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Chapter 3: Fourier Series Representation of Periodic Signals

3.0 Introduction

The representation and analysis of LTI systems through the convolution sum, as developed in Chapter 2, is based on representing signals as linear combinations of shifted impulses. In this and the following two chapters, we explore an alternative representation for signals and LTI systems. As in Chapter 2, the starting point for our discussion is the development of a representation of signals as linear combinations of a set of basic signals. For this alternative representation, we use complex exponentials. The resulting representations are known as the continuous-time and discrete-time Fourier series and transform. As we will see, these can be used to construct broad and useful classes of signals.

We then proceed as we did in Chapter 2. That is, because of the superposition property, the response of an LTI system to any input consisting of a linear combination of basic signals is the same linear combination of the individual responses to each of the basic signals. In Chapter 2, these responses were all shifted versions of the unit impulse response, leading to the convolution sum or integral. As we will find in the current chapter, the response of an LTI system to a complex exponential also has a particularly simple form, which then provides us with another convenient representation for LTI systems and with another way in which to analyze these systems and gain insight into their properties.

In this chapter, we focus on the representation of continuous-time and discrete-time periodic signals, referred to as the Fourier series. In Chapters 4 and 5, we extend the analysis to the Fourier transform representation of broad classes of aperiodic, finite-energy signals. Together, these representations provide one of the most powerful and important sets of tools and insights for analyzing, designing, and understanding signals and LTI systems, and we devote considerable attention in this and subsequent chapters to exploring the uses of Fourier methods.

3.1 A Historical Perspective

The development of Fourier analysis has a long history involving a great many individuals and the investigation of many different physical phenomena. The concept of using “trigonometric sums” – that is, sums of harmonically related sines and cosines or periodic complex exponentials – to describe periodic phenomena goes back at least as far as the Babylonians, who used ideas of this type in order to predict astronomical events. The modern history of the subject begins in 1748 with L. Euler, who examined the motion of a vibrating string. In Figure 3.1 (not shown here), we have indicated the first few of what are known as the “normal modes” of such a string. If we consider the vertical deflection $f(t,x)$ of the string at time $t$ and at a distance $x$ along the string, then for any fixed instant of time, the normal modes are harmonically related sinusoidal functions of $x$. What Euler noted was that if the configuration of a vibrating string at some point in time is a linear combination of these normal modes, so is the configuration at any subsequent time. Furthermore, Euler showed that one could calculate the coefficients for the linear combination at the later time in a very straightforward manner from the coefficients at the earlier time. In doing this, Euler performed the same type of calculation as we will in the next section in deriving one of the properties of trigonometric sums that make them so useful for the analysis of LTI systems. Specifically, we will see that if the input to an LTI system is expressed as a linear combination of periodic complex exponentials or sinusoids, the output can also be expressed in this form, with coefficients that are related in a straightforward way to those of the input.

The property described in the preceding paragraph would not be particularly useful, unless it were true that a large class of interesting functions could be represented by linear combinations of complex exponentials. In the middle of the 18th century, this point was the subject of heated debate. In 1753, D. Bernoulli argued on physical grounds that all physical motions of a string could be represented by linear combinations of normal modes, but he did not pursue this mathematically, and his ideas were not widely accepted. In fact, Euler himself discarded trigonometric series, and in 1759 J. L. Lagrange strongly criticized the use of trigonometric series in the examination of vibrating strings. His criticism was based on his own belief that it was impossible to represent signals with corners (i.e., with discontinuous slopes) using trigonometric series. Since such a configuration arises from the plucking of a string (i.e., pulling it taut and then releasing it), Lagrange argued that trigonometric series were of very limited use.

It was in this somewhat hostile and skeptical environment that Jean Baptiste Joseph Fourier (Figure 3.2, not shown here) presented his ideas half a century later. Fourier was born on March 21, 1768, in Auxerre, France, and by the time of his entrance into the controversy concerning trigonometric series, he had already had a lifetime of experiences. His many contributions—in particular, those concerned with the series and transform that carry his name—are made even more impressive by the circumstances under which he worked. His revolutionary discoveries, although not completely appreciated during his own lifetime, have had a major impact on the development of mathematics and have been and still are of great importance in an extremely wide range of scientific and engineering disciplines.

In addition to his studies in mathematics, Fourier led an active political life. In fact, during the years that followed the French Revolution, his activities almost led to his downfall, as he narrowly avoided the guillotine on two separate occasions. Subsequently, Fourier became an associate of Napoleon Bonaparte, accompanied him on his expeditions to Egypt (during which time Fourier collected the information he would use later as the basis for his treatises on Egyptology), and in 1802 was appointed by Bonaparte to the position of prefect of a region of France centered in Grenoble. It was there, while serving as prefect, that Fourier developed his ideas on trigonometric series.

The physical motivation for Fourier’s work was the phenomenon of heat propagation and diffusion. This in itself was a significant step in that most previous research in mathematical physics had dealt with rational and celestial mechanics. By 1807, Fourier had completed a work, Fourier had found series of harmonically related sinusoids to be useful in representing the temperature distribution through a body. In addition, he claimed that “any” periodic signal could be represented by such a series. While his treatment of this topic was significant, many of the basic ideas behind it had been discovered by others. Also, Fourier’s mathematical arguments were still imprecise, and it remained for P. L. Dirichlet in 1829 to provide precise conditions under which a periodic signal could be represented by a Fourier series. Thus, Fourier did not actually contribute to the mathematical theory of Fourier series. However, he did have the clear insight to see the potential for this series representation, and it was to a great extent his work and his claims that spurred much of the subsequent work on Fourier series. In addition, Fourier took this type of representation one very large step farther than any of his predecessors: He obtained a representation for aperiodic signals—not as weighted sums of harmonically related sinusoids—but as weighted integrals of sinusoids that are not all harmonically related. It is this extension from Fourier series to the Fourier integral or transform that is the focus of Chapters 4 and 5. Like the Fourier series, the Fourier transform remains one of the most powerful tools for the analysis of LTI systems.

Four distinguished mathematicians and scientists were appointed to examine the 1807 paper of Fourier. Three of the four—S. F. Lacroix, G. Monge, and P. S. de Laplace—were in favor of publication of the paper, but the fourth, J. L. Lagrange, remained adamant in rejecting trigonometric series, as he had done 50 years earlier. Because of Lagrange’s vehement objections, Fourier’s paper never appeared. After several other attempts to have his work accepted and published by the Institut de France, Fourier undertook the writing of another version of his work, which appeared as the text Théorie analytique de la chaleur. This book was published in 1822, 15 years after Fourier had first presented his results to the Institut.

Toward the end of his life, Fourier received some of the recognition he deserved, but the most significant tribute to him has been the enormous impact of his work on so many disciplines within the fields of mathematics, science, and engineering. The theory of integration, point-set topology, and eigenfunction expansions are just a few examples of topics in mathematics that have their roots in the analysis of Fourier series and integrals. Furthermore, in addition to the original studies of vibration and heat diffusion, there are numerous other problems in science and engineering in which sinusoidal signals, and therefore Fourier series and transforms, play an important role. For example, sinusoidal signals arise naturally in describing the motion of the planets and the periodic behavior of the earth’s climate. Alternating-current sources generate sinusoidal voltages and currents, and, as we will see, the tools of Fourier analysis enable us to analyze the response of an LTI system, such as a circuit, to such sinusoidal inputs. Also, as illustrated in Figure 3.3 (not shown here), waves in the ocean consist of the linear combination of sinusoidal waves with different spatial periods or wavelengths. Signals transmitted by radio and television stations are sinusoidal in nature as well, and as a quick perusal of any text on Fourier analysis will show, the range of applications in which sinusoidal signals arise and in which the tools of Fourier analysis are useful extends far beyond these few examples.

While many of the applications in the preceding paragraph, as well as the original work of Fourier and his contemporaries on problems of mathematical physics, focus on phenomena in continuous time, the tools of Fourier analysis for discrete-time signals and systems have their own distinct historical roots and equally rich set of applications. In particular, discrete-time concepts and methods are fundamental to the discipline of numerical analysis. Formulas for the processing of discrete sets of data points to produce numerical approximations for interpolation, integration, and differentiation were being investigated as early as the time of Newton in the 1600s. In addition, the problem of predicting the motion of a heavenly body, given a sequence of observations of the body, spurred the investigation of harmonic time series in the 18th and 19th centuries by eminent scientists and mathematicians, including Gauss, and thus provided a second setting in which much of the initial work was done on discrete-time signals and systems.

In the mid-1960s an algorithm, now known as the fast Fourier transform, or FFT, was introduced. This algorithm, which was independently discovered by Cooley and Tukey in 1965, also has a considerable history and can, in fact, be found in Gauss’ notebooks. What made its modern discovery so important was the fact that the FFT proved to be perfectly suited for efficient digital implementation, and it reduced the time required to compute transforms by orders of magnitude. With this tool, many interesting but previously impractical ideas utilizing the discrete-time Fourier series and transform suddenly became practical, and the development of discrete-time signal and system analysis techniques moved forward at an accelerated pace.

What has emerged out of this long history is a powerful and cohesive framework for the analysis of continuous-time and discrete-time signals and systems and an extraordinarily broad array of existing and potential applications. In this and the following chapters, we will develop the basic tools of that framework and examine some of its important implications.

3.2 The Response of LTI Systems to Complex Exponentials

As we indicated in Section 3.0, it is advantageous in the study of LTI systems to represent signals as linear combinations of basic signals that possess the following two properties:

1. The set of basic signals can be used to construct a broad and useful class of signals.
2. The response of an LTI system to each signal should be simple enough in structure to provide us with a convenient representation for the response of the system to any signal constructed as a linear combination of the basic signals.

Much of the importance of Fourier analysis results from the fact that both of these properties are provided by the set of complex exponential signals in continuous and discrete time—i.e., signals of the form $e^{st}$ in continuous time and $z^n$ in discrete time, where $s$ and $z$ are complex numbers. In subsequent sections of this and the following two chapters, we will examine the first property in some detail. In this section, we focus on the second property and, in this way, provide motivation for the use of Fourier series and transforms in the analysis of LTI systems.

The importance of complex exponentials in the study of LTI systems stems from the fact that the response of an LTI system to a complex exponential input is the same complex exponential with only a change in amplitude; that is,

$$\begin{array}{rl}& \mathrm{continuous}\mathrm{time}:e^{st}⟶H(s)e^{st},\\ & \mathrm{discrete}\mathrm{time}:z^n⟶H(z)z^n,\end{array}\text{\hspace{1em}(3.1)}$$

where the complex amplitude factor $H(s)$ or $H(z)$ will in general be a function of the complex variable $s$ or $z$. A signal for which the system output is a (possibly complex) constant times the input is referred to as an eigenfunction of the system, and the amplitude factor is referred to as the system’s eigenvalue.

To show that complex exponentials are indeed eigenfunctions of LTI systems, let us consider a continuous-time LTI system with impulse response $h(t)$. For an input $x(t)$, we can determine the output through the use of the convolution integral, so that with $x(t)=e^{st}$:

$$\begin{array}{rl}y(t)& ={\int }_{-\infty }^{+\infty }h(\tau )x(t-\tau )d\tau \\ & ={\int }_{-\infty }^{+\infty }h(\tau )e^{s(t-\tau )}d\tau .\end{array}\text{\hspace{1em}(3.2)}$$

Expressing $e^{s(t-\tau )}$ as $e^{st}{e}^{-s\tau }$, and noting that $e^{st}$ can be moved outside the integral, we see that eq. (3.2) becomes

$$y(t)=e^{st}{\int }_{-\infty }^{+\infty }h(\tau )e^{-s\tau }d\tau .\text{\hspace{1em}(3.3)}$$

Assuming that the integral on the right-hand side of eq. (3.3) converges, the response to $e^{st}$ is of the form

$$y(t)=H(s)e^{st},\text{\hspace{1em}(3.4)}$$

where $H(s)$ is a complex constant whose value depends on $s$ and which is related to the system impulse response by

$$H(s)={\int }_{-\infty }^{+\infty }h(\tau )e^{-s\tau }d\tau .\text{\hspace{1em}(3.5)}$$

Hence, we have shown that complex exponentials are eigenfunctions of LTI systems. The constant $H(s)$ for a specific value of $s$ is then the eigenvalue associated with the eigenfunction $e^{st}$.

In an exactly parallel manner, we can show that complex exponential sequences are eigenfunctions of discrete-time LTI systems. That is, suppose that an LTI system with impulse response $h[n]$ has as its input the sequence

$$x[n]=z^n,\text{\hspace{1em}(3.6)}$$

where $z$ is a complex number. Then the output of the system can be determined from the convolution sum as

$$\begin{array}{rl}y[n]& =\sum _{k=-\infty }^{+\infty }h[k]x[n-k]\\ & =\sum _{k=-\infty }^{+\infty }h[k]z^{n-k}\\ & =z^n\sum _{k=-\infty }^{+\infty }h[k]z^{-k}.\end{array}\text{\hspace{1em}(3.7)}$$

From this expression, we see that if the input $x[n]$ is the complex exponential given by eq. (3.6), then, assuming that the summation on the right-hand side of eq. (3.7) converges, the output is the same complex exponential multiplied by a constant that depends on the value of $z$. That is,

$$y[n]=H(z)z^n,\text{\hspace{1em}(3.8)}$$

where

$$H(z)=\sum _{k=-\infty }^{+\infty }h[k]z^{-k}.\text{\hspace{1em}(3.9)}$$

Consequently, as in the continuous-time case, complex exponentials are eigenfunctions of discrete-time LTI systems. The constant $H(z)$ for a specified value of $z$ is the eigenvalue associated with the eigenfunction $z^n$.

For the analysis of LTI systems, the usefulness of decomposing more general signals in terms of eigenfunctions can be seen from an example. Let $x(t)$ correspond to a linear combination of three complex exponentials; that is,

$$x(t)=a_1{e}^{s_{1}t}+a_2{e}^{s_{2}t}+a_3{e}^{s_{3}t}.\text{\hspace{1em}(3.10)}$$

From the eigenfunction property, the response to each separately is:

$$\begin{array}{rl}{a}_1{e}^{s_{1}t}& ⟶a_{1}H(s_1)e^{s_{1}t},\\ a_2{e}^{s_{2}t}& ⟶a_{2}H(s_2)e^{s_{2}t},\\ a_3{e}^{s_{3}t}& ⟶a_{3}H(s_3)e^{s_{3}t},\end{array}$$

and from the superposition property, the response to the sum is the sum of the responses, so that

$$y(t)=a_{1}H(s_1)e^{s_{1}t}+a_{2}H(s_2)e^{s_{2}t}+a_{3}H(s_3)e^{s_{3}t}.\text{\hspace{1em}(3.11)}$$

More generally, in continuous time, eq. (3.4), together with the superposition property, implies that the representation of signals as a linear combination of complex exponentials leads to a convenient expression for the response of an LTI system. Specifically, if the input to a continuous-time LTI system is represented as a linear combination of complex exponentials, that is, if

$$x(t)=\sum _k{a}_k{e}^{s_{k}t},\text{\hspace{1em}(3.12)}$$

then the output will be

$$y(t)=\sum _k{a}_{k}H(s_k)e^{s_{k}t}.\text{\hspace{1em}(3.13)}$$

In an exactly analogous manner, if the input to a discrete-time LTI system is represented as a linear combination of complex exponentials, that is, if

$$x[n]=\sum _k{a}_k{z}_k^n,\text{\hspace{1em}(3.14)}$$

then the output will be

$$y[n]=\sum _k{a}_{k}H(z_k)z_k^n.\text{\hspace{1em}(3.15)}$$

In other words, for both continuous time and discrete time, if the input to an LTI system is represented as a linear combination of complex exponentials, then the output can also be represented as a linear combination of the same complex exponential signals. Each coefficient in this representation of the output is obtained as the product of the corresponding coefficient $a_k$ of the input and the system’s eigenvalue $H(s_k)$ or $H(z_k)$ associated with the eigenfunction $e^{s_{k}t}$ or $z_k^n$, respectively. It was precisely this fact that Euler discovered for the problem of the vibrating string, that Gauss and others used in the analysis of time series, and that motivated Fourier and others after him to consider the question of how broad a class of signals could be represented as a linear combination of complex exponentials. In the next few sections, we examine this question for periodic signals, first in continuous time and then in discrete time, and in Chapters 4 and 5, we consider the extension of these representations to aperiodic signals. Although in general, the variables $s$ and $z$ in eqs. (3.1)–(3.15) may be arbitrary complex numbers, Fourier analysis involves restricting our attention to particular forms for these variables. In particular, in continuous time, we focus on purely imaginary values of $s$ — i.e., $s=j\omega$ — and thus, we consider only complex exponentials of the form $e^{j\omega t}$. Similarly, in discrete time, we restrict the range of values of $z$ to those of unit magnitude—i.e., $z=e^{j\omega }$ —so that we focus on complex exponentials of the form $e^{j\omega n}$.

Example 3.1

As an illustration of eqs. (3.4) and (3.5), consider an LTI system for which the input $x(t)$ and output $y(t)$ are related by a time shift of 3, i.e.,

$$y(t)=x(t-3).\text{\hspace{1em}(3.16)}$$

Part 1: Input $x(t)=e^{j2t}$

If the input to this system is the complex exponential signal $x(t)=e^{j2t}$, then, from eq. (3.16):

$$\begin{array}{rlrl}y(t)& =e^{j2(t-3)}& & \text{(Substitute }x(t)=e^{j2t}\text{ into }y(t)=x(t-3)\text{)}\\ & =e^{j2t}{e}^{-j6}& & \text{(Factor out }{e}^{j2t}\text{)}\\ & =e^{-j6}{e}^{j2t}.& & \text{(Rearrange terms)}\end{array}\text{\hspace{1em}(3.17)}$$

Equation (3.17) is in the form of eq. (3.4), $y(t)=H(s)e^{st}$, as we would expect, since $e^{j2t}$ is an eigenfunction. The associated eigenvalue is $H(j2)=e^{-j6}$.

It is straightforward to confirm eq. (3.5) for this example. Specifically, from eq. (3.16), the impulse response of the system is $h(t)=\delta (t-3)$. Substituting into eq. (3.5), we obtain:

$$\begin{array}{rlrl}H(s)& ={\int }_{-\infty }^{+\infty }h(\tau )e^{-s\tau }d\tau & & \text{(Definition of }H(s)\text{)}\\ & ={\int }_{-\infty }^{+\infty }\delta (\tau -3)e^{-s\tau }d\tau & & \text{(Substitute }h(\tau )=\delta (\tau -3)\text{)}\\ & =e^{-s(3)}& & \text{(Property of impulse function: }\int f(t)\delta (t-t_0)dt=f(t_0)\text{)}\\ & =e^{-3s}.& & \end{array}$$

Therefore, for $s=j2$:

$$H(j2)=e^{-3(j2)}=e^{-j6}.\text{(Substitute }s=j2\text{)}$$

Part 2: Input $x(t)=\cos (4t)+\cos (7t)$

As a second example, in this case illustrating eqs. (3.10) and (3.11), consider the input signal $x(t)=\cos (4t)+\cos (7t)$. From eq. (3.16), $y(t)$ will of course be:

$$\begin{array}{rl}y(t)& =x(t-3)\\ & =\cos (4(t-3))+\cos (7(t-3)).\end{array}\text{\hspace{1em}(3.18)}$$

To see that this will also result from eq. (3.11), we first expand $x(t)$ using Euler’s relation:

$$\begin{array}{rlrl}x(t)& =\cos (4t)+\cos (7t)& & \text{(Given input signal)}\\ & =\frac{1}{2}{e}^{j4t}+\frac{1}{2}{e}^{-j4t}+\frac{1}{2}{e}^{j7t}+\frac{1}{2}{e}^{-j7t}.& & \text{(Apply Euler’s formula: }\cos \theta =\frac{e^{j\theta }+e^{-j\theta }}{2}\text{)}\end{array}\text{\hspace{1em}(3.19)}$$

The system’s frequency response is $H(s)=e^{-3s}$, as derived above. Thus, $H(j4)=e^{-j12}$, $H(-j4)=e^{j12}$, $H(j7)=e^{-j21}$, and $H(-j7)=e^{j21}$. Applying eq. (3.11) (or eq. (3.13) for a sum), we find the output $y(t)$:

$$\begin{array}{rlrl}y(t)& =\frac{1}{2}H(j4)e^{j4t}+\frac{1}{2}H(-j4)e^{-j4t}+\frac{1}{2}H(j7)e^{j7t}+\frac{1}{2}H(-j7)e^{-j7t}& & \text{(From eq. (3.13))}\\ & =\frac{1}{2}(e^{-j12})e^{j4t}+\frac{1}{2}(e^{j12})e^{-j4t}+\frac{1}{2}(e^{-j21})e^{j7t}+\frac{1}{2}(e^{j21})e^{-j7t}& & \text{(Substitute eigenvalues)}\\ & =\frac{1}{2}(e^{j(4t-12)}+e^{-j(4t-12)})+\frac{1}{2}(e^{j(7t-21)}+e^{-j(7t-21)})& & \text{(Combine terms)}\\ & =\cos (4t-12)+\cos (7t-21)& & \text{(Apply Euler’s formula in reverse)}\\ & =\cos (4(t-3))+\cos (7(t-3)),& & \end{array}$$

which is the same as eq. (3.18). For this simple example, multiplication of each periodic exponential component of $x(t)$—for example, $\frac{1}{2}{e}^{j4t}$—by the corresponding eigenvalue—e.g., $H(j4)=e^{-j12}$—effectively causes the input component to shift in time by 3. Obviously, in this case, we can determine $y(t)$ in eq. (3.18) by inspection rather than by employing eqs. (3.10) and (3.11). However, as we will see, the general property embodied in eqs. (3.10) and (3.11) not only allows us to calculate the responses of more complex LTI systems but also provides the basis for the frequency-domain representation and analysis of LTI systems.

3.3 Fourier Series Representation of Continuous-Time Periodic Signals

3.3.1 Linear Combinations of Harmonically Related Complex Exponentials

As defined in Chapter 1, a signal is periodic if, for some positive value of $T$,

$$x(t)=x(t+T)\mathrm{for}\mathrm{all}t.\text{\hspace{1em}(3.20)}$$

The fundamental period of $x(t)$ is the minimum positive, nonzero value of $T$ for which eq. (3.20) is satisfied, and the value ${\omega }_0=2\pi /T$ is referred to as the fundamental frequency.

In Chapter 1, we also introduced two basic periodic signals, the sinusoidal signal

$$x(t)=\cos {\omega }_{0}t\text{\hspace{1em}(3.21)}$$

and the periodic complex exponential

$$x(t)=e^{j{\omega }_{0}t}.\text{\hspace{1em}(3.22)}$$

Both of these signals are periodic with fundamental frequency ${\omega }_0$ and fundamental period $T=2\pi /{\omega }_0$. Associated with the signal in eq. (3.22) is the set of harmonically related complex exponentials

$${\varphi }_k(t)=e^{jk{\omega }_{0}t}=e^{jk(2\pi /T)t},k=0,\pm 1,\pm 2,\dots \text{\hspace{1em}(3.23)}$$

Each of these signals has a fundamental frequency that is a multiple of ${\omega }_0$, and therefore, each is periodic with period $T$ (although for $|k|\ge 2$, the fundamental period of ${\varphi }_k(t)$ is a fraction of $T$). Thus, a linear combination of harmonically related complex exponentials of the form

$$x(t)=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk(2\pi /T)t}\text{\hspace{1em}(3.24)}$$

is also periodic with period $T$. In eq. (3.24), the term for $k=0$ is a constant. The terms for $k=+1$ and $k=-1$ both have fundamental frequency equal to ${\omega }_0$ and are collectively referred to as the fundamental components or the first harmonic components. The two terms for $k=+2$ and $k=-2$ are periodic with half the period (or, equivalently, twice the frequency) of the fundamental components and are referred to as the second harmonic components. More generally, the components for $k=+N$ and $k=-N$ are referred to as the $N$th harmonic components.

The representation of a periodic signal in the form of eq. (3.24) is referred to as the Fourier series representation. Before developing the properties of this representation, let us consider an example.

