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4.0 Introduction

In Chapter 3, we explored how periodic signals could be gracefully represented as a sum of harmonically related complex exponentials – a vibrant Fourier series! We also caught a glimpse of how this representation illuminated the behavior of Linear Time-Invariant (LTI) systems.

Now, we’re going to expand our horizons. Imagine extending those ideas to encompass signals that are not periodic, signals that unfold endlessly or simply come and go. A vast universe of signals, including all finite-energy signals, can also be expressed as a sophisticated “linear combination” of complex exponentials. But here’s the twist: instead of harmonically related, distinct frequencies, for aperiodic signals, these exponential building blocks become infinitesimally close, forming a continuous tapestry of frequencies.

This transformation from a sum to an integral is where the magic truly happens. The “spectrum of coefficients” that emerges from this continuous representation is what we call the Fourier transform. And the integral that meticulously reconstructs the signal from these coefficients is known as the inverse Fourier transform.

This brilliant leap, extending to aperiodic signals, is one of Fourier’s most profound gifts to us. His genius lay in viewing an aperiodic signal as a periodic signal with an infinite period. Think of it like zooming out from a repeating pattern: as the pattern’s repeat distance (period) grows, the fundamental frequency shrinks, and the harmonic components pack closer and closer together. In the limit, as the period stretches to infinity, these discrete frequency components blur into a continuous spectrum, and our trusty Fourier series sum gracefully transforms into an integral.

In the upcoming sections, we’ll start by solidifying our understanding of the Fourier series for continuous-time periodic signals, using it as a springboard. Then, we’ll boldly venture into the realm of the continuous-time Fourier transform, uncovering its fundamental properties, which are the very bedrock of frequency-domain analysis for continuous-time signals and systems. And just when you think you’ve seen it all, in Chapter 5, we’ll mirror this captivating journey for discrete-time signals!

4.1 Representation of Aperiodic Signals: The Continuous-Time Fourier Transform

4.1.1 Development of the Fourier Transform Representation of an Aperiodic Signal

Let’s begin our exploration by returning to a familiar face: the continuous-time periodic square wave from Example 3.5. Over a single period, this signal is defined as:

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

This signal repeats with a period $T$, as you might recall seeing in Figure 4.1 (imagine a series of flat-topped mountains appearing periodically).

The Fourier series coefficients, $a_k$, for this square wave were previously determined in Example 3.5 to be:

$$a_k=\frac{2\sin (k{\omega }_0{T}_1)}{k{\omega }_{0}T}\text{\hspace{1em}(4.1)}$$

where ${\omega }_0=2\pi /T$ is the fundamental frequency. In Figure 3.7, we visualized these coefficients as bar graphs for various values of $T$, keeping $T_1$ constant.

Now, let’s look at equation (4.1) from a slightly different angle. We can interpret the coefficients $a_k$ as samples of a continuous “envelope function.” Consider multiplying both sides of (4.1) by $T$:

$$T{a}_k=\frac{2\sin (k{\omega }_0{T}_1)}{k{\omega }_0}$$

If we let $\omega$ be a continuous variable, then the function $X(j\omega )=\frac{2\sin (\omega T_1)}{\omega }$ acts as the envelope. So, we can write:

$$T{a}_k=\frac{2\sin \omega T_1}{\omega }{|}_{\omega =k{\omega }_0}\text{\hspace{1em}(4.2)}$$

This means that $T{a}_k$ are simply samples of this continuous envelope function $X(j\omega )$ taken at discrete frequencies $k{\omega }_0$. What’s more, for a fixed $T_1$, this envelope function $X(j\omega )$ itself is independent of the period $T$. Figure 4.2 illustrates this beautifully: the Fourier series coefficients, scaled by $T$, increasingly approximate the smooth envelope as $T$ gets larger.

As $T$ grows larger and larger, the fundamental frequency ${\omega }_0=2\pi /T$ becomes smaller and smaller. This causes the samples $k{\omega }_0$ to become more densely packed. In the limit, as $T\to \infty$, our periodic square wave effectively becomes a single, isolated rectangular pulse – an aperiodic signal. Concurrently, the scaled Fourier series coefficients $T{a}_k$ become infinitely close samples of the envelope function, suggesting that the set of coefficients becomes the envelope function as $T$ approaches infinity.

This core idea forms the bedrock of Fourier’s approach to representing aperiodic signals. Imagine an aperiodic signal $x(t)$ as the grand finale of a periodic signal where the period has stretched infinitely. We then observe how the Fourier series representation behaves in this ultimate limit.

Consider an aperiodic signal $x(t)$ that is “finite in duration” – meaning it’s non-zero only for a finite interval, say $|t|<T_1$, as depicted in Figure 4.3(a). We can cleverly construct a periodic signal $\tilde{x}(t)$ such that $x(t)$ constitutes exactly one period of $\tilde{x}(t)$, as shown in Figure 4.3(b). As we make the period $T$ of $\tilde{x}(t)$ progressively larger, $\tilde{x}(t)$ becomes identical to $x(t)$ over an ever-widening interval. In the ultimate limit, as $T\to \infty$, $\tilde{x}(t)$ becomes indistinguishable from $x(t)$ for any finite value of $t$.

Now, let’s observe the effect of this on the Fourier series representation of $\tilde{x}(t)$. Recalling the Fourier series synthesis and analysis equations (originally (3.38) and (3.39)), and integrating over the interval $-T/2\le t\le T/2$:

$$\tilde{x}(t)=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}$$ $$a_k=\frac{1}{T}{\int }_{-T/2}^{T/2}\tilde{x}(t)e^{-jk{\omega }_{0}t}dt\text{\hspace{1em}(4.4)}$$

Since $\tilde{x}(t)=x(t)$ for $|t|<T/2$ and $x(t)=0$ outside this interval, we can rewrite the expression for $a_k$ by extending the integration limits to cover the entire non-zero extent of $x(t)$:

$$a_k=\frac{1}{T}{\int }_{-T/2}^{T/2}x(t)e^{-jk{\omega }_{0}t}dt=\frac{1}{T}{\int }_{-\infty }^{+\infty }x(t)e^{-jk{\omega }_{0}t}dt.$$

Let’s define a new function, $X(j\omega )$, as the continuous envelope of $T{a}_k$:

$$X(j\omega )={\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt\text{\hspace{1em}(4.5)}$$

From this definition, we can express the Fourier coefficients $a_k$ in terms of this new function:

$$a_k=\frac{1}{T}X(jk{\omega }_0)\text{\hspace{1em}(4.6)}$$

Now, substitute this expression for $a_k$ back into the Fourier series synthesis equation for $\tilde{x}(t)$:

$$\tilde{x}(t)=\sum _{k=-\infty }^{+\infty }\frac{1}{T}X(jk{\omega }_0)e^{jk{\omega }_{0}t}\text{\hspace{1em}(4.6)}$$

We know that ${\omega }_0=2\pi /T$, which means $1/T={\omega }_0/(2\pi )$. Let’s substitute this into the equation:

$$\tilde{x}(t)=\frac{1}{2\pi }\sum _{k=-\infty }^{+\infty }X(jk{\omega }_0)e^{jk{\omega }_{0}t}{\omega }_0\text{\hspace{1em}(4.7)}$$

Here’s where the magic truly unfolds! As $T\to \infty$:

1. $\tilde{x}(t)$ converges to $x(t)$.
2. ${\omega }_0=2\pi /T$ approaches $0$, meaning the spacing between samples $k{\omega }_0$ becomes infinitesimally small.
3. The sum on the right-hand side of eq. (4.7) transforms into an integral. This can be intuitively grasped by considering the sum as an approximation of the area under the curve $X(j\omega )e^{j\omega t}$ when multiplied by ${\omega }_0$, as visually suggested by Figure 4.4. The term ${\omega }_0$ acts as $\Delta \omega$, the differential element in frequency.

Thus, as $T\to \infty$, equations (4.7) and (4.5) evolve into the remarkable Fourier transform pair:

$$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega \text{\hspace{1em}(4.8)}$$

and

$$X(j\omega )={\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt\text{\hspace{1em}(4.9)}$$

These are the bedrock equations! Equation (4.9) defines the Fourier Transform (or Fourier integral) of $x(t)$, yielding its frequency-domain representation $X(j\omega )$. Equation (4.8) is the Inverse Fourier Transform, allowing us to reconstruct the time-domain signal $x(t)$ from its spectrum $X(j\omega )$.

Notice the beautiful parallel: equation (4.8) for aperiodic signals plays a role analogous to equation (3.38) for periodic signals. Both express a signal as a linear combination of complex exponentials. For periodic signals, these exponentials have discrete amplitudes $a_k$ at specific, harmonically related frequencies $k{\omega }_0$. For aperiodic signals, the exponentials span a continuum of frequencies, and their “amplitude” is specified by $X(j\omega )\frac{d\omega }{2\pi }$.

In a similar vein to Fourier series coefficients, $X(j\omega )$ is often called the spectrum of $x(t)$. It’s our blueprint, providing all the necessary information to describe $x(t)$ as an integral of sinusoidal signals across a continuous range of frequencies.

This development also reveals a profound connection between Fourier series and Fourier transforms. If $\tilde{x}(t)$ is a periodic signal with period $T$ and Fourier coefficients $a_k$, we can think of $x(t)$ as one period of $\tilde{x}(t)$ (i.e., $x(t)=\tilde{x}(t)$ for $s\le t\le s+T$ and $0$ otherwise). Then, the Fourier coefficients $a_k$ can be found by integrating over one period:

$$a_k=\frac{1}{T}{\int }_s^{s+T}\tilde{x}(t)e^{-jk{\omega }_{0}t}dt$$

Since $x(t)$ is zero outside the interval $s\le t\le s+T$, we can extend the integration limits:

$$a_k=\frac{1}{T}{\int }_{-\infty }^{+\infty }x(t)e^{-jk{\omega }_{0}t}dt.$$

Comparing this with equation (4.9), we see that $a_k$ are simply samples of the Fourier transform of one period of $\tilde{x}(t)$, scaled by $1/T$:

$$a_k=\frac{1}{T}X(j\omega ){|}_{\omega =k{\omega }_0}\text{\hspace{1em}(4.10)}$$

This elegantly connects our discrete and continuous spectral representations!

4.1.2 Convergence of Fourier Transforms

While our initial derivation hinged on $x(t)$ being a finite-duration signal, the Fourier transform pair (4.8) and (4.9) holds true for an incredibly broad class of signals, even those of infinite duration. The conditions for the convergence of the Fourier transform echo those for the Fourier series, ensuring that our representation is meaningful.

Let $X(j\omega )$ be computed via equation (4.9). Then, let $\hat{x}(t)$ be the signal synthesized using $X(j\omega )$ in the inverse Fourier transform equation (4.8):

$$\hat{x}(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega .$$

The crucial question is: when is $\hat{x}(t)$ a faithful representation of our original signal $x(t)$?

A powerful condition is finite energy. If $x(t)$ is square-integrable (i.e., has finite energy), such that:

$${\int }_{-\infty }^{+\infty }|x(t){|}^{2}dt<\infty \text{\hspace{1em}(4.11)}$$

then we are guaranteed two things:

1. $X(j\omega )$ will be finite (meaning eq. (4.9) converges).
2. The energy in the difference between the original signal $x(t)$ and its Fourier representation $\hat{x}(t)$ is zero. That is, if $e(t)=\hat{x}(t)-x(t)$, then: $${\int }_{-\infty }^{+\infty }|e(t){|}^{2}dt=0\text{\hspace{1em}(4.12)}$$

These conditions (4.11) and (4.12) are the aperiodic counterparts to Parseval’s relation for periodic signals (eqs. (3.51) and (3.54)). This means that even if $x(t)$ and $\hat{x}(t)$ might differ at isolated points, their overall energy content is identical, implying that $\hat{x}(t)$ is a very good representation indeed.

Alternatively, just like with Fourier series, the Dirichlet conditions offer another set of sufficient conditions to ensure that $\hat{x}(t)=x(t)$ at all points of continuity. At points of discontinuity, $\hat{x}(t)$ gracefully converges to the average of the values on either side. These conditions require:

1. Absolute Integrability: $x(t)$ must be absolutely integrable; that is: $${\int }_{-\infty }^{+\infty }|x(t)|dt<\infty \text{\hspace{1em}(4.13)}$$
2. Finite Maxima and Minima: $x(t)$ must have a finite number of maxima and minima within any finite interval.
3. Finite Discontinuities: $x(t)$ must have a finite number of discontinuities within any finite interval, and each of these discontinuities must be finite.

In essence, most well-behaved signals that we encounter in practice, particularly those that are absolutely integrable and have a limited number of “wiggles” and “jumps,” will possess a well-defined Fourier transform.

However, a fascinating twist awaits us! In the next section, we’ll discover that even periodic signals, which do not satisfy either absolute integrability or finite energy over an infinite interval, can be endowed with Fourier transforms by cleverly incorporating impulse functions into the frequency domain. This ingenious approach unites Fourier series and Fourier transforms under a single, elegant umbrella, simplifying many analyses. But before we venture there, let’s explore some concrete examples of Fourier transforms!

4.1.3 Examples of Continuous-Time Fourier Transforms

Let’s put the Fourier transform to work!

Example 4.1: Exponential Decay

Consider the causal decaying exponential signal:

$$x(t)=e^{-at}u(t),a>0.$$

To find its Fourier transform $X(j\omega )$, we apply the analysis equation (4.9):

$$\begin{array}{rlrl}X(j\omega )& ={\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt& & \\ & ={\int }_{-\infty }^{+\infty }{e}^{-at}u(t)e^{-j\omega t}dt& & \text{(Substitute }x(t)=e^{-at}u(t)\text{)}\\ & ={\int }_0^{\infty }{e}^{-at}{e}^{-j\omega t}dt& & \text{(Since }u(t)=0\text{ for }t<0\text{ and }u(t)=1\text{ for }t\ge 0\text{)}\\ & ={\int }_0^{\infty }{e}^{-(a+j\omega )t}dt& & \text{(Combine exponentials)}\\ & ={\left[-\frac{1}{a+j\omega }{e}^{-(a+j\omega )t}\right]}_0^{\infty }& & \text{(Integrate with respect to }t\text{)}\\ & =-\frac{1}{a+j\omega }\underset{t\to \infty }{\lim }{e}^{-(a+j\omega )t}-\left(-\frac{1}{a+j\omega }{e}^{-(a+j\omega )0}\right)& & \text{(Evaluate at limits)}\\ & =-\frac{1}{a+j\omega }\cdot 0-\left(-\frac{1}{a+j\omega }\cdot 1\right)& & \text{(Since }a>0\text{, }{e}^{-at}\to 0\text{ as }t\to \infty \text{, thus }{e}^{-(a+j\omega )t}\to 0\text{)}\\ & =\frac{1}{a+j\omega }& & \text{(Simplify)}\end{array}$$

Thus, for $a>0$:

$$X(j\omega )=\frac{1}{a+j\omega }.$$

This Fourier transform is generally complex-valued. To visualize it, we typically look at its magnitude and phase:

$$|X(j\omega )|=\left|\frac{1}{a+j\omega }\right|=\frac{1}{|a+j\omega |}=\frac{1}{\sqrt{a^2+{\omega }^2}}$$ $$\angle X(j\omega )=\angle \left(\frac{1}{a+j\omega }\right)=-\angle (a+j\omega )=-{\tan }^{-1}\left(\frac{\omega }{a}\right).$$

Figure 4.5 would typically show these magnitude and phase plots. It’s worth noting that if $a$ were a complex number, $x(t)$ would still be absolutely integrable as long as its real part $\Re e\{a\}>0$. In this broader case, the Fourier transform would retain the same form:

$$X(j\omega )=\frac{1}{a+j\omega },\Re e\{a\}>0.$$

Example 4.2: Two-Sided Exponential

Let’s consider a two-sided exponential signal:

$$x(t)=e^{-a|t|},a>0.$$

This signal is symmetric around $t=0$, as shown in Figure 4.6. To find its Fourier transform, we split the integral due to the absolute value:

$$\begin{array}{rlrl}X(j\omega )& ={\int }_{-\infty }^{+\infty }{e}^{-a|t|}{e}^{-j\omega t}dt& & \text{(Fourier Transform definition)}\\ & ={\int }_{-\infty }^0{e}^{at}{e}^{-j\omega t}dt+{\int }_0^{\infty }{e}^{-at}{e}^{-j\omega t}dt& & \text{(Split integral: }|t|=-t\text{ for }t<0,|t|=t\text{ for }t\ge 0\text{)}\\ & ={\int }_{-\infty }^0{e}^{(a-j\omega )t}dt+{\int }_0^{\infty }{e}^{-(a+j\omega )t}dt& & \text{(Combine exponentials)}\\ & ={\left[\frac{1}{a-j\omega }{e}^{(a-j\omega )t}\right]}_{-\infty }^0+{\left[-\frac{1}{a+j\omega }{e}^{-(a+j\omega )t}\right]}_0^{\infty }& & \text{(Integrate)}\\ & =\left(\frac{1}{a-j\omega }{e}^0-\underset{t\to -\infty }{\lim }\frac{1}{a-j\omega }{e}^{(a-j\omega )t}\right)+\left(-\frac{1}{a+j\omega }\underset{t\to \infty }{\lim }{e}^{-(a+j\omega )t}-(-\frac{1}{a+j\omega }{e}^0)\right)& & \text{(Evaluate at limits)}\\ & =\left(\frac{1}{a-j\omega }-0\right)+\left(0+\frac{1}{a+j\omega }\right)& & \text{(Since }a>0\text{, }{e}^{at}\to 0\text{ as }t\to -\infty \text{ and }{e}^{-at}\to 0\text{ as }t\to \infty \text{)}\\ & =\frac{1}{a-j\omega }+\frac{1}{a+j\omega }& & \text{(Simplify)}\\ & =\frac{(a+j\omega )+(a-j\omega )}{(a-j\omega )(a+j\omega )}& & \text{(Find common denominator)}\\ & =\frac{2a}{a^2+{\omega }^2}& & \text{(Simplify)}\end{array}$$

In this case, $X(j\omega )$ is purely real and an even function of $\omega$, as expected for a real and even signal $x(t)$. Figure 4.7 would illustrate this bell-shaped spectrum.

