7.3 Some Properties of the Fourier Transform 701

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7.3 Some Properties of the Fourier Transform 701

The Fourier transform is given by

$$u(t) \Longleftrightarrow \frac{1}{j\omega} + \pi \delta(\omega)$$

Clearly, X(jω) = X(ω) in this case.

To understand this puzzle, consider the fact that we obtain X(jω) by setting s = jω in Eq. (7.24). This implies that the integral on the right-hand side of Eq. (7.24) converges for s = jω, meaning that s = jω (the imaginary axis) lies in the ROC for X(s). The general rule is that only when the ROC for X(s) includes the ω axis, does setting s = jω in X(s) yield the Fourier transform X(ω), that is, X(jω) = X(ω). This is the case of absolutely integrable x(t). If the ROC of X(s) excludes the ω axis, X(jω) = X(ω). This is the case for exponentially growing x(t) and also x(t) that is constant or is oscillating with constant amplitude.

The reason for this peculiar behavior has something to do with the nature of convergence of the Laplace and the Fourier integrals when x(t) is not absolutely integrable.†

This discussion shows that although the Fourier transform may be considered as a special case of the Laplace transform, we need to circumscribe such a view. This fact can also be confirmed by noting that a periodic signal has the Fourier transform, but the Laplace transform does not exist.

7.3 SOME PROPERTIES OF THE FOURIER TRANSFORM

We now study some of the important properties of the Fourier transform and their implications as well as applications. We have already encountered two important properties, linearity [Eq. (7.15)] and the conjugation property [Eq. (7.11)].

Before embarking on this study, we shall explain an important and pervasive aspect of the Fourier transform: the time-frequency duality.

^† To explain this point, consider the unit step function and its transforms. Both the Laplace and the Fourier transform synthesize x(t), using everlasting exponentials of the form est. The frequency s can be anywhere in the complex plane for the Laplace transform, but it must be restricted to the ω axis in the case of the Fourier transform. The unit step function is readily synthesized in the Laplace transform by a relatively simple spectrum X(s) = 1/s, in which the frequencies s are chosen in the RHP [the region of convergence for u(t) is Re s > 0]. In the Fourier transform, however, we are restricted to values of s on the ω axis only. The function u(t) can still be synthesized by frequencies along the ω axis, but the spectrum is more complicated than it is when we are free to choose the frequencies in the RHP. In contrast, when x(t) is absolutely integrable, the region of convergence for the Laplace transform includes the ω axis, and we can synthesize x(t) by using frequencies along the ω axis in both transforms. This leads to X(jω) = X(ω).

We may explain this concept by an example of two countries, X and Y. Suppose these countries want to construct similar dams in their respective territories. Country X has financial resources but not much manpower. In contrast, Y has considerable manpower but few financial resources. The dams will still be constructed in both countries, although the methods used will be different. Country X will use expensive but efficient equipment to compensate for its lack of manpower, whereas Y will use the cheapest possible equipment in a labor-intensive approach to the project. Similarly, both Fourier and Laplace integrals converge for u(t), but the makeup of the components used to synthesize u(t) will be very different for two cases because of the constraints of the Fourier transform, which are not present for the Laplace transform.

TIME-FREQUENCY DUALITY IN THE TRANSFORM OPERATIONS

Equations (7.9) and (7.10) show an interesting fact: the direct and the inverse transform operations are remarkably similar. These operations, required to go from x(t) to X(ω) and then from X(ω) to x(t), are depicted graphically in Fig. 7.18. The inverse transform equation can be obtained from the direct transform equation by replacing x(t) with X(ω), t with ω, and ω with t. In a similar way, we can obtain the direct from the inverse. There are only two minor differences in these operations: the factor 2π appears only in the inverse operator, and the exponential indices in the two operations have opposite signs. Otherwise the two equations are duals of each other.†

This observation has far-reaching consequences in the study of the Fourier transform. It is the basis of the so-called duality of time and frequency. The duality principle may be compared with a photograph and its negative. A photograph can be obtained from its negative, and by using an identical procedure, a negative can be obtained from the photograph. For any result or relationship between x(t) and X(ω), there exists a dual result or relationship, obtained by interchanging the roles of x(t) and X(ω) in the original result (along with some minor modifications arising because of the factor 2π and a sign change). For example, the time-shifting property, to be proved later, states that if x(t) ⇐⇒ X(ω), then

$$x(t-t_0) \Longleftrightarrow X(\omega)e^{-j\omega t_0}$$

Figure 7.18 A near symmetry between the direct and the inverse Fourier transforms.

$$X(2\pi f) = \int_{-\infty}^{\infty} x(t)e^{-j2\pi ft} dt \quad \text{and} \quad x(t) = \int_{-\infty}^{\infty} X(2\pi f)e^{j2\pi ft} df$$

This leaves only one significant difference, that of sign change in the exponential index.

^† Of the two differences, the former can be eliminated by change of variable from ω to f (in hertz). In this case ω = 2πf and dω = 2π df .