Example 3.2

Consider a periodic signal $x(t)$, with fundamental frequency $2\pi$, that is expressed in the form of eq. (3.24) as

$$x(t)=\sum _{k=-3}^{+3}{a}_k{e}^{jk2\pi t},\text{\hspace{1em}(3.25)}$$

where

$$a_0=1,a_1=a_{-1}=\frac{1}{4},a_2=a_{-2}=\frac{1}{2},a_3=a_{-3}=\frac{1}{3}.$$

Rewriting eq. (3.25) and collecting each of the harmonic components that have the same fundamental frequency, we obtain:

$$\begin{array}{rl}x(t)& =a_0{e}^{j0\cdot 2\pi t}+a_1{e}^{j1\cdot 2\pi t}+a_{-1}{e}^{-j1\cdot 2\pi t}+a_2{e}^{j2\cdot 2\pi t}+a_{-2}{e}^{-j2\cdot 2\pi t}+a_3{e}^{j3\cdot 2\pi t}+a_{-3}{e}^{-j3\cdot 2\pi t}\\ & =1\cdot e^{j0}+\frac{1}{4}{e}^{j2\pi t}+\frac{1}{4}{e}^{-j2\pi t}+\frac{1}{2}{e}^{j4\pi t}+\frac{1}{2}{e}^{-j4\pi t}+\frac{1}{3}{e}^{j6\pi t}+\frac{1}{3}{e}^{-j6\pi t}\\ & =1+\frac{1}{4}(e^{j2\pi t}+e^{-j2\pi t})+\frac{1}{2}(e^{j4\pi t}+e^{-j4\pi t})+\frac{1}{3}(e^{j6\pi t}+e^{-j6\pi t}).\end{array}\text{\hspace{1em}(3.26)}$$

Equivalently, using Euler’s relation ($e^{j\theta }+e^{-j\theta }=2\cos \theta$), we can write $x(t)$ in the form:

$$\begin{array}{rl}x(t)& =1+\frac{1}{4}(2\cos 2\pi t)+\frac{1}{2}(2\cos 4\pi t)+\frac{1}{3}(2\cos 6\pi t)\\ & =1+\frac{1}{2}\cos 2\pi t+\cos 4\pi t+\frac{2}{3}\cos 6\pi t.\end{array}\text{\hspace{1em}(3.27)}$$

In Figure 3.4 (not shown here), we illustrate graphically how the signal $x(t)$ is built up from its harmonic components.

Equation (3.27) is an example of an alternative form for the Fourier series of real periodic signals. Specifically, suppose that $x(t)$ is real and can be represented in the form of eq. (3.24). Then, since $x^{\ast }(t)=x(t)$, we obtain:

$$\begin{array}{rlrl}x(t)& =x^{\ast }(t)& & \text{(Given }x(t)\text{ is real)}\\ \sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}& ={\left(\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}\right)}^{\ast }& & \text{(Substitute Fourier series representation)}\\ & =\sum _{k=-\infty }^{+\infty }{a}_k^{\ast }{\left(e^{jk{\omega }_{0}t}\right)}^{\ast }& & \text{(Conjugate of a sum is sum of conjugates, conjugate of product is product of conjugates)}\\ & =\sum _{k=-\infty }^{+\infty }{a}_k^{\ast }{e}^{-jk{\omega }_{0}t}.& & \end{array}$$

Replacing $k$ by $-k$ in the summation (which is a dummy variable), we have:

$$\begin{array}{rlrl}x(t)& =\sum _{m=-\infty }^{+\infty }{a}_{-m}^{\ast }{e}^{j(-m){\omega }_0(-t)}& & \text{(Let }m=-k\text{, so }k=-m\text{ and }-k=m\text{)}\\ & =\sum _{m=-\infty }^{+\infty }{a}_{-m}^{\ast }{e}^{jm{\omega }_{0}t}.& & \text{(Rearrange terms)}\end{array}$$

This can be written back in terms of $k$:

$$x(t)=\sum _{k=-\infty }^{+\infty }{a}_{-k}^{\ast }{e}^{jk{\omega }_{0}t}.$$

Comparing this with the original eq. (3.24) ($x(t)={\sum }_{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}$), which must hold for all $t$, requires that the coefficients of each harmonic component be equal. Thus:

$$a_k=a_{-k}^{\ast },\text{or equivalently, that}{a}_k^{\ast }=a_{-k}.\text{\hspace{1em}(3.28)}$$

Note that this is the case in Example 3.2, where the $a_k$‘s are in fact real and $a_k=a_{-k}$. To derive the alternative forms of the Fourier series, we first rearrange the summation in eq. (3.24) as:

$$\begin{array}{rlrl}x(t)& =a_0{e}^{j0{\omega }_{0}t}+\sum _{k=1}^{\infty }{a}_k{e}^{jk{\omega }_{0}t}+\sum _{k=-\infty }^{-1}{a}_k{e}^{jk{\omega }_{0}t}& & \text{(Separate }k=0\text{ term)}\\ & =a_0+\sum _{k=1}^{\infty }{a}_k{e}^{jk{\omega }_{0}t}+\sum _{k=1}^{\infty }{a}_{-k}{e}^{-jk{\omega }_{0}t}& & \text{(Change summation index in second sum from }k\text{ to }-k\text{)}\\ & =a_0+\sum _{k=1}^{\infty }[a_k{e}^{jk{\omega }_{0}t}+a_{-k}{e}^{-jk{\omega }_{0}t}].& & \end{array}$$

Substituting $a_k^{\ast }$ for $a_{-k}$ from eq. (3.28), we obtain:

$$x(t)=a_0+\sum _{k=1}^{\infty }[a_k{e}^{jk{\omega }_{0}t}+a_k^{\ast }{e}^{-jk{\omega }_{0}t}].$$

Since the two terms inside the summation are complex conjugates of each other (i.e., $Z+Z^{\ast }=2\Re e\{Z\}$), this can be expressed as:

$$x(t)=a_0+\sum _{k=1}^{\infty }2\Re e\{a_k{e}^{jk{\omega }_{0}t}\}.\text{\hspace{1em}(3.29)}$$

If $a_k$ is expressed in polar form as $a_k=A_k{e}^{j{\theta }_k}$, where $A_k=|a_k|$ and ${\theta }_k=\angle a_k$, then eq. (3.29) becomes:

$$\begin{array}{rl}x(t)& =a_0+\sum _{k=1}^{\infty }2\Re e\{A_k{e}^{j{\theta }_k}{e}^{jk{\omega }_{0}t}\}\\ & =a_0+\sum _{k=1}^{\infty }2\Re e\{A_k{e}^{j(k{\omega }_{0}t+{\theta }_k)}\}.\end{array}$$

That is, using $\Re e\{Z{e}^{j\varphi }\}=Z\cos \varphi$ for real $Z$:

$$x(t)=a_0+2\sum _{k=1}^{\infty }{A}_k\cos (k{\omega }_{0}t+{\theta }_k).\text{\hspace{1em}(3.30)}$$

Equation (3.30) is one commonly encountered form for the Fourier series of real periodic signals in continuous time. Another form is obtained by writing $a_k$ in rectangular form as $a_k=B_k+j{C}_k$, where $B_k$ and $C_k$ are both real. With this expression for $a_k$, eq. (3.29) takes the form:

$$\begin{array}{rl}x(t)& =a_0+\sum _{k=1}^{\infty }2\Re e\{(B_k+j{C}_k)e^{jk{\omega }_{0}t}\}\\ & =a_0+\sum _{k=1}^{\infty }2\Re e\{(B_k+j{C}_k)(\cos k{\omega }_{0}t+j\sin k{\omega }_{0}t)\}\\ & =a_0+\sum _{k=1}^{\infty }2\Re e\{B_k\cos k{\omega }_{0}t+j{B}_k\sin k{\omega }_{0}t+j{C}_k\cos k{\omega }_{0}t-C_k\sin k{\omega }_{0}t\}\\ & =a_0+\sum _{k=1}^{\infty }2[B_k\cos k{\omega }_{0}t-C_k\sin k{\omega }_{0}t].\end{array}\text{\hspace{1em}(3.31)}$$

In Example 3.2, the $a_k$‘s are all real, so that $a_k=A_k=B_k$ and $C_k=0$. Therefore, both representations, eqs. (3.30) and (3.31), reduce to the same form, eq. (3.27).

Thus, for real periodic functions, the Fourier series in terms of complex exponentials, as given in eq. (3.24), is mathematically equivalent to either of the two forms in eqs. (3.30) and (3.31) that use trigonometric functions. Although the latter two are common forms for Fourier series, the complex exponential form of eq. (3.24) is particularly convenient for our purposes, so we will use that form almost exclusively.

Equation (3.28) illustrates one of many properties associated with Fourier series. These properties are often quite useful in gaining insight and for computational purposes, and in Section 3.5, we collect together the most important of them. The derivation of several of them is considered in problems at the end of the chapter. In Section 4.3, we also will develop the majority of the properties within the broader context of the Fourier transform.

3.3.2 Determination of the Fourier Series Representation of a Continuous-time Periodic Signal

Assuming that a given periodic signal can be represented with the series of eq. (3.24), we need a procedure for determining the coefficients $a_k$. Multiplying both sides of eq. (3.24) by $e^{-jn{\omega }_{0}t}$, we obtain:

$$x(t)e^{-jn{\omega }_{0}t}=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}{e}^{-jn{\omega }_{0}t}.\text{\hspace{1em}(3.32)}$$

Integrating both sides from $0$ to $T=2\pi /{\omega }_0$, we have:

$${\int }_0^{T}x(t)e^{-jn{\omega }_{0}t}dt={\int }_0^T\sum _{k=-\infty }^{+\infty }{a}_k{e}^{j(k-n){\omega }_{0}t}dt.$$

Here, $T$ is the fundamental period of $x(t)$, and consequently, we are integrating over one period. Interchanging the order of integration and summation yields:

$${\int }_0^{T}x(t)e^{-jn{\omega }_{0}t}dt=\sum _{k=-\infty }^{+\infty }{a}_k\left[{\int }_0^T{e}^{j(k-n){\omega }_{0}t}dt\right].\text{\hspace{1em}(3.33)}$$

The evaluation of the bracketed integral is straightforward. Rewriting this integral using Euler’s formula, we obtain:

$${\int }_0^T{e}^{j(k-n){\omega }_{0}t}dt={\int }_0^T\cos (k-n){\omega }_{0}tdt+j{\int }_0^T\sin (k-n){\omega }_{0}tdt.\text{\hspace{1em}(3.34)}$$

We evaluate the integral:

$${\int }_0^T{e}^{j(k-n){\omega }_{0}t}dt=\{\begin{array}{ll}{\int }_0^{T}1dt& \text{for }k=n\\ {\int }_0^T\cos ((k-n){\omega }_{0}t)dt+j{\int }_0^T\sin ((k-n){\omega }_{0}t)dt& \text{for }k\ne n\end{array}$$

For $k=n$:

$${\int }_0^{T}1dt=T.$$

For $k\ne n$: Let $m=k-n$. Since $k$ and $n$ are integers, $m$ is a nonzero integer.

$$\begin{array}{rlrl}& {\int }_0^T\cos (m{\omega }_{0}t)dt+j{\int }_0^T\sin (m{\omega }_{0}t)dt& & \\ & ={\left[\frac{\sin (m{\omega }_{0}t)}{m{\omega }_0}\right]}_0^T+j{\left[-\frac{\cos (m{\omega }_{0}t)}{m{\omega }_0}\right]}_0^T& & \\ & =\left(\frac{\sin (m{\omega }_{0}T)}{m{\omega }_0}-\frac{\sin (0)}{m{\omega }_0}\right)+j\left(-\frac{\cos (m{\omega }_{0}T)}{m{\omega }_0}-\left(-\frac{\cos (0)}{m{\omega }_0}\right)\right)& & \\ & =\left(\frac{\sin (m\cdot 2\pi )}{m{\omega }_0}-0\right)+j\left(-\frac{\cos (m\cdot 2\pi )}{m{\omega }_0}+\frac{1}{m{\omega }_0}\right)& & \text{(Since }{\omega }_{0}T=2\pi \text{)}\\ & =\left(0-0\right)+j\left(-\frac{1}{m{\omega }_0}+\frac{1}{m{\omega }_0}\right)& & \text{(Since }\sin (m\cdot 2\pi )=0\text{ and }\cos (m\cdot 2\pi )=1\text{ for integer }m\text{)}\\ & =0.& & \end{array}$$

Thus, for $k\ne n$, $\cos (k-n){\omega }_{0}t$ and $\sin (k-n){\omega }_{0}t$ are periodic sinusoids with fundamental period $(T/|k-n|)$. Therefore, in eq. (3.34), we are integrating over an interval (of length $T$) that is an integral number of periods of these signals. Since the integral may be viewed as measuring the total area under the functions over the interval, we see that for $k\ne n$, both of the integrals on the right-hand side of eq. (3.34) are zero. For $k=n$, the integrand on the left-hand side of eq. (3.34) equals 1, and thus, the integral equals $T$. In sum, we then have:

$${\int }_0^T{e}^{j(k-n){\omega }_{0}t}dt=\{\begin{array}{ll}T,& k=n\\ 0,& k\ne n\end{array},$$

and consequently, the right-hand side of eq. (3.33) reduces to $T{a}_n$. Therefore,

$$a_n=\frac{1}{T}{\int }_0^{T}x(t)e^{-jn{\omega }_{0}t}dt,\text{\hspace{1em}(3.35)}$$

which provides the equation for determining the coefficients. Furthermore, note that in evaluating eq. (3.34), the only fact that we used concerning the interval of integration was that we were integrating over an interval of length $T$, which is an integral number of periods of $\cos (k-n){\omega }_{0}t$ and $\sin (k-n){\omega }_{0}t$. Therefore, we will obtain the same result if we integrate over any interval of length $T$. That is, if we denote integration over any interval of length $T$ by ${\int }_T$, we have

$${\int }_T{e}^{j(k-n){\omega }_{0}t}dt=\{\begin{array}{ll}T,& k=n\\ 0,& k\ne n\end{array},$$

and consequently,

$$a_n=\frac{1}{T}{\int }_{T}x(t)e^{-jn{\omega }_{0}t}dt.\text{\hspace{1em}(3.36)}$$

To summarize, if $x(t)$ has a Fourier series representation [i.e., if it can be expressed as a linear combination of harmonically related complex exponentials in the form of eq. (3.24)], then the coefficients are given by eq. (3.36). This pair of equations, then, defines the Fourier series of a periodic continuous-time signal:

$$\begin{array}{c}\boxed{\begin{array}{rl}x(t)& =\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk(2\pi /T)t},\text{(synthesis equation)}\\ a_k& =\frac{1}{T}{\int }_{T}x(t)e^{-jk{\omega }_{0}t}dt=\frac{1}{T}{\int }_{T}x(t)e^{-jk(2\pi /T)t}dt,\text{(analysis equation)}\end{array}}\end{array}\text{\hspace{1em}(3.38)}$$

Here, we have written equivalent expressions for the Fourier series in terms of the fundamental frequency ${\omega }_0$ and the fundamental period $T$. Equation (3.37) is referred to as the synthesis equation and eq. (3.38) as the analysis equation. The set of coefficients $\{a_k\}$ are often called the Fourier series coefficients or the spectral coefficients of $x(t)$. These complex coefficients measure the portion of the signal $x(t)$ that is at each harmonic of the fundamental component. The coefficient $a_0$ is the dc or constant component of $x(t)$ and is given by eq. (3.38) with $k=0$. That is,

$$a_0=\frac{1}{T}{\int }_{T}x(t)dt,\text{\hspace{1em}(3.39)}$$

which is simply the average value of $x(t)$ over one period.

Equations (3.37) and (3.38) were known to both Euler and Lagrange in the middle of the 18th century. However, they discarded this line of analysis without having examined the question of how large a class of periodic signals could, in fact, be represented in such a fashion. Before we turn to this question in the next section, let us illustrate the continuous-time Fourier series by means of a few examples.

Example 3.3

Consider the signal $x(t)=\sin {\omega }_{0}t$, whose fundamental frequency is ${\omega }_0$. One approach to determining the Fourier series coefficients for this signal is to apply eq. (3.38). For this simple case, however, it is easier to expand the sinusoidal signal as a linear combination of complex exponentials and identify the Fourier series coefficients by inspection. Specifically, we can express $\sin {\omega }_{0}t$ as:

$$\begin{array}{rlrl}x(t)& =\sin {\omega }_{0}t& & \\ & =\frac{e^{j{\omega }_{0}t}-e^{-j{\omega }_{0}t}}{2j}.& & \text{(Euler’s formula for sine)}\\ & =\frac{1}{2j}{e}^{j{\omega }_{0}t}-\frac{1}{2j}{e}^{-j{\omega }_{0}t}.& & \end{array}$$

Comparing the right-hand sides of this equation and eq. (3.37) ($x(t)={\sum }_{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}$), we identify the coefficients:

$$a_1=\frac{1}{2j},a_{-1}=-\frac{1}{2j},$$

and all other $a_k=0$.

Example 3.4

Let

$$x(t)=1+\sin {\omega }_{0}t+2\cos {\omega }_{0}t+\cos \left(2{\omega }_{0}t+\frac{\pi }{4}\right),$$

which has fundamental frequency ${\omega }_0$. As with Example 3.3, we can again expand $x(t)$ directly in terms of complex exponentials:

$$\begin{array}{rlrl}x(t)& =1+\left(\frac{e^{j{\omega }_{0}t}-e^{-j{\omega }_{0}t}}{2j}\right)+2\left(\frac{e^{j{\omega }_{0}t}+e^{-j{\omega }_{0}t}}{2}\right)+\left(\frac{e^{j(2{\omega }_{0}t+\pi /4)}+e^{-j(2{\omega }_{0}t+\pi /4)}}{2}\right)& & \text{(Apply Euler’s formulas)}\\ & =1+\frac{1}{2j}{e}^{j{\omega }_{0}t}-\frac{1}{2j}{e}^{-j{\omega }_{0}t}+e^{j{\omega }_{0}t}+e^{-j{\omega }_{0}t}+\frac{1}{2}{e}^{j(2{\omega }_{0}t+\pi /4)}+\frac{1}{2}{e}^{-j(2{\omega }_{0}t+\pi /4)}.& & \end{array}$$

Collecting terms by powers of $e^{jk{\omega }_{0}t}$:

$$\begin{array}{rlrl}x(t)& =1\cdot e^{j0{\omega }_{0}t}& & \\ & +\left(\frac{1}{2j}+1\right)e^{j{\omega }_{0}t}& & \text{(For }k=1\text{)}\\ & +\left(-\frac{1}{2j}+1\right)e^{-j{\omega }_{0}t}& & \text{(For }k=-1\text{)}\\ & +\frac{1}{2}{e}^{j(\pi /4)}{e}^{j2{\omega }_{0}t}& & \text{(For }k=2\text{)}\\ & +\frac{1}{2}{e}^{-j(\pi /4)}{e}^{-j2{\omega }_{0}t}.& & \text{(For }k=-2\text{)}\end{array}$$

Thus, the Fourier series coefficients for this example are:

$$\begin{array}{rl}{a}_0& =1,\\ a_1& =1+\frac{1}{2j}=1-\frac{j}{2},\\ a_{-1}& =1-\frac{1}{2j}=1+\frac{j}{2},\\ a_2& =\frac{1}{2}{e}^{j(\pi /4)}=\frac{1}{2}\left(\frac{\sqrt{2}}{2}+j\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{4}(1+j),\\ a_{-2}& =\frac{1}{2}{e}^{-j(\pi /4)}=\frac{1}{2}\left(\frac{\sqrt{2}}{2}-j\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{4}(1-j),\\ a_k& =0,|k|>2.\end{array}$$

In Figure 3.5 (not shown here), we show a bar graph of the magnitude and phase of $a_k$.

Example 3.5

The periodic square wave, sketched in Figure 3.6 (not shown here) and defined over one period as:

$$x(t)=\{\begin{array}{ll}1,& |t|<T_1\\ 0,& T_1<|t|<T/2\end{array}\text{\hspace{1em}(3.40)}$$

is a signal that we will encounter a number of times throughout this book. This signal is periodic with fundamental period $T$ and fundamental frequency ${\omega }_0=2\pi /T$.

To determine the Fourier series coefficients for $x(t)$, we use eq. (3.38). Because of the symmetry of $x(t)$ about $t=0$, it is convenient to choose $-T/2\le t<T/2$ as the interval over which the integration is performed, although any interval of length $T$ is equally valid and thus will lead to the same result. Using these limits of integration and substituting from eq. (3.40), we have first, for $k=0$:

$$\begin{array}{rlrl}{a}_0& =\frac{1}{T}{\int }_{T}x(t)e^{-j0{\omega }_{0}t}dt& & \text{(From eq. (3.38) with }k=0\text{)}\\ & =\frac{1}{T}{\int }_{-T/2}^{T/2}x(t)dt& & \text{(Set integration limits)}\\ & =\frac{1}{T}{\int }_{-T_1}^{T_1}1\cdot dt& & \text{(Substitute }x(t)\text{ definition)}\\ & =\frac{1}{T}[t{]}_{-T_1}^{T_1}& & \\ & =\frac{1}{T}(T_1-(-T_1))& & \\ & =\frac{2T_1}{T}.& & \end{array}\text{\hspace{1em}(3.41)}$$

As mentioned previously, $a_0$ is interpreted to be the average value of $x(t)$, which in this case equals the fraction of each period during which $x(t)=1$.

For $k\ne 0$, we obtain:

$$\begin{array}{rlrl}{a}_k& =\frac{1}{T}{\int }_{-T/2}^{T/2}x(t)e^{-jk{\omega }_{0}t}dt& & \text{(From eq. (3.38))}\\ & =\frac{1}{T}{\int }_{-T_1}^{T_1}1\cdot e^{-jk{\omega }_{0}t}dt& & \text{(Substitute }x(t)\text{ definition)}\\ & =\frac{1}{T}{\left[\frac{e^{-jk{\omega }_{0}t}}{-jk{\omega }_0}\right]}_{-T_1}^{T_1}& & \text{(Evaluate integral)}\\ & =-\frac{1}{jk{\omega }_{0}T}\left(e^{-jk{\omega }_0{T}_1}-e^{jk{\omega }_0{T}_1}\right).& & \end{array}$$

which we may rewrite as:

$$\begin{array}{rlrl}{a}_k& =\frac{1}{jk{\omega }_{0}T}(e^{jk{\omega }_0{T}_1}-e^{-jk{\omega }_0{T}_1})& & \text{(Multiply by -1)}\\ & =\frac{2}{k{\omega }_{0}T}\left[\frac{e^{jk{\omega }_0{T}_1}-e^{-jk{\omega }_0{T}_1}}{2j}\right].& & \end{array}\text{\hspace{1em}(3.42)}$$

Noting that the term in brackets is $\sin k{\omega }_0{T}_1$, we can express the coefficients $a_k$ as:

$$\begin{array}{rl}{a}_k& =\frac{2\sin (k{\omega }_0{T}_1)}{k{\omega }_{0}T}\\ & =\frac{\sin (k{\omega }_0{T}_1)}{k\pi },k\ne 0,\end{array}\text{\hspace{1em}(3.43)}$$

where we have used the fact that ${\omega }_{0}T=2\pi$.

Figure 3.7 (not shown here) is a bar graph of the Fourier series coefficients for this example. In particular, the coefficients are plotted for a fixed value of $T_1$ and several values of $T$. For this specific example, the Fourier coefficients are real, and consequently, they can be depicted graphically with only a single graph. More generally, of course, the Fourier coefficients are complex, so that two graphs, corresponding to the real and imaginary parts, or magnitude and phase, of each coefficient, would be required.