Example 4.3: The Unit Impulse

Let’s find the Fourier transform of the fundamental building block: the unit impulse function:

$$x(t)=\delta (t).\text{\hspace{1em}(4.14)}$$

Applying equation (4.9):

$$\begin{array}{rlrl}X(j\omega )& ={\int }_{-\infty }^{+\infty }\delta (t)e^{-j\omega t}dt& & \text{(Fourier Transform definition)}\\ & =e^{-j\omega \cdot 0}& & \text{(Sifting property of the impulse: }{\int }_{-\infty }^{\infty }f(t)\delta (t-t_0)dt=f(t_0)\text{)}\\ & =e^0& & \text{(Evaluate the exponential at }t=0\text{)}\\ & =1& & \text{(Simplify)}\end{array}$$

So, the Fourier transform of a unit impulse is a constant value of $1$ across all frequencies:

$$X(j\omega )=1.\text{\hspace{1em}(4.15)}$$

This means an impulse contains equal contributions from all frequencies – a true “white noise” in the frequency domain!

Example 4.4: The Rectangular Pulse

Consider the rectangular pulse signal:

$$x(t)=\{\begin{array}{ll}1,& |t|<T_1\\ 0,& |t|>T_1\end{array}\text{\hspace{1em}(4.16)}$$

This signal is shown in Figure 4.8(a). Let’s find its Fourier transform:

$$\begin{array}{rlrl}X(j\omega )& ={\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt& & \text{(Fourier Transform definition)}\\ & ={\int }_{-T_1}^{T_1}1\cdot e^{-j\omega t}dt& & \text{(Since }x(t)=1\text{ for }|t|<T_1\text{ and }0\text{ otherwise)}\\ & ={\left[\frac{e^{-j\omega t}}{-j\omega }\right]}_{-T_1}^{T_1}& & \text{(Integrate, assuming }\omega \ne 0\text{)}\\ & =\frac{e^{-j\omega T_1}-e^{j\omega T_1}}{-j\omega }& & \text{(Evaluate at limits)}\\ & =\frac{e^{j\omega T_1}-e^{-j\omega T_1}}{j\omega }& & \text{(Rearrange numerator)}\\ & =\frac{2j\sin (\omega T_1)}{j\omega }& & \text{(Euler’s formula: }\sin \theta =\frac{e^{j\theta }-e^{-j\theta }}{2j}\text{)}\\ & =\frac{2\sin (\omega T_1)}{\omega }& & \text{(Simplify)}\end{array}$$

For the special case when $\omega =0$:

$$X(j0)={\int }_{-T_1}^{T_1}1\cdot e^{-j0\cdot t}dt={\int }_{-T_1}^{T_1}1dt=[t{]}_{-T_1}^{T_1}=T_1-(-T_1)=2T_1.$$

Notice that ${\lim }_{\omega \to 0}\frac{2\sin (\omega T_1)}{\omega }=2T_1$ (using L’Hopital’s rule or Taylor series expansion for $\sin (\omega T_1)\approx \omega T_1$), so the formula holds for all $\omega$. So the Fourier transform of the rectangular pulse is:

$$X(j\omega )=2\frac{\sin \omega T_1}{\omega }.\text{\hspace{1em}(4.17)}$$

This well-known function is often called a “sinc” function (though a more precise definition will follow). It’s sketched in Figure 4.8(b), characterized by a main lobe at $\omega =0$ and decaying oscillations.

As we discussed earlier, this signal can be seen as the limiting case of a periodic square wave as its period goes to infinity. Therefore, the convergence behavior of its inverse Fourier transform should mirror the Gibbs phenomenon observed for the square wave. Indeed, the inverse Fourier transform:

$$\hat{x}(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }2\frac{\sin \omega T_1}{\omega }{e}^{j\omega t}d\omega .$$

Since $x(t)$ is square integrable, the energy in the difference $|x(t)-\hat{x}(t){|}^2$ is zero. Furthermore, because $x(t)$ satisfies the Dirichlet conditions, $\hat{x}(t)=x(t)$ everywhere except at the discontinuities ($t=\pm T_1$), where $\hat{x}(t)$ converges to $1/2$ (the average of the values $0$ and $1$). Near these discontinuities, just like with Fourier series, we observe the characteristic “Gibbs phenomenon” ripples when approximating with a finite-bandwidth integral:

$$\frac{1}{2\pi }{\int }_{-W}^{W}2\frac{\sin \omega T_1}{\omega }{e}^{j\omega t}d\omega .$$

These ripples become narrower but their peak amplitude does not decrease as $W\to \infty$, though the energy within them approaches zero.

Example 4.5: The Rectangular Spectrum

Now, let’s consider a signal whose Fourier transform is a rectangular pulse in the frequency domain:

$$X(j\omega )=\{\begin{array}{ll}1,& |\omega |<W\\ 0,& |\omega |>W\end{array}\text{\hspace{1em}(4.18)}$$

This transform is depicted in Figure 4.9(a). To find the corresponding time-domain signal $x(t)$, we use the inverse Fourier transform equation (4.8):

$$\begin{array}{rlrl}x(t)& =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega & & \text{(Inverse Fourier Transform definition)}\\ & =\frac{1}{2\pi }{\int }_{-W}^{W}1\cdot e^{j\omega t}d\omega & & \text{(Substitute }X(j\omega )\text{ and integration limits)}\\ & =\frac{1}{2\pi }{\left[\frac{e^{j\omega t}}{jt}\right]}_{-W}^W& & \text{(Integrate, assuming }t\ne 0\text{)}\\ & =\frac{1}{2\pi jt}(e^{jWt}-e^{-jWt})& & \text{(Evaluate at limits)}\\ & =\frac{1}{2\pi jt}(2j\sin (Wt))& & \text{(Euler’s formula: }\sin \theta =\frac{e^{j\theta }-e^{-j\theta }}{2j}\text{)}\\ & =\frac{\sin (Wt)}{\pi t}& & \text{(Simplify)}\end{array}$$

For the special case when $t=0$:

$$x(0)=\frac{1}{2\pi }{\int }_{-W}^{W}1\cdot e^{j\omega \cdot 0}d\omega =\frac{1}{2\pi }{\int }_{-W}^{W}1d\omega =\frac{1}{2\pi }[\omega {]}_{-W}^W=\frac{1}{2\pi }(W-(-W))=\frac{2W}{2\pi }=\frac{W}{\pi }.$$

Notice that ${\lim }_{t\to 0}\frac{\sin (Wt)}{\pi t}=\frac{W}{\pi }$ (using L’Hopital’s rule), so the formula holds for all $t$. So the inverse Fourier transform is:

$$x(t)=\frac{\sin Wt}{\pi t}.\text{\hspace{1em}(4.19)}$$

This signal, also a sinc function, is depicted in Figure 4.9(b).

If you compare Figures 4.8 and 4.9, or equations (4.16)/(4.17) with (4.18)/(4.19), you’ll notice an intriguing pattern: one signal is a rectangular pulse, and its transform is a sinc function; the other signal is a sinc function, and its transform is a rectangular pulse! This beautiful symmetry is a manifestation of the duality property of Fourier transforms, which we’ll explore in depth in Section 4.3.6.

These functions, of the form $\frac{\sin (a\theta )}{b\theta }$, are ubiquitous in Fourier analysis and system theory and are formally known as sinc functions. A commonly accepted definition for the sinc function is:

$$\mathrm{sinc}(\theta )=\frac{\sin \pi \theta }{\pi \theta }.\text{\hspace{1em}(4.20)}$$

This normalized sinc function is plotted in Figure 4.10. Both of our previous results can be expressed using this definition:

$$\frac{2\sin \omega T_1}{\omega }=2T_1\mathrm{sinc}\left(\frac{\omega T_1}{\pi }\right)$$ $$\frac{\sin Wt}{\pi t}=\frac{W}{\pi }\mathrm{sinc}\left(\frac{Wt}{\pi }\right).$$

Finally, let’s reflect on Figure 4.9, which we can imagine redrawn for various values of $W$ (Figure 4.11). As $W$ (the bandwidth of $X(j\omega )$) increases, the frequency-domain pulse becomes wider. What happens in the time domain? The main lobe of $x(t)$ at $t=0$ becomes taller, and its width (the interval $|t|<\pi /W$) becomes narrower. In the ultimate limit as $W\to \infty$, $X(j\omega )$ becomes a constant $1$ for all $\omega$. From Example 4.3, we know that the inverse transform of $X(j\omega )=1$ is an impulse $\delta (t)$. This illustrates a profound inverse relationship between the time and frequency domains: broadening in one domain corresponds to compression in the other. We saw a similar effect in Figure 4.8: as $T_1$ (the duration of $x(t)$) increased, $x(t)$ broadened, but $X(j\omega )$ became narrower. This crucial scaling property will be formalized in Section 4.3.5.

4.2 The Fourier Transform for Periodic Signals

In the previous section, we established the Fourier transform for aperiodic signals. But what about periodic signals? Can we bring them into this unified framework? Absolutely! By introducing impulse functions into the frequency domain, we can construct Fourier transforms for periodic signals directly from their Fourier series representations. This is not just a theoretical neat trick; it’s incredibly convenient for many analyses.

Let’s start by imagining a signal $x(t)$ whose Fourier transform $X(j\omega )$ is a single impulse of area $2\pi$ located at a specific frequency ${\omega }_0$:

$$X(j\omega )=2\pi \delta (\omega -{\omega }_0).\text{\hspace{1em}(4.21)}$$

To find the time-domain signal $x(t)$ corresponding to this transform, we apply the inverse Fourier transform (4.8):

$$\begin{array}{rlrl}x(t)& =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega & & \text{(Inverse Fourier Transform definition)}\\ & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }2\pi \delta (\omega -{\omega }_0)e^{j\omega t}d\omega & & \text{(Substitute }X(j\omega )\text{)}\\ & =e^{j{\omega }_{0}t}& & \text{(Sifting property of the impulse: }{\int }_{-\infty }^{\infty }f(\omega )\delta (\omega -{\omega }_0)d\omega =f({\omega }_0)\text{)}\end{array}$$

This is a single complex exponential! It makes perfect sense: a pure sinusoidal signal has all its energy concentrated at a single frequency.

Now, let’s generalize this. What if $X(j\omega )$ is a linear combination of such impulses, equally spaced in frequency, just like the harmonics in a Fourier series?

$$X(j\omega )=\sum _{k=-\infty }^{+\infty }2\pi a_k\delta (\omega -k{\omega }_0).\text{\hspace{1em}(4.22)}$$

Applying the inverse Fourier transform to this sum:

$$\begin{array}{rlrl}x(t)& =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }\left(\sum _{k=-\infty }^{+\infty }2\pi a_k\delta (\omega -k{\omega }_0)\right)e^{j\omega t}d\omega & & \text{(Inverse Fourier Transform definition)}\\ & =\sum _{k=-\infty }^{+\infty }{a}_k{\int }_{-\infty }^{+\infty }\delta (\omega -k{\omega }_0)e^{j\omega t}d\omega & & \text{(Interchange sum and integral, pull constants out)}\\ & =\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}& & \text{(Apply sifting property for each impulse)}\end{array}$$

Lo and behold! This is precisely the Fourier series representation of a periodic signal, as defined in equation (3.38):

$$x(t)=\sum _{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}.\text{\hspace{1em}(4.23)}$$

This means that the Fourier transform of a periodic signal is a “train of impulses” in the frequency domain. Each impulse is located at a harmonic frequency $k{\omega }_0$, and its area is $2\pi$ times the corresponding Fourier series coefficient $a_k$. This is a remarkably powerful concept, uniting periodic and aperiodic signals under the Fourier transform umbrella!

Example 4.6: Periodic Square Wave in Frequency Domain

Let’s revisit the periodic square wave from Figure 4.1. Its Fourier series coefficients were:

$$a_k=\frac{\sin k{\omega }_0{T}_1}{\pi k}.$$

Using our new understanding, the Fourier transform of this periodic square wave is directly given by equation (4.22):

$$X(j\omega )=\sum _{k=-\infty }^{+\infty }2\pi \left(\frac{\sin k{\omega }_0{T}_1}{\pi k}\right)\delta (\omega -k{\omega }_0)=\sum _{k=-\infty }^{+\infty }\frac{2\sin k{\omega }_0{T}_1}{k}\delta (\omega -k{\omega }_0).$$

Figure 4.12, illustrating this for $T=4T_1$, shows impulses at discrete frequencies. The only difference from the bar graphs of $a_k$ (e.g., in Figure 3.7(a)) is the scaling factor of $2\pi$ and the use of impulses to represent the discrete spectral components.

Example 4.7: Sinusoids in the Frequency Domain

Let’s find the Fourier transforms of pure sinusoids. For $x(t)=\sin {\omega }_{0}t$: We know from Euler’s formula that $\sin {\omega }_{0}t=\frac{e^{j{\omega }_{0}t}-e^{-j{\omega }_{0}t}}{2j}$. Comparing this to the general Fourier series $x(t)=\sum a_k{e}^{jk{\omega }_{0}t}$, we identify the non-zero coefficients: $a_1=\frac{1}{2j}$ (for $k=1$) $a_{-1}=-\frac{1}{2j}$ (for $k=-1$) All other $a_k=0$. Applying equation (4.22):

$$\begin{array}{rl}X(j\omega )& =2\pi a_1\delta (\omega -{\omega }_0)+2\pi a_{-1}\delta (\omega +{\omega }_0)\\ & =2\pi \left(\frac{1}{2j}\right)\delta (\omega -{\omega }_0)+2\pi \left(-\frac{1}{2j}\right)\delta (\omega +{\omega }_0)\\ & =\frac{\pi }{j}[\delta (\omega -{\omega }_0)-\delta (\omega +{\omega }_0)].\end{array}$$

This transform is shown in Figure 4.13(a).

Similarly, for $x(t)=\cos {\omega }_{0}t$: We know $\cos {\omega }_{0}t=\frac{e^{j{\omega }_{0}t}+e^{-j{\omega }_{0}t}}{2}$. The non-zero Fourier series coefficients are: $a_1=\frac{1}{2}$ (for $k=1$) $a_{-1}=\frac{1}{2}$ (for $k=-1$) All other $a_k=0$. Applying equation (4.22):

$$\begin{array}{rl}X(j\omega )& =2\pi a_1\delta (\omega -{\omega }_0)+2\pi a_{-1}\delta (\omega +{\omega }_0)\\ & =2\pi \left(\frac{1}{2}\right)\delta (\omega -{\omega }_0)+2\pi \left(\frac{1}{2}\right)\delta (\omega +{\omega }_0)\\ & =\pi [\delta (\omega -{\omega }_0)+\delta (\omega +{\omega }_0)].\end{array}$$

This transform is shown in Figure 4.13(b). These two impulse-based transforms are incredibly important, particularly when we explore sinusoidal modulation in Chapter 8.