Therefore, the direct and the inverse transforms are given by

The dual of this property (the frequency-shifting property) states that

$$x(t)e^{j\omega_0 t} \Longleftrightarrow X(\omega - \omega_0)$$

Observe the role reversal of time and frequency in these two equations (with the minor difference of the sign change in the exponential index). The value of this principle lies in the fact that whenever we derive any result, we can be sure that it has a dual. This possibility can give valuable insights about many unsuspected properties or results in signal processing.

The properties of the Fourier transform are useful not only in deriving the direct and inverse transforms of many functions, but also in obtaining several valuable results in signal processing. The reader should not fail to observe the ever-present duality in this discussion.

LINEARITY

The linearity property, already introduced as Eq. (7.15), states that if x1(t) ⇐⇒ X1(ω) and x2(t) ⇐⇒ X2(ω), then a1x1(t)+a2x2(t) ⇐⇒ a1X1(ω)+a2X2(ω).

CONJUGATION AND CONJUGATE SYMMETRY

The conjugation property, which has already been introduced, states that if x(t) ⇐⇒ X(ω), then

$$x^*(t) \Longleftrightarrow X^*(-\omega)$$

From this property follows the conjugate symmetry property, also introduced earlier, which states that if x(t) is real, then

$$X(-\omega) = X^*(\omega)$$

DUALITY

The duality property states that if

$$x(t) \Longleftrightarrow X(\omega)$$

then

$$X(t) \Longleftrightarrow 2\pi x(-\omega) \tag{7.25}$$

Proof. From Eq. (7.10) we can write

$$x(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} X(u)e^{iut} du$$

Hence,

$$2\pi x(-t) = \int_{-\infty}^{\infty} X(u)e^{-jut} du$$

Changing t to ω yields Eq. (7.25).

t

From Eq. (7.19) we have

1

$$\underbrace{\text{rect}\left(\frac{t}{\tau}\right)}_{x(t)} \Longleftrightarrow \underbrace{\tau \text{ sinc}\left(\frac{\omega \tau}{2}\right)}_{X(\omega)}$$

Also, X(t) is the same as X(ω) with ω replaced by t, and x(−ω) is the same as x(t) with t replaced by −ω. Therefore, the duality property of Eq. (7.25) yields

$$\underbrace{\tau \text{ sinc}\left(\frac{\tau t}{2}\right)}_{X(t)} \Longleftrightarrow \underbrace{2\pi \text{ rect}\left(\frac{-\omega}{\tau}\right)}_{2\pi x(-\omega)} = 2\pi \text{ rect}\left(\frac{\omega}{\tau}\right)$$

In this result, we used the fact that rect(−x) = rect(x) because rect is an even function. Figure 7.19b shows this pair graphically. Observe the interchange of the roles of t and ω (with the minor adjustment of the factor 2π). This result appears as pair 18 in Table 7.1 (with τ/2 = W).

As an interesting exercise, the reader should generate the dual of every pair in Table 7.1 by applying the duality property.

DR ILL 7.4 Applying the Duality Property of the Fourier Transform

Apply the duality property to pairs 1, 3, and 9 (Table 7.1) to show that

• (a) 1/(jt +a) ⇐⇒ 2πea^ωu(−ω)
• (b) 2a/(t ² +a2) ⇐⇒ 2πe−a|ω^|
• (c) δ(t +t0)+δ(t −t0) ⇐⇒ 2 cos t0ω

THE SCALING PROPERTY If

x(t) ⇐⇒ X(ω)

then, for any real constant a,

$$x(at) \Longleftrightarrow \frac{1}{|a|}X\left(\frac{\omega}{a}\right) \tag{7.26}$$

Proof. For a positive real constant a,

$$\mathcal{F}[x(at)] = \int_{-\infty}^{\infty} x(at)e^{-j\omega t}dt = \frac{1}{a}\int_{-\infty}^{\infty} x(u)e^{(-j\omega/a)u}du = \frac{1}{a}X\left(\frac{\omega}{a}\right)$$

Similarly, we can demonstrate that if a < 0,

$$x(at) \Longleftrightarrow \frac{-1}{a}X\left(\frac{\omega}{a}\right)$$

Hence follows Eq. (7.26).

SIGNIFICANCE OF THE SCALING PROPERTY

The function x(at) represents the function x(t) compressed in time by a factor a (see Sec. 1.2-2). Similarly, a function X(ω/a) represents the function X(ω) expanded in frequency by the same factor a. The scaling property states that time compression of a signal results in its spectral expansion, and time expansion of the signal results in its spectral compression. Intuitively, compression in time by factor a means that the signal is varying faster by factor a. † To synthesize such a signal, the frequencies of its sinusoidal components must be increased by the factor a, implying that its frequency spectrum is expanded by the factor a. Similarly, a signal expanded in time varies more slowly; hence the frequencies of its components are lowered, implying that its frequency spectrum is compressed. For instance, the signal cos 2ω0t is the same as the signal cosω0t time-compressed by a factor of 2. Clearly, the spectrum of the former (impulse at ±2ω0) is an expanded version of the spectrum of the latter (impulse at ±ω0). The effect of this scaling is demonstrated in Fig. 7.20.