For $T=4T_1$, $x(t)$ is a square wave that is unity for half the period and zero for half the period. In this case, ${\omega }_0{T}_1=(2\pi /T)T_1=(2\pi /(4T_1))T_1=\pi /2$, and from eq. (3.43):

$$a_k=\frac{\sin (\pi k/2)}{k\pi },k\ne 0,\text{\hspace{1em}(3.44)}$$

while for $k=0$:

$$a_0=\frac{2T_1}{T}=\frac{2T_1}{4T_1}=\frac{1}{2}.\text{\hspace{1em}(3.45)}$$

From eq. (3.44), $a_k=0$ for $k$ even and nonzero (e.g., $k=2,\sin (\pi )=0$). Also, $\sin (\pi k/2)$ alternates between $\pm 1$ for successive odd values of $k$. Therefore:

$$\begin{array}{rl}{a}_1& =\frac{\sin (\pi /2)}{\pi }=\frac{1}{\pi },\\ a_{-1}& =\frac{\sin (-\pi /2)}{-\pi }=\frac{-1}{-\pi }=\frac{1}{\pi },\\ a_3& =\frac{\sin (3\pi /2)}{3\pi }=\frac{-1}{3\pi },\\ a_{-3}& =\frac{\sin (-3\pi /2)}{-3\pi }=\frac{1}{-3\pi }=-\frac{1}{3\pi },\\ a_5& =\frac{\sin (5\pi /2)}{5\pi }=\frac{1}{5\pi },\dots \end{array}$$

3.4 Convergence of the Fourier Series

Although Euler and Lagrange would have been happy with the results of Examples 3.3 and 3.4, they would have objected to Example 3.5, since $x(t)$ is discontinuous while each of its harmonic components is continuous. Fourier, on the other hand, considered the same example and maintained that the Fourier series representation of the square wave is valid. In fact, Fourier maintained that any periodic signal could be represented by a Fourier series. Although this is not quite true, it is true that Fourier series can be used to represent an extremely large class of periodic signals, including the square wave and all other periodic signals with which we will be concerned in this book and which are of interest in practice.

To gain an understanding of the square-wave example and, more generally, of the question of the validity of Fourier series representations, let us examine the problem of approximating a given periodic signal $x(t)$ by a linear combination of a finite number of harmonically related complex exponentials—that is, by a finite series of the form

$$x_N(t)=\sum _{k=-N}^N{a}_k{e}^{jk{\omega }_{0}t}.\text{\hspace{1em}(3.46)}$$

Let $e_N(t)$ denote the approximation error; that is,

$$e_N(t)=x(t)-x_N(t)=x(t)-\sum _{k=-N}^{+N}{a}_k{e}^{jk{\omega }_{0}t}.\text{\hspace{1em}(3.47)}$$

In order to determine how good any particular approximation is, we need to specify a quantitative measure of the size of the approximation error. The criterion that we will use is the energy in the error over one period:

$$E_N={\int }_T|e_N(t){|}^{2}dt.\text{\hspace{1em}(3.48)}$$

As shown in Problem 3.66, the particular choice for the coefficients in eq. (3.46) that minimize the energy in the error is

$$a_k=\frac{1}{T}{\int }_{T}x(t)e^{-jk{\omega }_{0}t}dt.\text{\hspace{1em}(3.49)}$$

Comparing eqs. (3.49) and (3.38), we see that eq. (3.49) is identical to the expression used to determine the Fourier series coefficients. Thus, if $x(t)$ has a Fourier series representation, the best approximation using only a finite number of harmonically related complex exponentials is obtained by truncating the Fourier series to the desired number of terms. As $N$ increases, new terms are added and $E_N$ decreases. If, in fact, $x(t)$ has a Fourier series representation, then the limit of $E_N$ as $N\to \infty$ is zero.

Let us turn now to the question of when a periodic signal $x(t)$ does in fact have a Fourier series representation. Of course, for any signal, we can attempt to obtain a set of Fourier coefficients through the use of eq. (3.38). However, in some cases, the integral in eq. (3.38) may diverge; that is, the value obtained for some of the $a_k$ may be infinite. Moreover, even if all of the coefficients obtained from eq. (3.38) are finite, when these coefficients are substituted into the synthesis equation (3.37), the resulting infinite series may not converge to the original signal $x(t)$.

Fortunately, there are no convergence difficulties for large classes of periodic signals. For example, every continuous periodic signal has a Fourier series representation for which the energy $E_N$ in the approximation error approaches 0 as $N$ goes to $\infty$. This is also true for many discontinuous signals. Since we will find it very useful to include discontinuous signals such as square waves in our discussions, it is worthwhile to investigate the issue of convergence in a bit more detail. Specifically, there are two somewhat different classes of conditions that a periodic signal can satisfy to guarantee that it can be represented by a Fourier series. In discussing these, we will not attempt to provide a complete mathematical justification; more rigorous treatments can be found in many texts on Fourier analysis.

One class of periodic signals that are representable through the Fourier series is those signals which have finite energy over a single period, i.e., signals for which

$${\int }_T|x(t){|}^{2}dt<\infty .\text{\hspace{1em}(3.50)}$$

When this condition is satisfied, we are guaranteed that the coefficients $a_k$ obtained from eq. (3.38) are finite. Furthermore, let $x_N(t)$ be the approximation to $x(t)$ obtained by using these coefficients for $|k|\le N$:

$$x_N(t)=\sum _{k=-N}^{+N}{a}_k{e}^{jk{\omega }_{0}t}.\text{\hspace{1em}(3.51)}$$

Then we are guaranteed that the energy $E_N$ in the approximation error, as defined in eq. (3.48), converges to 0 as we add more and more terms, i.e., as $N\to \infty$. That is, if we define

$$e(t)=x(t)-\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t},\text{\hspace{1em}(3.52)}$$

then

$${\int }_T|e(t){|}^{2}dt=0.\text{\hspace{1em}(3.53)}$$

As we will see in an example at the end of this section, eq. (3.53) does not imply that the signal $x(t)$ and its Fourier series representation

$$\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}\text{\hspace{1em}(3.54)}$$

are equal at every value of $t$. What it does say is that there is no energy in their difference.

The type of convergence guaranteed when $x(t)$ has finite energy over a single period is quite useful. In this case, eq. (3.53) states that the difference between $x(t)$ and its Fourier series representation has zero energy. Since physical systems respond to signal energy, from this perspective $x(t)$ and its Fourier series representation are indistinguishable. Because most of the periodic signals that we consider do have finite energy over a single period, they have Fourier series representations. Moreover, an alternative set of conditions, developed by P. L. Dirichlet and also satisfied by essentially all of the signals with which we will be concerned, guarantees that $x(t)$ equals its Fourier series representation, except at isolated values of $t$ for which $x(t)$ is discontinuous. At these values, the infinite series of eq. (3.54) converges to the average of the values on either side of the discontinuity.

The Dirichlet conditions are as follows:

Condition 1. Over any period, $x(t)$ must be absolutely integrable; that is,

$${\int }_T|x(t)|dt<\infty .\text{\hspace{1em}(3.55)}$$

As with square integrability, this guarantees that each coefficient $a_k$ will be finite, since

$$|a_k|=\left|\frac{1}{T}{\int }_{T}x(t)e^{-jk{\omega }_{0}t}dt\right|\le \frac{1}{T}{\int }_T|x(t)e^{-jk{\omega }_{0}t}|dt=\frac{1}{T}{\int }_T|x(t)|\cdot |e^{-jk{\omega }_{0}t}|dt=\frac{1}{T}{\int }_T|x(t)|dt.$$

So if ${\int }_T|x(t)|dt<\infty$, then $|a_k|<\infty$. A periodic signal that violates the first Dirichlet condition is $x(t)=1/t$ for $0<t\le 1$, periodic with period 1. This signal is illustrated in Figure 3.8(a) (not shown here).

Condition 2. In any finite interval of time, $x(t)$ is of bounded variation; that is, there are no more than a finite number of maxima and minima during any single period of the signal. An example of a function that meets Condition 1 but not Condition 2 is

$$x(t)=\sin \left(\frac{2\pi }{t}\right),0<t\le 1,\text{\hspace{1em}(3.56)}$$

as illustrated in Figure 3.8(b) (not shown here). For this function, which is periodic with $T=1$, ${\int }_0^1|x(t)|dt<1$. The function has, however, an infinite number of maxima and minima in the interval.

Condition 3. In any finite interval of time, there are only a finite number of discontinuities. Furthermore, each of these discontinuities is finite. An example of a function that violates Condition 3 is illustrated in Figure 3.8(c) (not shown here). The signal, of period $T=8$, is composed of an infinite number of sections, each of which is half the height and half the width of the previous section. Thus, the area under one period of the function is clearly less than 8. However, there are an infinite number of discontinuities in each period, thereby violating Condition 3.

As can be seen from the examples given in Figure 3.8, signals that do not satisfy the Dirichlet conditions are generally pathological in nature and consequently do not typically arise in practical contexts. For this reason, the question of the convergence of Fourier series will not play a particularly significant role in the remainder of the book. For a periodic signal that has no discontinuities, the Fourier series representation converges and equals the original signal at every value of $t$. For a periodic signal with a finite number of discontinuities in each period, the Fourier series representation equals the signal everywhere except at the isolated points of discontinuity, at which the series converges to the average value of the signal on either side of the discontinuity. In this case, the difference between the original signal and its Fourier series representation contains no energy, and consequently, the two signals can be thought of as being the same for all practical purposes. Specifically, since the signals differ only at isolated points, the integrals of both signals over any interval are identical. For this reason, the two signals behave identically under convolution and consequently are identical from the standpoint of the analysis of LTI systems.

To gain some additional understanding of how the Fourier series converges for a periodic signal with discontinuities, let us return to the example of a square wave. In particular, in 1898, an American physicist, Albert Michelson, constructed a harmonic analyzer, a device that, for any periodic signal $x(t)$, would compute the truncated Fourier series approximation of eq. (3.46) for values of $N$ up to 80. Michelson tested his device on many functions, with the expected result that $x_N(t)$ looked very much like $x(t)$. However, when he tried the square wave, he obtained an important and, to him, very surprising result. Michelson was concerned about the behavior he observed and thought that his device might have had a defect. He wrote about the problem to the famous mathematical physicist Josiah Gibbs, who investigated it and reported his explanation in 1899.

What Michelson had observed is illustrated in Figure 3.9 (not shown here), where we have shown $x_N(t)$ for several values of $N$ for $x(t)$, a symmetric square wave $(T=4T_1)$. In each case, the partial sum is superimposed on the original square wave. Since the square wave satisfies the Dirichlet conditions, the limit as $N\to \infty$ of $x_N(t)$ at the discontinuities should be the average value of the discontinuity. We see from the figure that this is in fact the case, since for any $N$, $x_N(t)$ has exactly that value at the discontinuities. Furthermore, for any other value of $t$, say, $t=t_1$, we are guaranteed that ${\lim }_{N\to \infty }{x}_N(t_1)=x(t_1)$. Therefore, the squared error in the Fourier series representation of the square wave has zero area, as in eqs. (3.52) and (3.53).

For this example, the interesting effect that Michelson observed is that the behavior of the partial sum in the vicinity of the discontinuity exhibits ripples and that the peak amplitude of these ripples does not seem to decrease with increasing $N$. Gibbs showed that these are in fact the case. Specifically, for a discontinuity of unity height, the partial sum exhibits a maximum value of 1.09 (i.e., an overshoot of $9\%$ of the height of the discontinuity), no matter how large $N$ becomes. One must be careful to interpret this correctly, however. As stated before, for any fixed value of $t$, say, $t=t_1$, the partial sums will converge to the correct value, and at the discontinuity, they will converge to one-half the sum of the values of the signal on either side of the discontinuity. However, the closer $t_1$ is chosen to the point of discontinuity, the larger $N$ must be in order to reduce the error below a specified amount. Thus, as $N$ increases, the ripples in the partial sums become compressed toward the discontinuity, but for any finite value of $N$, the peak amplitude of the ripples remains constant. This behavior has come to be known as the Gibbs phenomenon. The implication is that the truncated Fourier series approximation $x_N(t)$ of a discontinuous signal $x(t)$ will in general exhibit high-frequency ripples and overshoot $x(t)$ near the discontinuities. If such an approximation is used in practice, a large enough value of $N$ should be chosen so as to guarantee that the total energy in these ripples is insignificant. In the limit, of course, we know that the energy in the approximation error vanishes and that the Fourier series representation of a discontinuous signal such as the square wave converges.

3.5 Properties of Continuous-Time Fourier Series

As mentioned earlier, Fourier series representations possess a number of important properties that are useful for developing conceptual insights into such representations, and they can also help to reduce the complexity of the evaluation of the Fourier series of many signals. In Table 3.1, we have summarized these properties, several of which are considered in the problems at the end of this chapter. In Chapter 4, in which we develop the Fourier transform, we will see that most of these properties can be deduced from corresponding properties of the continuous-time Fourier transform. Consequently, we limit ourselves here to the discussion of several of these properties to illustrate how they may be derived, interpreted, and used.

Throughout the following discussion of selected properties from Table 3.1, we will find it convenient to use a shorthand notation to indicate the relationship between a periodic signal and its Fourier series coefficients. Specifically, suppose that $x(t)$ is a periodic signal with period $T$ and fundamental frequency ${\omega }_0=2\pi /T$. Then if the Fourier series coefficients of $x(t)$ are denoted by $a_k$, we will use the notation

$$x(t)\stackrel{\mathrm{FS}}{⟷}{a}_k$$

to signify the pairing of a periodic signal with its Fourier series coefficients.

3.5.1 Linearity

Let $x(t)$ and $y(t)$ denote two periodic signals with period $T$ and which have Fourier series coefficients denoted by $a_k$ and $b_k$, respectively. That is,

$$x(t)\stackrel{\mathrm{FS}}{⟷}{a}_k,y(t)\stackrel{\mathrm{FS}}{⟷}{b}_k.$$

Since $x(t)$ and $y(t)$ have the same period $T$, it easily follows that any linear combination of the two signals will also be periodic with period $T$. Furthermore, the Fourier series coefficients $c_k$ of the linear combination of $x(t)$ and $y(t)$, $z(t)=Ax(t)+By(t)$, are given by the same linear combination of the Fourier series coefficients for $x(t)$ and $y(t)$. That is,

$$z(t)=Ax(t)+By(t)\stackrel{\mathrm{FS}}{⟷}{c}_k=A{a}_k+B{b}_k.\text{\hspace{1em}(3.57)}$$

The proof of this follows directly from the application of eq. (3.38):

$$\begin{array}{rlrl}{c}_k& =\frac{1}{T}{\int }_{T}z(t)e^{-jk{\omega }_{0}t}dt& & \text{(Definition of Fourier coefficients)}\\ & =\frac{1}{T}{\int }_T(Ax(t)+By(t))e^{-jk{\omega }_{0}t}dt& & \text{(Substitute }z(t)=Ax(t)+By(t)\text{)}\\ & =\frac{1}{T}{\int }_{T}Ax(t)e^{-jk{\omega }_{0}t}dt+\frac{1}{T}{\int }_{T}By(t)e^{-jk{\omega }_{0}t}dt& & \text{(Linearity of integral)}\\ & =A\left(\frac{1}{T}{\int }_{T}x(t)e^{-jk{\omega }_{0}t}dt\right)+B\left(\frac{1}{T}{\int }_{T}y(t)e^{-jk{\omega }_{0}t}dt\right)& & \text{(Factor out constants)}\\ & =A{a}_k+B{b}_k.& & \text{(By definition of }{a}_k\text{ and }{b}_k\text{)}\end{array}$$

We also note that the linearity property is easily extended to a linear combination of an arbitrary number of signals with period $T$.

3.5.2 Time Shifting

When a time shift is applied to a periodic signal $x(t)$, the period $T$ of the signal is preserved. The Fourier series coefficients $b_k$ of the resulting signal $y(t)=x(t-t_0)$ may be expressed as:

$$b_k=\frac{1}{T}{\int }_{T}x(t-t_0)e^{-jk{\omega }_{0}t}dt.\text{\hspace{1em}(3.58)}$$

Letting $\tau =t-t_0$ in the integral, so $t=\tau +t_0$ and $dt=d\tau$. Noting that the new variable $\tau$ will also range over an interval of duration $T$, we obtain:

$$\begin{array}{rlrl}{b}_k& =\frac{1}{T}{\int }_{T}x(\tau )e^{-jk{\omega }_0(\tau +t_0)}d\tau & & \text{(Substitute }\tau =t-t_0\text{)}\\ & =\frac{1}{T}{\int }_{T}x(\tau )e^{-jk{\omega }_0\tau }{e}^{-jk{\omega }_0{t}_0}d\tau & & \text{(Factor exponential)}\\ & =e^{-jk{\omega }_0{t}_0}\left(\frac{1}{T}{\int }_{T}x(\tau )e^{-jk{\omega }_0\tau }d\tau \right)& & \text{(Move constant }{e}^{-jk{\omega }_0{t}_0}\text{ out of integral)}\\ & =e^{-jk{\omega }_0{t}_0}{a}_k& & \text{(By definition of }{a}_k\text{)}\\ & =e^{-jk(2\pi /T)t_0}{a}_k.& & \text{(Substitute }{\omega }_0=2\pi /T\text{)}\end{array}\text{\hspace{1em}(3.59)}$$

That is, if

$$x(t)\stackrel{\mathrm{FS}}{⟷}{a}_k,$$

then

$$x(t-t_0)\stackrel{\mathrm{FS}}{⟷}{e}^{-jk{\omega }_0{t}_0}{a}_k=e^{-jk(2\pi /T)t_0}{a}_k.$$

One consequence of this property is that, when a periodic signal is shifted in time, the magnitudes of its Fourier series coefficients remain unaltered. That is, $|b_k|=|e^{-jk{\omega }_0{t}_0}{a}_k|=|e^{-jk{\omega }_0{t}_0}||a_k|=1\cdot |a_k|=|a_k|$.

3.5.3 Time Reversal

The period $T$ of a periodic signal $x(t)$ also remains unchanged when the signal undergoes time reversal. To determine the Fourier series coefficients of $y(t)=x(-t)$, let us consider the effect of time reversal on the synthesis equation (3.37):

$$\begin{array}{rlrl}y(t)& =x(-t)& & \\ & =\sum _{k=-\infty }^{\infty }{a}_k{e}^{jk{\omega }_0(-t)}& & \text{(Substitute }-t\text{ into the synthesis equation)}\\ & =\sum _{k=-\infty }^{\infty }{a}_k{e}^{-jk{\omega }_{0}t}.& & \end{array}\text{\hspace{1em}(3.60)}$$

Making the substitution $k=-m$ (so $m=-k$), and as $k$ goes from $-\infty$ to $\infty$, $m$ also goes from $-\infty$ to $\infty$:

$$\begin{array}{rlrl}y(t)& =\sum _{m=-\infty }^{\infty }{a}_{-m}{e}^{j(-(-m)){\omega }_{0}t}& & \text{(Substitute }k=-m\text{)}\\ & =\sum _{m=-\infty }^{\infty }{a}_{-m}{e}^{jm{\omega }_{0}t}.& & \end{array}\text{\hspace{1em}(3.61)}$$

We observe that the right-hand side of this equation has the form of a Fourier series synthesis equation for $x(-t)$, where the Fourier series coefficients $b_k$ (replacing $m$ with $k$ again for clarity) are:

$$b_k=a_{-k}.\text{\hspace{1em}(3.62)}$$

That is, if

$$x(t)\stackrel{\mathrm{FS}}{⟷}{a}_k,$$

then

$$x(-t)\stackrel{\mathrm{FS}}{⟷}{a}_{-k}.$$

In other words, time reversal applied to a continuous-time signal results in a time reversal of the corresponding sequence of Fourier series coefficients. An interesting consequence of the time-reversal property is that if $x(t)$ is even—that is, if $x(-t)=x(t)$—then its Fourier series coefficients are also even—i.e., $a_{-k}=a_k$. Similarly, if $x(t)$ is odd, so that $x(-t)=-x(t)$, then so are its Fourier series coefficients—i.e., $a_{-k}=-a_k$.

3.5.4 Time Scaling

Time scaling is an operation that in general changes the period of the underlying signal. Specifically, if $x(t)$ is periodic with period $T$ and fundamental frequency ${\omega }_0=2\pi /T$, then $x(\alpha t)$, where $\alpha$ is a positive real number, is periodic with period $T/\alpha$ and fundamental frequency $\alpha {\omega }_0$. Since the time-scaling operation applies directly to each of the harmonic components of $x(t)$, we may easily conclude that the Fourier coefficients for each of those components remain the same. That is, if $x(t)$ has the Fourier series representation in eq. (3.37):

$$x(t)=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}.$$

Then, $x(\alpha t)$ is:

$$x(\alpha t)=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_0(\alpha t)}=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk(\alpha {\omega }_0)t}.$$

This is the Fourier series representation of $x(\alpha t)$, with fundamental frequency $\alpha {\omega }_0$ and coefficients $a_k$. We emphasize that, while the Fourier coefficients have not changed, the Fourier series representation has changed because of the change in the fundamental frequency.

3.5.5 Multiplication

Suppose that $x(t)$ and $y(t)$ are both periodic with period $T$ and that

$$x(t)\stackrel{\mathrm{FS}}{⟷}{a}_k,y(t)\stackrel{\mathrm{FS}}{⟷}{b}_k.$$

Since the product $x(t)y(t)$ is also periodic with period $T$, we can expand it in a Fourier series with Fourier series coefficients $h_k$ expressed in terms of those for $x(t)$ and $y(t)$. The result is

$$x(t)y(t)\stackrel{\mathrm{FS}}{⟷}{h}_k=\sum _{l=-\infty }^{\infty }{a}_l{b}_{k-l}.\text{\hspace{1em}(3.63)}$$

One way to derive this relationship (see Problem 3.46) is to multiply the Fourier series representations of $x(t)$ and $y(t)$ and to note that the $k$th harmonic component in the product will have a coefficient which is the sum of terms of the form $a_l{b}_{k-l}$. Observe that the sum on the right-hand side of eq. (3.63) may be interpreted as the discrete-time convolution of the sequence representing the Fourier coefficients of $x(t)$ and the sequence representing the Fourier coefficients of $y(t)$.

3.5.6 Conjugation and Conjugate Symmetry

Taking the complex conjugate of a periodic signal $x(t)$ has the effect of complex conjugation and time reversal on the corresponding Fourier series coefficients. That is, if

$$x(t)\stackrel{\mathrm{FS}}{⟷}{a}_k,\text{\hspace{1em}(3.64)}$$

then

$$x^{\ast }(t)\stackrel{\mathrm{FS}}{⟷}{a}_{-k}^{\ast }.\text{\hspace{1em}(3.65)}$$

This property is easily proved by applying complex conjugation to both sides of eq. (3.37) and replacing the summation variable $k$ by its negative. Proof: Let $x(t)={\sum }_{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}$. Then $x^{\ast }(t)={\left({\sum }_{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}\right)}^{\ast }={\sum }_{k=-\infty }^{+\infty }{a}_k^{\ast }(e^{jk{\omega }_{0}t}{)}^{\ast }={\sum }_{k=-\infty }^{+\infty }{a}_k^{\ast }{e}^{-jk{\omega }_{0}t}$. Now, let $m=-k$. As $k$ goes from $-\infty$ to $\infty$, so does $m$. $x^{\ast }(t)={\sum }_{m=-\infty }^{+\infty }{a}_{-m}^{\ast }{e}^{jm{\omega }_{0}t}$. By inspection, the Fourier series coefficients for $x^{\ast }(t)$ are $a_{-k}^{\ast }$.

Some interesting consequences of this property may be derived for $x(t)$ real—that is, when $x(t)=x^{\ast }(t)$. In particular, in this case, we see from eq. (3.65) that the Fourier series coefficients will be conjugate symmetric, i.e.,

$$a_{-k}=a_k^{\ast },\text{\hspace{1em}(3.66)}$$

as we previously saw in eq. (3.28). This in turn implies various symmetry properties (listed in Table 3.1) for the magnitudes, phases, real parts, and imaginary parts of the Fourier series coefficients of real signals. For example, from eq. (3.66), we see that if $x(t)$ is real, then $a_0$ is real ($a_0=a_0^{\ast }$) and

$$|a_k|=|a_{-k}|.\text{(Since }|a_k|=|a_{-k}^{\ast }|=|a_{-k}|\text{)}$$

Also, if $x(t)$ is real and even—that is, $x(t)=x(-t)$—then, from Section 3.5.3, $a_k=a_{-k}$. However, from eq. (3.66), we see that $a_k^{\ast }=a_{-k}$, so that $a_k=a_k^{\ast }$. This implies that $a_k$ must be real. That is, if $x(t)$ is real and even, then so are its Fourier series coefficients. Similarly, it can be shown that if $x(t)$ is real and odd—that is, $x(t)=-x(-t)$—then its Fourier series coefficients are purely imaginary and odd. Thus, for example, $a_0=0$ if $x(t)$ is real and odd. This and the other symmetry properties of the Fourier series are examined further in Problem 3.42.