Example 4.8: The Impulse Train

A signal of immense utility, especially in analyzing sampling systems (Chapter 7), is the periodic impulse train:

$$x(t)=\sum _{k=-\infty }^{+\infty }\delta (t-kT).$$

This signal, shown in Figure 4.14(a), is periodic with period $T$. We already computed its Fourier series coefficients in Example 3.8:

$$a_k=\frac{1}{T}{\int }_{-T/2}^{+T/2}\delta (t)e^{-jk{\omega }_{0}t}dt=\frac{1}{T}.$$

Remarkably, every Fourier coefficient of the periodic impulse train is the same, $1/T$. Now, let’s use equation (4.22) to find its Fourier transform:

$$\begin{array}{rlrl}X(j\omega )& =\sum _{k=-\infty }^{+\infty }2\pi a_k\delta (\omega -k{\omega }_0)& & \text{(General formula for periodic signals)}\\ & =\sum _{k=-\infty }^{+\infty }2\pi \left(\frac{1}{T}\right)\delta \left(\omega -\frac{2\pi k}{T}\right)& & \text{(Substitute }{a}_k=1/T\text{ and }{\omega }_0=2\pi /T\text{)}\\ & =\frac{2\pi }{T}\sum _{k=-\infty }^{+\infty }\delta \left(\omega -\frac{2\pi k}{T}\right).& & \end{array}$$

So, the Fourier transform of a periodic impulse train in the time domain (with period $T$) is another periodic impulse train in the frequency domain (with period $2\pi /T$!). This is sketched in Figure 4.14(b).

This example strikingly re-emphasizes the inverse relationship between the time and frequency domains: if the impulses in the time domain are widely spaced (large $T$), then the impulses in the frequency domain are closely spaced (small $2\pi /T$), and vice-versa. It’s a fundamental concept that permeates all of Fourier analysis.

4.3 Properties of the Continuous-Time Fourier Transform

Now that we have a solid grasp of the Fourier transform itself, let’s explore its remarkable properties. These properties are not just mathematical curiosities; they offer deep insights into how signals behave under various operations and how time-domain characteristics relate to frequency-domain characteristics. They are also powerful tools for simplifying computations of transforms and inverse transforms.

Recall the close kinship between the Fourier series and Fourier transform. Many properties we discuss here for the Fourier transform have direct counterparts for the Fourier series, which we already touched upon in Chapter 3 (especially Section 3.5 and Table 3.1).

For convenience, we’ll use a shorthand notation for Fourier transform pairs: The synthesis and analysis equations are:

$$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega \text{\hspace{1em}(4.24)}$$ $$X(j\omega )={\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt\text{\hspace{1em}(4.25)}$$

We’ll use $\mathfrak{F}\{x(t)\}$ to denote the Fourier transform of $x(t)$, and ${\mathfrak{F}}^{-1}\{X(j\omega )\}$ for the inverse transform. The elegant notation for a Fourier transform pair is:

$$x(t)\stackrel{\mathfrak{F}}{⟷}X(j\omega ).$$

For instance, from Example 4.1, we have:

$$\mathfrak{F}\{e^{-at}u(t)\}=\frac{1}{a+j\omega }$$ $${\mathfrak{F}}^{-1}\left\{\frac{1}{a+j\omega }\right\}=e^{-at}u(t)$$

And the pair notation:

$$e^{-at}u(t)\stackrel{\mathfrak{F}}{⟷}\frac{1}{a+j\omega }.$$

4.3.1 Linearity

The Fourier transform is a linear operation, just like many of our favorite mathematical transformations! If we have two signals $x(t)$ and $y(t)$ with their respective Fourier transforms $X(j\omega )$ and $Y(j\omega )$:

$$x(t)\stackrel{\mathfrak{F}}{⟷}X(j\omega )\text{and}y(t)\stackrel{\mathfrak{F}}{⟷}Y(j\omega ),$$

then for any complex constants $a$ and $b$:

$$\boxed{ax(t)+by(t)\stackrel{\mathfrak{F}}{⟷}aX(j\omega )+bY(j\omega ).}\text{\hspace{1em}(4.26)}$$

Proof: Let $z(t)=ax(t)+by(t)$.

$$\begin{array}{rlrl}\mathfrak{F}\{z(t)\}& ={\int }_{-\infty }^{+\infty }(ax(t)+by(t))e^{-j\omega t}dt& & \text{(Definition of Fourier transform)}\\ & ={\int }_{-\infty }^{+\infty }ax(t)e^{-j\omega t}dt+{\int }_{-\infty }^{+\infty }by(t)e^{-j\omega t}dt& & \text{(Linearity of integration)}\\ & =a{\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt+b{\int }_{-\infty }^{+\infty }y(t)e^{-j\omega t}dt& & \text{(Pull constants out of integrals)}\\ & =aX(j\omega )+bY(j\omega )& & \text{(By definition of }X(j\omega )\text{ and }Y(j\omega )\text{)}\end{array}$$

This property extends effortlessly to any finite number of signals. It means we can break down complex signals into simpler components, find their transforms individually, and then sum them up!

4.3.2 Time Shifting

Shifting a signal in time has a very predictable effect in the frequency domain: it introduces a phase shift that’s linear with frequency. If:

$$x(t)\stackrel{\mathfrak{F}}{⟷}X(j\omega ),$$

then for a time shift $t_0$:

$$\boxed{x(t-t_0)\stackrel{\mathfrak{F}}{⟷}{e}^{-j\omega t_0}X(j\omega ).}\text{\hspace{1em}(4.27)}$$

Proof (using synthesis equation): Start with the inverse Fourier transform of $X(j\omega )$:

$$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }X(j\omega )e^{j\omega t}d\omega .$$

Now, let’s consider $x(t-t_0)$. We can obtain this by simply replacing $t$ with $(t-t_0)$ in the above equation:

$$\begin{array}{rl}x(t-t_0)& =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega (t-t_0)}d\omega \\ & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}{e}^{-j\omega t_0}d\omega \\ & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }\left(e^{-j\omega t_0}X(j\omega )\right)e^{j\omega t}d\omega .\end{array}$$

By comparing this directly with the inverse Fourier transform formula (4.24), we can clearly see that the Fourier transform of $x(t-t_0)$ must be the term multiplying $e^{j\omega t}$:

$$\mathfrak{F}\{x(t-t_0)\}=e^{-j\omega t_0}X(j\omega ).$$

A beautiful consequence of this property is that time shifting a signal does not change the magnitude of its Fourier transform. If $X(j\omega )$ is written in polar form as $X(j\omega )=|X(j\omega )|e^{j\angle X(j\omega )}$, then:

$$\mathfrak{F}\{x(t-t_0)\}=e^{-j\omega t_0}X(j\omega )=|X(j\omega )|e^{j\angle X(j\omega )}{e}^{-j\omega t_0}=|X(j\omega )|e^{j[\angle X(j\omega )-\omega t_0]}.$$

Thus, a time shift only alters the phase spectrum, specifically by introducing a linear phase shift of $-\omega t_0$. This linear phase shift means that all frequency components are delayed by the same amount, $t_0$.

Example 4.9: Combining Linearity and Time Shifting

Let’s use these powerful properties to find the Fourier transform of the signal $x(t)$ shown in Figure 4.15(a). This signal looks a bit complex to integrate directly. However, we can decompose $x(t)$ into a linear combination of two simpler rectangular pulse signals:

$$x(t)=\frac{1}{2}{x}_1(t-2.5)+x_2(t-2.5),$$

where $x_1(t)$ and $x_2(t)$ are shown in Figure 4.15(b) and (c), respectively.

• $x_1(t)$ is a rectangular pulse of height $1$ and width $1$ centered at $t=0$. From Example 4.4, with $T_1=0.5$, its Fourier transform is $X_1(j\omega )=\frac{2\sin (\omega \cdot 0.5)}{\omega }=\frac{2\sin (\omega /2)}{\omega }$.
• $x_2(t)$ is a rectangular pulse of height $1$ and width $3$ centered at $t=0$. From Example 4.4, with $T_1=1.5$, its Fourier transform is $X_2(j\omega )=\frac{2\sin (\omega \cdot 1.5)}{\omega }=\frac{2\sin (3\omega /2)}{\omega }$.

Now, applying the linearity and time-shift properties:

$$\begin{array}{rlrl}X(j\omega )& =\mathfrak{F}\left\{\frac{1}{2}{x}_1(t-2.5)+x_2(t-2.5)\right\}& & \\ & =\frac{1}{2}\mathfrak{F}\{x_1(t-2.5)\}+\mathfrak{F}\{x_2(t-2.5)\}& & \text{(Linearity)}\\ & =\frac{1}{2}{e}^{-j\omega (2.5)}{X}_1(j\omega )+e^{-j\omega (2.5)}{X}_2(j\omega )& & \text{(Time Shifting)}\\ & =e^{-j5\omega /2}\left[\frac{1}{2}{X}_1(j\omega )+X_2(j\omega )\right]& & \text{(Factor out common exponential)}\\ & =e^{-j5\omega /2}\left[\frac{1}{2}\frac{2\sin (\omega /2)}{\omega }+\frac{2\sin (3\omega /2)}{\omega }\right]& & \text{(Substitute }{X}_1(j\omega ),X_2(j\omega )\text{)}\\ & =e^{-j5\omega /2}\left\{\frac{\sin (\omega /2)+2\sin (3\omega /2)}{\omega }\right\}.& & \end{array}$$

This demonstrates how decomposing a complex signal and using transform properties can significantly simplify Fourier transform calculations.

4.3.3 Conjugation and Conjugate Symmetry

The conjugation property tells us how taking the complex conjugate of a signal in the time domain affects its Fourier transform. If:

$$x(t)\stackrel{\mathfrak{F}}{⟷}X(j\omega ),$$

then:

$$x^{\ast }(t)\stackrel{\mathfrak{F}}{⟷}{X}^{\ast }(-j\omega ).\text{\hspace{1em}(4.28)}$$

Proof: Let’s start by taking the complex conjugate of the Fourier transform definition (4.25):

$$\begin{array}{rlrl}{X}^{\ast }(j\omega )& ={\left[{\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt\right]}^{\ast }& & \\ & ={\int }_{-\infty }^{+\infty }{x}^{\ast }(t)(e^{-j\omega t}{)}^{\ast }dt& & \text{(Property of complex conjugate of integral: }(\int f(t)dt{)}^{\ast }=\int f^{\ast }(t)dt\text{)}\\ & ={\int }_{-\infty }^{+\infty }{x}^{\ast }(t)e^{j\omega t}dt.& & \end{array}$$

Now, to match the form of a Fourier transform, we need an $e^{-j(\cdot )t}$ term. Let’s replace $\omega$ with $-\omega$ in the expression we just derived for $X^{\ast }(j\omega )$:

$$X^{\ast }(-j\omega )={\int }_{-\infty }^{+\infty }{x}^{\ast }(t)e^{j(-\omega )t}dt={\int }_{-\infty }^{+\infty }{x}^{\ast }(t)e^{-j\omega t}dt.\text{\hspace{1em}(4.29)}$$

The right-hand side of equation (4.29) is precisely the definition of the Fourier transform of $x^{\ast }(t)$. Thus, we have proved the property:

$$\mathfrak{F}\{x^{\ast }(t)\}=X^{\ast }(-j\omega ).$$

A profoundly important consequence arises when $x(t)$ is a real signal. If $x(t)$ is real, then $x^{\ast }(t)=x(t)$. In this case, the conjugation property implies conjugate symmetry for its Fourier transform:

$$X(-j\omega )=X^{\ast }(j\omega )[x(t)\text{ real}].\text{\hspace{1em}(4.30)}$$

Derivation of (4.30) from (4.29): If $x(t)$ is real, then $x^{\ast }(t)=x(t)$. Substituting this into eq. (4.29):

$$X^{\ast }(-j\omega )={\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt=X(j\omega ).$$

Now, take the complex conjugate of both sides:

$$(X^{\ast }(-j\omega ){)}^{\ast }=(X(j\omega ){)}^{\ast }⟹X(-j\omega )=X^{\ast }(j\omega ).$$

This property has several implications for real signals:

1. Real and Imaginary Parts: If $X(j\omega )=\Re e\{X(j\omega )\}+j\Im m\{X(j\omega )\}$, then for a real $x(t)$:
   • $\Re e\{X(j\omega )\}$ is an even function of $\omega$: $\Re e\{X(j\omega )\}=\Re e\{X(-j\omega )\}$.
   • $\Im m\{X(j\omega )\}$ is an odd function of $\omega$: $\Im m\{X(j\omega )\}=-\Im m\{X(-j\omega )\}$.
2. Magnitude and Phase: If $X(j\omega )=|X(j\omega )|e^{j\angle X(j\omega )}$, then for a real $x(t)$:
   • $|X(j\omega )|$ is an even function of $\omega$: $|X(j\omega )|=|X(-j\omega )|$.
   • $\angle X(j\omega )$ is an odd function of $\omega$: $\angle X(j\omega )=-\angle X(-j\omega )$.

These symmetry properties are extremely useful! When dealing with real-valued signals, you only need to compute or plot the Fourier transform for positive frequencies ($\omega >0$). The values for negative frequencies can be directly inferred from these symmetries. For example, in Example 4.1 ($x(t)=e^{-at}u(t)$), $X(j\omega )=1/(a+j\omega )$, and indeed $X(-j\omega )=1/(a-j\omega )=X^{\ast }(j\omega )$.

Let’s dig deeper into symmetries for real signals:

• If $x(t)$ is real and even: Then $x(t)=x(-t)$. From $X(-j\omega )=X^{\ast }(j\omega )$ and $\mathfrak{F}\{x(-t)\}=X(-j\omega )$ (which we’ll prove in Section 4.3.5), we have $X(j\omega )=X(-j\omega )$. Combining $X(-j\omega )=X^{\ast }(j\omega )$ and $X(j\omega )=X(-j\omega )$, it implies $X(j\omega )=X^{\ast }(j\omega )$, meaning $X(j\omega )$ is purely real. Since $X(j\omega )=X(-j\omega )$, it is also an even function. So, if $x(t)$ is real and even, $X(j\omega )$ is real and even. Example 4.2 ($e^{-a|t|}$) perfectly illustrates this.
• If $x(t)$ is real and odd: Then $x(t)=-x(-t)$. Following a similar line of reasoning, it can be shown that $X(j\omega )$ will be purely imaginary and an odd function of $\omega$.

Recall that any real function $x(t)$ can be decomposed into its even and odd parts: $x(t)=x_e(t)+x_o(t)$, where $x_e(t)=\mathcal{E}v\{x(t)\}=\frac{x(t)+x(-t)}{2}$ and $x_o(t)=\mathcal{O}d\{x(t)\}=\frac{x(t)-x(-t)}{2}$. From the linearity of the Fourier transform:

$$\mathfrak{F}\{x(t)\}=\mathfrak{F}\{x_e(t)\}+\mathfrak{F}\{x_o(t)\}.$$

Given our discussion above:

• $\mathfrak{F}\{x_e(t)\}$ is real and even.
• $\mathfrak{F}\{x_o(t)\}$ is purely imaginary and odd. Therefore, for a real signal $x(t)$:

$$x(t)\stackrel{\mathfrak{F}}{⟷}X(j\omega )$$ $$\mathcal{E}v\{x(t)\}\stackrel{\mathfrak{F}}{⟷}\Re e\{X(j\omega )\}$$ $$\mathcal{O}d\{x(t)\}\stackrel{\mathfrak{F}}{⟷}j\Im m\{X(j\omega )\}.$$

These relationships provide a powerful way to relate the time-domain symmetries of a signal to the symmetries of its frequency-domain components.

Example 4.10: Utilizing Symmetry for Fourier Transform Calculation

Let’s re-evaluate the Fourier transform of $x(t)=e^{-a|t|}$ (from Example 4.2), this time leveraging symmetry properties. We already know from Example 4.1 that:

$$e^{-at}u(t)\stackrel{\mathfrak{F}}{⟷}\frac{1}{a+j\omega }.$$

Now, observe the signal $x(t)=e^{-a|t|}$. It is a real and even function. It can be expressed as:

$$x(t)=e^{-at}u(t)+e^{at}u(-t).$$

Alternatively, for $t>0$, $x(t)$ is $e^{-at}$, which is the positive-time part of $e^{-at}u(t)$. Since $x(t)$ is even, $x(t)$ is effectively twice the even part of $e^{-at}u(t)$:

$$x(t)=2\mathcal{E}v\{e^{-at}u(t)\}.$$

Since $e^{-at}u(t)$ is a real-valued signal, its even part’s Fourier transform corresponds to the real part of its Fourier transform. That is:

$$\mathcal{E}v\{e^{-at}u(t)\}\stackrel{\mathfrak{F}}{⟷}\Re e\left\{\frac{1}{a+j\omega }\right\}.$$

Let’s find the real part of $\frac{1}{a+j\omega }$:

$$\begin{array}{rl}\Re e\left\{\frac{1}{a+j\omega }\right\}& =\Re e\left\{\frac{1}{a+j\omega }\cdot \frac{a-j\omega }{a-j\omega }\right\}\\ & =\Re e\left\{\frac{a-j\omega }{a^2+{\omega }^2}\right\}\\ & =\frac{a}{a^2+{\omega }^2}.\end{array}$$

Therefore, using linearity, the Fourier transform $X(j\omega )$ of $x(t)=e^{-a|t|}$ is:

$$X(j\omega )=\mathfrak{F}\{2\mathcal{E}v\{e^{-at}u(t)\}\}=2\Re e\left\{\frac{1}{a+j\omega }\right\}=2\left(\frac{a}{a^2+{\omega }^2}\right)=\frac{2a}{a^2+{\omega }^2}.$$

This result precisely matches what we found in Example 4.2, showcasing the elegance and efficiency of using symmetry properties!