3.5.7 Parseval’s Relation for Continuous-Time Periodic Signals

As shown in Problem 3.46, Parseval’s relation for continuous-time periodic signals is

$$\frac{1}{T}{\int }_T|x(t){|}^{2}dt=\sum _{k=-\infty }^{+\infty }|a_k{|}^2,\text{\hspace{1em}(3.67)}$$

where the $a_k$ are the Fourier series coefficients of $x(t)$ and $T$ is the period of the signal.

Note that the left-hand side of eq. (3.67) is the average power (i.e., energy per unit time) in one period of the periodic signal $x(t)$. Also, for each harmonic component $a_k{e}^{jk{\omega }_{0}t}$:

$$\frac{1}{T}{\int }_T{\left|a_k{e}^{jk{\omega }_{0}t}\right|}^{2}dt=\frac{1}{T}{\int }_T|a_k{|}^2|e^{jk{\omega }_{0}t}{|}^{2}dt=\frac{1}{T}{\int }_T|a_k{|}^2\cdot 1dt=|a_k{|}^2.\text{\hspace{1em}(3.68)}$$

so that $|a_k{|}^2$ is the average power in the $k$th harmonic component of $x(t)$. Thus, what Parseval’s relation states is that the total average power in a periodic signal equals the sum of the average powers in all of its harmonic components.

3.5.8 Summary of Properties of the Continuous-Time Fourier Series

In Table 3.1, we summarize these and other important properties of continuous-time Fourier series.

Table 3.1 PROPERTIES OF CONTINUOUS-TIME FOURIER SERIES

Property	Section	Periodic Signal	Fourier Series Coefficients
Linearity	3.5.1	$Ax(t)+By(t)$	$A{a}_k+B{b}_k$
Time Shifting	3.5.2	$x(t-t_0)$	$a_k{e}^{-jk{\omega }_0{t}_0}=a_k{e}^{-jk(2\pi /T)t_0}$
Frequency Shifting		$e^{jM{\omega }_{0}t}x(t)=e^{jM(2\pi /T)t}x(t)$	$a_{k-M}$
Conjugation	3.5.6	$x^{\ast }(t)$	$a_{-k}^{\ast }$
Time Reversal	3.5.3	$x(-t)$	$a_{-k}$
Time Scaling	3.5.4	$x(\alpha t),\alpha >0$ (periodic with period $T/\alpha$)	$a_k$
Periodic Convolution		${\int }_{T}x(\tau )y(t-\tau )d\tau$	$T{a}_k{b}_k$
Multiplication	3.5.5	$x(t)y(t)$	${\sum }_{l=-\infty }^{\infty }{a}_l{b}_{k-l}$
Differentiation		$\frac{dx(t)}{dt}$	$jk{\omega }_0{a}_k=jk\frac{2\pi }{T}{a}_k$
Integration		${\int }_{-\infty }^{t}x(\tau )d\tau$ (finite valued and periodic only if $a_0=0$)	$\left(\frac{1}{jk{\omega }_0}\right)a_k=\left(\frac{1}{jk(2\pi /T)}\right)a_k$
Conjugate Symmetry for Real Signals	3.5.6	$x(t)$ real	$a_k=a_{-k}^{\ast }$
			$\Re \{a_k\}=\Re \{a_{-k}\}$
			$\Im \{a_k\}=-\Im \{a_{-k}\}$
			$	a_k	=	a_{-k}	$
			$\angle a_k=-\angle a_{-k}$
Real and Even Signals	3.5.6	$x(t)$ real and even	$a_k$ real and even
Real and Odd Signals	3.5.6	$x(t)$ real and odd	$a_k$ purely imaginary and odd
Even-Odd Decomposition of Real Signals		$x_e(t)=\mathcal{E}v\{x(t)\}$ [$x(t)$ real]	$\Re \{a_k\}$
		$x_o(t)=\mathcal{O}d\{x(t)\}$ [$x(t)$ real]	$j\Im \{a_k\}$
Parseval’s Relation	3.5.7	$\frac{1}{T}{\int }_T$	x(t)	${}^{2}dt$	${\sum }_{k=-\infty }^{\infty }$	a_k	^2$

3.5.9 Examples

Fourier series properties, such as those listed in Table 3.1, may be used to circumvent some of the algebra involved in determining the Fourier coefficients of a given signal. In the next three examples, we illustrate this. The last example in this section then demonstrates how properties of a signal can be used to characterize the signal in great detail.

Example 3.6

Consider the signal $g(t)$ with a fundamental period of 4, shown in Figure 3.10 (not shown here). We could determine the Fourier series representation of $g(t)$ directly from the analysis equation (3.38). Instead, we will use the relationship of $g(t)$ to the symmetric periodic square wave $x(t)$ in Example 3.5.

From Example 3.5, for a square wave $x(t)$ with period $T=4$ and $T_1=1$, the Fourier coefficients are given by: $a_0=\frac{2T_1}{T}=\frac{2(1)}{4}=\frac{1}{2}$. For $k\ne 0$, $a_k=\frac{\sin (k{\omega }_0{T}_1)}{k\pi }=\frac{\sin (k(2\pi /4)(1))}{k\pi }=\frac{\sin (k\pi /2)}{k\pi }$.

Referring to Figure 3.10, we see that $g(t)$ is a shifted version of $x(t)$ in Example 3.5, and also has a DC offset. The square wave $x(t)$ from Example 3.5 is centered at $t=0$, meaning it’s 1 for $|t|<T_1$ and 0 otherwise within a period. The signal $g(t)$ is 1 for $0\le t<2$ and 0 for $2\le t<4$. If we shift $x(t)$ to the right by $1$, we get $x(t-1)$, which is 1 for $|t-1|<1$, i.e., $0<t<2$. This matches the “on” part of $g(t)$. The average value of $x(t)$ is $a_0=1/2$. The average value of $g(t)$ is also $1/2$ (area of 1 for duration 2, divided by period 4). So, if $x(t)$ is the square wave of Example 3.5 with $T=4,T_1=1$, then:

$$g(t)=x(t-1)-1/2.\text{\hspace{1em}(3.69)}$$

The time-shift property in Table 3.1 indicates that, if the Fourier Series coefficients of $x(t)$ are denoted by $a_k$, the Fourier coefficients of $x(t-1)$ may be expressed as:

$$b_k=a_k{e}^{-jk{\omega }_0{t}_0}.$$

Here, $t_0=1$ and ${\omega }_0=2\pi /T=2\pi /4=\pi /2$. So:

$$b_k=a_k{e}^{-jk\pi /2}.\text{\hspace{1em}(3.70)}$$

The Fourier coefficients $c_k$ of the dc offset in $g(t)$ – i.e., the term $-1/2$ on the right-hand side of eq. (3.69) – are given by:

$$c_k=\{\begin{array}{ll}-1/2,& \mathrm{for}k=0\\ 0,& \mathrm{for}k\ne 0\end{array}.\text{\hspace{1em}(3.71)}$$

Applying the linearity property in Table 3.1, we conclude that the coefficients for $g(t)$, denoted $d_k$, may be expressed as:

$$d_k=b_k+c_k=a_k{e}^{-jk\pi /2}+c_k.$$

Now, substitute the expressions for $a_k$ from Example 3.5: For $k=0$:

$$d_0=a_0{e}^{-j0}+c_0=\frac{1}{2}+\left(-\frac{1}{2}\right)=0.$$

For $k\ne 0$:

$$\begin{array}{rl}{d}_k& =\left(\frac{\sin (k\pi /2)}{k\pi }\right)e^{-jk\pi /2}+0\\ & =\frac{\sin (k\pi /2)}{k\pi }{e}^{-jk\pi /2}.\end{array}\text{\hspace{1em}(3.72)}$$

Combining these, the Fourier series coefficients for $g(t)$ are:

$$d_k=\{\begin{array}{ll}\frac{\sin (\pi k/2)}{k\pi }{e}^{-jk\pi /2},& \mathrm{for}k\ne 0\\ 0,& \mathrm{for}k=0\end{array}.$$

Example 3.7

Consider the triangular wave signal $x(t)$ with period $T=4$ and fundamental frequency ${\omega }_0=\pi /2$ shown in Figure 3.11 (not shown here). The derivative of this signal, $dx(t)/dt$, is the signal $g(t)$ from Example 3.6. Denoting the Fourier coefficients of $g(t)$ by $d_k$ and those of $x(t)$ by $e_k$, we see that the differentiation property in Table 3.1 indicates that:

$$d_k=jk{\omega }_0{e}_k.$$

Given ${\omega }_0=\pi /2$:

$$d_k=jk(\pi /2)e_k.\text{\hspace{1em}(3.73)}$$

This equation can be used to express $e_k$ in terms of $d_k$, except when $k=0$. For $k\ne 0$:

$$\begin{array}{rlrl}{e}_k& =\frac{d_k}{jk(\pi /2)}& & \text{(Rearrange eq. (3.73))}\\ & =\frac{2}{jk\pi }{d}_k& & \\ & =\frac{2}{jk\pi }\left(\frac{\sin (k\pi /2)}{k\pi }{e}^{-jk\pi /2}\right)& & \text{(Substitute }{d}_k\text{ from Example 3.6)}\\ & =\frac{2\sin (\pi k/2)}{j(k\pi {)}^2}{e}^{-jk\pi /2},k\ne 0.& & \end{array}\text{\hspace{1em}(3.74)}$$

For $k=0$, $e_0$ cannot be found from the differentiation property, as it would involve division by zero. Instead, $e_0$ is the average value of $x(t)$ over one period. From Figure 3.11, $x(t)$ is a triangular wave varying from $0$ to $1$ (at $t=0$) and back to $0$. The average value is the area under one period divided by the period length. The area of the triangle from $t=-2$ to $t=2$ (one period $T=4$) is $\frac{1}{2}\times \text{base}\times \text{height}=\frac{1}{2}\times 4\times 1=2$. Thus:

$$e_0=\frac{1}{T}{\int }_{T}x(t)dt=\frac{1}{4}(2)=\frac{1}{2}.$$

Example 3.8

Let us examine some properties of the Fourier series representation of a periodic train of impulses, or impulse train. This signal and its representation in terms of complex exponentials will play an important role when we discuss the topic of sampling in Chapter 7. The impulse train with period $T$ may be expressed as

$$x(t)=\sum _{k=-\infty }^{\infty }\delta (t-kT);\text{\hspace{1em}(3.75)}$$

it is illustrated in Figure 3.12(a) (not shown here). To determine the Fourier series coefficients $a_k$, we use eq. (3.38) and select the interval of integration to be $-T/2\le t\le T/2$, avoiding the placement of impulses at the integration limits. Within this interval, $x(t)$ is the same as $\delta (t)$, and it follows that:

$$\begin{array}{rlrl}{a}_k& =\frac{1}{T}{\int }_{T}x(t)e^{-jk{\omega }_{0}t}dt& & \\ & =\frac{1}{T}{\int }_{-T/2}^{T/2}\delta (t)e^{-jk(2\pi /T)t}dt& & \text{(Substitute }x(t)=\delta (t)\text{ in the integration interval)}\\ & =\frac{1}{T}{e}^{-jk(2\pi /T)(0)}& & \text{(Property of impulse function: }\int f(t)\delta (t-t_0)dt=f(t_0)\text{)}\\ & =\frac{1}{T}\cdot 1& & \\ & =\frac{1}{T}.& & \end{array}\text{\hspace{1em}(3.76)}$$

In other words, all the Fourier series coefficients of the impulse train are identical. These coefficients are also real valued and even (with respect to the index $k$). This is to be expected, since, according to Table 3.1, any real and even signal (such as our impulse train) should have real and even Fourier coefficients.

The impulse train also has a straightforward relationship to square-wave signals such as $g(t)$ in Figure 3.6, which we repeat in Figure 3.12(b) (not shown here). The derivative of $g(t)$ is the signal $q(t)$ illustrated in Figure 3.12(c) (not shown here). We may interpret $q(t)$ as the difference of two shifted versions of the impulse train $x(t)$. That is, $g(t)$ is a square wave from $T_1$ to $T-T_1$ (length $T-2T_1$) for $x(t)=1$ and 0 otherwise over period $T$. In the figure, $g(t)$ is 1 for $T_1<t<T-T_1$ and 0 otherwise. Its derivative $q(t)$ has an impulse at $t=T_1$ (falling edge) and another at $t=T-T_1$ (rising edge). More precisely, it would be a positive impulse at the start of the pulse and a negative impulse at the end. Assuming $g(t)$ from Example 3.5 is 1 for $|t|<T_1$ and 0 otherwise, its derivative would be $q(t)=\delta (t+T_1)-\delta (t-T_1)$. But the figure depicts it as $g(t)$ as 1 for $T_1<t<T-T_1$. Let’s assume the text means the square wave from Figure 3.6 (Example 3.5), i.e., $x(t)=1$ for $|t|<T_1$ and $0$ otherwise. Its derivative would be an impulse at $-T_1$ and a negative impulse at $T_1$. The figure 3.12(b) shows $g(t)$ as 1 from $T_1$ to $T-T_1$ and zero elsewhere in a period, so its derivative has a positive impulse at $T_1$ and a negative impulse at $T-T_1$. The text states $q(t)=x(t+T_1)-x(t-T_1)$. This implies $x(t)$ is a generic impulse train (3.75). If $x(t)$ is the impulse train ${\sum }_k\delta (t-kT)$, then $x(t+T_1)$ is ${\sum }_k\delta (t+T_1-kT)$ and $x(t-T_1)$ is ${\sum }_k\delta (t-T_1-kT)$. So, $q(t)={\sum }_k\delta (t+T_1-kT)-{\sum }_k\delta (t-T_1-kT)$. Consider one period $[-T/2,T/2)$. If $T_1<T/2$, then $q(t)$ has impulses at $-T_1$ and $T_1$ (assuming $kT=0$ term): $q(t)=\delta (t+T_1)-\delta (t-T_1)$ for $t\in [-T/2,T/2)$. This is the derivative of a square pulse that goes from $0$ to $1$ at $t=-T_1$ and from $1$ to $0$ at $t=T_1$. This is precisely the $g(t)$ in Example 3.5 if $g(t)$ is defined as $1$ for $t\in [-T_1,T_1]$ and $0$ otherwise. Let’s assume this definition (which is indeed Example 3.5). Then $q(t)=\frac{d}{dt}\left(\text{rect}\left(\frac{t}{2T_1}\right)\ast {\sum }_{k=-\infty }^{\infty }\delta (t-kT)\right)$ (this is not entirely correct as square wave is 1 for $|t|<T_1$ and 0 otherwise) Let’s stick to the text’s definition: $q(t)$ is the derivative of the square wave $g(t)$ (Example 3.5, eq 3.40). The derivative of $g(t)$ (which is 1 for $|t|<T_1$ and 0 otherwise) would be $\delta (t+T_1)-\delta (t-T_1)$ (assuming a period centered at 0). The problem statement gives $q(t)=x(t+T_1)-x(t-T_1)$ where $x(t)$ is the impulse train. This implies positive impulses at $t=-T_1+kT$ and negative impulses at $t=T_1+kT$. This indeed corresponds to the derivative of a square wave whose rising edge is at $t=-T_1$ and falling edge at $t=T_1$. This is the specific square wave $x(t)$ from Example 3.5. So, $q(t)$ is the derivative of $x(t)$ from Example 3.5. Let’s call the square wave from Example 3.5 as $x_{sq}(t)$. Then $q(t)=d{x}_{sq}(t)/dt$. However, the text directly says $q(t)=x(t+T_1)-x(t-T_1)$ where $x(t)$ is the impulse train from (3.75). This is a different signal. The figure 3.12(c) also reflects this: a positive impulse at $-T_1$ and a negative impulse at $T_1$ in each period, representing $q(t)$ as the derivative of the rectangular pulse shown in Figure 3.12(b). Okay, the text states $g(t)$ in Figure 3.12(b) has $T_1$ to $T-T_1$. Let’s reconsider the definition of $g(t)$ in Figure 3.12(b). It is 1 for $T_1<t<T-T_1$ and 0 for $-T_1<t<T_1$. This definition is consistent if $T_1$ is small relative to $T/2$. The derivative $q(t)$ would then be $\delta (t-T_1)-\delta (t-(T-T_1))$ (in one period $[0,T)$). Or if centered, $\delta (t-T_1)-\delta (t+T_1)$ (in $[-T/2,T/2)$). The text says: $q(t)=x(t+T_1)-x(t-T_1)$. If $x(t)$ is the impulse train (3.75), then $x(t+T_1)$ has impulses at $-T_1,-T_1+T,\dots$ and $x(t-T_1)$ has impulses at $T_1,T_1+T,\dots$. So $q(t)$ has positive impulses at $-T_1$ (and $-T_1+T$, etc.) and negative impulses at $T_1$ (and $T_1+T$, etc.). This is consistent with $q(t)$ being the derivative of a square wave that goes from 0 to 1 at $-T_1$ and from 1 to 0 at $T_1$. This is exactly the square wave from Example 3.5 (eq. 3.40). So $g(t)$ in eq. (3.40) is the integral of $q(t)$. Let’s use the symbols as defined in the text, so $g(t)$ in eq (3.40) is what we are calling $x_{sq}(t)$ and $q(t)=d{x}_{sq}(t)/dt$. Let $a_k$ be the Fourier coefficients of the impulse train $x(t)$. Let $b_k$ be the Fourier coefficients of $q(t)$. Let $c_k$ be the Fourier coefficients of $g(t)$ from Example 3.5.

From $q(t)=x(t+T_1)-x(t-T_1)$. Using the time-shifting and linearity properties, we see that the Fourier series coefficients $b_k$ of $q(t)$ may be expressed in terms of the Fourier series coefficients $a_k$ of $x(t)$:

$$\begin{array}{rlrl}q(t)\stackrel{\mathrm{FS}}{⟷}{b}_k& =a_k{e}^{-jk{\omega }_0(-T_1)}-a_k{e}^{-jk{\omega }_0(T_1)}& & \text{(Time-shifting and linearity)}\\ & =a_k{e}^{jk{\omega }_0{T}_1}-a_k{e}^{-jk{\omega }_0{T}_1}& & \\ & =a_k(e^{jk{\omega }_0{T}_1}-e^{-jk{\omega }_0{T}_1}).& & \end{array}$$

where ${\omega }_0=2\pi /T$. Using eq. (3.76), $a_k=1/T$:

$$\begin{array}{rlrl}{b}_k& =\frac{1}{T}[e^{jk{\omega }_0{T}_1}-e^{-jk{\omega }_0{T}_1}]& & \text{(Substitute }{a}_k=1/T\text{)}\\ & =\frac{1}{T}[2j\sin (k{\omega }_0{T}_1)]& & \text{(Euler’s formula for sine)}\\ & =\frac{2j\sin (k{\omega }_0{T}_1)}{T}.& & \end{array}$$

Finally, since $q(t)$ is the derivative of $g(t)$ (the square wave from Example 3.5), we can use the differentiation property in Table 3.1 to write:

$$b_k=jk{\omega }_0{c}_k,\text{\hspace{1em}(3.78)}$$

where the $c_k$ are the Fourier series coefficients of $g(t)$. Thus, for $k\ne 0$:

$$\begin{array}{rlrl}{c}_k& =\frac{b_k}{jk{\omega }_0}& & \text{(Rearrange eq. (3.78))}\\ & =\frac{2j\sin (k{\omega }_0{T}_1)/T}{jk{\omega }_0}& & \text{(Substitute }{b}_k\text{)}\\ & =\frac{2\sin (k{\omega }_0{T}_1)}{k{\omega }_{0}T}& & \\ & =\frac{\sin (k{\omega }_0{T}_1)}{k\pi },k\ne 0,& & \end{array}\text{\hspace{1em}(3.79)}$$

where we have used the fact that ${\omega }_{0}T=2\pi$. Note that eq. (3.79) is valid for $k\ne 0$, since we cannot solve for $c_0$ from eq. (3.78) with $k=0$. However, since $c_0$ is just the average value of $g(t)$ over one period, we can determine it by inspection from Figure 3.12(b) (which shows $g(t)$ is 1 for $T_1<t<T-T_1$): No, the figure 3.12(b) shows a square wave that is 1 for $2T_1$ duration (from $T_1$ to $T-T_1$) and 0 for $T-2T_1$ duration. The total period is $T$. No, this is confusing. Let’s assume Figure 3.12(b) is the same square wave from Example 3.5, i.e., $g(t)=1$ for $|t|<T_1$ and $0$ otherwise. Its average value is $2T_1/T$.

$$c_0=\frac{2T_1}{T}.\text{\hspace{1em}(3.80)}$$

Eqs. (3.80) and (3.79) are identical to eqs. (3.41) and (3.43), respectively, for the Fourier series coefficients of the square wave derived in Example 3.5.

The next example is chosen to illustrate the use of many of the properties in Table 3.1.

Example 3.9

Suppose we are given the following facts about a signal $x(t)$:

1. $x(t)$ is a real signal.
2. $x(t)$ is periodic with period $T=4$, and it has Fourier series coefficients $a_k$.
3. $a_k=0$ for $|k|>1$.
4. The signal with Fourier coefficients $b_k=e^{-j\pi k/2}{a}_{-k}$ is odd.
5. $\frac{1}{4}{\int }_4|x(t){|}^{2}dt=1/2$.

Let us show that this information is sufficient to determine the signal $x(t)$ to within a sign factor. Analysis of Fact 2 and 3: From Fact 2, $T=4$, so the fundamental frequency is ${\omega }_0=2\pi /T=2\pi /4=\pi /2$. From Fact 3, $a_k=0$ for $|k|>1$. This means the only non-zero coefficients are $a_{-1},a_0,a_1$. So, $x(t)$ can be written as:

$$x(t)=a_0{e}^{j0{\omega }_{0}t}+a_1{e}^{j1{\omega }_{0}t}+a_{-1}{e}^{-j1{\omega }_{0}t}=a_0+a_1{e}^{j\pi t/2}+a_{-1}{e}^{-j\pi t/2}.$$

Analysis of Fact 1: Since $x(t)$ is real (Fact 1), we use the conjugate symmetry property (Table 3.1, eq. 3.66): $a_k=a_{-k}^{\ast }$. This implies:

• $a_0=a_0^{\ast }$, so $a_0$ must be real.
• $a_1=a_{-1}^{\ast }$, so $a_{-1}=a_1^{\ast }$. Substituting $a_{-1}=a_1^{\ast }$ into the expression for $x(t)$:

$$x(t)=a_0+a_1{e}^{j\pi t/2}+a_1^{\ast }{e}^{-j\pi t/2}=a_0+2\Re e\{a_1{e}^{j\pi t/2}\}.\text{\hspace{1em}(3.81)}$$

Analysis of Fact 4: The signal $y(t)$ corresponding to coefficients $b_k=e^{-j\pi k/2}{a}_{-k}$ is odd. Let’s analyze $b_k$ using properties from Table 3.1:

• Time reversal: If $x(t)\stackrel{\mathrm{FS}}{⟷}{a}_k$, then $x(-t)\stackrel{\mathrm{FS}}{⟷}{a}_{-k}$.
• Time shifting: If $z(t)\stackrel{\mathrm{FS}}{⟷}{c}_k$, then $z(t-t_0)\stackrel{\mathrm{FS}}{⟷}{c}_k{e}^{-jk{\omega }_0{t}_0}$. The term $e^{-j\pi k/2}$ can be written as $e^{-jk(2\pi /4)(1)}=e^{-jk{\omega }_0{t}_0}$ with $t_0=1$. So, $e^{-j\pi k/2}{a}_{-k}$ corresponds to a signal $x(-t-1)=x(-(t+1))$. No, this is $x(-(t-1))=x(-t+1)$. Let $z(t)\stackrel{\mathrm{FS}}{⟷}{a}_{-k}$. This means $z(t)=x(-t)$. Then $e^{-jk{\omega }_0{t}_0}z(t)$ corresponds to $z(t-t_0)$, i.e., $x(-(t-t_0))$. Here $t_0=1$. So $b_k$ corresponds to $x(-(t-1))=x(-t+1)$. So, the signal $y(t)=x(-t+1)$ is odd (from Fact 4). Also, since $x(t)$ is real, $x(-t+1)$ must also be real. From Table 3.1, if a signal is real and odd, its Fourier coefficients are purely imaginary and odd. Thus, $b_k$ must be purely imaginary and odd. This means:
• $b_0=0$.
• $b_{-k}=-b_k$.

Let’s find $b_0$ in terms of $a_k$: $b_0=e^{-j\pi (0)/2}{a}_{-0}=1\cdot a_0=a_0$. Since $b_0=0$, we conclude that $a_0=0$.

Now let’s use $b_k$ being purely imaginary and odd for $k=1$: $b_{-1}=-b_1$. Also, for $k=1$: $b_1=e^{-j\pi /2}{a}_{-1}=(-j)a_{-1}$. And for $k=-1$: $b_{-1}=e^{-j\pi (-1)/2}{a}_{-(-1)}=e^{j\pi /2}{a}_1=(j)a_1$. Since $b_1$ must be purely imaginary, $a_{-1}$ must be purely real. (From $b_1=-j{a}_{-1}$, if $a_{-1}$ is purely real, then $b_1$ is purely imaginary). From $b_{-1}=j{a}_1$, if $a_1$ is purely real, then $b_{-1}$ is purely imaginary. These are consistent with $a_1=a_{-1}^{\ast }$. If $a_{-1}$ is purely real, then $a_1=\Re \{a_{-1}{\}}^{\ast }=\Re \{a_{-1}\}$, so $a_1$ is also purely real. In this case, $a_1$ and $a_{-1}$ are real and $a_1=a_{-1}$. Then $b_1=-j{a}_1$ and $b_{-1}=j{a}_1$. And we also have $b_{-1}=-b_1$, so $j{a}_1=-(-j{a}_1)=j{a}_1$. This is consistent. So, $a_1$ and $a_{-1}$ are real and equal. Analysis of Fact 5: The average power of $x(t)$ is $\frac{1}{4}{\int }_4|x(t){|}^{2}dt=1/2$. Using Parseval’s relation (Table 3.1):

$$\frac{1}{T}{\int }_T|x(t){|}^{2}dt=\sum _{k=-\infty }^{+\infty }|a_k{|}^2.$$

Substituting given values:

$$\frac{1}{2}=\sum _{k=-1}^1|a_k{|}^2.\text{(Since }{a}_k=0\text{ for }|k|>1$$

Since $a_0=0$, we have:

$$\frac{1}{2}=|a_{-1}{|}^2+|a_1{|}^2.$$

As $a_1$ and $a_{-1}$ are real and equal, let $a_1=a_{-1}=A$. Then $\frac{1}{2}=A^2+A^2=2A^2$. So $A^2=1/4$, which implies $A=\pm 1/2$. Therefore, $a_1=a_{-1}=1/2$ or $a_1=a_{-1}=-1/2$.

Let’s reconstruct $x(t)$ using $a_0=0$ and the possible values for $a_1,a_{-1}$: Case 1: $a_1=1/2$ and $a_{-1}=1/2$. Then $x(t)=0+\frac{1}{2}{e}^{j\pi t/2}+\frac{1}{2}{e}^{-j\pi t/2}=\cos (\pi t/2)$. Case 2: $a_1=-1/2$ and $a_{-1}=-1/2$. Then $x(t)=0-\frac{1}{2}{e}^{j\pi t/2}-\frac{1}{2}{e}^{-j\pi t/2}=-\cos (\pi t/2)$.

Thus, this information is sufficient to determine $x(t)$ to within a sign factor.

3.6 Fourier Series Representation of Discrete-Time Periodic Signals

In this section, we consider the Fourier series representation of discrete-time periodic signals. While the discussion closely parallels that of Section 3.3, there are some important differences. In particular, the Fourier series representation of a discrete-time periodic signal is a finite series, as opposed to the infinite series representation required for continuous-time periodic signals. As a consequence, there are no mathematical issues of convergence such as those discussed in Section 3.4.

3.6.1 Linear Combinations of Harmonically Related Complex Exponentials

As defined in Chapter 1, a discrete-time signal $x[n]$ is periodic with period $N$ if

$$x[n]=x[n+N].\text{\hspace{1em}(3.84)}$$

The fundamental period is the smallest positive integer $N$ for which eq. (3.84) holds, and ${\omega }_0=2\pi /N$ is the fundamental frequency. For example, the complex exponential $e^{j(2\pi /N)n}$ is periodic with period $N$. Furthermore, the set of all discrete-time complex exponential signals that are periodic with period $N$ is given by

$${\varphi }_k[n]=e^{jk{\omega }_{0}n}=e^{jk(2\pi /N)n},k=0,\pm 1,\pm 2,\dots \text{\hspace{1em}(3.85)}$$

All of these signals have fundamental frequencies that are multiples of $2\pi /N$ and thus are harmonically related.

As mentioned in Section 1.3.3, there are only $N$ distinct signals in the set given by eq. (3.85). This is a consequence of the fact that discrete-time complex exponentials which differ in frequency by a multiple of $2\pi$ are identical. Specifically, ${\varphi }_0[n]={\varphi }_N[n],{\varphi }_1[n]={\varphi }_{N+1}[n]$, and, in general,

$${\varphi }_k[n]={\varphi }_{k+rN}[n].\text{\hspace{1em}(3.86)}$$

That is, when $k$ is changed by any integer multiple of $N$, the identical sequence is generated. This differs from the situation in continuous time in which the signals ${\varphi }_k(t)$ defined in eq. (3.23) are all different from one another.

We now wish to consider the representation of more general periodic sequences in terms of linear combinations of the sequences ${\varphi }_k[n]$ in eq. (3.85). Such a linear combination has the form

$$x[n]=\sum _k{a}_k{\varphi }_k[n]=\sum _k{a}_k{e}^{jk{\omega }_{0}n}=\sum _k{a}_k{e}^{jk(2\pi /N)n}.\text{\hspace{1em}(3.87)}$$

Since the sequences ${\varphi }_k[n]$ are distinct only over a range of $N$ successive values of $k$, the summation in eq. (3.87) need only include terms over this range. Thus, the summation is on $k$ as $k$ varies over a range of $N$ successive integers, beginning with any value of $k$. We indicate this by expressing the limits of the summation as $k=⟨N⟩$. That is,

$$x[n]=\sum _{k=⟨N⟩}{a}_k{\varphi }_k[n]=\sum _{k=⟨N⟩}{a}_k{e}^{jk{\omega }_{0}n}=\sum _{k=⟨N⟩}{a}_k{e}^{jk(2\pi /N)n}.\text{\hspace{1em}(3.88)}$$

For example, $k$ could take on the values $k=0,1,\dots ,N-1$, or $k=3,4,\dots ,N+2$. In either case, by virtue of eq. (3.86), exactly the same set of complex exponential sequences appears in the summation on the right-hand side of eq. (3.88). Equation (3.88) is referred to as the discrete-time Fourier series and the coefficients $a_k$ as the Fourier series coefficients.

3.6.2 Determination of the Fourier Series Representation of a Periodic Signal

Suppose now that we are given a sequence $x[n]$ that is periodic with fundamental period $N$. We would like to determine whether a representation of $x[n]$ in the form of eq. (3.88) exists and, if so, what the values of the coefficients $a_k$ are. This question can be phrased in terms of finding a solution to a set of linear equations. Specifically, if we evaluate eq. (3.88) for $N$ successive values of $n$ corresponding to one period of $x[n]$, we obtain

$$\begin{array}{rl}x[0]& =\sum _{k=⟨N⟩}{a}_k,\\ x[1]& =\sum _{k=⟨N⟩}{a}_k{e}^{j2\pi k/N},\\ ⋮& \\ x[N-1]& =\sum _{k=⟨N⟩}{a}_k{e}^{j2\pi k(N-1)/N}.\end{array}\text{\hspace{1em}(3.89)}$$

Thus, eq. (3.89) represents a set of $N$ linear equations for the $N$ unknown coefficients $a_k$ as $k$ ranges over a set of $N$ successive integers. It can be shown that this set of equations is linearly independent and consequently can be solved to obtain the coefficients $a_k$ in terms of the given values of $x[n]$. In Problem 3.32, we consider an example in which the Fourier series coefficients are obtained by explicitly solving the set of $N$ equations given in eq. (3.89). However, by following steps parallel to those used in continuous time, it is possible to obtain a closed-form expression for the coefficients $a_k$ in terms of the values of the sequence $x[n]$.

The basis for this result is the fact, shown in Problem 3.54, that

$$\sum _{n=⟨N⟩}{e}^{jk(2\pi /N)n}=\{\begin{array}{ll}N,& k=0,\pm N,\pm 2N,\dots \\ 0,& \mathrm{otherwise}\end{array}.\text{\hspace{1em}(3.90)}$$

Equation (3.90) states that the sum over one period of the values of a periodic complex exponential is zero, unless that complex exponential is a constant.

Now consider the Fourier series representation of eq. (3.88). Multiplying both sides by $e^{-jr(2\pi /N)n}$ and summing over $N$ terms (i.e., over one period, $⟨N⟩$), we obtain:

$$\sum _{n=⟨N⟩}x[n]e^{-jr(2\pi /N)n}=\sum _{n=⟨N⟩}\left(\sum _{k=⟨N⟩}{a}_k{e}^{jk(2\pi /N)n}\right)e^{-jr(2\pi /N)n}.\text{\hspace{1em}(3.91)}$$

Interchanging the order of summation on the right-hand side, we have:

$$\sum _{n=⟨N⟩}x[n]e^{-jr(2\pi /N)n}=\sum _{k=⟨N⟩}{a}_k\sum _{n=⟨N⟩}{e}^{j(k-r)(2\pi /N)n}.\text{\hspace{1em}(3.92)}$$

From the identity in eq. (3.90), the innermost sum on $n$ on the right-hand side of eq. (3.92) is zero, unless $k-r$ is zero or an integer multiple of $N$. Since the outer summation for $k$ is taken over a range of $N$ consecutive values, and we choose $r$ to also be in this range, $k-r$ can only be $0$ within this range. (If $k-r$ was a multiple of $N$, say $N$, then $k=r+N$. But $k$ and $r$ are in the same range of $N$ consecutive integers, so this is not possible unless $N=0$ which is not the case for period). Therefore, if we choose values for $r$ over the same range as that over which $k$ varies in the outer summation, the innermost sum on the right-hand side of eq. (3.92) equals $N$ if $k=r$ and $0$ if $k\ne r$. The right-hand side of eq. (3.92) then reduces to $N{a}_r$:

$$\sum _{n=⟨N⟩}x[n]e^{-jr(2\pi /N)n}=N{a}_r.$$

And thus, solving for $a_r$:

$$a_r=\frac{1}{N}\sum _{n=⟨N⟩}x[n]e^{-jr(2\pi /N)n}.\text{\hspace{1em}(3.93)}$$

This provides a closed-form expression for obtaining the Fourier series coefficients, and we have the discrete-time Fourier series pair:

$$\begin{array}{c}\begin{array}{rlr}x[n]& =\sum _{k=⟨N⟩}{a}_k{e}^{jk{\omega }_{0}n}=\sum _{k=⟨N⟩}{a}_k{e}^{jk(2\pi /N)n},& \text{(synthesis equation)}\\ a_k& =\frac{1}{N}\sum _{n=⟨N⟩}x[n]e^{-jk{\omega }_{0}n}=\frac{1}{N}\sum _{n=⟨N⟩}x[n]e^{-jk(2\pi /N)n}.& \text{(analysis equation)}\end{array}\end{array}\text{\hspace{1em}(3.94–3.95)}$$

These equations play the same role for discrete-time periodic signals that eqs. (3.37) and (3.38) play for continuous-time periodic signals, with eq. (3.94) the synthesis equation and eq. (3.95) the analysis equation. As in continuous time, the discrete-time Fourier series coefficients $a_k$ are often referred to as the spectral coefficients of $x[n]$. These coefficients specify a decomposition of $x[n]$ into a sum of $N$ harmonically related complex exponentials.

Referring to eq. (3.88), we see that if we take $k$ in the range from $0$ to $N-1$, we have

$$x[n]=a_0{\varphi }_0[n]+a_1{\varphi }_1[n]+\dots +a_{N-1}{\varphi }_{N-1}[n].\text{\hspace{1em}(3.96)}$$

Similarly, if $k$ ranges from $1$ to $N$, we obtain

$$x[n]=a_1{\varphi }_1[n]+a_2{\varphi }_2[n]+\dots +a_N{\varphi }_N[n].\text{\hspace{1em}(3.97)}$$

From eq. (3.86), ${\varphi }_0[n]={\varphi }_N[n]$, and therefore, upon comparing eqs. (3.96) and (3.97), we conclude that $a_0=a_N$. Similarly, by letting $k$ range over any set of $N$ consecutive integers and using eq. (3.86), we can conclude that

$$a_k=a_{k+N}.\text{\hspace{1em}(3.98)}$$

That is, if we consider more than $N$ sequential values of $k$, the values $a_k$ repeat periodically with period $N$. It is important that this fact be interpreted carefully. In particular, since there are only $N$ distinct complex exponentials that are periodic with period $N$, the discrete-time Fourier series representation is a finite series with $N$ terms. Therefore, if we fix the $N$ consecutive values of $k$ over which we define the Fourier series in eq. (3.94), we will obtain a set of exactly $N$ Fourier coefficients from eq. (3.95). On the other hand, at times it will be convenient to use different sets of $N$ values of $k$, and consequently, it is useful to regard eq. (3.94) as a sum over any arbitrary set of $N$ successive values of $k$. For this reason, it is sometimes convenient to think of $a_k$ as a sequence defined for all values of $k$, but where only $N$ successive elements in the sequence will be used in the Fourier series representation. Furthermore, since the ${\varphi }_k[n]$ repeat periodically with period $N$ as we vary $k$ [eq. (3.86)], so must the $a_k$ [eq. (3.98)]. This viewpoint is illustrated in the next example.

Example 3.10

Consider the signal

$$x[n]=\sin {\omega }_{0}n,\text{\hspace{1em}(3.99)}$$

which is the discrete-time counterpart of the signal $x(t)=\sin {\omega }_{0}t$ of Example 3.3. $x[n]$ is periodic only if $2\pi /{\omega }_0$ is an integer or a ratio of integers.

Case 1: ${\omega }_0=2\pi /N$ (so $2\pi /{\omega }_0$ is an integer $N$) In this case, $x[n]$ is periodic with fundamental period $N$, and we obtain a result that is exactly analogous to the continuous-time case. Expanding the signal as a sum of two complex exponentials, we get:

$$\begin{array}{rl}x[n]& =\sin \left(\frac{2\pi }{N}n\right)\\ & =\frac{e^{j(2\pi /N)n}-e^{-j(2\pi /N)n}}{2j}.\end{array}\text{\hspace{1em}(3.100)}$$

Comparing eq. (3.100) with eq. (3.94) ($x[n]={\sum }_{k=⟨N⟩}{a}_k{e}^{jk(2\pi /N)n}$), we see by inspection that:

$$a_1=\frac{1}{2j},a_{-1}=-\frac{1}{2j},$$

and the remaining coefficients over the interval of summation are zero. As described previously, these coefficients repeat with period $N$; thus, $a_{N+1}$ is also equal to $(1/2j)$ and $a_{N-1}$ equals $(-1/2j)$. The Fourier series coefficients for this example with $N=5$ are illustrated in Figure 3.13 (not shown here). The fact that they repeat periodically is indicated. However, only one period is utilized in the synthesis equation (3.94).

Case 2: ${\omega }_0=2\pi M/N$ (so $2\pi /{\omega }_0=N/M$ is a ratio of integers) Assuming that $M$ and $N$ do not have any common factors, $x[n]$ has a fundamental period of $N$. Again expanding $x[n]$ as a sum of two complex exponentials, we have:

$$x[n]=\frac{e^{jM(2\pi /N)n}-e^{-jM(2\pi /N)n}}{2j},$$

from which we can determine by inspection that $a_M=(1/2j)$, $a_{-M}=(-1/2j)$, and the remaining coefficients over one period of length $N$ are zero. The Fourier coefficients for this example with $M=3$ and $N=5$ are depicted in Figure 3.14 (not shown here). Again, we have indicated the periodicity of the coefficients. For example, for $N=5$, $a_2=a_{-3}$ (since $-3$ is equivalent to $2$ modulo $5$), which in our example equals $(-1/2j)$. Note, however, that over any period of length 5, there are only two nonzero Fourier coefficients, and therefore there are only two nonzero terms in the synthesis equation.

Example 3.12

In this example, we consider the discrete-time periodic square wave shown in Figure 3.16 (not shown here). It is defined as $x[n]=1$ for $|n|\le N_1$ and $x[n]=0$ for $N_1<|n|<N/2$, and is periodic with period $N$. We can evaluate the Fourier series for this signal using eq. (3.95). Because $x[n]=1$ for $-N_1\le n\le N_1$, it is particularly convenient to choose the length-$N$ interval of summation in eq. (3.95) so that it includes the range $-N_1\le n\le N_1$. In this case, we can express eq. (3.95) as:

$$\begin{array}{rl}{a}_k& =\frac{1}{N}\sum _{n=⟨N⟩}x[n]e^{-jk(2\pi /N)n}\\ & =\frac{1}{N}\sum _{n=-N_1}^{N_1}1\cdot e^{-jk(2\pi /N)n}.\end{array}\text{\hspace{1em}(3.102)}$$

Letting $m=n+N_1$, so $n=m-N_1$. When $n=-N_1$, $m=0$. When $n=N_1$, $m=2N_1$. The sum ranges from $m=0$ to $2N_1$. We observe that eq. (3.102) becomes:

$$\begin{array}{rlrl}{a}_k& =\frac{1}{N}\sum _{m=0}^{2N_1}{e}^{-jk(2\pi /N)(m-N_1)}& & \text{(Substitute }n=m-N_1\text{)}\\ & =\frac{1}{N}{e}^{jk(2\pi /N)N_1}\sum _{m=0}^{2N_1}{e}^{-jk(2\pi /N)m}.& & \text{(Factor out constant term)}\end{array}\text{\hspace{1em}(3.103)}$$

The summation in eq. (3.103) consists of the sum of a geometric series ${\sum }_{m=0}^L{r}^m=\frac{1-r^{L+1}}{1-r}$, where $r=e^{-jk(2\pi /N)}$ and $L=2N_1$. This yields:

$$\begin{array}{rlrl}{a}_k& =\frac{1}{N}{e}^{jk(2\pi /N)N_1}\left(\frac{1-e^{-jk(2\pi /N)(2N_1+1)}}{1-e^{-jk(2\pi /N)}}\right)& & \text{(Sum of geometric series)}\\ & =\frac{1}{N}{e}^{jk(2\pi /N)N_1}\frac{e^{-jk(2\pi /N)(N_1+1/2)}\left[e^{jk(2\pi /N)(N_1+1/2)}-e^{-jk(2\pi /N)(N_1+1/2)}\right]}{e^{-jk(2\pi /2N)}\left[e^{jk(2\pi /2N)}-e^{-jk(2\pi /2N)}\right]}& & \text{(Factor out half angles)}\\ & =\frac{1}{N}{e}^{jk(2\pi /N)N_1}\frac{e^{-jk(2\pi /N)(N_1+1/2)}(2j\sin [k(2\pi /N)(N_1+1/2)])}{e^{-jk(2\pi /2N)}(2j\sin [k(2\pi /2N)])}& & \text{(Use }\sin \theta =\frac{e^{j\theta }-e^{-j\theta }}{2j}\text{)}\\ & =\frac{1}{N}{e}^{jk(2\pi /N)N_1}\frac{e^{-jk(2\pi /N)(N_1+1/2)}}{e^{-jk(2\pi /2N)}}\frac{\sin [k(2\pi /N)(N_1+1/2)]}{\sin [k(\pi /N)]}& & \\ & =\frac{1}{N}{e}^{jk(2\pi /N)(N_1-(N_1+1/2)+1/2)}\frac{\sin [k(2\pi /N)(N_1+1/2)]}{\sin [k(\pi /N)]}& & \\ & =\frac{1}{N}{e}^{j0}\frac{\sin [k(2\pi /N)(N_1+1/2)]}{\sin [k(\pi /N)]}& & \\ & =\frac{1}{N}\frac{\sin [2\pi k(N_1+1/2)/N]}{\sin (\pi k/N)},k\ne 0,\pm N,\pm 2N,\dots & & \end{array}\text{\hspace{1em}(3.104)}$$

For values of $k$ that are integer multiples of $N$ (i.e., $k=rN$ for some integer $r$), the denominator $\sin (\pi k/N)=\sin (r\pi )$ is $0$. In this case, we evaluate the original sum directly for these values of $k$. If $k=rN$:

$$\begin{array}{rlrl}{a}_{rN}& =\frac{1}{N}\sum _{n=-N_1}^{N_1}{e}^{-j(rN)(2\pi /N)n}=\frac{1}{N}\sum _{n=-N_1}^{N_1}{e}^{-jr2\pi n}& & \\ & =\frac{1}{N}\sum _{n=-N_1}^{N_1}1& & \text{(Since }{e}^{-jr2\pi n}=(e^{-j2\pi }{)}^{rn}=1^{rn}=1\text{ for integer }r,n\text{)}\\ & =\frac{1}{N}(N_1-(-N_1)+1)& & \\ & =\frac{2N_1+1}{N},k=0,\pm N,\pm 2N,\dots & & \end{array}\text{\hspace{1em}(3.105)}$$

The coefficients $a_k$ for $2N_1+1=5$ are sketched for $N=10,20$, and $40$ in Figures 3.17(a), (b), and (c), respectively (not shown here).

In discussing the convergence of the continuous-time Fourier series in Section 3.4, we considered the example of a symmetric square wave and observed how the finite sum in eq. (3.46) converged to the square wave as the number of terms approached infinity. In contrast, for the discrete-time Fourier series there is no issue of convergence because the synthesis equation (3.94) tells us how to recover the values of the original sequence in terms of a finite series. Thus, if $N$ is odd and we take $M=(N-1)/2$ in the summation from $-M$ to $M$ (which is $N$ terms: from $-(N-1)/2$ to $(N-1)/2$), then this sum includes exactly $N$ terms, and consequently, from the synthesis equations, we have $\hat{x}[n]=x[n]$. Similarly, if $N$ is even and we let:

$$\hat{x}[n]=\sum _{k=-M+1}^M{a}_k{e}^{jk(2\pi /N)n},\text{\hspace{1em}(3.107)}$$

then with $M=N/2$, this sum consists of $N$ terms ($k$ ranges from $-(N/2)+1$ to $N/2$), and again, we can conclude from eq. (3.94) that $\hat{x}[n]=x[n]$.

In contrast, a continuous-time periodic signal takes on a continuum of values over a single period, and an infinite number of Fourier coefficients are required to represent it.

3.7 Properties of Discrete-Time Fourier Series

The derivations of many of these properties are very similar to those of the corresponding properties for continuous-time Fourier series, and several such derivations are considered in the problems at the end of the chapter. In addition, in Chapter 5, we will see that most of the properties can be inferred from corresponding properties of the discrete-time Fourier transform. Consequently, we limit the discussion in the following subsections to only a few of these properties, including several that have important differences relative to those for continuous time. We also provide examples illustrating the usefulness of various discrete-time Fourier series properties for developing conceptual insights and helping to reduce the complexity of the evaluation of the Fourier series of many periodic sequences.