4.3.4 Differentiation and Integration

These properties are absolutely crucial for analyzing LTI systems described by differential equations. They show how operations in one domain transform into simpler operations in the other.

Differentiation in Time: If $x(t)$ has Fourier transform $X(j\omega )$, then differentiating $x(t)$ with respect to time $t$ corresponds to multiplying its Fourier transform by $j\omega$.

$$\frac{dx(t)}{dt}\stackrel{\mathfrak{F}}{⟷}j\omega X(j\omega ).\text{\hspace{1em}(4.31)}$$

Proof: Start with the Fourier transform synthesis equation:

$$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega .$$

Differentiating both sides with respect to $t$ (assuming we can interchange differentiation and integration, which is usually valid under conditions for Fourier transform existence):

$$\begin{array}{rl}\frac{dx(t)}{dt}& =\frac{d}{dt}\left(\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega \right)\\ & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )\frac{d}{dt}(e^{j\omega t})d\omega \\ & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )(j\omega e^{j\omega t})d\omega \\ & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }(j\omega X(j\omega ))e^{j\omega t}d\omega .\end{array}$$

Comparing this with the synthesis equation form, we see that the Fourier transform of $\frac{dx(t)}{dt}$ is indeed $j\omega X(j\omega )$. This property transforms a differential operator in the time domain into a simple algebraic multiplication in the frequency domain, a cornerstone of frequency-domain analysis!

Integration in Time: If differentiation leads to multiplication by $j\omega$, then integration should involve division by $j\omega$. This is mostly true, but we must account for any DC (average) component that might arise from integration.

$$\boxed{{\int }_{-\infty }^{t}x(\tau )d\tau \stackrel{\mathfrak{F}}{⟷}\frac{1}{j\omega }X(j\omega )+\pi X(0)\delta (\omega ).}\text{\hspace{1em}(4.32)}$$

Here, $X(0)$ represents the DC value of the signal $x(t)$ (i.e., ${\int }_{-\infty }^{+\infty }x(t)dt$). The impulse term $\pi X(0)\delta (\omega )$ accounts for any constant offset that the integral might introduce. If the integral of $x(t)$ over all time is zero, then $X(0)=0$, and this term vanishes.

Let’s illustrate these with some examples.

Example 4.11: Fourier Transform of the Unit Step

Let’s find the Fourier transform of the unit step function $x(t)=u(t)$ using the integration property (4.32). We know that the derivative of the unit step is the unit impulse: $\frac{d}{dt}u(t)=\delta (t)$. Let $g(t)=\delta (t)$. From Example 4.3, its Fourier transform is $G(j\omega )=1$. Now, $u(t)$ can be expressed as the integral of $\delta (t)$:

$$u(t)={\int }_{-\infty }^t\delta (\tau )d\tau .$$

Applying the integration property (4.32) to $g(t)=\delta (t)$:

$$\mathfrak{F}\left\{{\int }_{-\infty }^{t}g(\tau )d\tau \right\}=\frac{1}{j\omega }G(j\omega )+\pi G(0)\delta (\omega ).$$

Substitute $G(j\omega )=1$:

$$X(j\omega )=\frac{1}{j\omega }(1)+\pi (1)\delta (\omega ).$$

Thus, the Fourier transform of the unit step is:

$$X(j\omega )=\frac{1}{j\omega }+\pi \delta (\omega ).\text{\hspace{1em}(4.33)}$$

The $\pi \delta (\omega )$ term accounts for the DC value of $u(t)$, which effectively has an average value of $1/2$ (a constant of $1/2$ would be $2\pi (1/2)\delta (\omega )=\pi \delta (\omega )$).

Let’s do a quick check using the differentiation property to recover the transform of the impulse. If we differentiate $u(t)$:

$$\frac{du(t)}{dt}=\delta (t).$$

Then, its Fourier transform should be $j\omega X(j\omega )$:

$$\mathfrak{F}\{\delta (t)\}=j\omega \left[\frac{1}{j\omega }+\pi \delta (\omega )\right]=j\omega \frac{1}{j\omega }+j\omega \pi \delta (\omega )=1+\pi (j\omega \delta (\omega )).$$

Recall a crucial property of the impulse function: $\omega \delta (\omega )=0$. (This is because $\delta (\omega )$ is zero everywhere except at $\omega =0$, where the multiplication by $\omega$ makes the product zero.) Therefore, $j\omega \pi \delta (\omega )=0$, and:

$$\mathfrak{F}\{\delta (t)\}=1.$$

This is consistent with Example 4.3!

Example 4.12: Using Differentiation for Complex Signals

Suppose we want to find the Fourier transform $X(j\omega )$ of the signal $x(t)$ shown in Figure 4.16(a). Directly integrating this piecewise linear signal might be tedious. Instead, let’s consider its derivative: $g(t)=\frac{d}{dt}x(t)$. Looking at Figure 4.16(b), $g(t)$ is much simpler! It’s composed of:

• An impulse at $t=-2$ with area $-1$.
• A rectangular pulse from $t=-1$ to $t=1$ with height $1$.
• An impulse at $t=2$ with area $-1$. So, we can write $g(t)$ as:

$$g(t)=-\delta (t+2)+(\text{rect}(t/2))-\delta (t-2).$$

The Fourier transforms of these components are:

• $\mathfrak{F}\{-\delta (t+2)\}=-e^{j\omega 2}$ (using time-shift property on $\mathfrak{F}\{\delta (t)\}=1$)
• $\mathfrak{F}\{\text{rect}(t/2)\}$: This is a rectangular pulse from $-1$ to $1$. From Example 4.4, with $T_1=1$, its transform is $\frac{2\sin (\omega \cdot 1)}{\omega }=\frac{2\sin \omega }{\omega }$.
• $\mathfrak{F}\{-\delta (t-2)\}=-e^{-j\omega 2}$ (using time-shift property). So, using linearity:

$$G(j\omega )=-e^{j2\omega }+\frac{2\sin \omega }{\omega }-e^{-j2\omega }=\frac{2\sin \omega }{\omega }-(e^{j2\omega }+e^{-j2\omega })=\frac{2\sin \omega }{\omega }-2\cos (2\omega ).$$

Notice that $G(j0)={\lim }_{\omega \to 0}\left(\frac{2\sin \omega }{\omega }-2\cos (2\omega )\right)=2-2(1)=0$. This is expected, as the net change in $x(t)$ is zero (it starts at 0 and ends at 0), so its derivative must have zero area, meaning $G(j0)=0$. Now, using the integration property (4.32) in reverse (since $x(t)={\int }_{-\infty }^{t}g(\tau )d\tau$):

$$X(j\omega )=\frac{G(j\omega )}{j\omega }+\pi G(0)\delta (\omega ).$$

Since $G(0)=0$, the impulse term vanishes:

$$\begin{array}{rl}X(j\omega )& =\frac{1}{j\omega }\left(\frac{2\sin \omega }{\omega }-2\cos (2\omega )\right)\\ & =\frac{2\sin \omega }{j{\omega }^2}-\frac{2\cos (2\omega )}{j\omega }.\end{array}$$

This expression for $X(j\omega )$ is purely imaginary and odd. This aligns perfectly with the fact that $x(t)$ is a real and odd function, reinforcing our understanding of symmetry properties!

4.3.5 Time and Frequency Scaling

This property formalizes the inverse relationship between the time and frequency domains that we’ve observed repeatedly. If:

$$x(t)\stackrel{\mathfrak{F}}{⟷}X(j\omega ),$$

then for a nonzero real constant $a$:

$$\boxed{x(at)\stackrel{\mathfrak{F}}{⟷}\frac{1}{|a|}X\left(\frac{j\omega }{a}\right).}\text{\hspace{1em}(4.34)}$$

Proof: We start with the Fourier transform definition:

$$\mathfrak{F}\{x(at)\}={\int }_{-\infty }^{+\infty }x(at)e^{-j\omega t}dt.$$

Let’s use a substitution: $\tau =at$. Then $t=\tau /a$, and $dt=\frac{1}{a}d\tau$. We need to consider two cases for $a$:

Case 1: $a>0$ If $a>0$, the limits of integration remain the same (from $-\infty$ to $+\infty$).

$$\begin{array}{rl}\mathfrak{F}\{x(at)\}& ={\int }_{-\infty }^{+\infty }x(\tau )e^{-j\omega (\tau /a)}\frac{1}{a}d\tau \\ & =\frac{1}{a}{\int }_{-\infty }^{+\infty }x(\tau )e^{-j(\omega /a)\tau }d\tau .\end{array}$$

The integral on the right is simply the Fourier transform of $x(\tau )$ with the frequency variable replaced by $\omega /a$, i.e., $X(j\omega /a)$. So, for $a>0$: $\mathfrak{F}\{x(at)\}=\frac{1}{a}X(j\omega /a)$.

Case 2: $a<0$ If $a<0$, the limits of integration reverse. When $t\to -\infty$, $\tau =at\to +\infty$. When $t\to +\infty$, $\tau =at\to -\infty$. Also, $dt=\frac{1}{a}d\tau$, and since $a$ is negative, $\frac{1}{a}$ is negative.

$$\begin{array}{rlrl}\mathfrak{F}\{x(at)\}& ={\int }_{+\infty }^{-\infty }x(\tau )e^{-j\omega (\tau /a)}\frac{1}{a}d\tau & & \\ & =-\frac{1}{a}{\int }_{-\infty }^{+\infty }x(\tau )e^{-j(\omega /a)\tau }d\tau & & \text{(Swap limits and negate, }-1/a\text{ is positive)}\\ & =-\frac{1}{a}X(j\omega /a).& & \end{array}$$

Since $a<0$, we have $-1/a=1/|a|$. So, for $a<0$: $\mathfrak{F}\{x(at)\}=\frac{1}{|a|}X(j\omega /a)$.

Combining both cases, we get the general formula:

$$\mathfrak{F}\{x(at)\}=\frac{1}{|a|}X\left(\frac{j\omega }{a}\right).$$

This property shows that:

• Time Compression ($|a|>1$): If a signal is compressed in time ($x(at)$ for $a>1$), its Fourier transform is stretched in frequency by the factor $a$ (the argument becomes $\omega /a$) and its amplitude is scaled down by $1/a$.
• Time Expansion ($|a|<1$): If a signal is stretched in time ($x(at)$ for $0<a<1$), its Fourier transform is compressed in frequency by the factor $a$ (the argument becomes $\omega /a$) and its amplitude is scaled up by $1/a$.

This is precisely the inverse relationship we observed! A fast-changing signal (compressed in time) requires a broader range of frequencies to represent it. A slow-changing signal (expanded in time) has a narrower, more concentrated frequency content. This is why playing an audiotape at a higher speed (time compression) makes the sound higher pitched (frequency expansion).

A special case of the scaling property is when $a=-1$:

$$x(-t)\stackrel{\mathfrak{F}}{⟷}\frac{1}{|-1|}X\left(\frac{j\omega }{-1}\right)=X(-j\omega ).\text{\hspace{1em}(4.35)}$$

This means that reversing a signal in time also reverses its Fourier transform.

The inverse relationship between time and frequency is a recurring theme in signal processing. It’s reflected in the uncertainty principle in quantum mechanics (you can’t simultaneously know a particle’s exact position and momentum) and has profound implications in areas like filter design and communication systems.

4.3.6 Duality

The Fourier transform pair equations (4.24) and (4.25) exhibit a beautiful structural symmetry. This symmetry gives rise to the duality property. We hinted at this when comparing Examples 4.4 and 4.5. From Example 4.4, we have the pair:

$$x_1(t)=\{\begin{array}{ll}1,& |t|<T_1\\ 0,& |t|>T_1\end{array}\stackrel{\mathfrak{F}}{⟷}{X}_1(j\omega )=\frac{2\sin \omega T_1}{\omega }.\text{\hspace{1em}(4.36)}$$

And from Example 4.5, we have the pair:

$$x_2(t)=\frac{\sin Wt}{\pi t}\stackrel{\mathfrak{F}}{⟷}{X}_2(j\omega )=\{\begin{array}{ll}1,& |\omega |<W\\ 0,& |\omega |>W\end{array}.\text{\hspace{1em}(4.37)}$$

Figure 4.17 visually highlights this stunning relationship. The time function of one pair resembles the frequency function of the other, and vice versa!

The duality property states that if $x(t)\stackrel{\mathfrak{F}}{⟷}X(j\omega )$, then a signal whose time-domain shape is $X(t)$ will have a Fourier transform whose frequency-domain shape is related to $x(-\omega )$. Specifically, it is $2\pi x(-j\omega )$.

Let’s derive the general form of duality. We have:

$$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega .$$

Now, let’s swap the roles of $t$ and $\omega$. Let’s call the new time variable ${\omega }^{\prime }$ and the new frequency variable $t^{\prime }$. This sounds confusing, so let’s stick to the text’s method for now: Multiply the synthesis equation by $2\pi$:

$$2\pi x(t)={\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega .$$

Now, replace $t$ with $-t$:

$$2\pi x(-t)={\int }_{-\infty }^{+\infty }X(j\omega )e^{-j\omega t}d\omega .$$

The right-hand side is now exactly in the form of a Fourier transform analysis equation, but with $X(j\omega )$ acting as the time-domain signal and $\omega$ as the time variable, and $t$ as the frequency variable. So, if we define a new time-domain signal $g(t)=X(jt)$, its Fourier transform $G(j\omega )$ would be $2\pi x(-\omega )$. More generally, if $x(t)\stackrel{\mathfrak{F}}{⟷}X(j\omega )$, then $X(jt)\stackrel{\mathfrak{F}}{⟷}2\pi x(-j\omega )$.

Example 4.13: Duality in Action

Let’s use duality to find the Fourier transform $G(j\omega )$ of the signal:

$$g(t)=\frac{2}{1+t^2}.$$

We recognize this functional form from Example 4.2. There, we found the pair:

$$e^{-|t|}\stackrel{\mathfrak{F}}{⟷}\frac{2}{1+{\omega }^2}.$$

Let’s call $x(t)=e^{-|t|}$ and $X(j\omega )=\frac{2}{1+{\omega }^2}$. The synthesis equation for this pair is:

$$e^{-|t|}=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }\left(\frac{2}{1+{\omega }^2}\right)e^{j\omega t}d\omega .$$

Now, we want to find the transform of $g(t)=\frac{2}{1+t^2}$. Notice that $g(t)$ has the same functional form as $X(j\omega )$, but with $t$ replacing $\omega$. To utilize duality, we want to match the integral form of the Fourier transform definition (4.25). Multiply the synthesis equation by $2\pi$:

$$2\pi e^{-|t|}={\int }_{-\infty }^{\infty }\left(\frac{2}{1+{\omega }^2}\right)e^{j\omega t}d\omega .$$

Now, replace $t$ with $-t$:

$$2\pi e^{-|-t|}={\int }_{-\infty }^{\infty }\left(\frac{2}{1+{\omega }^2}\right)e^{-j\omega t}d\omega .$$

Since $|-t|=|t|$, we have:

$$2\pi e^{-|t|}={\int }_{-\infty }^{\infty }\left(\frac{2}{1+{\omega }^2}\right)e^{-j\omega t}d\omega .$$

Now, here’s the clever part. Let’s swap the dummy variables: let $\omega$ become $t$, and $t$ become $\omega$. (This is just relabeling, so it’s perfectly valid).

$$2\pi e^{-|\omega |}={\int }_{-\infty }^{\infty }\left(\frac{2}{1+t^2}\right)e^{-j\omega t}dt.\text{\hspace{1em}(4.38)}$$

The right-hand side of equation (4.38) is exactly the definition of the Fourier transform of the signal $\frac{2}{1+t^2}$. Therefore, we conclude that:

$$\mathfrak{F}\left\{\frac{2}{1+t^2}\right\}=2\pi e^{-|\omega |}.$$

This demonstrates the power of duality: if you know one transform pair, you can often deduce a related dual pair.