As with continuous time, it is often convenient to use a shorthand notation to indicate the relationship between a periodic signal and its Fourier series coefficients. Specifically, if $x[n]$ is a periodic signal with period $N$ and with Fourier series coefficients denoted by $a_k$, then we will write

$$x[n]\stackrel{\mathrm{FS}}{⟷}{a}_k.$$

3.7.1 Multiplication

The multiplication property of the Fourier series representation is one example of a property that reflects the difference between continuous time and discrete time. From Table 3.1, the product of two continuous-time signals of period $T$ results in a periodic signal with period $T$ whose sequence of Fourier series coefficients is the convolution of the sequences of Fourier series coefficients of the two signals being multiplied. In discrete time, suppose that

$$x[n]\stackrel{\mathrm{FS}}{⟷}{a}_k$$

and

$$y[n]\stackrel{\mathrm{FS}}{⟷}{b}_k$$

are both periodic with period $N$. Then the product $x[n]y[n]$ is also periodic with period $N$ and, as shown in Problem 3.57, its Fourier coefficients, $d_k$, are given by

$$x[n]y[n]\stackrel{\mathrm{FS}}{⟷}{d}_k=\sum _{l=⟨N⟩}{a}_l{b}_{k-l}.\text{\hspace{1em}(3.108)}$$

Equation (3.108) is analogous to the definition of convolution, except that the summation variable is now restricted to an interval of $N$ consecutive samples. As shown in Problem 3.57, the summation can be taken over any set of $N$ consecutive values of $l$. We refer to this type of operation as a periodic convolution between the two periodic sequences of Fourier coefficients. The usual form of the convolution sum (where the summation variable ranges from $-\infty$ to $\infty$) is sometimes referred to as aperiodic convolution, to distinguish it from periodic convolution.

3.7.2 First Difference

The discrete-time parallel to the differentiation property of the continuous-time Fourier series involves the use of the first-difference operation, which is defined as $x[n]-x[n-1]$. If $x[n]$ is periodic with period $N$, then so is $y[n]$, since shifting $x[n]$ or linearly combining $x[n]$ with another periodic signal whose period is $N$ always results in a periodic signal with period $N$. Also, if

$$x[n]\stackrel{\mathrm{FS}}{⟷}{a}_k,$$

then the Fourier coefficients corresponding to the first difference of $x[n]$ may be expressed as:

$$\begin{array}{rlrl}\mathcal{F}\{x[n]-x[n-1]\}& =\mathcal{F}\{x[n]\}-\mathcal{F}\{x[n-1]\}& & \text{(Linearity property)}\\ & \stackrel{\mathrm{FS}}{⟷}{a}_k-a_k{e}^{-jk(2\pi /N)}& & \text{(Time-shifting property)}\\ & \stackrel{\mathrm{FS}}{⟷}(1-e^{-jk(2\pi /N)})a_k.& & \end{array}\text{\hspace{1em}(3.109)}$$

which is easily obtained by applying the time-shifting and linearity properties in Table 3.2 (to be summarized later). A common use of this property is in situations where evaluation of the Fourier series coefficients is easier for the first difference than for the original sequence. (See Problem 3.31.)

3.7.3 Parseval’s Relation for Discrete-Time Periodic Signals

As shown in Problem 3.57, Parseval’s relation for discrete-time periodic signals is given by

$$\frac{1}{N}\sum _{n=⟨N⟩}|x[n]{|}^2=\sum _{k=⟨N⟩}|a_k{|}^2,\text{\hspace{1em}(3.110)}$$

where the $a_k$ are the Fourier series coefficients of $x[n]$ and $N$ is the period. As in the continuous-time case, the left-hand side of Parseval’s relation is the average power in one period for the periodic signal $x[n]$. Similarly, $|a_k{|}^2$ is the average power in the $k$th harmonic component of $x[n]$. Thus, once again, Parseval’s relation states that the average power in a periodic signal equals the sum of the average powers in all of its harmonic components. In discrete time, of course, there are only $N$ distinct harmonic components, and since the $a_k$ are periodic with period $N$, the sum on the right-hand side of eq. (3.110) can be taken over any $N$ consecutive values of $k$.

3.7.4 Examples

In this subsection, we present several examples illustrating how properties of the discrete-time Fourier series can be used to characterize discrete-time periodic signals and to compute their Fourier series representations. Specifically, Fourier series properties, such as those listed in Table 3.2 (to be summarized later), may be used to simplify the process of determining the Fourier series coefficients of a given signal. This involves first expressing the given signal in terms of other signals whose Fourier series coefficients are already known or are simpler to compute. Then, using Table 3.2, we can express the Fourier series coefficients of the given signal in terms of the Fourier series coefficients of the other signals. This is illustrated in Example 3.13. Example 3.14 then illustrates the determination of a sequence from some partial information. In Example 3.15, we illustrate the use of the periodic convolution property in Table 3.2.

Example 3.13

Let us consider the problem of finding the Fourier series coefficients $a_k$ of the sequence $x[n]$ shown in Figure 3.19(a) (not shown here). This sequence has a fundamental period of 5. We observe that $x[n]$ may be viewed as the sum of the square wave $x_1[n]$ in Figure 3.19(b) and the dc sequence $x_2[n]$ in Figure 3.19(c). Specifically, $x[n]$ is 2 for $n=-1,0,1$ and 1 for $n=2,3$. Period is $N=5$. $x_1[n]$ is 1 for $n=-1,0,1$ and 0 for $n=2,3$. Period $N=5$. $x_2[n]$ is 1 for all $n$. Period $N=5$. Thus, $x[n]=x_1[n]+x_2[n]$. Denoting the Fourier series coefficients of $x_1[n]$ by $b_k$ and those of $x_2[n]$ by $c_k$, we use the linearity property to conclude that:

$$a_k=b_k+c_k.\text{\hspace{1em}(3.111)}$$

From Example 3.12 (with $N_1=1$ and $N=5$), the square wave $x_1[n]$ is 1 for $|n|\le N_1=1$ and 0 otherwise. Its Fourier series coefficients $b_k$ can be expressed as:

$$b_k=\frac{1}{N}\frac{\sin [2\pi k(N_1+1/2)/N]}{\sin (\pi k/N)}.$$

Substituting $N=5$ and $N_1=1$:

$$b_k=\frac{1}{5}\frac{\sin [2\pi k(1+1/2)/5]}{\sin (\pi k/5)}=\frac{1}{5}\frac{\sin (3\pi k/5)}{\sin (\pi k/5)},\mathrm{for}k\ne 0,\pm 5,\pm 10,\dots$$

And for $k=0,\pm N,\dots$:

$$b_k=\frac{2N_1+1}{N}=\frac{2(1)+1}{5}=\frac{3}{5},\mathrm{for}k=0,\pm 5,\pm 10,\dots$$

So:

$$b_k=\{\begin{array}{ll}\frac{1}{5}\frac{\sin (3\pi k/5)}{\sin (\pi k/5)},& \mathrm{for}k\ne 0,\pm 5,\pm 10,\dots \\ \frac{3}{5},& \mathrm{for}k=0,\pm 5,\pm 10,\dots \end{array}.\text{\hspace{1em}(3.112)}$$

The sequence $x_2[n]$ is $x_2[n]=1$ for all $n$. It is a constant signal. Its Fourier series coefficients $c_k$ are simply its average value over one period.

$$c_k=\frac{1}{N}\sum _{n=⟨N⟩}{x}_2[n]e^{-jk(2\pi /N)n}.$$

For $k=0$:

$$c_0=\frac{1}{5}\sum _{n=0}^{4}1=\frac{1}{5}(5)=1.\text{\hspace{1em}(3.113)}$$

For $k\ne 0,\pm N,\dots$: $c_k=\frac{1}{N}{\sum }_{n=0}^{N-1}{e}^{-jk(2\pi /N)n}$. This sum is zero because $e^{-jk(2\pi /N)n}$ is a periodic complex exponential with an integer number of cycles over the summation length $N$. (From eq. 3.90). Thus, $c_k=0$ for $k\ne 0,\pm N,\dots$. Since the discrete-time Fourier series coefficients are periodic, it follows that $c_k=1$ whenever $k$ is an integer multiple of 5. The remaining coefficients of $x_2[n]$ must be zero, because $x_2[n]$ contains only a dc component. We can now substitute the expressions for $b_k$ and $c_k$ into eq. (3.111) to obtain $a_k$: For $k=0,\pm 5,\pm 10,\dots$: $a_k=b_k+c_k=\frac{3}{5}+1=\frac{8}{5}$. For $k\ne 0,\pm 5,\pm 10,\dots$: $a_k=b_k+c_k=\frac{1}{5}\frac{\sin (3\pi k/5)}{\sin (\pi k/5)}+0=\frac{1}{5}\frac{\sin (3\pi k/5)}{\sin (\pi k/5)}$. So:

$$a_k=\{\begin{array}{ll}\frac{1}{5}\frac{\sin (3\pi k/5)}{\sin (\pi k/5)},& \mathrm{for}k\ne 0,\pm 5,\pm 10,\dots \\ \frac{8}{5},& \mathrm{for}k=0,\pm 5,\pm 10,\dots \end{array}.\text{\hspace{1em}(3.114)}$$

Example 3.14

Suppose we are given the following facts about a sequence $x[n]$:

1. $x[n]$ is periodic with period $N=6$.
2. ${\sum }_{n=0}^{5}x[n]=2$.
3. ${\sum }_{n=2}^7(-1{)}^{n}x[n]=1$.
4. $x[n]$ has the minimum power per period among the set of signals satisfying the preceding three conditions.

Let us determine the sequence $x[n]$. We denote the Fourier series coefficients of $x[n]$ by $a_k$. Analysis of Fact 2: The sum ${\sum }_{n=0}^{5}x[n]$ is the sum of $x[n]$ over one period. From the Fourier series analysis equation (3.95) with $k=0$:

$$a_0=\frac{1}{N}\sum _{n=⟨N⟩}x[n]e^{-j0(2\pi /N)n}=\frac{1}{N}\sum _{n=0}^{N-1}x[n].$$

Given $N=6$ and ${\sum }_{n=0}^{5}x[n]=2$:

$$a_0=\frac{1}{6}(2)=\frac{1}{3}.$$

Analysis of Fact 3: The sum ${\sum }_{n=2}^7(-1{)}^{n}x[n]=1$. The sum is over $N=6$ consecutive terms, so it’s a sum over one period. We need to relate $(-1{)}^n$ to a complex exponential. $(-1{)}^n=(e^{j\pi }{)}^n=e^{j\pi n}$. We know that $a_k=\frac{1}{N}{\sum }_{n=⟨N⟩}x[n]e^{-jk(2\pi /N)n}$. To get a sum with $e^{j\pi n}$, we need to choose $k$ such that $-k(2\pi /N)=\pi$. $-k(2\pi /6)=\pi \Rightarrow -k(\pi /3)=\pi \Rightarrow -k/3=1\Rightarrow k=-3$. However, Fourier coefficients are periodic with period $N=6$. So $a_{-3}$ is equivalent to $a_3$ for coefficients in the range $0\le k<N$. So, let’s look at the expression for $N{a}_k$: $N{a}_k={\sum }_{n=⟨N⟩}x[n]e^{-jk(2\pi /N)n}$. Let $k=3$. Then $N{a}_3={\sum }_{n=2}^{7}x[n]e^{-j3(2\pi /6)n}={\sum }_{n=2}^{7}x[n]e^{-j\pi n}$. The sum given is ${\sum }_{n=2}^7(-1{)}^{n}x[n]={\sum }_{n=2}^7{e}^{j\pi n}x[n]$. So, we have: ${\sum }_{n=2}^{7}x[n]e^{j\pi n}=1$. This expression is $N\cdot a_k$ if $e^{j\pi n}=e^{-jk(2\pi /N)n}$. So $\pi n=-k(2\pi /6)n\Rightarrow \pi =-k\pi /3\Rightarrow k=-3$. So, $N{a}_{-3}=1$. Since $N=6$, $a_{-3}=1/6$. Due to the periodicity of coefficients ($a_k=a_{k+N}$), $a_{-3}=a_{-3+6}=a_3$. Therefore, $a_3=1/6$.

Analysis of Fact 4: $x[n]$ has the minimum power per period among the set of signals satisfying the preceding three conditions. From Parseval’s relation (3.110), the average power in $x[n]$ is:

$$P=\frac{1}{N}\sum _{n=⟨N⟩}|x[n]{|}^2=\sum _{k=⟨N⟩}|a_k{|}^2.\text{\hspace{1em}(3.115)}$$

We know $a_0=1/3$ and $a_3=1/6$. The summation range $⟨N⟩$ can be taken as $k=0,1,2,3,4,5$. So, $P=|a_0{|}^2+|a_1{|}^2+|a_2{|}^2+|a_3{|}^2+|a_4{|}^2+|a_5{|}^2$. Substituting the known coefficients: $P=(1/3{)}^2+|a_1{|}^2+|a_2{|}^2+(1/6{)}^2+|a_4{|}^2+|a_5{|}^2$. To minimize $P$, we must choose the unknown coefficients to be zero. Therefore, $a_1=a_2=a_4=a_5=0$.

Now we have all Fourier coefficients for one period: $a_0=1/3$ $a_1=0$ $a_2=0$ $a_3=1/6$ $a_4=0$ $a_5=0$

Reconstruct $x[n]$ using the synthesis equation (3.94):

$$\begin{array}{rlrl}x[n]& =\sum _{k=0}^{N-1}{a}_k{e}^{jk(2\pi /N)n}& & \\ & =a_0{e}^{j0(2\pi /6)n}+a_1{e}^{j1(2\pi /6)n}+a_2{e}^{j2(2\pi /6)n}+a_3{e}^{j3(2\pi /6)n}+a_4{e}^{j4(2\pi /6)n}+a_5{e}^{j5(2\pi /6)n}& & \\ & =\frac{1}{3}\cdot 1+0+0+\frac{1}{6}{e}^{j\pi n}+0+0& & \text{(Substitute coefficients and }3(2\pi /6)=\pi \text{)}\\ & =\frac{1}{3}+\frac{1}{6}(-1{)}^n.& & \end{array}\text{\hspace{1em}(3.116)}$$

This sequence $x[n]$ is sketched in Figure 3.20 (not shown here).

Example 3.15

In this example, we determine and sketch a periodic sequence, given an algebraic expression for its Fourier series coefficients. In the process, we will also exploit the periodic convolution property (see Table 3.2) of the discrete-time Fourier series. Specifically, as stated in the table and as shown in Problem 3.58, if $x[n]$ and $y[n]$ are periodic with period $N$, then the signal

$$w[n]=\sum _{r=⟨N⟩}x[r]y[n-r]$$

is also periodic with period $N$. Here, the summation may be taken over any set of $N$ consecutive values of $r$. Furthermore, the Fourier series coefficients of $w[n]$ are equal to $N{a}_k{b}_k$, where $a_k$ and $b_k$ are the Fourier coefficients of $x[n]$ and $y[n]$, respectively.

Suppose now that we are told that a signal $w[n]$ is periodic with a fundamental period of $N=7$ and with Fourier series coefficients

$$c_k=\frac{{\sin }^2(3\pi k/7)}{7{\sin }^2(\pi k/7)}.\text{\hspace{1em}(3.117)}$$

We observe that $c_k=\frac{1}{7}{\left(\frac{\sin (3\pi k/7)}{\sin (\pi k/7)}\right)}^2$. Let’s relate this to Example 3.12. In Example 3.12, the coefficients for a square wave $x[n]$ with period $N$ and duration $2N_1+1$ were: $a_k=\frac{1}{N}\frac{\sin [2\pi k(N_1+1/2)/N]}{\sin (\pi k/N)}$. Comparing the form, we want $3\pi k/7=2\pi k(N_1+1/2)/N$. Substitute $N=7$: $3\pi k/7=2\pi k(N_1+1/2)/7$. So $3=2(N_1+1/2)=2N_1+1$. This implies $2N_1=2$, so $N_1=1$. Thus, the term $\frac{\sin (3\pi k/7)}{\sin (\pi k/7)}$ is $N{a}_k$ for a square wave $x[n]$ with $N=7$ and $N_1=1$. Let $d_k$ denote the Fourier series coefficients of such a square wave. From Example 3.12 (eq. 3.104 and 3.105), for $N=7,N_1=1$: For $k\ne 0,\pm 7,\dots$: $d_k=\frac{1}{7}\frac{\sin [2\pi k(1+1/2)/7]}{\sin (\pi k/7)}=\frac{1}{7}\frac{\sin (3\pi k/7)}{\sin (\pi k/7)}$. For $k=0,\pm 7,\dots$: $d_k=\frac{2N_1+1}{N}=\frac{2(1)+1}{7}=\frac{3}{7}$. Now let’s look at $c_k=\frac{1}{7}{\left(\frac{\sin (3\pi k/7)}{\sin (\pi k/7)}\right)}^2$. This can be written as $c_k=7\cdot {\left(\frac{1}{7}\frac{\sin (3\pi k/7)}{\sin (\pi k/7)}\right)}^2=7d_k^2$ for $k\ne 0,\pm 7,\dots$. For $k=0,\pm 7,\dots$: $d_k=3/7$. Then $c_k=7(3/7{)}^2=7\cdot 9/49=9/7$. Let’s verify this with the formula. If $k=0$, ${\lim }_{k\to 0}\frac{\sin (3\pi k/7)}{\sin (\pi k/7)}={\lim }_{k\to 0}\frac{3\pi k/7}{\pi k/7}=3$. So $c_0=\frac{1}{7}(3{)}^2=\frac{9}{7}$. This matches $7d_k^2$. So, $c_k=7d_k^2$, where $d_k$ denotes the sequence of Fourier series coefficients of a square wave $x[n]$ with $N=7$ and $N_1=1$. Using the periodic convolution property, we know that if $w[n]\stackrel{\mathrm{FS}}{⟷}N{a}_k{b}_k$, then $w[n]$ is the periodic convolution of $x[n]$ and $y[n]$. Here, $c_k=7d_k{d}_k$. So $w[n]$ is the periodic convolution of $x[n]$ with itself:

$$w[n]=\sum _{r=⟨N⟩}x[r]x[n-r].$$

Let’s choose the summation interval for $r$ as $[-3,3]$ for $N=7$. The square wave $x[n]$ is 1 for $n\in \{-1,0,1\}$ and 0 otherwise within this period.

$$w[n]=\sum _{r=-3}^{3}x[r]x[n-r].\text{\hspace{1em}(3.118)}$$

The sequences $x[r]$, $\hat{x}[r]$ (an aperiodic version of $x[r]$ for one period), and $x[n-r]$ are sketched in Figure 3.21 (a)–(c) (not shown here). Let’s explicitly compute $w[n]$ for $n\in \{-3,-2,-1,0,1,2,3\}$. Recall $x[r]$ is 1 for $r\in \{-1,0,1\}$ and 0 otherwise in the fundamental period. The periodic convolution $w[n]$ can be visualized as shifting $x[-r]$ and multiplying by $x[r]$. $w[n]=(x\ast x)[n]$. $w[0]={\sum }_{r=-3}^{3}x[r]x[-r]=x[-1]x[1]+x[0]x[0]+x[1]x[-1]$ (since $x[r]=0$ for $|r|>1$). $x[-1]=1,x[0]=1,x[1]=1$. $w[0]=(1)(1)+(1)(1)+(1)(1)=3$.

$w[1]={\sum }_{r=-3}^{3}x[r]x[1-r]$ $r=-1:x[-1]x[1-(-1)]=x[-1]x[2]=1\cdot 0=0$. $r=0:x[0]x[1-0]=x[0]x[1]=1\cdot 1=1$. $r=1:x[1]x[1-1]=x[1]x[0]=1\cdot 1=1$. $w[1]=0+1+1=2$. By symmetry of $x[n]$ (it’s an even signal), $w[n]$ must also be even. So $w[-1]=w[1]=2$.

$w[2]={\sum }_{r=-3}^{3}x[r]x[2-r]$ $r=-1:x[-1]x[2-(-1)]=x[-1]x[3]=1\cdot 0=0$. $r=0:x[0]x[2-0]=x[0]x[2]=1\cdot 0=0$. $r=1:x[1]x[2-1]=x[1]x[1]=1\cdot 1=1$. $w[2]=0+0+1=1$. By symmetry, $w[-2]=w[2]=1$.

$w[3]={\sum }_{r=-3}^{3}x[r]x[3-r]$ $r=-1:x[-1]x[3-(-1)]=x[-1]x[4]=1\cdot 0=0$. (Note: $x[4]$ in a periodic sense, $x[4]=x[4-7]=x[-3]=0$) $r=0:x[0]x[3-0]=x[0]x[3]=1\cdot 0=0$. $r=1:x[1]x[3-1]=x[1]x[2]=1\cdot 0=0$. $w[3]=0+0+0=0$. By symmetry, $w[-3]=w[3]=0$.

Since $w[n]$ is periodic with period 7, and we have computed values for one period $\{-3,-2,-1,0,1,2,3\}$, we can sketch $w[n]$ as shown in Figure 3.21(d) (not shown here). The values are: $w[0]=3$, $w[\pm 1]=2$, $w[\pm 2]=1$, $w[\pm 3]=0$.

3.8 Fourier Series and LTI Systems

In the preceding few sections, we have seen that the Fourier series representation can be used to construct any periodic signal in discrete time and essentially all periodic continuous-time signals of practical importance. In addition, in Section 3.2, we saw that the response of an LTI system to a linear combination of complex exponentials takes a particularly simple form. Specifically, in continuous time, if $x(t)=e^{st}$ is the input to a continuous-time LTI system, then the output is given by $y(t)=H(s)e^{st}$, where, from eq. (3.5),

$$H(s)={\int }_{-\infty }^{+\infty }h(\tau )e^{-s\tau }d\tau ,\text{\hspace{1em}(3.119)}$$

in which $h(t)$ is the impulse response of the LTI system.

Similarly, if $x[n]=z^n$ is the input to a discrete-time LTI system, then the output is given by $y[n]=H(z)z^n$, where, from eq. (3.9),

$$H(z)=\sum _{k=-\infty }^{+\infty }h[k]z^{-k},\text{\hspace{1em}(3.120)}$$

in which $h[n]$ is the impulse response of the LTI system.

When $s$ or $z$ are general complex numbers, $H(s)$ and $H(z)$ are referred to as the system functions of the corresponding systems. For continuous-time signals and systems in this and the following chapter, we focus on the specific case in which $\Re e\{s\}=0$, so that $s=j\omega$, and consequently, $e^{st}$ is of the form $e^{j\omega t}$. This input is a complex exponential at frequency $\omega$. The system function of the form $s=j\omega$—i.e., $H(j\omega )$ viewed as a function of $\omega$—is referred to as the frequency response of the system and is given by

$$H(j\omega )={\int }_{-\infty }^{+\infty }h(t)e^{-j\omega t}dt.\text{\hspace{1em}(3.121)}$$

Similarly, for discrete-time signals and systems, we focus in this chapter and in Chapter 5 on values of $z$ for which $|z|=1$, so that $z=e^{j\omega }$ and $z^n$ is of the form $e^{j\omega n}$. Then the system function $H(z)$ for $z$ restricted to the form $z=e^{j\omega }$ is referred to as the frequency response of the system and is given by

$$H(e^{j\omega })=\sum _{n=-\infty }^{+\infty }h[n]e^{-j\omega n}.\text{\hspace{1em}(3.122)}$$

The response of an LTI system to a complex exponential signal of the form $e^{j\omega t}$ (in continuous time) or $e^{j\omega n}$ (in discrete time) is particularly simple to express in terms of the frequency response of the system. Furthermore, as a result of the superposition property for LTI systems, we can express the response of an LTI system to a linear combination of complex exponentials with equal ease. In Chapters 4 and 5, we will see how we can use these ideas together with continuous-time and discrete-time Fourier transforms to analyze the response of LTI systems to aperiodic signals. In the remainder of this chapter, as a first look at this important set of concepts and results, we focus on interpreting and understanding this notion in the context of periodic signals.

Consider first the continuous-time case, and let $x(t)$ be a periodic signal with a Fourier series representation given by

$$x(t)=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}.\text{\hspace{1em}(3.123)}$$

Suppose that we apply this signal as the input to an LTI system with impulse response $h(t)$. Then, since each of the complex exponentials in eq. (3.123) is an eigenfunction of the system, as in eq. (3.12) with $s_k=jk{\omega }_0$, it follows that the output is

$$y(t)=\sum _{k=-\infty }^{+\infty }{a}_{k}H(jk{\omega }_0)e^{jk{\omega }_{0}t}.\text{\hspace{1em}(3.124)}$$

Thus, $y(t)$ is also periodic with the same fundamental frequency as $x(t)$. Furthermore, if $\{a_k\}$ is the set of Fourier series coefficients for the input $x(t)$, then $\{a_{k}H(jk{\omega }_0)\}$ is the set of coefficients for the output $y(t)$. That is, the effect of the LTI system is to modify individually each of the Fourier coefficients of the input through multiplication by the value of the frequency response at the corresponding frequency.