Duality also helps us predict other properties. Since differentiation in the time domain corresponds to multiplication by $j\omega$ in the frequency domain, duality suggests that multiplication by $jt$ in the time domain should correspond to differentiation in the frequency domain. Let’s verify this! Differentiate the Fourier transform analysis equation (4.25) with respect to $\omega$:

$$\begin{array}{rlrl}\frac{dX(j\omega )}{d\omega }& =\frac{d}{d\omega }{\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt& & \\ & ={\int }_{-\infty }^{+\infty }x(t)\frac{d}{d\omega }(e^{-j\omega t})dt& & \text{(Interchange differentiation and integration)}\\ & ={\int }_{-\infty }^{+\infty }x(t)(-jt{e}^{-j\omega t})dt& & \\ & ={\int }_{-\infty }^{+\infty }(-jtx(t))e^{-j\omega t}dt.& & \end{array}$$

This directly gives us the property:

$$-jtx(t)\stackrel{\mathfrak{F}}{⟷}\frac{dX(j\omega )}{d\omega }.\text{\hspace{1em}(4.39)}$$

Or, more commonly written:

$$tx(t)\stackrel{\mathfrak{F}}{⟷}j\frac{dX(j\omega )}{d\omega }.\text{\hspace{1em}(4.40)}$$

Similarly, we can derive the duals of the time-shift and integration properties:

• Frequency Shifting (Dual of Time Shifting): $$e^{j{\omega }_{0}t}x(t)\stackrel{\mathfrak{F}}{⟷}X(j(\omega -{\omega }_0)).\text{\hspace{1em}(4.41)}$$ Multiplying a signal by a complex exponential $e^{j{\omega }_{0}t}$ shifts its entire spectrum by ${\omega }_0$. This is foundational for modulation in communications!
• Integration in Frequency (Dual of Differentiation in Time, with a caveat): $$\mathfrak{F}\left\{\frac{1}{jt}x(t)\right\}={\int }_{-\infty }^{\omega }X(j\eta )d\eta +\pi x(0)\delta (\omega ).$$ However, the text presents a slightly different form, the inverse of which would correspond to integration in frequency: $$-\frac{1}{jt}x(t)+\pi x(0)\delta (t)\stackrel{\mathfrak{F}}{⟷}{\int }_{-\infty }^{\omega }X(j\eta )d\eta .\text{\hspace{1em}(4.42)}$$ This is less commonly used than the other properties, but shows the comprehensive nature of duality.

4.3.7 Parseval’s Relation

Parseval’s relation is a beautiful statement about the conservation of energy between the time and frequency domains. If $x(t)$ and $X(j\omega )$ form a Fourier transform pair, then:

$$\boxed{{\int }_{-\infty }^{+\infty }|x(t){|}^{2}dt=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }|X(j\omega ){|}^{2}d\omega .}\text{\hspace{1em}(4.43)}$$

Proof: We start with the left-hand side, representing the total energy of $x(t)$:

$${\int }_{-\infty }^{+\infty }|x(t){|}^{2}dt={\int }_{-\infty }^{+\infty }x(t)x^{\ast }(t)dt.$$

Now, substitute the inverse Fourier transform for $x^{\ast }(t)$. To do this, we first need to find $x^{\ast }(t)$. If $x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega$, then $x^{\ast }(t)={\left(\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega \right)}^{\ast }=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{X}^{\ast }(j\omega )e^{-j\omega t}d\omega$.

$${\int }_{-\infty }^{+\infty }|x(t){|}^{2}dt={\int }_{-\infty }^{+\infty }x(t)\left[\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{X}^{\ast }(j\omega )e^{-j\omega t}d\omega \right]dt.$$

Now, we can interchange the order of integration (assuming conditions allow for this, which they do for finite-energy signals):

$${\int }_{-\infty }^{+\infty }|x(t){|}^{2}dt=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{X}^{\ast }(j\omega )\left[{\int }_{-\infty }^{+\infty }x(t)e^{-j\omega t}dt\right]d\omega .$$

The bracketed term is precisely the definition of the Fourier transform of $x(t)$, which is $X(j\omega )$:

$${\int }_{-\infty }^{+\infty }|x(t){|}^{2}dt=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{X}^{\ast }(j\omega )X(j\omega )d\omega .$$

Since $X^{\ast }(j\omega )X(j\omega )=|X(j\omega ){|}^2$:

$${\int }_{-\infty }^{+\infty }|x(t){|}^{2}dt=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }|X(j\omega ){|}^{2}d\omega .$$

This truly profound relationship states that the total energy of a signal can be calculated either by integrating its squared magnitude in the time domain or by integrating the squared magnitude of its Fourier transform (scaled by $1/(2\pi )$) in the frequency domain. For this reason, $|X(j\omega ){|}^2$ is often referred to as the energy-density spectrum of the signal $x(t)$, telling us how the signal’s energy is distributed across different frequencies.

Parseval’s relation for aperiodic signals is the direct counterpart of Parseval’s relation for periodic signals (eq. 3.67), which stated that the average power of a periodic signal is the sum of the average powers of its harmonic components.

Example 4.14: Applying Parseval’s Relation and Differentiation Property

Let’s use these properties to evaluate some time-domain characteristics directly from given Fourier transforms. We want to find:

$$E={\int }_{-\infty }^{\infty }|x(t){|}^{2}dt\text{and}D={\frac{d}{dt}x(t)|}_{t=0}.$$

for the Fourier transforms shown in Figure 4.18.

Part 1: Evaluating $E$ (Total Energy) We use Parseval’s relation (4.43):

$$E=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }|X(j\omega ){|}^{2}d\omega .\text{\hspace{1em}(4.44)}$$

• For Figure 4.18(a): $X(j\omega )$ is a rectangular pulse with height $1$ from $\omega =-1$ to $\omega =1$, and height $1/2$ from $\omega =-2$ to $\omega =-1$ and $\omega =1$ to $\omega =2$. The magnitude squared $|X(j\omega ){|}^2$ will be: $|X(j\omega ){|}^2=1^2=1$ for $|\omega |<1$. $|X(j\omega ){|}^2=(1/2{)}^2=1/4$ for $1<|\omega |<2$.

  $$\begin{array}{rl}{E}_a& =\frac{1}{2\pi }\left[{\int }_{-1}^1{1}^{2}d\omega +{\int }_{-2}^{-1}{\left(\frac{1}{2}\right)}^{2}d\omega +{\int }_1^2{\left(\frac{1}{2}\right)}^{2}d\omega \right]\\ & =\frac{1}{2\pi }\left[[\omega {]}_{-1}^1+[\frac{1}{4}\omega {]}_{-2}^{-1}+[\frac{1}{4}\omega {]}_1^2\right]\\ & =\frac{1}{2\pi }\left[(1-(-1))+\left(\frac{-1}{4}-\frac{-2}{4}\right)+\left(\frac{2}{4}-\frac{1}{4}\right)\right]\\ & =\frac{1}{2\pi }\left[2+\frac{1}{4}+\frac{1}{4}\right]\\ & =\frac{1}{2\pi }\left[2+\frac{1}{2}\right]=\frac{1}{2\pi }\left[\frac{5}{2}\right]\\ & =\frac{5}{4\pi }.\end{array}$$

  Correction: The original problem description states the result should be $5/8$. Let’s re-read the figure description. Ah, I see: Figure 4.18(a) has constant real magnitude and zero imaginary component. The function is $X(j\omega )=1$ for $|\omega |<1$ and $X(j\omega )=1/2$ for $1<|\omega |<2$. The calculation is correct for this interpretation: $E=\frac{5}{4\pi }$. The problem text’s result of $5/8$ is peculiar unless there’s an implied $2\pi$ factor missing somewhere or a different normalization, but adhering strictly to (4.44) yields $5/(4\pi )$. Let’s assume the $5/8$ is a target value that might involve different constants for now. I’ll stick to my derivation.

• For Figure 4.18(b): $X(j\omega )$ is a triangular pulse from $\omega =-\sqrt{\pi }$ to $\omega =\sqrt{\pi }$, with maximum height $1$ at $\omega =0$. The equation for the triangular pulse can be written as: $X(j\omega )=1-\frac{|\omega |}{\sqrt{\pi }}$ for $|\omega |<\sqrt{\pi }$, and $0$ otherwise.

  $$\begin{array}{rlrl}{E}_b& =\frac{1}{2\pi }{\int }_{-\sqrt{\pi }}^{\sqrt{\pi }}{\left(1-\frac{|\omega |}{\sqrt{\pi }}\right)}^{2}d\omega & & \\ & =\frac{1}{2\pi }\cdot 2{\int }_0^{\sqrt{\pi }}{\left(1-\frac{\omega }{\sqrt{\pi }}\right)}^{2}d\omega & & \text{(Symmetry: }X(j\omega )\text{ is real and even)}\\ & =\frac{1}{\pi }{\int }_0^{\sqrt{\pi }}\left(1-\frac{2\omega }{\sqrt{\pi }}+\frac{{\omega }^2}{\pi }\right)d\omega & & \\ & =\frac{1}{\pi }{\left[\omega -\frac{{\omega }^2}{\sqrt{\pi }}+\frac{{\omega }^3}{3\pi }\right]}_0^{\sqrt{\pi }}& & \\ & =\frac{1}{\pi }\left[\sqrt{\pi }-\frac{(\sqrt{\pi }{)}^2}{\sqrt{\pi }}+\frac{(\sqrt{\pi }{)}^3}{3\pi }-0\right]& & \\ & =\frac{1}{\pi }\left[\sqrt{\pi }-\frac{\pi }{\sqrt{\pi }}+\frac{\pi \sqrt{\pi }}{3\pi }\right]& & \\ & =\frac{1}{\pi }\left[\sqrt{\pi }-\sqrt{\pi }+\frac{\sqrt{\pi }}{3}\right]& & \\ & =\frac{1}{\pi }\left[\frac{\sqrt{\pi }}{3}\right]=\frac{1}{3\sqrt{\pi }}.& & \end{array}$$

  Again, the problem states the result is $1$. This suggests my interpretation of the plots or the formula is off by a constant. The question asks to use Parseval’s relation and equation (4.44). I’ve done that. Let me re-verify if perhaps $X(j\omega )$ in 4.18(b) is meant to be a simple rectangle of width $2\sqrt{\pi }$ and height $1$? No, it’s clearly triangular. The area under $|X(j\omega ){|}^2$ must be $2\pi$ for $E=1$. The area is ${\int }_{-\sqrt{\pi }}^{\sqrt{\pi }}(1-|\omega |/\sqrt{\pi }{)}^{2}d\omega =2{\int }_0^{\sqrt{\pi }}(1-\omega /\sqrt{\pi }{)}^{2}d\omega =2[\omega -{\omega }^2/\sqrt{\pi }+{\omega }^3/(3\pi ){]}_0^{\sqrt{\pi }}=2[\sqrt{\pi }-\sqrt{\pi }+\sqrt{\pi }/3]=2\sqrt{\pi }/3$. So $E=(1/(2\pi ))(2\sqrt{\pi }/3)=1/(3\sqrt{\pi })$. I will proceed with my derived results for $E$.

Part 2: Evaluating $D$ (Derivative at $t=0$) We know from the differentiation property (4.31) that if $g(t)=\frac{d}{dt}x(t)$, then $G(j\omega )=j\omega X(j\omega )$. We also know that the value of a signal at $t=0$ can be found from its Fourier transform using the inverse transform formula at $t=0$:

$$g(0)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }G(j\omega )e^{j\omega \cdot 0}d\omega =\frac{1}{2\pi }{\int }_{-\infty }^{\infty }G(j\omega )d\omega .$$

Therefore, $D=g(0)$ can be found as:

$$D=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }j\omega X(j\omega )d\omega .\text{\hspace{1em}(4.45)}$$

• For Figure 4.18(a): $X(j\omega )$ is real and even. Therefore, $j\omega X(j\omega )$ is purely imaginary and odd. The integral of an odd function over a symmetric interval ($-\infty$ to $+\infty$) is always zero.

  $$D_a=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }j\omega X(j\omega )d\omega =0.$$

  This matches the problem text. If $x(t)$ is real and even, its derivative $dx(t)/dt$ is real and odd. For an odd function $g(t)$, $g(0)$ must be $0$.

• For Figure 4.18(b): $X(j\omega )$ is real and even. Again, $j\omega X(j\omega )$ is purely imaginary and odd, so the integral should be zero. Let’s verify this using the explicit function for $X(j\omega )=1-\frac{|\omega |}{\sqrt{\pi }}$. Then $j\omega X(j\omega )=j\omega \left(1-\frac{|\omega |}{\sqrt{\pi }}\right)$. This function is indeed odd, as for $\omega >0$, $j\omega (1-\omega /\sqrt{\pi })$, and for $\omega <0$, $j\omega (1+\omega /\sqrt{\pi })$. Let $F(\omega )=j\omega X(j\omega )$. We need to calculate ${\int }_{-\infty }^{\infty }F(\omega )d\omega$. Since $F(\omega )$ is an odd function and the limits are symmetric, the integral is $0$.

  $$D_b=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }j\omega X(j\omega )d\omega =0.$$

  The problem text gives $D_b=\frac{-1}{2\sqrt{\pi }}$. This implies $X(j\omega )$ is not entirely real or not entirely even in its structure, or my interpretation is again different. If $X(j\omega )$ had an imaginary odd component, or a real odd component, then $j\omega X(j\omega )$ would have a real even component, which would integrate to a non-zero value. For $D_b=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }j\omega X(j\omega )d\omega =\frac{-1}{2\sqrt{\pi }}$, this would mean ${\int }_{-\infty }^{\infty }j\omega X(j\omega )d\omega =-\pi /\sqrt{\pi }=-\sqrt{\pi }$. Since $X(j\omega )$ is real and even from the graph, $j\omega X(j\omega )$ is purely imaginary and odd. The integral of a purely imaginary odd function over symmetric limits is $0$. Perhaps $X(j\omega )$ in the figure refers to $|X(j\omega )|$? No, it’s explicitly labelled $X(j\omega )$. This suggests that the solution provided in the notes might have made an assumption about $X(j\omega )$‘s phase or odd part that isn’t immediately obvious from the drawing, or the question’s target value is based on a different problem setup. I will again stick to my derivation based on the given figures representing real, even functions.

These properties are indispensable for gaining intuition and solving problems. Next, we explore two fundamental properties central to LTI systems: convolution and multiplication.

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4.4 The Convolution Property

In Chapter 3, we discovered a beautiful truth about LTI systems: if a periodic signal (a sum of complex exponentials) enters an LTI system, the output is also a sum of complex exponentials, each merely scaled by the system’s frequency response $H(jk{\omega }_0)$. This scaling is simply the Fourier series coefficient of the input multiplied by the system’s response at that specific harmonic frequency.

Now, we extend this powerful idea to aperiodic signals using the continuous-time Fourier transform. We’ll start with an intuitive argument, building upon what we know, and then provide a formal derivation.

Recall the Fourier transform synthesis equation (4.8), which expresses $x(t)$ as a continuous superposition of complex exponentials. We can visualize this as the limit of a sum, as we developed in equation (4.7):

$$x(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )e^{j\omega t}d\omega =\underset{{\omega }_0\to 0}{\lim }\frac{1}{2\pi }\sum _{k=-\infty }^{+\infty }X(jk{\omega }_0)e^{jk{\omega }_{0}t}{\omega }_0.\text{\hspace{1em}(4.47)}$$

As we learned in Sections 3.2 and 3.8, complex exponentials are the cherished eigenfunctions of LTI systems. This means if an LTI system with impulse response $h(t)$ is fed a complex exponential $e^{jk{\omega }_{0}t}$, its output will simply be a scaled version: $H(jk{\omega }_0)e^{jk{\omega }_{0}t}$. Here, $H(jk{\omega }_0)$ is the system’s frequency response, which, if you recall, is defined as the Fourier transform of the impulse response $h(t)$ evaluated at $\omega =k{\omega }_0$:

$$H(j\omega )={\int }_{-\infty }^{+\infty }h(t)e^{-j\omega t}dt.\text{\hspace{1em}(4.48)}$$

Thanks to the linearity of LTI systems (superposition!), if the input is a sum of these exponentials, the output will be a sum of their individually scaled responses:

$$\frac{1}{2\pi }\sum _{k=-\infty }^{+\infty }X(jk{\omega }_0)e^{jk{\omega }_{0}t}{\omega }_0\stackrel{\text{LTI System }h(t)}{\to }\frac{1}{2\pi }\sum _{k=-\infty }^{+\infty }X(jk{\omega }_0)H(jk{\omega }_0)e^{jk{\omega }_{0}t}{\omega }_0.$$

Now, taking the limit as ${\omega }_0\to 0$ (i.e., $T\to \infty$), the sum transforms back into an integral. The output $y(t)$ of the LTI system is then:

$$y(t)=\underset{{\omega }_0\to 0}{\lim }\frac{1}{2\pi }\sum _{k=-\infty }^{+\infty }X(jk{\omega }_0)H(jk{\omega }_0)e^{jk{\omega }_{0}t}{\omega }_0=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\omega )H(j\omega )e^{j\omega t}d\omega .\text{\hspace{1em}(4.49)}$$

We also know that the output $y(t)$ and its Fourier transform $Y(j\omega )$ are related by the synthesis equation:

$$y(t)=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }Y(j\omega )e^{j\omega t}d\omega .\text{\hspace{1em}(4.50)}$$

Comparing equations (4.49) and (4.50), we can directly identify the Fourier transform of the output $Y(j\omega )$:

$$Y(j\omega )=X(j\omega )H(j\omega ).\text{\hspace{1em}(4.51)}$$

This is a monumental result! It states that convolution in the time domain becomes simple multiplication in the frequency domain.

For a more formal derivation, let’s start with the definition of convolution itself:

$$y(t)={\int }_{-\infty }^{+\infty }x(\tau )h(t-\tau )d\tau .\text{\hspace{1em}(4.52)}$$

We want to find $Y(j\omega )$, the Fourier transform of $y(t)$:

$$Y(j\omega )=\mathfrak{F}\{y(t)\}={\int }_{-\infty }^{+\infty }\left[{\int }_{-\infty }^{+\infty }x(\tau )h(t-\tau )d\tau \right]e^{-j\omega t}dt.\text{\hspace{1em}(4.53)}$$

Assuming we can interchange the order of integration (Fubini’s theorem), and noting that $x(\tau )$ doesn’t depend on $t$:

$$Y(j\omega )={\int }_{-\infty }^{+\infty }x(\tau )\left[{\int }_{-\infty }^{+\infty }h(t-\tau )e^{-j\omega t}dt\right]d\tau .\text{\hspace{1em}(4.54)}$$

Now, look closely at the bracketed term: ${\int }_{-\infty }^{+\infty }h(t-\tau )e^{-j\omega t}dt$. This is the Fourier transform of a time-shifted signal $h(t-\tau )$. Using the time-shift property (4.27), we know this is equal to $e^{-j\omega \tau }H(j\omega )$. Substitute this back into equation (4.54):

$$Y(j\omega )={\int }_{-\infty }^{+\infty }x(\tau )e^{-j\omega \tau }H(j\omega )d\tau .$$

Since $H(j\omega )$ is independent of $\tau$, we can pull it out of the integral:

$$Y(j\omega )=H(j\omega ){\int }_{-\infty }^{+\infty }x(\tau )e^{-j\omega \tau }d\tau .\text{\hspace{1em}(4.55)}$$

The remaining integral is precisely the definition of the Fourier transform of $x(\tau )$, which is $X(j\omega )$. Hence, we arrive at the same crucial result:

$$Y(j\omega )=H(j\omega )X(j\omega ).$$

So, the convolution property states:

$$\boxed{y(t)=h(t)\ast x(t)\stackrel{\mathfrak{F}}{⟷}Y(j\omega )=H(j\omega )X(j\omega ).}\text{\hspace{1em}(4.56)}$$

Equation (4.56) is one of the most fundamental relationships in signal and system analysis. It elegantly transforms the often-complex operation of convolution in the time domain into a simple algebraic multiplication in the frequency domain. $H(j\omega )$, the frequency response (the Fourier transform of the impulse response), acts as a spectral filter, modifying the amplitude and phase of each frequency component of the input signal $X(j\omega )$.

For instance, in frequency-selective filtering (like a lowpass filter), we desire $H(j\omega )\approx 1$ for a certain range of frequencies (the passband) to let them pass unchanged, and $H(j\omega )\approx 0$ for other frequencies (the stopband) to eliminate them.

The frequency response $H(j\omega )$ is as important as the impulse response $h(t)$ in characterizing an LTI system. Since $h(t)$ completely describes the system, so must $H(j\omega )$. Moreover, many LTI system properties become clear in the frequency domain. For example, the impulse response of two cascaded LTI systems is the convolution of their individual impulse responses. From the convolution property, this means that the overall frequency response of cascaded systems is simply the product of their individual frequency responses, as depicted in Figure 4.19. This immediately shows that the order of cascading does not matter, as multiplication is commutative.

A crucial point about the existence of $H(j\omega )$: the Fourier transform is guaranteed to exist only under certain conditions. For an LTI system to have a well-defined frequency response $H(j\omega )$, its impulse response $h(t)$ must generally satisfy the Dirichlet conditions, particularly absolute integrability:

$${\int }_{-\infty }^{+\infty }|h(t)|dt<\infty .\text{\hspace{1em}(4.57)}$$

This condition is precisely what ensures the stability of an LTI system (as we learned in Section 2.3.7 and Problem 2.49). Therefore, a stable LTI system is guaranteed to have a Fourier transformable impulse response and thus a well-defined frequency response $H(j\omega )$. In our Fourier analysis, we’ll focus on systems whose impulse responses possess Fourier transforms. For unstable LTI systems, we’ll need a generalization of the Fourier transform called the Laplace transform, which we’ll explore in Chapter 9. But for now, the Fourier transform suffices for a vast array of practical and theoretical problems.

4.4.1 Examples

Let’s see the convolution property in action!

Example 4.15: A Pure Time Shift

Consider an LTI system with the simplest possible impulse response:

$$h(t)=\delta (t-t_0).\text{\hspace{1em}(4.58)}$$

This system merely delays the input by $t_0$. Its frequency response $H(j\omega )$ is the Fourier transform of $h(t)$. Using the time-shift property (4.27) and knowing $\mathfrak{F}\{\delta (t)\}=1$:

$$H(j\omega )=\mathfrak{F}\{\delta (t-t_0)\}=e^{-j\omega t_0}\cdot 1=e^{-j\omega t_0}.\text{\hspace{1em}(4.59)}$$

Now, if an input $x(t)$ with Fourier transform $X(j\omega )$ passes through this system, the output $Y(j\omega )$ is:

$$Y(j\omega )=H(j\omega )X(j\omega )=e^{-j\omega t_0}X(j\omega ).\text{\hspace{1em}(4.60)}$$

Taking the inverse Fourier transform of $Y(j\omega )$ confirms $y(t)=x(t-t_0)$, which is exactly what a system with impulse response $\delta (t-t_0)$ does. Note that the magnitude of this frequency response is $|H(j\omega )|=|e^{-j\omega t_0}|=1$ for all frequencies, meaning it passes all frequency components without attenuation. The phase response is $\angle H(j\omega )=-\omega t_0$, which is a linear phase characteristic.

Example 4.16: The Differentiator

An LTI system that performs differentiation has an input-output relationship:

$$y(t)=\frac{dx(t)}{dt}.$$

From the differentiation property (4.31) we just discussed:

$$Y(j\omega )=j\omega X(j\omega ).\text{\hspace{1em}(4.61)}$$

Comparing this with the convolution property $Y(j\omega )=H(j\omega )X(j\omega )$, we immediately see that the frequency response of a differentiator is:

$$H(j\omega )=j\omega .\text{\hspace{1em}(4.62)}$$

This is a powerful simplification! Differentiation, a complex operation in the time domain, becomes simple multiplication by $j\omega$ in the frequency domain.

Example 4.17: The Integrator

Consider an integrator, an LTI system whose output is the running integral of its input:

$$y(t)={\int }_{-\infty }^{t}x(\tau )d\tau .$$

The impulse response of this system is the unit step $h(t)=u(t)$. From Example 4.11, the Fourier transform of $u(t)$ is:

$$H(j\omega )=\frac{1}{j\omega }+\pi \delta (\omega ).$$

Then, using the convolution property $Y(j\omega )=H(j\omega )X(j\omega )$:

$$\begin{array}{rl}Y(j\omega )& =\left(\frac{1}{j\omega }+\pi \delta (\omega )\right)X(j\omega )\\ & =\frac{1}{j\omega }X(j\omega )+\pi X(j\omega )\delta (\omega ).\end{array}$$

Recall that $f(\omega )\delta (\omega )=f(0)\delta (\omega )$. So $X(j\omega )\delta (\omega )=X(j0)\delta (\omega )=X(0)\delta (\omega )$.

$$Y(j\omega )=\frac{1}{j\omega }X(j\omega )+\pi X(0)\delta (\omega ).$$

This is precisely consistent with the integration property (4.32) we derived earlier!

Example 4.18: Ideal Lowpass Filter Characteristics

As discussed in Section 3.9.2, frequency-selective filters are LTI systems designed to pass certain frequency ranges and attenuate others. The ideal lowpass filter is a perfect example, with a frequency response shown in Figure 4.20:

$$H(j\omega )=\{\begin{array}{ll}1,& |\omega |<{\omega }_c\\ 0,& |\omega |>{\omega }_c\end{array}.\text{\hspace{1em}(4.63)}$$

Here, ${\omega }_c$ is the cutoff frequency. Now that we have the Fourier transform, we can find the impulse response $h(t)$ of this ideal filter by taking the inverse Fourier transform of $H(j\omega )$. This is exactly the scenario of Example 4.5! Using that result, with $W={\omega }_c$:

$$h(t)=\frac{\sin {\omega }_{c}t}{\pi t}.\text{\hspace{1em}(4.64)}$$

This impulse response is plotted in Figure 4.21.

Looking at $h(t)$ for the ideal lowpass filter, we see some interesting characteristics:

1. Non-causal: $h(t)$ is non-zero for $t<0$. This means the ideal lowpass filter is non-causal, producing an output that depends on future inputs. In real-world applications where causality is a requirement (e.g., real-time processing), an ideal lowpass filter cannot be perfectly realized.
2. Oscillatory Behavior (Ringing): The sinc function exhibits significant oscillations that decay slowly. In some applications (like audio or image processing), this “ringing” in the time domain can be undesirable.

These observations highlight a fundamental trade-off in filter design: ideal frequency-domain characteristics (perfect brick-wall cutoff) often lead to undesirable time-domain characteristics (non-causality, ringing). Real-world filters involve compromises.

Consider, for example, a simple RC circuit from Section 3.10. Its impulse response is:

$$h(t)=e^{-t}u(t).\text{\hspace{1em}(4.65)}$$

Its frequency response, from Example 4.1, is:

$$H(j\omega )=\frac{1}{j\omega +1}.\text{\hspace{1em}(4.66)}$$

The impulse response and the magnitude of the frequency response are shown in Figure 4.22. This system is causal ($h(t)=0$ for $t<0$) and has an impulse response that decays monotonically without oscillation. While its frequency selectivity isn’t as sharp as the ideal lowpass filter, its desirable time-domain properties, ease of implementation, and causality make it (or variations of it) often preferable in practical applications. This interplay between time and frequency characteristics is a rich area of study in Chapter 6.

Example 4.19: Solving Convolution using Fourier Transforms

Let’s find the output $y(t)=x(t)\ast h(t)$ for an LTI system with:

$$h(t)=e^{-at}u(t),a>0$$

and input:

$$x(t)=e^{-bt}u(t),b>0.$$

Instead of directly evaluating the convolution integral (which can be messy!), we’ll use the convolution property. First, find the Fourier transforms of $x(t)$ and $h(t)$. From Example 4.1:

$$X(j\omega )=\frac{1}{b+j\omega }$$ $$H(j\omega )=\frac{1}{a+j\omega }.$$

Now, apply the convolution property $Y(j\omega )=H(j\omega )X(j\omega )$:

$$Y(j\omega )=\frac{1}{(a+j\omega )(b+j\omega )}.\text{\hspace{1em}(4.67)}$$

To find $y(t)$, we need the inverse Fourier transform of $Y(j\omega )$. This is a common task, and partial-fraction expansion is the go-to technique. Assuming $b\ne a$, we express $Y(j\omega )$ as:

$$Y(j\omega )=\frac{A}{a+j\omega }+\frac{B}{b+j\omega }.\text{\hspace{1em}(4.68)}$$

To find $A$ and $B$:

• Multiply both sides by $(a+j\omega )$ and set $j\omega =-a$: $$A={\left[\frac{1}{b+j\omega }\right]}_{j\omega =-a}=\frac{1}{b-a}.$$
• Multiply both sides by $(b+j\omega )$ and set $j\omega =-b$: $$B={\left[\frac{1}{a+j\omega }\right]}_{j\omega =-b}=\frac{1}{a-b}=-\frac{1}{b-a}.$$

So, $A=\frac{1}{b-a}$ and $B=-\frac{1}{b-a}$. Substituting these back into (4.68):

$$Y(j\omega )=\frac{1}{b-a}\left[\frac{1}{a+j\omega }-\frac{1}{b+j\omega }\right].\text{\hspace{1em}(4.69)}$$

Now, we can find the inverse transform of each term by inspection (again using Example 4.1 and linearity):

$${\mathfrak{F}}^{-1}\left\{\frac{1}{a+j\omega }\right\}=e^{-at}u(t)$$ $${\mathfrak{F}}^{-1}\left\{\frac{1}{b+j\omega }\right\}=e^{-bt}u(t).$$

Therefore, the output $y(t)$ is:

$$y(t)=\frac{1}{b-a}[e^{-at}u(t)-e^{-bt}u(t)].$$

What if $b=a$? The partial-fraction expansion (4.69) is not valid. In this case, equation (4.67) becomes:

$$Y(j\omega )=\frac{1}{(a+j\omega {)}^2}.$$

To find its inverse transform, we can use the differentiation-in-frequency property (4.40), or look it up in Table 4.2. We know that $e^{-at}u(t)\stackrel{\mathfrak{F}}{⟷}\frac{1}{a+j\omega }$. From (4.40), $tx(t)\stackrel{\mathfrak{F}}{⟷}j\frac{dX(j\omega )}{d\omega }$. Let $x(t)=e^{-at}u(t)$. Then $X(j\omega )=\frac{1}{a+j\omega }$.

$$\mathfrak{F}\{t{e}^{-at}u(t)\}=j\frac{d}{d\omega }\left[\frac{1}{a+j\omega }\right]=j\left[-(a+j\omega {)}^{-2}\cdot j\right]=j^2\left[-\frac{1}{(a+j\omega {)}^2}\right](-1)=\frac{1}{(a+j\omega {)}^2}.$$

Thus, if $b=a$:

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

This example beautifully illustrates how Fourier transforms turn convolution problems into algebraic ones, making solutions much more tractable.

Example 4.20: Convolution of Sinc Functions

Let’s find the response of an ideal lowpass filter to an input that is also a sinc function:

$$x(t)=\frac{\sin {\omega }_{i}t}{\pi t}.$$

The impulse response of an ideal lowpass filter is:

$$h(t)=\frac{\sin {\omega }_{c}t}{\pi t}.$$

We want to find $y(t)=x(t)\ast h(t)$. Again, we turn to the frequency domain! We know the Fourier transforms of these sinc functions are rectangular pulses (from Example 4.5):

$$X(j\omega )=\{\begin{array}{ll}1,& |\omega |<{\omega }_i\\ 0,& |\omega |>{\omega }_i\end{array}$$ $$H(j\omega )=\{\begin{array}{ll}1,& |\omega |<{\omega }_c\\ 0,& |\omega |>{\omega }_c\end{array}.$$

Now, by the convolution property, $Y(j\omega )=X(j\omega )H(j\omega )$. This is simply the product of two rectangular pulses:

$$Y(j\omega )=\{\begin{array}{ll}1\cdot 1,& |\omega |<\min ({\omega }_i,{\omega }_c)\\ 0,& \text{otherwise}\end{array}.$$

Let ${\omega }_0=\min ({\omega }_i,{\omega }_c)$. Then:

$$Y(j\omega )=\{\begin{array}{ll}1,& |\omega |<{\omega }_0\\ 0,& |\omega |>{\omega }_0\end{array}.$$

This is another rectangular pulse in the frequency domain! Taking its inverse Fourier transform (again, Example 4.5):

$$y(t)=\frac{\sin {\omega }_{0}t}{\pi t}.$$

So, the convolution of two sinc functions results in another sinc function, whose bandwidth is the smaller of the two original bandwidths. This is a very elegant result!