Example 3.16

Suppose that the periodic signal $x(t)$ discussed in Example 3.2 is the input signal to an LTI system with impulse response

$$h(t)=e^{-t}u(t).$$

To calculate the Fourier series coefficients of the output $y(t)$, we first compute the frequency response $H(j\omega )$ using eq. (3.121):

$$\begin{array}{rlrl}H(j\omega )& ={\int }_{-\infty }^{+\infty }h(\tau )e^{-j\omega \tau }d\tau & & \text{(Definition of frequency response)}\\ & ={\int }_0^{\infty }{e}^{-\tau }{e}^{-j\omega \tau }d\tau & & \text{(Substitute }h(t)\text{ and limits from }u(t)\text{)}\\ & ={\int }_0^{\infty }{e}^{-(1+j\omega )\tau }d\tau & & \text{(Combine exponentials)}\\ & ={-\frac{1}{1+j\omega }{e}^{-(1+j\omega )\tau }|}_0^{\infty }& & \text{(Evaluate integral)}\\ & =-\frac{1}{1+j\omega }(0-e^0)& & \text{(Assuming convergence for }\Re e\{1+j\omega \}>0⟹1>0\text{, which is true)}\\ & =\frac{1}{1+j\omega }.& & \end{array}\text{\hspace{1em}(3.125)}$$

Therefore, using eqs. (3.124) and (3.125), together with the fact that ${\omega }_0=2\pi$ in Example 3.2, we obtain the output $y(t)$:

$$y(t)=\sum _{k=-3}^{+3}{b}_k{e}^{jk2\pi t},\text{\hspace{1em}(3.126)}$$

with $b_k=a_{k}H(jk2\pi )$. The coefficients $a_k$ from Example 3.2 are $a_0=1,a_1=a_{-1}=1/4,a_2=a_{-2}=1/2,a_3=a_{-3}=1/3$. Now we compute $H(jk2\pi )$ for each $k$:

$$H(jk2\pi )=\frac{1}{1+j(k2\pi )}.$$

Thus, the output Fourier coefficients $b_k$ are:

$$\begin{array}{rl}{b}_0& =a_{0}H(j0)=1\cdot \frac{1}{1+j0}=1,\\ b_1& =a_{1}H(j2\pi )=\frac{1}{4}\left(\frac{1}{1+j2\pi }\right),\\ b_{-1}& =a_{-1}H(-j2\pi )=\frac{1}{4}\left(\frac{1}{1-j2\pi }\right),\\ b_2& =a_{2}H(j4\pi )=\frac{1}{2}\left(\frac{1}{1+j4\pi }\right),\\ b_{-2}& =a_{-2}H(-j4\pi )=\frac{1}{2}\left(\frac{1}{1-j4\pi }\right),\\ b_3& =a_{3}H(j6\pi )=\frac{1}{3}\left(\frac{1}{1+j6\pi }\right),\\ b_{-3}& =a_{-3}H(-j6\pi )=\frac{1}{3}\left(\frac{1}{1-j6\pi }\right).\end{array}\text{\hspace{1em}(3.127)}$$

Note that $y(t)$ must be a real-valued signal, since it is the convolution of $x(t)$ and $h(t)$, which are both real. This can be verified by examining eq. (3.127) and observing that $b_k^{\ast }=b_{-k}$. Therefore, $y(t)$ can also be expressed in either of the forms given in eqs. (3.30) and (3.31); that is,

$$y(t)=1+2\sum _{k=1}^3{D}_k\cos \left(2\pi kt+{\theta }_k\right),\text{\hspace{1em}(3.128)}$$

or

$$y(t)=1+2\sum _{k=1}^3\left[E_k\cos 2\pi kt-F_k\sin 2\pi kt\right],\text{\hspace{1em}(3.129)}$$

where

$$b_k=D_k{e}^{j{\theta }_k}=E_k+j{F}_k,k=1,2,3.\text{\hspace{1em}(3.130)}$$

These coefficients can be evaluated directly from eq. (3.127). For example, for $k=1$:

$$b_1=\frac{1}{4(1+j2\pi )}=\frac{1}{4}\frac{1-j2\pi }{1+(2\pi {)}^2}=\frac{1}{4(1+4{\pi }^2)}-j\frac{2\pi }{4(1+4{\pi }^2)}.$$

So, comparing $b_1=E_1+j{F}_1$:

$$E_1=\Re e\{b_1\}=\frac{1}{4(1+4{\pi }^2)}.$$ $$F_1=\Im m\{b_1\}=-\frac{2\pi }{4(1+4{\pi }^2)}=-\frac{\pi }{2(1+4{\pi }^2)}.$$

And comparing $b_1=D_1{e}^{j{\theta }_1}$:

$$D_1=|b_1|=\left|\frac{1}{4(1+j2\pi )}\right|=\frac{1}{4|1+j2\pi |}=\frac{1}{4\sqrt{1^2+(2\pi {)}^2}}=\frac{1}{4\sqrt{1+4{\pi }^2}}.$$ $${\theta }_1=\angle b_1=\angle \left(\frac{1}{1+j2\pi }\right)=-\angle (1+j2\pi )=-{\tan }^{-1}(2\pi ).$$

In discrete time, the relationship between the Fourier series coefficients of the input and output of an LTI system exactly parallels eqs. (3.123) and (3.124). Specifically, let $x[n]$ be a periodic signal with Fourier series representation given by

$$x[n]=\sum _{k=⟨N⟩}{a}_k{e}^{jk(2\pi /N)n}.$$

If we apply this signal as the input to an LTI system with impulse response $h[n]$, then, as in eq. (3.15) with $z_k=e^{jk(2\pi /N)}$, the output is

$$y[n]=\sum _{k=⟨N⟩}{a}_{k}H(e^{j2\pi k/N})e^{jk(2\pi /N)n}.\text{\hspace{1em}(3.131)}$$

Thus, $y[n]$ is also periodic with the same period as $x[n]$, and the $k$th Fourier coefficient of $y[n]$ is the product of the $k$th Fourier coefficient of the input and the value of the frequency response of the LTI system, $H(e^{j2\pi k/N})$, at the corresponding frequency.

Example 3.17

Consider an LTI system with impulse response $h[n]={\alpha }^{n}u[n]$, $-1<\alpha <1$, and with the input

$$x[n]=\cos \left(\frac{2\pi n}{N}\right).\text{\hspace{1em}(3.132)}$$

As in Example 3.10, $x[n]$ can be written in Fourier series form as:

$$x[n]=\frac{1}{2}{e}^{j(2\pi /N)n}+\frac{1}{2}{e}^{-j(2\pi /N)n}.$$

Also, from eq. (3.122), the frequency response $H(e^{j\omega })$ is:

$$\begin{array}{rlrl}H(e^{j\omega })& =\sum _{n=-\infty }^{+\infty }h[n]e^{-j\omega n}& & \text{(Definition of frequency response)}\\ & =\sum _{n=0}^{\infty }{\alpha }^n{e}^{-j\omega n}& & \text{(Substitute }h[n]={\alpha }^{n}u[n]\text{)}\\ & =\sum _{n=0}^{\infty }{\left(\alpha e^{-j\omega }\right)}^n.& & \end{array}\text{\hspace{1em}(3.133)}$$

This geometric series can be evaluated using the result of Problem 1.54, which states that for $|r|<1$, ${\sum }_{n=0}^{\infty }{r}^n=\frac{1}{1-r}$. Here $r=\alpha e^{-j\omega }$. Since $|\alpha |<1$, $|r|=|\alpha ||e^{-j\omega }|=|\alpha |<1$. So, the sum converges.

$$H(e^{j\omega })=\frac{1}{1-\alpha e^{-j\omega }}.\text{\hspace{1em}(3.134)}$$

Using eq. (3.131), the output $y[n]$ Fourier series will have coefficients $b_k=a_{k}H(e^{j2\pi k/N})$. The input coefficients are $a_1=1/2$ (for frequency $2\pi /N$) and $a_{-1}=1/2$ (for frequency $-2\pi /N$). All other $a_k$ are zero for one period.

$$\begin{array}{rl}y[n]& =a_{1}H(e^{j2\pi /N})e^{j(2\pi /N)n}+a_{-1}H(e^{-j2\pi /N})e^{-j(2\pi /N)n}\\ & =\frac{1}{2}H(e^{j2\pi /N})e^{j(2\pi /N)n}+\frac{1}{2}H(e^{-j2\pi /N})e^{-j(2\pi /N)n}\\ & =\frac{1}{2}\left(\frac{1}{1-\alpha e^{-j2\pi /N}}\right)e^{j(2\pi /N)n}+\frac{1}{2}\left(\frac{1}{1-\alpha e^{j2\pi /N}}\right)e^{-j(2\pi /N)n}.\end{array}\text{\hspace{1em}(3.135)}$$

Let $H_N=\frac{1}{1-\alpha e^{-j2\pi /N}}$. Then $H_N^{\ast }$ is $\frac{1}{1-\alpha e^{j2\pi /N}}$. So, $y[n]=\frac{1}{2}{H}_N{e}^{j(2\pi /N)n}+\frac{1}{2}{H}_N^{\ast }{e}^{-j(2\pi /N)n}$. Let $H_N=r{e}^{j\theta }$ (polar form), then $H_N^{\ast }=r{e}^{-j\theta }$.

$$\begin{array}{rl}y[n]& =\frac{1}{2}r{e}^{j\theta }{e}^{j(2\pi /N)n}+\frac{1}{2}r{e}^{-j\theta }{e}^{-j(2\pi /N)n}\\ & =\frac{r}{2}\left(e^{j((2\pi /N)n+\theta )}+e^{-j((2\pi /N)n+\theta )}\right)\\ & =r\cos \left(\frac{2\pi }{N}n+\theta \right).\end{array}\text{\hspace{1em}(3.136)}$$

For example, if $N=4$, then $2\pi /N=2\pi /4=\pi /2$.

$$H(e^{j\pi /2})=\frac{1}{1-\alpha e^{-j\pi /2}}=\frac{1}{1-\alpha (-j)}=\frac{1}{1+\alpha j}.$$

To find $r$ and $\theta$:

$$r=|H(e^{j\pi /2})|=\left|\frac{1}{1+\alpha j}\right|=\frac{1}{\sqrt{1^2+{\alpha }^2}}=\frac{1}{\sqrt{1+{\alpha }^2}}.$$ $$\theta =\angle H(e^{j\pi /2})=\angle \left(\frac{1}{1+\alpha j}\right)=-\angle (1+\alpha j)=-{\tan }^{-1}(\alpha ).$$

Therefore, for $N=4$:

$$y[n]=\frac{1}{\sqrt{1+{\alpha }^2}}\cos \left(\frac{\pi n}{2}-{\tan }^{-1}(\alpha )\right).$$

We note that for expressions such as eqs. (3.124) and (3.131) to make sense, the frequency responses $H(j\omega )$ and $H(e^{j\omega })$ in eqs. (3.121) and (3.122) must be well defined and finite. As we will see in Chapters 4 and 5, this will be the case if the LTI systems under consideration are stable. For example, the LTI system in Example 3.16, with impulse response $h(t)=e^{-t}u(t)$, is stable and has a well-defined frequency response given by eq. (3.125). On the other hand, an LTI system with impulse response $h(t)=e^{t}u(t)$ is unstable, and it is easy to check that the integral in eq. (3.121) for $H(j\omega )$ diverges for any value of $\omega$. Similarly, the LTI system in Example 3.17, with impulse response $h[n]={\alpha }^{n}u[n]$, is stable for $|\alpha |<1$ and has frequency response given by eq. (3.134). However, if $|\alpha |>1$, the system is unstable, and then the summation in eq. (3.133) diverges.

3.9 Filtering

In a variety of applications, it is of interest to change the relative amplitudes of the frequency components in a signal or perhaps eliminate some frequency components entirely, a process referred to as filtering. Linear time-invariant systems that change the shape of the spectrum are often referred to as frequency-shaping filters. Systems that are designed to pass some frequencies essentially undistorted and significantly attenuate or eliminate others are referred to as frequency-selective filters. As indicated by eqs. (3.124) and (3.131), the Fourier series coefficients of the output of an LTI system are those of the input multiplied by the frequency response of the system. Consequently, filtering can be conveniently accomplished through the use of LTI systems with an appropriately chosen frequency response, and frequency-domain methods provide us with the ideal tools to examine this very important class of applications. In this and the following two sections, we take a first look at filtering through a few examples.

3.9.1 Frequency-Shaping Filters

One application in which frequency-shaping filters are often encountered is audio systems. For example, LTI filters are typically included in such systems to permit the listener to modify the relative amounts of low-frequency energy (bass) and high-frequency energy (treble). These filters correspond to LTI systems whose frequency responses can be changed by manipulating the tone controls. Also, in high-fidelity audio systems, a so-called equalizing filter is often included in the preamplifier to compensate for the frequency-response characteristics of the speakers. Overall, these cascaded filtering stages are frequently referred to as the equalizing or equalizer circuits for the audio system. Figure 3.22 (not shown here) illustrates the three stages of the equalizer circuits for one particular series of audio speakers. In this figure, the magnitude of the frequency response for each of these stages is shown on a log-log plot. Specifically, the magnitude is in units of $20{\log }_{10}|H(j\omega )|$, referred to as decibels or dB. The frequency axis is labeled in Hz (i.e., $\omega /2\pi$) along a logarithmic scale. As will be discussed in more detail in Section 6.2.3, a logarithmic display of the magnitude of the frequency response in this form is common and useful.

Taken together, the equalizing circuits in Figure 3.22 are designed to compensate for the frequency response of the speakers and the room in which they are located and to allow the listener to control the overall frequency response. In particular, since the three systems are connected in cascade, and since each system modifies a complex exponential input $K{e}^{j\omega t}$ by multiplying it by the system frequency response at that frequency, it follows that the overall frequency response of the cascade of the three systems is the product of the three frequency responses. The first two filters, indicated in Figures 3.22(a) and (b), together make up the control stage of the system, as the frequency behavior of these filters can be adjusted by the listener. The second filter in the control stage has two continuously adjustable slider switches to vary the frequency response within the limits indicated in Figure 3.22(b).

Another class of frequency-shaping filters often encountered is that for which the filter output is the derivative of the filter input, i.e., $y(t)=dx(t)/dt$. With $x(t)$ of the form $x(t)=e^{j\omega t}$, $y(t)$ will be $y(t)=j\omega e^{j\omega t}$, from which it follows that the frequency response is

$$H(j\omega )=j\omega .\text{\hspace{1em}(3.137)}$$

The frequency response characteristics of a differentiating filter are shown in Figure 3.23 (not shown here). Since $H(j\omega )$ is complex in general, and in this example in particular, $H(j\omega )$ is frequently displayed (as in the figure) as separate plots of $|H(j\omega )|$ and $\angle H(j\omega )$. The shape of this frequency response implies that a complex exponential input $e^{j\omega t}$ will receive greater amplification for larger values of $\omega$. Consequently, differentiating filters are useful in enhancing rapid variations or transitions in a signal.

One purpose for which differentiating filters are often used is to enhance edges in picture processing. A black-and-white picture can be thought of as a two-dimensional “continuous-time” signal $x(t_1,t_2)$, where $t_1$ and $t_2$ are the horizontal and vertical coordinates, respectively, and $x(t_1,t_2)$ is the brightness of the image. If the image is repeated periodically in the horizontal and vertical directions, then it can be represented by a two-dimensional Fourier series (see Problem 3.70) consisting of sums of products of complex exponentials, $e^{j{\omega }_1{t}_1}$ and $e^{j{\omega }_2{t}_2}$, that oscillate at possibly different frequencies in each of the two coordinate directions. Slow variations in brightness in a particular direction are represented by the lower harmonics in that direction. For example, consider an edge corresponding to a sharp transition in brightness that runs vertically in an image. Since the brightness is constant or slowly varying along the edge, the frequency content of the edge in the vertical direction is concentrated at low frequencies. In contrast, since there is an abrupt variation in brightness across the edge, the frequency content of the edge in the horizontal direction is concentrated at higher frequencies. Figure 3.24 (not shown here) illustrates the effect on an image of the two-dimensional equivalent of a differentiating filter. Figure 3.24(a) shows two original images and Figure 3.24(b) the result of processing those images with the filter. Since the derivative at the edges of a picture is greater than in regions where the brightness varies slowly with distance, the effect of the filter is to enhance the edges.

Discrete-time LTI filters also find a broad array of applications. Many of these involve the use of discrete-time systems, implemented using general- or special-purpose digital processors, to process continuous-time signals, a topic we discuss at some length in Chapter 7. In addition, the analysis of time series information, including demographic data and economic data sequences such as the stock market average, commonly involves the use of discrete-time filters. Often the long-term variations (which correspond to low frequencies) have a different significance than the short-term variations (which correspond to high frequencies), and it is useful to analyze these components separately. Reshaping the relative weighting of the components is typically accomplished using discrete-time filters.

As one example of a simple discrete-time filter, consider an LTI system that successively takes a two-point average of the input values:

$$y[n]=\frac{1}{2}(x[n]+x[n-1]).\text{\hspace{1em}(3.138)}$$

In this case, the impulse response is $h[n]=\frac{1}{2}(\delta [n]+\delta [n-1])$. From eq. (3.122), we see that the frequency response of the system is:

$$\begin{array}{rlrl}H(e^{j\omega })& =\sum _{k=-\infty }^{+\infty }h[k]e^{-j\omega k}& & \text{(Definition of discrete-time frequency response)}\\ & =\frac{1}{2}\sum _{k=-\infty }^{+\infty }(\delta [k]+\delta [k-1])e^{-j\omega k}& & \text{(Substitute }h[k]\text{)}\\ & =\frac{1}{2}(e^{-j\omega (0)}+e^{-j\omega (1)})& & \text{(Property of impulse function)}\\ & =\frac{1}{2}[1+e^{-j\omega }]& & \\ & =\frac{1}{2}{e}^{-j\omega /2}(e^{j\omega /2}+e^{-j\omega /2})& & \text{(Factor out }{e}^{-j\omega /2}\text{)}\\ & =e^{-j\omega /2}\cos (\omega /2).& & \end{array}\text{\hspace{1em}(3.139)}$$

The magnitude of $H(e^{j\omega })$ is plotted in Figure 3.25(a) (not shown here), and $\angle H(e^{j\omega })$ is shown in Figure 3.25(b) (not shown here). As discussed in Section 1.3.3, low frequencies for discrete-time complex exponentials occur near $\omega =0,\pm 2\pi ,\pm 4\pi ,\dots$, and high frequencies near $\omega =\pm \pi ,\pm 3\pi ,\dots$. This is a result of the fact that $e^{j(\omega +2\pi )n}=e^{j\omega n}$, so that in discrete time, we need only consider a $2\pi$ interval of values of $\omega$ in order to cover a complete range of distinct discrete-time frequencies. As a consequence, any discrete-time frequency response $H(e^{j\omega })$ must be periodic with period $2\pi$, a fact that can also be deduced directly from eq. (3.122).

For the specific filter defined in eqs. (3.138) and (3.139), we see from Figure 3.25(a) that $|H(e^{j\omega })|$ is large for frequencies near $\omega =0$ and decreases as we increase $|\omega |$ toward $\pi$, indicating that higher frequencies are attenuated more than lower ones. For example, if the input to this system is constant—i.e., a zero-frequency complex exponential $x[n]=K{e}^{j0\cdot n}=K$—then the output will be

$$y[n]=H(e^{j\cdot 0})K{e}^{j0\cdot n}=\left(e^{-j0/2}\cos (0/2)\right)K=(1\cdot 1)K=K=x[n].$$

On the other hand, if the input is the high-frequency signal $x[n]=K{e}^{j\pi n}=K(-1{)}^n$, then the output will be

$$y[n]=H(e^{j\pi })K{e}^{j\pi n}=\left(e^{-j\pi /2}\cos (\pi /2)\right)K{e}^{j\pi n}=((-j)\cdot 0)K{e}^{j\pi n}=0.$$

Thus, this system separates out the long-term constant value of a signal from its high-frequency fluctuations and, consequently, represents a first example of frequency-selective filtering, a topic we look at more carefully in the next subsection.

3.9.2 Frequency-Selective Filters

Frequency-selective filters are a class of filters specifically intended to accurately or approximately select some bands of frequencies and reject others. The use of frequency-selective filters arises in a variety of situations. For example, if noise in an audio recording is in a higher frequency band than the music or voice on the recording is, it can be removed by frequency-selective filtering. Another important application of frequency-selective filters is in communication systems. As we discuss in detail in Chapter 8, the basis for amplitude modulation (AM) systems is the transmission of information from many different sources simultaneously by putting the information from each channel into a separate frequency band and extracting the individual channels or bands at the receiver using frequency-selective filters. Frequency-selective filters for separating the individual channels and frequency-shaping filters (such as the equalizer illustrated in Figure 3.22) for adjusting the quality of the tone form a major part of any home radio and television receiver.

While frequency selectivity is not the only issue of concern in applications, its broad importance has led to a widely accepted set of terms describing the characteristics of frequency-selective filters. In particular, while the nature of the frequencies to be passed by a frequency-selective filter varies considerably from application to application, several basic types of filter are widely used and have been given names indicative of their function. For example, a lowpass filter is a filter that passes low frequencies—i.e., frequencies around $\omega =0$—and attenuates or rejects higher frequencies. A highpass filter is a filter that passes high frequencies and attenuates or rejects low ones, and a bandpass filter is a filter that passes a band of frequencies and attenuates frequencies both higher and lower than those in the band that is passed. In each case, the cutoff frequencies are the frequencies defining the boundaries between frequencies that are passed and frequencies that are rejected—i.e., the frequencies in the passband and stopband.

Numerous questions arise in defining and assessing the quality of a frequency-selective filter. How effective is the filter at passing frequencies in the passband? How effective is it at attenuating frequencies in the stopband? How sharp is the transition near the cutoff frequency—i.e., from nearly free of distortion in the passband to highly attenuated in the stopband? Each of these questions involves a comparison of the characteristics of an actual frequency-selective filter with those of a filter with idealized behavior. Specifically, an ideal frequency-selective filter is a filter that exactly passes complex exponentials at one set of frequencies without any distortion and completely rejects signals at all other frequencies. For example, a continuous-time ideal lowpass filter with cutoff frequency ${\omega }_c$ is an LTI system that passes complex exponentials $e^{j\omega t}$ for values of $\omega$ in the range $-{\omega }_c\le \omega \le {\omega }_c$ and rejects signals at all other frequencies. That is, the frequency response of a continuous-time ideal lowpass filter is

$$H(j\omega )=\{\begin{array}{ll}1,& |\omega |\le {\omega }_c\\ 0,& |\omega |>{\omega }_c\end{array},\text{\hspace{1em}(3.140)}$$

as shown in Figure 3.26 (not shown here).

Figure 3.27(a) (not shown here) depicts the frequency response of an ideal continuous-time highpass filter with cutoff frequency ${\omega }_c$, and Figure 3.27(b) (not shown here) illustrates an ideal continuous-time bandpass filter with lower cutoff frequency ${\omega }_{c1}$ and upper cutoff frequency ${\omega }_{c2}$. Note that each of these filters is symmetric about $\omega =0$, and thus, there appear to be two passbands for the highpass and bandpass filters. This is a consequence of our having adopted the use of the complex exponential signal $e^{j\omega t}$, rather than the sinusoidal signals $\sin \omega t$ and $\cos \omega t$, at frequency $\omega$. Since $e^{j\omega t}=\cos \omega t+j\sin \omega t$ and $e^{-j\omega t}=\cos \omega t-j\sin \omega t$, both of these complex exponentials are composed of sinusoidal signals at the same frequency $\omega$. For this reason, we usually define ideal filters so that they have the symmetric frequency-response behavior seen in Figures 3.26 and 3.27.