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4.5 The Multiplication Property

We’ve seen that convolution in the time domain corresponds to multiplication in the frequency domain. Thanks to the beautiful duality between time and frequency, we should expect the reverse: multiplication in the time domain corresponds to convolution in the frequency domain! Specifically, if $r(t)=s(t)p(t)$, then its Fourier transform $R(j\omega )$ is given by:

$$\boxed{R(j\omega )=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }S(j\theta )P(j(\omega -\theta ))d\theta .}\text{\hspace{1em}(4.70)}$$

This integral is recognized as a convolution integral, but in the frequency domain! It’s denoted as $R(j\omega )=\frac{1}{2\pi }S(j\omega )\ast P(j\omega )$. The factor of $1/(2\pi )$ arises from the definition of the inverse Fourier transform and the way the transform is normalized.

This property can be derived either by exploiting duality from the convolution property (as mentioned in Section 4.3.6) or by directly substituting the Fourier transform relations, much like how we derived the convolution property.

Multiplying one signal by another is often called amplitude modulation, as one signal’s amplitude is scaled or “modulated” by the other. Thus, equation (4.70) is sometimes referred to as the modulation property. This property is absolutely fundamental in many communication systems and in sampling theory, as we’ll explore in Chapters 7 and 8. Let’s see it in action.

Example 4.21: Sinusoidal Amplitude Modulation

Consider a signal $s(t)$ whose spectrum $S(j\omega )$ is depicted in Figure 4.23(a) (a band-limited signal). Now, let’s multiply $s(t)$ by a sinusoidal signal:

$$p(t)=\cos {\omega }_{0}t.$$

From Example 4.7, the Fourier transform of $p(t)$ is:

$$P(j\omega )=\pi \delta (\omega -{\omega }_0)+\pi \delta (\omega +{\omega }_0).$$

This is shown in Figure 4.23(b). Now, let $r(t)=s(t)p(t)$. We apply the multiplication property (4.70) to find $R(j\omega )$:

$$\begin{array}{rlrl}R(j\omega )& =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }S(j\theta )P(j(\omega -\theta ))d\theta & & \text{(Multiplication Property)}\\ & =\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }S(j\theta )[\pi \delta (\omega -\theta -{\omega }_0)+\pi \delta (\omega -\theta +{\omega }_0)]d\theta & & \text{(Substitute }P(j(\omega -\theta ))\text{)}\\ & =\frac{1}{2}{\int }_{-\infty }^{+\infty }S(j\theta )\delta (\theta -(\omega -{\omega }_0))d\theta +\frac{1}{2}{\int }_{-\infty }^{+\infty }S(j\theta )\delta (\theta -(\omega +{\omega }_0))d\theta & & \text{(Split integral, pull out }\pi \text{)}\\ & =\frac{1}{2}S(j(\omega -{\omega }_0))+\frac{1}{2}S(j(\omega +{\omega }_0)).& & \text{(Sifting Property of Impulse)}\end{array}$$

So, the Fourier transform of $r(t)$ is:

$$R(j\omega )=\frac{1}{2}S(j(\omega -{\omega }_0))+\frac{1}{2}S(j(\omega +{\omega }_0)).\text{\hspace{1em}(4.71)}$$

This result is sketched in Figure 4.23(c), assuming ${\omega }_0>{\omega }_1$ (so the two shifted spectra don’t overlap). What has happened? The original spectrum $S(j\omega )$ has been shifted to be centered around $+{\omega }_0$ and $-{\omega }_0$, and its amplitude has been scaled by $1/2$. All the information from $s(t)$ is still present, but it’s now located at higher frequencies. This “frequency shifting” is the essence of sinusoidal amplitude modulation, a cornerstone of radio communication.

Example 4.22: Demodulation

Now, let’s say we have the modulated signal $r(t)$ from Example 4.21, and we want to recover the original signal $s(t)$. This process is called demodulation. Let’s multiply $r(t)$ by $p(t)=\cos {\omega }_{0}t$ again:

$$g(t)=r(t)p(t).$$

The Fourier transforms $R(j\omega )$ and $P(j\omega )$ are shown in Figure 4.24(a) and (b), respectively. Applying the multiplication property again to find $G(j\omega )$:

$$G(j\omega )=\frac{1}{2\pi }R(j\omega )\ast P(j\omega ).$$

This involves convolving $R(j\omega )$ with $P(j\omega )$. Remember $R(j\omega )$ has two copies of $S(j\omega )$, and $P(j\omega )$ has two impulses. Each impulse in $P(j\omega )$ will shift a copy of $R(j\omega )$.

$$\begin{array}{rl}G(j\omega )& =\frac{1}{2\pi }\left[\frac{1}{2}S(j(\omega -{\omega }_0))+\frac{1}{2}S(j(\omega +{\omega }_0))\right]\ast [\pi \delta (\omega -{\omega }_0)+\pi \delta (\omega +{\omega }_0)]\\ & =\frac{1}{2}[\frac{1}{2}S(j(\omega -{\omega }_0))\ast \delta (\omega -{\omega }_0)+\frac{1}{2}S(j(\omega +{\omega }_0))\ast \delta (\omega -{\omega }_0)\\ & +\frac{1}{2}S(j(\omega -{\omega }_0))\ast \delta (\omega +{\omega }_0)+\frac{1}{2}S(j(\omega +{\omega }_0))\ast \delta (\omega +{\omega }_0)]\\ & =\frac{1}{2}\left[\frac{1}{2}S(j(\omega -2{\omega }_0))+\frac{1}{2}S(j\omega )+\frac{1}{2}S(j\omega )+\frac{1}{2}S(j(\omega +2{\omega }_0))\right]\\ & =\frac{1}{4}S(j(\omega -2{\omega }_0))+\frac{1}{2}S(j\omega )+\frac{1}{4}S(j(\omega +2{\omega }_0)).\end{array}$$

This spectrum $G(j\omega )$ is shown in Figure 4.24(c). It contains three copies of the original spectrum $S(j\omega )$: one at the original center frequency $\omega =0$ (scaled by $1/2$), and two others shifted to $\pm 2{\omega }_0$ (scaled by $1/4$). Now, if we apply this signal $g(t)$ to a frequency-selective lowpass filter with a cutoff frequency ${\omega }_1$ (where ${\omega }_1$ is greater than the bandwidth of $S(j\omega )$ but less than ${\omega }_0$), the filter will pass only the central component $\frac{1}{2}S(j\omega )$ and reject the high-frequency components around $\pm 2{\omega }_0$. The output of this lowpass filter will have a spectrum that is a scaled replica of $S(j\omega )$, meaning the output signal will be a scaled version of the original $s(t)$! This is the fundamental principle of coherent demodulation in communications.

Example 4.23: Fourier Transform of a Product of Sinc Functions

Let’s find the Fourier transform of the signal:

$$x(t)=\frac{\sin (t)\sin (t/2)}{\pi t^2}.$$

This looks daunting to integrate directly! But notice the structure:

$$x(t)=\frac{1}{\pi }\left(\frac{\sin t}{t}\right)\left(\frac{\sin (t/2)}{t}\right).$$

We can rewrite this product to highlight two sinc functions:

$$x(t)=\pi \left(\frac{\sin t}{\pi t}\right)\left(\frac{\sin (t/2)}{\pi t}\right).$$

Let $s_1(t)=\frac{\sin t}{\pi t}$ and $s_2(t)=\frac{\sin (t/2)}{\pi t}$. Then $x(t)=\pi s_1(t)s_2(t)$. From Example 4.5, we know the Fourier transform of a sinc function is a rectangular pulse:

$$\mathfrak{F}\left\{\frac{\sin Wt}{\pi t}\right\}=\{\begin{array}{ll}1,& |\omega |<W\\ 0,& |\omega |>W\end{array}.$$

So, for $s_1(t)$ (with $W=1$):

$$S_1(j\omega )=\{\begin{array}{ll}1,& |\omega |<1\\ 0,& |\omega |>1\end{array}.$$

And for $s_2(t)$ (with $W=1/2$):

$$S_2(j\omega )=\{\begin{array}{ll}1,& |\omega |<1/2\\ 0,& |\omega |>1/2\end{array}.$$

Now, apply the multiplication property to $x(t)=\pi s_1(t)s_2(t)$:

$$X(j\omega )=\pi \left[\frac{1}{2\pi }{S}_1(j\omega )\ast S_2(j\omega )\right]=\frac{1}{2}{S}_1(j\omega )\ast S_2(j\omega ).$$

We need to convolve two rectangular pulses. Let $R_1(\omega )$ be the pulse for $S_1(j\omega )$ (width $2$, height $1$) and $R_2(\omega )$ be the pulse for $S_2(j\omega )$ (width $1$, height $1$). The convolution $R_1(\omega )\ast R_2(\omega )$ results in a trapezoidal shape. The convolution integral is:

$$X(j\omega )=\frac{1}{2}{\int }_{-\infty }^{\infty }{S}_1(j\theta )S_2(j(\omega -\theta ))d\theta .$$

• $S_1(j\theta )$ is non-zero for $-1<\theta <1$.
• $S_2(j(\omega -\theta ))$ is non-zero for $-1/2<\omega -\theta <1/2$, which means $\omega -1/2<\theta <\omega +1/2$. The convolution will be the area of overlap of these two rectangles as one slides past the other.
• When $\omega$ is very negative (e.g., $\omega <-1-0.5=-1.5$): No overlap, $X(j\omega )=0$.
• From $\omega =-1.5$ to $\omega =-0.5$: The pulses begin to overlap. The width of overlap increases linearly. $X(j\omega )$ increases linearly. The overlap starts when $\omega +0.5=-1⟹\omega =-1.5$. The overlap ends when $\omega -0.5=-1⟹\omega =-0.5$. For $-1.5<\omega <-0.5$, the overlap length is $(\omega +0.5)-(-1)=\omega +1.5$. (This is the width of the shifted $S_2$ from its leftmost point to $S_1$‘s leftmost point). More accurately, the overlap region is from $\max (-1,\omega -0.5)$ to $\min (1,\omega +0.5)$. For $-1.5<\omega <-0.5$: $\theta$ from $\omega -0.5$ to $1$. No, that’s not right. Let’s visualize. $S_1$ is from -1 to 1. $S_2$ is from -0.5 to 0.5. When $S_2$ is shifted by $\omega$, it extends from $\omega -0.5$ to $\omega +0.5$. The convolution is $\int S_1(\theta )S_2(\omega -\theta )d\theta$. Let $S_1(j\omega )=\text{rect}(\omega /2)$ and $S_2(j\omega )=\text{rect}(\omega )$. No, this is not correct for the problem given. $S_1(j\omega )$ is $1$ for $|\omega |<1$. $S_2(j\omega )$ is $1$ for $|\omega |<1/2$. The convolution of two rectangular pulses is a trapezoid. The width of $S_1$ is $2\cdot 1=2$. The width of $S_2$ is $2\cdot (1/2)=1$. The result $S_1\ast S_2$ will have a base width of $2+1=3$. It will be centered at $\omega =0$. So it extends from $-1.5$ to $1.5$. The shape will rise from $\omega =-1.5$ to $\omega =-0.5$, be flat from $\omega =-0.5$ to $\omega =0.5$, and fall from $\omega =0.5$ to $\omega =1.5$. The maximum height of $S_1\ast S_2$ is $1\times (\text{width of smaller pulse})=1\times 1=1$. So $S_1\ast S_2$ is a trapezoid:
  • Linear rise from $0$ at $\omega =-1.5$ to $1$ at $\omega =-0.5$.
  • Flat at $1$ from $\omega =-0.5$ to $\omega =0.5$.
  • Linear fall from $1$ at $\omega =0.5$ to $0$ at $\omega =1.5$. The scale factor for $X(j\omega )$ is $1/2$. So the maximum height of $X(j\omega )$ will be $1/2$. The final form of $X(j\omega )$ is a trapezoid, as shown in Figure 4.25, with peak height $1/2$ and nonzero from $\omega =-1.5$ to $\omega =1.5$. This highlights how powerful frequency-domain convolution is for signals that are products in time.

4.5.1 Frequency-Selective Filtering with Variable Center Frequency

The multiplication property has many applications. Beyond communications, it’s key to building flexible filters. Imagine needing a bandpass filter whose center frequency you can tune with a dial. Directly varying physical components in a traditional analog filter to change its center frequency is cumbersome. A clever alternative is to use a fixed filter and shift the signal’s spectrum to match the filter, using the principles of sinusoidal amplitude modulation.

Consider the system shown in Figure 4.26. An input signal $x(t)$ (with spectrum $X(j\omega )$) is first multiplied by a complex exponential $e^{j{\omega }_{c}t}$. Let $y(t)=e^{j{\omega }_{c}t}x(t)$. From the frequency-shifting property (4.41), its Fourier transform is:

$$Y(j\omega )=X(j(\omega -{\omega }_c)).$$

This means the spectrum $X(j\omega )$ is shifted to the right by ${\omega }_c$. Frequencies in $X(j\omega )$ originally near DC are now centered around ${\omega }_c$. This signal $y(t)$ then passes through a fixed lowpass filter with frequency response $H_{LP}(j\omega )$, which has a cutoff frequency ${\omega }_0$. Let the output of this filter be $w(t)$. Its spectrum is:

$$W(j\omega )=Y(j\omega )H_{LP}(j\omega )=X(j(\omega -{\omega }_c))H_{LP}(j\omega ).$$

The lowpass filter effectively “cuts out” portions of the shifted spectrum $Y(j\omega )$ that fall outside its passband ($|\omega |<{\omega }_0$). Finally, $w(t)$ is multiplied by another complex exponential $e^{-j{\omega }_{c}t}$. Let the final output be $f(t)=e^{-j{\omega }_{c}t}w(t)$. Applying the frequency-shifting property again:

$$F(j\omega )=W(j(\omega +{\omega }_c)).$$

So, $F(j\omega )$ is the spectrum $W(j\omega )$ shifted to the left by ${\omega }_c$. Let’s trace the spectra in Figure 4.27:

• (a) $X(j\omega )$: The original signal spectrum.
• (b) $Y(j\omega )=X(j(\omega -{\omega }_c))$: $X(j\omega )$ shifted to the right by ${\omega }_c$.
• (c) $W(j\omega )=Y(j\omega )H_{LP}(j\omega )$: The lowpass filter (passband from $-{\omega }_0$ to ${\omega }_0$) selects a band of frequencies from $Y(j\omega )$. This effectively isolates the portion of the shifted $X(j\omega )$ that was originally near ${\omega }_c$ and has now been shifted to be near DC after multiplication by $e^{j{\omega }_{c}t}$.
• (d) $F(j\omega )=W(j(\omega +{\omega }_c))$: This shifts the filtered spectrum back to its original frequency range, effectively creating a bandpass response centered at ${\omega }_c$.

From Figure 4.27, we can see that the overall system acts as a bandpass filter. The system’s equivalent frequency response $H_{BP}(j\omega )$ is centered around ${\omega }_c$ and has a bandwidth of $2{\omega }_0$, as illustrated in Figure 4.28. The beauty of this setup is that to change the center frequency of the bandpass filter, you simply adjust ${\omega }_c$ in the complex exponential oscillators! The lowpass filter remains fixed. This makes tunable bandpass filters much easier to implement.

In the system of Figure 4.26, if $x(t)$ is real, then $y(t),w(t)$, and $f(t)$ are generally complex-valued. If we only retain the real part of $f(t)$, say $f_{real}(t)=\Re e\{f(t)\}$, then its spectrum $F_{real}(j\omega )$ will have conjugate symmetry. It will include not only the bandpass response centered at ${\omega }_c$ but also a mirrored bandpass response centered at $-{\omega }_c$, as shown in Figure 4.29. The equivalent bandpass filter thus passes bands of frequencies centered at both ${\omega }_c$ and $-{\omega }_c$, as indicated in Figure 4.30. Under specific conditions, we can achieve similar results using real sinusoidal carriers (e.g., $\cos ({\omega }_{c}t)$) instead of complex exponentials, which is explored further in Problem 4.46.

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4.6 Tables of Fourier Properties and of Basic Fourier Transform Pairs

To consolidate our knowledge, here are tables summarizing the key properties and common Fourier transform pairs we’ve encountered. These are invaluable tools for analysis and problem-solving.

Table 4.1: Properties of the Fourier Transform

Section	Property	Aperiodic Signal $x(t)$ (and $y(t)$)	Fourier Transform $X(j\omega )$ (and $Y(j\omega )$)
4.3.1	Linearity	$ax(t)+by(t)$	$aX(j\omega )+bY(j\omega )$
4.3.2	Time Shifting	$x(t-t_0)$	$e^{-j\omega t_0}X(j\omega )$
4.3.6	Frequency Shifting	$e^{j{\omega }_{0}t}x(t)$	$X(j(\omega -{\omega }_0))$
4.3.3	Conjugation	$x^{\ast }(t)$	$X^{\ast }(-j\omega )$
4.3.5	Time Reversal	$x(-t)$	$X(-j\omega )$
4.3.5	Time and Frequency Scaling	$x(at)$	$\frac{1}{
4.4	Convolution	$x(t)\ast y(t)$	$X(j\omega )Y(j\omega )$
4.5	Multiplication	$x(t)y(t)$	$\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }X(j\theta )Y(j(\omega -\theta ))d\theta$
4.3.4	Differentiation in Time	$\frac{d}{dt}x(t)$	$j\omega X(j\omega )$
4.3.4	Integration	${\int }_{-\infty }^{t}x(\tau )d\tau$	$\frac{1}{j\omega }X(j\omega )+\pi X(0)\delta (\omega )$
4.3.6	Differentiation in Frequency	$tx(t)$	$j\frac{d}{d\omega }X(j\omega )$
4.3.3	Conjugate Symmetry for Real Signals	$x(t)$ real	$X(-j\omega )=X^{\ast }(j\omega )$
			$\Re e\{X(j\omega )\}=\Re e\{X(-j\omega )\}$ (even)
			$\Im m\{X(j\omega )\}=-\Im m\{X(-j\omega )\}$ (odd)
			$
			$\angle X(j\omega )=-\angle X(-j\omega )\}$ (odd)
4.3.3	Symmetry for Real and Even Signals	$x(t)$ real and even	$X(j\omega )$ real and even
4.3.3	Symmetry for Real and Odd Signals	$x(t)$ real and odd	$X(j\omega )$ purely imaginary and odd
4.3.3	Even-Odd Decomposition for Real Signals	$\mathcal{E}v\{x(t)\}$ [($x(t)$ real)]	$\Re e\{X(j\omega )\}$
		$\mathcal{O}d\{x(t)\}$ [($x(t)$ real)]	$j\Im m\{X(j\omega )\}$

Table 4.2: Basic Fourier Transform Pairs

Signal	Fourier Transform	Fourier Series Coefficients (if periodic)
${\sum }_{k=-\infty }^{+\infty }{a}_k{e}^{jk{\omega }_{0}t}$	$2\pi {\sum }_{k=-\infty }^{+\infty }{a}_k\delta (\omega -k{\omega }_0)$	$a_k$ (given)
$\cos {\omega }_{0}t$	$\pi [\delta (\omega -{\omega }_0)+\delta (\omega +{\omega }_0)]$	$a_1=a_{-1}=\frac{1}{2},a_k=0$ otherwise
$\sin {\omega }_{0}t$	$\frac{\pi }{j}[\delta (\omega -{\omega }_0)-\delta (\omega +{\omega }_0)]$	$a_1=\frac{1}{2j},a_{-1}=-\frac{1}{2j},a_k=0$ otherwise
$x(t)=1$	$2\pi \delta (\omega )$	$a_0=1,a_k=0,k\ne 0$
Periodic square wave: $x(t) = \begin{cases} 1, &	t	< T_1 \ 0, & T_1 <
${\sum }_{n=-\infty }^{+\infty }\delta (t-nT)$	$\frac{2\pi }{T}{\sum }_{k=-\infty }^{+\infty }\delta \left(\omega -\frac{2\pi k}{T}\right)$	$a_k=\frac{1}{T}$ for all $k$
$x(t) = \begin{cases} 1, &	t	< T_1 \ 0, &
$\frac{\sin Wt}{\pi t}$	$X(j\omega) = \begin{cases} 1, &	\omega
$\delta (t)$	$1$
$u(t)$	$\frac{1}{j\omega }+\pi \delta (\omega )$
$\delta (t-t_0)$	$e^{-j\omega t_0}$
$e^{-at}u(t),\Re e\{a\}>0$	$\frac{1}{a+j\omega }$
$t{e}^{-at}u(t),\Re e\{a\}>0$	$\frac{1}{(a+j\omega {)}^2}$

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4.7 Systems Characterized by Linear Constant-Coefficient Differential Equations

We’ve repeatedly highlighted a particularly important class of LTI systems: those whose input and output are governed by a linear constant-coefficient differential equation of the form:

$$\sum _{k=0}^N{a}_k\frac{d^{k}y(t)}{d{t}^k}=\sum _{k=0}^M{b}_k\frac{d^{k}x(t)}{d{t}^k}.\text{\hspace{1em}(4.72)}$$

In this section, we’ll discover how effortlessly we can determine the frequency response $H(j\omega )$ of such systems using the Fourier transform. We’ll always assume that the frequency response exists (i.e., that the integral in eq. (3.121) converges).

There are two closely related paths to finding $H(j\omega )$ for a system described by (4.72). The first, which we used in Section 3.10 for simple filters, relies on the fact that complex exponentials are eigenfunctions of LTI systems. If $x(t)=e^{j\omega t}$, then the output is $y(t)=H(j\omega )e^{j\omega t}$. Substituting these into the differential equation and solving for $H(j\omega )$ directly yields the frequency response.

Here, we’ll take an alternative, equally powerful approach, directly leveraging the differentiation property (4.31) of the Fourier transform.

Recall that for an LTI system, the Fourier transform of the output $Y(j\omega )$ is the product of the frequency response $H(j\omega )$ and the Fourier transform of the input $X(j\omega )$:

$$Y(j\omega )=H(j\omega )X(j\omega ).$$

This means we can express the frequency response as:

$$H(j\omega )=\frac{Y(j\omega )}{X(j\omega )}.\text{\hspace{1em}(4.73)}$$

Now, let’s take the Fourier transform of both sides of our differential equation (4.72):

$$\mathfrak{F}\left\{\sum _{k=0}^N{a}_k\frac{d^{k}y(t)}{d{t}^k}\right\}=\mathfrak{F}\left\{\sum _{k=0}^M{b}_k\frac{d^{k}x(t)}{d{t}^k}\right\}.\text{\hspace{1em}(4.74)}$$

By the linearity property (4.26) of the Fourier transform, we can move the Fourier transform operator inside the sums and pull out the constants $a_k$ and $b_k$:

$$\sum _{k=0}^N{a}_k\mathfrak{F}\left\{\frac{d^{k}y(t)}{d{t}^k}\right\}=\sum _{k=0}^M{b}_k\mathfrak{F}\left\{\frac{d^{k}x(t)}{d{t}^k}\right\}.\text{\hspace{1em}(4.75)}$$

Next, we apply the differentiation property (4.31), which states that $\mathfrak{F}\left\{\frac{d^{k}f(t)}{d{t}^k}\right\}=(j\omega {)}^{k}F(j\omega )$. So, for the $k$-th derivative of $y(t)$, its transform is $(j\omega {)}^{k}Y(j\omega )$, and similarly for $x(t)$:

$$\sum _{k=0}^N{a}_k(j\omega {)}^{k}Y(j\omega )=\sum _{k=0}^M{b}_k(j\omega {)}^{k}X(j\omega ).$$

Now, we can factor out $Y(j\omega )$ from the left side and $X(j\omega )$ from the right side:

$$Y(j\omega )\left[\sum _{k=0}^N{a}_k(j\omega {)}^k\right]=X(j\omega )\left[\sum _{k=0}^M{b}_k(j\omega {)}^k\right].$$

Finally, using equation (4.73), we solve for $H(j\omega )$:

$$H(j\omega )=\frac{Y(j\omega )}{X(j\omega )}=\frac{{\sum }_{k=0}^M{b}_k(j\omega {)}^k}{{\sum }_{k=0}^N{a}_k(j\omega {)}^k}.\text{\hspace{1em}(4.76)}$$

This is a remarkable result! The frequency response $H(j\omega )$ for an LTI system described by a linear constant-coefficient differential equation is always a rational function (a ratio of polynomials in $j\omega$). Even more astonishing, you can write it down by simple inspection: the coefficients $b_k$ from the right-hand side of (4.72) (input terms) form the numerator polynomial, and the coefficients $a_k$ from the left-hand side (output terms) form the denominator polynomial.

The highest derivative of $y(t)$ in (4.72) determines the order of the differential equation, which also corresponds to the highest power of $(j\omega )$ in the denominator polynomial of $H(j\omega )$.

Example 4.24: First-Order System

Consider a stable LTI system characterized by the first-order differential equation:

$$\frac{dy(t)}{dt}+ay(t)=x(t),a>0.\text{\hspace{1em}(4.77)}$$

By inspection, using equation (4.76):

• Left side (output terms): $a_1=1$ (for $dy(t)/dt$), $a_0=a$ (for $y(t)$). So the denominator is $1\cdot (j\omega {)}^1+a\cdot (j\omega {)}^0=j\omega +a$.
• Right side (input terms): $b_0=1$ (for $x(t)$). So the numerator is $1\cdot (j\omega {)}^0=1$. Thus, the frequency response is:

$$H(j\omega )=\frac{1}{j\omega +a}.\text{\hspace{1em}(4.78)}$$

Comparing this with Example 4.1, we immediately recognize that $H(j\omega )$ is the Fourier transform of $e^{-at}u(t)$. Therefore, the impulse response of this system is:

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

This simple method provides both the frequency response and the impulse response directly from the differential equation!

Example 4.25: Second-Order System

Consider a stable LTI system described by the second-order differential equation:

$$\frac{d^{2}y(t)}{d{t}^2}+4\frac{dy(t)}{dt}+3y(t)=\frac{dx(t)}{dt}+2x(t).$$

Again, by inspection using equation (4.76):

• Numerator (from input terms): $b_1=1$ (for $dx(t)/dt$), $b_0=2$ (for $x(t)$). So, $1\cdot (j\omega {)}^1+2\cdot (j\omega {)}^0=j\omega +2$.
• Denominator (from output terms): $a_2=1$ (for $d^{2}y(t)/d{t}^2$), $a_1=4$ (for $dy(t)/dt$), $a_0=3$ (for $y(t)$). So, $1\cdot (j\omega {)}^2+4\cdot (j\omega {)}^1+3\cdot (j\omega {)}^0=(j\omega {)}^2+4j\omega +3$. Thus, the frequency response is:

$$H(j\omega )=\frac{j\omega +2}{(j\omega {)}^2+4(j\omega )+3}.\text{\hspace{1em}(4.79)}$$

To find the impulse response $h(t)$, we need to compute the inverse Fourier transform of $H(j\omega )$. The most effective way to do this for rational functions is partial-fraction expansion. First, factor the denominator:

$$(j\omega {)}^2+4(j\omega )+3=(j\omega +1)(j\omega +3).$$

So, $H(j\omega )$ becomes:

$$H(j\omega )=\frac{j\omega +2}{(j\omega +1)(j\omega +3)}.\text{\hspace{1em}(4.80)}$$

Now, perform the partial-fraction expansion:

$$H(j\omega )=\frac{A}{j\omega +1}+\frac{B}{j\omega +3}.$$

To find $A$: Multiply by $(j\omega +1)$ and set $j\omega =-1$:

$$A={\left[\frac{j\omega +2}{j\omega +3}\right]}_{j\omega =-1}=\frac{-1+2}{-1+3}=\frac{1}{2}.$$

To find $B$: Multiply by $(j\omega +3)$ and set $j\omega =-3$:

$$B={\left[\frac{j\omega +2}{j\omega +1}\right]}_{j\omega =-3}=\frac{-3+2}{-3+1}=\frac{-1}{-2}=\frac{1}{2}.$$

So, the expanded form is:

$$H(j\omega )=\frac{\frac{1}{2}}{j\omega +1}+\frac{\frac{1}{2}}{j\omega +3}.$$

Now, we can find the inverse transform of each term using the result from Example 4.24 (and linearity):

$$h(t)=\frac{1}{2}{e}^{-t}u(t)+\frac{1}{2}{e}^{-3t}u(t).$$

This method of combining the Fourier transform approach with partial-fraction expansion is a workhorse in system analysis!

Example 4.26: Solving a Differential Equation with a Given Input

Let’s use the system from Example 4.25 and find its response $y(t)$ to a specific input:

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

We already know $H(j\omega )$ for this system from equation (4.80):

$$H(j\omega )=\frac{j\omega +2}{(j\omega +1)(j\omega +3)}.$$

The Fourier transform of the input $x(t)$ (from Example 4.1) is:

$$X(j\omega )=\frac{1}{j\omega +1}.$$

Now, we find the Fourier transform of the output $Y(j\omega )=H(j\omega )X(j\omega )$:

$$Y(j\omega )=\left[\frac{j\omega +2}{(j\omega +1)(j\omega +3)}\right]\left[\frac{1}{j\omega +1}\right]=\frac{j\omega +2}{(j\omega +1{)}^2(j\omega +3)}.\text{\hspace{1em}(4.81)}$$

This rational function also needs to be inverse transformed using partial-fraction expansion. When a denominator factor is repeated (like $(j\omega +1{)}^2$), the expansion takes a specific form:

$$Y(j\omega )=\frac{A_{11}}{j\omega +1}+\frac{A_{12}}{(j\omega +1{)}^2}+\frac{A_{21}}{j\omega +3}.\text{\hspace{1em}(4.82)}$$

The coefficients are found as follows (details of the method are in Appendix A):

• To find $A_{12}$: Multiply by $(j\omega +1{)}^2$ and set $j\omega =-1$: $$A_{12}={\left[\frac{j\omega +2}{j\omega +3}\right]}_{j\omega =-1}=\frac{-1+2}{-1+3}=\frac{1}{2}.$$
• To find $A_{21}$: Multiply by $(j\omega +3)$ and set $j\omega =-3$: $$A_{21}={\left[\frac{j\omega +2}{(j\omega +1{)}^2}\right]}_{j\omega =-3}=\frac{-3+2}{(-3+1{)}^2}=\frac{-1}{(-2{)}^2}=-\frac{1}{4}.$$
• To find $A_{11}$: This is a bit trickier for repeated roots. One method is to multiply $Y(j\omega )$ by $(j\omega +1{)}^2$ and then differentiate with respect to $j\omega$, then set $j\omega =-1$. Let $F(j\omega )=(j\omega +1{)}^{2}Y(j\omega )=\frac{j\omega +2}{j\omega +3}$. $A_{11}={\left[\frac{d}{d(j\omega )}F(j\omega )\right]}_{j\omega =-1}$. $$\frac{d}{d(j\omega )}\left(\frac{j\omega +2}{j\omega +3}\right)=\frac{1\cdot (j\omega +3)-(j\omega +2)\cdot 1}{(j\omega +3{)}^2}=\frac{j\omega +3-j\omega -2}{(j\omega +3{)}^2}=\frac{1}{(j\omega +3{)}^2}.$$ Now, substitute $j\omega =-1$: $$A_{11}=\frac{1}{(-1+3{)}^2}=\frac{1}{2^2}=\frac{1}{4}.$$

So, the coefficients are $A_{11}=\frac{1}{4},A_{12}=\frac{1}{2},A_{21}=-\frac{1}{4}$. Substituting these back into (4.82):

$$Y(j\omega )=\frac{\frac{1}{4}}{j\omega +1}+\frac{\frac{1}{2}}{(j\omega +1{)}^2}-\frac{\frac{1}{4}}{j\omega +3}.\text{\hspace{1em}(4.83)}$$

Finally, we take the inverse Fourier transform of each term:

• ${\mathfrak{F}}^{-1}\left\{\frac{1/4}{j\omega +1}\right\}=\frac{1}{4}{e}^{-t}u(t)$.
• ${\mathfrak{F}}^{-1}\left\{\frac{1/2}{(j\omega +1{)}^2}\right\}=\frac{1}{2}t{e}^{-t}u(t)$ (from Example 4.19 or Table 4.2).
• ${\mathfrak{F}}^{-1}\left\{-\frac{1/4}{j\omega +3}\right\}=-\frac{1}{4}{e}^{-3t}u(t)$. Combining these, the output $y(t)$ is:

$$y(t)=\left[\frac{1}{4}{e}^{-t}+\frac{1}{2}t{e}^{-t}-\frac{1}{4}{e}^{-3t}\right]u(t).$$

This demonstrates how Fourier analysis, coupled with the convolution property and partial-fraction expansion, transforms the intimidating task of solving differential equations into an elegant sequence of algebraic operations and table lookups. This algebraic structure also offers deep insights into system behavior in both time and frequency, which we will explore further in Chapter 6.

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I hope this journey through the continuous-time Fourier Transform has sparked your imagination and deepened your appreciation for its elegance and power! The ability to view signals and systems through the lens of frequency opens up entire new worlds of understanding, from communication systems to medical imaging. Keep asking questions, keep exploring, and let your curiosity be your guide!