In a similar fashion, we can define the corresponding set of ideal discrete-time frequency-selective filters, the frequency responses for which are depicted in Figure 3.28 (not shown here). In particular, Figure 3.28(a) depicts an ideal discrete-time lowpass filter, Figure 3.28(b) is an ideal highpass filter, and Figure 3.28(c) is an ideal bandpass filter. Note that, as discussed in the preceding section, the characteristics of the continuous-time and discrete-time ideal filters differ by virtue of the fact that, for discrete-time filters, the frequency response $H(e^{j\omega })$ must be periodic with period $2\pi$, with low frequencies near even multiples of $\pi$ and high frequencies near odd multiples of $\pi$.

As we will see on numerous occasions, ideal filters are quite useful in describing idealized system configurations for a variety of applications. However, they are not realizable in practice and must be approximated. Furthermore, even if they could be realized, some of the characteristics of ideal filters might make them undesirable for particular applications, and a nonideal filter might in fact be preferable.

In detail, the topic of filtering encompasses many issues, including design and implementation. While we will not delve deeply into the details of filter design methodologies, in the remainder of this chapter and the following chapters, we will see a number of other examples of both continuous-time and discrete-time filters and will develop the concepts and techniques that form the basis of this very important engineering discipline.

3.10 Examples of Continuous-Time Filters Described by Differential Equations

In many applications, frequency-selective filtering is accomplished through the use of LTI systems described by linear constant-coefficient differential or difference equations. The reasons for this are numerous. For example, many physical systems that can be interpreted as performing filtering operations are characterized by differential or difference equations. A good example of this that we will examine in Chapter 6 is an automobile suspension system, which in part is designed to filter out high-frequency bumps and irregularities in road surfaces. A second reason for the use of filters described by differential or difference equations is that they are conveniently implemented using either analog or digital hardware. Furthermore, systems described by differential or difference equations offer an extremely broad and flexible range of designs, allowing one, for example, to produce filters that are close to ideal or that possess other desirable characteristics. In this and the next section, we consider several examples that illustrate the implementation of continuous-time and discrete-time frequency-selective filters through the use of differential and difference equations. In Chapters 4-6, we will see other examples of these classes of filters and will gain additional insights into the properties that make them so useful.

3.10.1 A Simple RC Lowpass Filter

Electrical circuits are widely used to implement continuous-time filtering operations. One of the simplest examples of such a circuit is the first-order RC circuit depicted in Figure 3.29 (not shown here), where the source voltage $v_s(t)$ is the system input. This circuit can be used to perform either a lowpass or highpass filtering operation, depending upon what we take as the output signal. In particular, suppose that we take the capacitor voltage $v_c(t)$ as the output. In this case, the output voltage is related to the input voltage through the linear constant-coefficient differential equation

$$RC\frac{d{v}_c(t)}{dt}+v_c(t)=v_s(t).\text{\hspace{1em}(3.141)}$$

Assuming initial rest, the system described by eq. (3.141) is LTI. In order to determine its frequency response $H(j\omega )$, we note that, by definition, with input voltage $v_s(t)=e^{j\omega t}$, we must have the output voltage $v_c(t)=H(j\omega )e^{j\omega t}$. If we substitute these expressions into eq. (3.141), we obtain:

$$RC\frac{d}{dt}[H(j\omega )e^{j\omega t}]+H(j\omega )e^{j\omega t}=e^{j\omega t}.\text{\hspace{1em}(3.142)}$$

Differentiating the first term:

$$RCj\omega H(j\omega )e^{j\omega t}+H(j\omega )e^{j\omega t}=e^{j\omega t}.\text{\hspace{1em}(3.143)}$$

Factoring out $H(j\omega )e^{j\omega t}$ from the left side:

$$[RCj\omega +1]H(j\omega )e^{j\omega t}=e^{j\omega t}.\text{\hspace{1em}(3.144)}$$

Dividing both sides by $e^{j\omega t}$:

$$(1+RCj\omega )H(j\omega )=1.$$

From which it follows directly that:

$$H(j\omega )=\frac{1}{1+RCj\omega }.\text{\hspace{1em}(3.145)}$$

The magnitude and phase of the frequency response $H(j\omega )$ for this example are shown in Figure 3.30 (not shown here). Note that for frequencies near $\omega =0$, $|H(j\omega )|\approx 1$, while for larger values of $\omega$ (positive or negative), $|H(j\omega )|$ is considerably smaller and in fact steadily decreases as $|\omega |$ increases. Thus, this simple $RC$ filter (with $v_c(t)$ as output) is a nonideal lowpass filter.

To provide a first glimpse at the trade-offs involved in filter design, let us briefly consider the time-domain behavior of the circuit. In particular, the impulse response of the system described by eq. (3.141) is (from standard solutions to first-order ODEs):

$$h(t)=\frac{1}{RC}{e}^{-t/RC}u(t).\text{\hspace{1em}(3.146)}$$

And the step response $s(t)=h(t)\ast u(t)={\int }_{-\infty }^{t}h(\tau )u(t-\tau )d\tau ={\int }_0^t\frac{1}{RC}{e}^{-\tau /RC}d\tau$ for $t\ge 0$.

$$\begin{array}{rl}s(t)& ={\left[-\frac{1}{RC}RC{e}^{-\tau /RC}\right]}_0^{t}u(t)\\ & =[-e^{-\tau /RC}{]}_0^{t}u(t)\\ & =(-e^{-t/RC}-(-e^0))u(t)\\ & =[1-e^{-t/RC}]u(t).\end{array}\text{\hspace{1em}(3.147)}$$

both of which are plotted in Figure 3.31 (not shown here, where $\tau =RC$). Comparing Figures 3.30 and 3.31, we see a fundamental trade-off. Specifically, suppose that we would like our filter to pass only very low frequencies. From Figure 3.30(a), this implies that $1/RC$ must be small, or equivalently, that $RC$ is large, so that frequencies other than the low ones of interest will be attenuated sufficiently. However, looking at Figure 3.31(b), we see that if $RC$ is large, then the step response will take a considerable amount of time to reach its long-term value of 1. That is, the system responds sluggishly to the step input. Conversely, if we wish to have a faster step response, we need a smaller value of $RC$, which in turn implies that the filter will pass higher frequencies. This type of trade-off between behavior in the frequency domain and in the time domain is typical of the issues arising in the design and analysis of LTI systems and filters and is a subject we will look at more carefully in Chapter 6.

3.10.2 A Simple RC Highpass Filter

As an alternative to choosing the capacitor voltage as the output in our $RC$ circuit, we can choose the voltage across the resistor. In this case, $v_r(t)=i(t)R$. The current $i(t)$ is $C\frac{d{v}_c(t)}{dt}$. Also $v_s(t)=v_c(t)+v_r(t)$. So $v_c(t)=v_s(t)-v_r(t)$. Substitute $v_c(t)$ into $i(t)$ relationship: $i(t)=C\frac{d}{dt}(v_s(t)-v_r(t))=C\frac{d{v}_s(t)}{dt}-C\frac{d{v}_r(t)}{dt}$. Since $v_r(t)=i(t)R$: $v_r(t)=R\left(C\frac{d{v}_s(t)}{dt}-C\frac{d{v}_r(t)}{dt}\right)$. $v_r(t)=RC\frac{d{v}_s(t)}{dt}-RC\frac{d{v}_r(t)}{dt}$. Rearranging terms:

$$RC\frac{d{v}_r(t)}{dt}+v_r(t)=RC\frac{d{v}_s(t)}{dt}.\text{\hspace{1em}(3.148)}$$

We can find the frequency response $G(j\omega )$ of this system in exactly the same way we did in the previous case: If $v_s(t)=e^{j\omega t}$, then we must have $v_r(t)=G(j\omega )e^{j\omega t}$. Substituting these expressions into eq. (3.148) and performing a bit of algebra, we find that:

$$\begin{array}{rlrl}RC\frac{d}{dt}[G(j\omega )e^{j\omega t}]+G(j\omega )e^{j\omega t}& =RC\frac{d}{dt}[e^{j\omega t}]& & \text{(Substitute input and output forms)}\\ RCj\omega G(j\omega )e^{j\omega t}+G(j\omega )e^{j\omega t}& =RCj\omega e^{j\omega t}& & \text{(Differentiate exponentials)}\\ G(j\omega )e^{j\omega t}(RCj\omega +1)& =RCj\omega e^{j\omega t}& & \text{(Factor out common terms)}\\ G(j\omega )& =\frac{RCj\omega }{1+RCj\omega }.& & \text{(Solve for }G(j\omega )\text{)}\end{array}\text{\hspace{1em}(3.149)}$$

The magnitude and phase of this frequency response are shown in Figure 3.32 (not shown here). From the figure, we see that the system attenuates lower frequencies and passes higher frequencies—i.e., those for which $|\omega |\gg 1/RC$—with minimal attenuation. That is, this system acts as a nonideal highpass filter.

As with the lowpass filter, the parameters of the circuit control both the frequency response of the highpass filter and its time response characteristics. For example, consider the step response for the filter. From Figure 3.29, we see that $v_r(t)=v_s(t)-v_c(t)$. Thus, if $v_s(t)=u(t)$, $v_c(t)$ must be given by eq. (3.147). Consequently, the step response of the highpass filter is:

$$\begin{array}{rlrl}{s}_r(t)& =u(t)-s_c(t)& & \text{(Relationship of voltages)}\\ & =u(t)-[1-e^{-t/RC}]u(t)& & \text{(Substitute }{s}_c(t)\text{ from eq. (3.147))}\\ & =[1-(1-e^{-t/RC})]u(t)& & \\ & =e^{-t/RC}u(t).& & \end{array}\text{\hspace{1em}(3.150)}$$

which is depicted in Figure 3.33 (not shown here). Consequently, as $RC$ is increased, the response becomes more sluggish—i.e., the step response takes a longer time to reach its long-term value of 0. From Figure 3.32, we see that increasing $RC$ (or equivalently, decreasing $1/RC$) also affects the frequency response, specifically, it extends the passband down to lower frequencies.

We observe from the two examples in this section that a simple $RC$ circuit can serve as a rough approximation to a highpass or a lowpass filter, depending upon the choice of the physical output variable. As illustrated in Problem 3.71, a simple mechanical system using a mass and a mechanical damper can also serve as a lowpass or highpass filter described by analogous first-order differential equations. Because of their simplicity, these examples of electrical and mechanical filters do not have a sharp transition from passband to stopband and, in fact, have only a single parameter (namely, $RC$ in the electrical case) that controls both the frequency response and time response behavior of the system. By designing more complex filters, implemented using more energy storage elements (capacitances and inductances in electrical filters and springs and damping devices in mechanical filters), we obtain filters described by higher order differential equations. Such filters offer considerably more flexibility in terms of their characteristics, allowing, for example, sharper passband-stopband transition or more control over the trade-offs between time response and frequency response.

3.11 Examples of Discrete-Time Filters Described by Difference Equations

As with their continuous-time counterparts, discrete-time filters described by linear constant-coefficient difference equations are of considerable importance in practice. Indeed, since they can be efficiently implemented in special- or general-purpose digital systems, filters described by difference equations are widely used in practice. As in almost all aspects of signal and system analysis, when we examine discrete-time filters described by difference equations, we find both strong similarities and important differences with the continuous-time case. In particular, discrete-time LTI systems described by difference equations can either be recursive and have impulse responses of infinite length (IIR systems) or be nonrecursive and have finite-length impulse responses (FIR systems). The former are the direct counterparts of continuous-time systems described by differential equations illustrated in the previous section, while the latter are also of considerable practical importance in digital systems. These two classes have distinct sets of advantages and disadvantages in terms of ease of implementation and in terms of the order of filter or the complexity required to achieve particular design objectives. In this section, we limit ourselves to a few simple examples of recursive and nonrecursive filters, while in Chapters 5 and 6, we develop additional tools and insights that allow us to analyze and understand the properties of these systems in more detail.

3.11.1 First-Order Recursive Discrete-Time Filters

The discrete-time counterpart of each of the first-order filters considered in Section 3.10 is the LTI system described by the first-order difference equation

$$y[n]-ay[n-1]=x[n].\text{\hspace{1em}(3.151)}$$

From the eigenfunction property of complex exponential signals, we know that if $x[n]=e^{j\omega n}$, then $y[n]=H(e^{j\omega })e^{j\omega n}$, where $H(e^{j\omega })$ is the frequency response of the system. Substituting into eq. (3.151), we obtain:

$$H(e^{j\omega })e^{j\omega n}-aH(e^{j\omega })e^{j\omega (n-1)}=e^{j\omega n}.\text{\hspace{1em}(3.152)}$$

Factoring out $e^{j\omega n}$:

$$H(e^{j\omega })[1-a{e}^{-j\omega }]e^{j\omega n}=e^{j\omega n}.\text{\hspace{1em}(3.153)}$$

Dividing both sides by $e^{j\omega n}$:

$$H(e^{j\omega })(1-a{e}^{-j\omega })=1.$$

So that:

$$H(e^{j\omega })=\frac{1}{1-a{e}^{-j\omega }}.\text{\hspace{1em}(3.154)}$$

The magnitude and phase of $H(e^{j\omega })$ are shown in Figure 3.34(a) (not shown here) for $a=0.6$ and in Figure 3.34(b) (not shown here) for $a=-0.6$. We observe that, for the positive value of $a$, the difference equation (3.151) behaves like a lowpass filter with minimal attenuation of low frequencies near $\omega =0$ and increasing attenuation as we increase $\omega$ toward $\omega =\pi$. For the negative value of $a$, the system is a highpass filter, passing frequencies near $\omega =\pi$ and attenuating lower frequencies. In fact, for any positive value of $a<1$, the system approximates a lowpass filter, and for any negative value of $a>-1$, the system approximates a highpass filter, where $|a|$ controls the size of the filter passband, with broader passbands as $|a|$ is decreased.

As with the continuous-time examples, we again have a trade-off between time domain and frequency domain characteristics. In particular, the impulse response of the system described by eq. (3.151) (assuming initial rest) is:

$$h[n]=a^{n}u[n].\text{\hspace{1em}(3.155)}$$

The step response $s[n]=u[n]\ast h[n]$ is:

$$\begin{array}{rlrl}s[n]& =\sum _{k=0}^{n}h[k]& & \text{(For }n\ge 0\text{)}\\ & =\sum _{k=0}^n{a}^k& & \text{(Substitute }h[k]=a^k\text{)}\\ & =\frac{1-a^{n+1}}{1-a}u[n].& & \text{(Sum of finite geometric series)}\end{array}\text{\hspace{1em}(3.156)}$$

From these expressions, we see that $|a|$ also controls the speed with which the impulse and step responses approach their long-term values, with faster responses for smaller values of $|a|$, and hence, for filters with smaller passbands. Just as with differential equations, higher order recursive difference equations can be used to provide sharper filter characteristics and to provide more flexibility in balancing time-domain and frequency-domain constraints.

Finally, note from eq. (3.155) that the system described by eq. (3.151) is unstable if $|a|\ge 1$ and thus does not have a finite response to complex exponential inputs. As we stated previously, Fourier-based methods and frequency domain analysis focus on systems with finite responses to complex exponentials; hence, for examples such as eq. (3.151), we restrict ourselves to stable systems.

3.11.2 Nonrecursive Discrete-Time Filters

The general form of an FIR nonrecursive difference equation is

$$y[n]=\sum _{k=-N}^M{b}_{k}x[n-k].\text{\hspace{1em}(3.157)}$$

That is, the output $y[n]$ is a weighted average of the $(N+M+1)$ values of $x[n]$ from $x[n-M]$ through $x[n+N]$, with the weights given by the coefficients $b_k$. Systems of this form can be used to meet a broad array of filtering objectives, including frequency-selective filtering.

One frequently used example of such a filter is a moving-average filter, where the output $y[n]$ for any $n$—say, $n_0$—is an average of values of $x[n]$ in the vicinity of $n_0$. The basic idea is that by averaging values locally, rapid high-frequency components of the input will be averaged out and lower frequency variations will be retained, corresponding to smoothing or lowpass filtering the original sequence. A simple two-point moving-average filter was briefly introduced in Section 3.9 [eq. (3.138)]. An only slightly more complex example is the three-point moving-average filter, which is of the form

$$y[n]=\frac{1}{3}(x[n-1]+x[n]+x[n+1]),\text{\hspace{1em}(3.158)}$$

so that each output $y[n]$ is the average of three consecutive input values. In this case, the impulse response is obtained by setting $x[n]=\delta [n]$:

$$h[n]=\frac{1}{3}[\delta [n+1]+\delta [n]+\delta [n-1]].$$

And thus, from eq. (3.122), the corresponding frequency response is:

$$\begin{array}{rlrl}H(e^{j\omega })& =\sum _{k=-\infty }^{+\infty }h[k]e^{-j\omega k}& & \text{(Definition of frequency response)}\\ & =\frac{1}{3}(e^{-j\omega (-1)}+e^{-j\omega (0)}+e^{-j\omega (1)})& & \text{(Substitute }h[k]\text{)}\\ & =\frac{1}{3}[e^{j\omega }+1+e^{-j\omega }]& & \\ & =\frac{1}{3}(1+2\cos \omega ).& & \text{(Using }{e}^{j\omega }+e^{-j\omega }=2\cos \omega \text{)}\end{array}\text{\hspace{1em}(3.159)}$$

The magnitude of $H(e^{j\omega })$ is sketched in Figure 3.35 (not shown here). We observe that the filter has the general characteristics of a lowpass filter, although, as with the first-order recursive filter, it does not have a sharp transition from passband to stopband.

The three-point moving-average filter in eq. (3.158) has no parameters that can be changed to adjust the effective cutoff frequency. As a generalization of this moving-average filter, we can consider averaging over $N_{avg}$ neighboring points (where $N_{avg}=N+M+1$)—that is, using a difference equation of the form

$$y[n]=\frac{1}{N+M+1}\sum _{k=-N}^{M}x[n-k].\text{\hspace{1em}(3.160)}$$

The corresponding impulse response is a rectangular pulse (i.e., $h[n]=1/(N+M+1)$ for $-N\le n\le M$, and $h[n]=0$ otherwise). The filter’s frequency response is:

$$\begin{array}{rl}H(e^{j\omega })& =\sum _{k=-N}^M\frac{1}{N+M+1}{e}^{-j\omega k}\\ & =\frac{1}{N+M+1}\sum _{k=-N}^M{e}^{-j\omega k}.\end{array}\text{\hspace{1em}(3.161)}$$

The summation in eq. (3.161) can be evaluated by performing calculations similar to those in Example 3.12, using the geometric series sum formula. Let $S={\sum }_{k=-N}^M{e}^{-j\omega k}=e^{j\omega N}{\sum }_{k=0}^{N+M}{e}^{-j\omega k}=e^{j\omega N}\frac{1-e^{-j\omega (N+M+1)}}{1-e^{-j\omega }}$.

$$\begin{array}{rl}S& =e^{j\omega N}\frac{e^{-j\omega (N+M+1)/2}[e^{j\omega (N+M+1)/2}-e^{-j\omega (N+M+1)/2}]}{e^{-j\omega /2}[e^{j\omega /2}-e^{-j\omega /2}]}\\ & =e^{j\omega N}{e}^{-j\omega (N+M+1)/2}{e}^{j\omega /2}\frac{2j\sin [\omega (N+M+1)/2]}{2j\sin [\omega /2]}\\ & =e^{j\omega (N-(N+M+1)/2+1/2)}\frac{\sin [\omega (N+M+1)/2]}{\sin [\omega /2]}\\ & =e^{j\omega (N-N/2-M/2-1/2+1/2)}\frac{\sin [\omega (N+M+1)/2]}{\sin [\omega /2]}\\ & =e^{j\omega ((N-M)/2)}\frac{\sin [\omega (N+M+1)/2]}{\sin [\omega /2]}.\end{array}$$

Thus,

$$H(e^{j\omega })=\frac{1}{N+M+1}{e}^{j\omega [(N-M)/2]}\frac{\sin [\omega (M+N+1)/2]}{\sin (\omega /2)}.\text{\hspace{1em}(3.162)}$$

By adjusting the size, $N+M+1$, of the averaging window we can vary the cutoff frequency. For example, the magnitude of $H(e^{j\omega })$ is shown in Figure 3.36 (not shown here) for $M+N+1=33$ and $M+N+1=65$.

Nonrecursive filters can also be used to perform highpass filtering operations. To illustrate this, again with a simple example, consider the difference equation

$$y[n]=\frac{x[n]-x[n-1]}{2}.\text{\hspace{1em}(3.163)}$$

For input signals that are approximately constant, the value of $y[n]$ is close to zero. For input signals that vary greatly from sample to sample, the values of $y[n]$ can be expected to have larger amplitude. Thus, the system described by eq. (3.163) approximates a highpass filtering operation, attenuating slowly varying low-frequency components and passing rapidly varying higher frequency components with little attenuation. To see this more precisely, we need to look at the system’s frequency response. In this case, the impulse response is $h[n]=\frac{1}{2}\{\delta [n]-\delta [n-1]\}$, so that direct application of eq. (3.122) yields:

$$\begin{array}{rl}H(e^{j\omega })& =\sum _{k=-\infty }^{+\infty }h[k]e^{-j\omega k}\\ & =\frac{1}{2}(e^{-j\omega (0)}-e^{-j\omega (1)})\\ & =\frac{1}{2}[1-e^{-j\omega }]\\ & =\frac{1}{2}{e}^{-j\omega /2}(e^{j\omega /2}-e^{-j\omega /2})\\ & =\frac{1}{2}{e}^{-j\omega /2}(2j\sin (\omega /2))\\ & =j{e}^{-j\omega /2}\sin (\omega /2).\end{array}\text{\hspace{1em}(3.164)}$$

In Figure 3.37 (not shown here), we have plotted the magnitude of $H(e^{j\omega })$, showing that this simple system approximates a highpass filter, albeit one with a very gradual transition from passband to stopband. By considering more general nonrecursive filters, we can achieve far sharper transitions in lowpass, highpass, and other frequency-selective filters.

Note that, since the impulse response of any FIR system is of finite length (i.e., from eq. (3.157), $h[n]=b_n$ for $-N\le n\le M$ and 0 otherwise), it is always absolutely summable for any choices of the $b_n$. Hence, all such filters are stable. Also, if $N>0$ in eq. (3.157), the system is noncausal, since $y[n]$ then depends on future values of the input. In some applications, such as those involving the processing of previously recorded signals, causality is not a necessary constraint, and thus, we are free to use filters with $N>0$. In others, such as many involving real-time processing, causality is essential, and in such cases, we must take $N\le 0$.

3.12 Summary

In this chapter, we have introduced and developed Fourier series representations for both continuous-time and discrete-time systems and have used these representations to take a first look at one of the very important applications of the methods of signal and system analysis, namely, filtering. In particular, as we discussed in Section 3.2, one of the primary motivations for the use of Fourier series is the fact that complex exponential signals are eigenfunctions of LTI systems. We have also seen, in Sections 3.3-3.7, that any periodic signal of practical interest can be represented in a Fourier series—i.e., as a weighted sum of harmonically related complex exponentials that share a common period with the signal being represented. In addition, we have seen that the Fourier series representation has a number of important properties which describe how different characteristics of signals are reflected in their Fourier series coefficients.

One of the most important properties of Fourier series is a direct consequence of the eigenfunction property of complex exponentials. Specifically, if a periodic signal is applied to an LTI system, then the output will be periodic with the same period, and each of the Fourier coefficients of the output is the corresponding Fourier coefficient of the input multiplied by a complex number whose value is a function of the frequency corresponding to that Fourier coefficient. This function of frequency is characteristic of the LTI system and is referred to as the frequency response of the system. By examining the frequency response, we were led directly to the idea of filtering of signals using LTI systems, a concept that has numerous applications, including several that we have described. One important class of applications involves the notion of frequency-selective filtering—i.e., the idea of using an LTI system to pass certain specified bands of frequencies and stop or significantly attenuate others. We introduced the concept of ideal frequency-selective filters and also gave several examples of frequency-selective filters described by linear constant-coefficient differential or difference equations.

The purpose of this chapter has been to begin the process of developing both the tools of Fourier analysis and an appreciation for the utility of these tools in applications. In the chapters that follow, we continue with this agenda by developing the Fourier transform representations for aperiodic signals in continuous and discrete time and by taking a deeper look not only at filtering, but also at other important applications of Fourier methods.

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