^† We are assuming a > 1, although the argument still holds if a < 1. In the latter case, compression becomes expansion by factor 1/a, and vice versa.

Figure 7.20 The scaling property of the Fourier transform.

RECIPROCITY OF SIGNAL DURATION AND ITS BANDWIDTH

The scaling property implies that if x(t) is wider, its spectrum is narrower, and vice versa. Doubling the signal duration halves its bandwidth, and vice versa. This suggests that the bandwidth of a signal is inversely proportional to the signal duration or width (in seconds).† We have already verified this fact for the gate pulse, where we found that the bandwidth of a gate pulse of width τ seconds is 1/τ Hz. More discussion of this interesting topic can be found in the literature [2].

By letting a = −1 in Eq. (7.26), we obtain the inversion (or reflection) property of time and frequency:

$$x(-t) \Longleftrightarrow X(-\omega) \tag{7.27}$$

EXAMPLE 7.12 Fourier Transform Reflection Property

Using the reflection property of the Fourier transform and Table 7.1, find the Fourier transforms of eatu(−t) and e−a|t^| .

Application of Eq. (7.27) to pair 1 of Table 7.1 yields

$$e^{at}u(-t) \Longleftrightarrow \frac{1}{a-j\omega} \qquad a > 0$$

Also,

$$e^{-a|t|} = e^{-at}u(t) + e^{at}u(-t)$$

^† When a signal has infinite duration, we must consider its effective or equivalent duration. There is no unique definition of effective signal duration. One possible definition is given in Eq. (2.47).

Therefore,

$$e^{-a|t|} \Longleftrightarrow \frac{1}{a+j\omega} + \frac{1}{a-j\omega} = \frac{2a}{a^2 + \omega^2} \qquad a > 0 \tag{7.28}$$

The signal e−a|t^| and its spectrum are illustrated in Fig. 7.21.

THE TIME-SHIFTING PROPERTY If

x(t) ⇐⇒ X(ω)

then

$$x(t - t_0) \Longleftrightarrow X(\omega)e^{-j\omega t_0} \tag{7.29}$$

Proof. By definition,

$$\mathcal{F}[x(t-t_0)] = \int_{-\infty}^{\infty} x(t-t_0)e^{-j\omega t} dt$$

Letting t −t⁰ = u, we have

$$\mathcal{F}[x(t-t_0)] = \int_{-\infty}^{\infty} x(u)e^{-j\omega(u+t_0)} du = e^{-j\omega t_0} \int_{-\infty}^{\infty} x(u)e^{-j\omega u} du = X(\omega)e^{-j\omega t_0}$$

This result shows that delaying a signal by t⁰ seconds does not change its amplitude spectrum. The phase spectrum, however, is changed by −ωt0.

PHYSICAL EXPLANATION OF THE LINEAR PHASE

Time delay in a signal causes a linear phase shift in its spectrum. This result can also be derived by heuristic reasoning. Imagine x(t) being synthesized by its Fourier components, which are sinusoids of certain amplitudes and phases. The delayed signal x(t − t0) can be synthesized by the same sinusoidal components, each delayed by t⁰ seconds. The amplitudes of the components remain unchanged. Therefore, the amplitude spectrum of x(t − t0) is identical to that of x(t). The time delay of t⁰ in each sinusoid, however, does change the phase of each component. Now, a sinusoid

Figure 7.22 Physical explanation of the time-shifting property.

cosωt delayed by t⁰ is given by

$$\cos \omega (t - t_0) = \cos (\omega t - \omega t_0)$$

Therefore a time delay t⁰ in a sinusoid of frequency ω manifests as a phase delay of ωt0. This is a linear function of ω, meaning that higher-frequency components must undergo proportionately higher phase shifts to achieve the same time delay. This effect is depicted in Fig. 7.22 with two sinusoids, the frequency of the lower sinusoid being twice that of the upper. The same time delay t⁰ amounts to a phase shift of π/2 in the upper sinusoid and a phase shift of π in the lower sinusoid. This verifies the fact that to achieve the same time delay, higher-frequency sinusoids must undergo proportionately higher phase shifts. The principle of linear phase shift is very important, and we shall encounter it again in distortionless signal transmission and filtering applications.

EXAMPLE 7.13 Fourier Transform Time-Shifting Property

Use the time-shifting property to find the Fourier transform of e−a|t−t0^| .

This function, shown in Fig. 7.23a, is a time-shifted version of e−a|t^| (depicted in Fig. 7.21a). From Eqs. (7.28) and (7.29), we have

$$e^{-a|t-t_0|} \Longleftrightarrow \frac{2a}{a^2 + \omega^2} e^{-j\omega t_0}$$

The spectrum of e−a|t−t0^| (Fig. 7.23b) is the same as that of e−a|t^| (Fig. 7.21b), except for an added phase shift of −ωt0.

EXAMPLE 7.14 Fourier Transform of a Time-Shifted Rectangular Pulse

Find the Fourier transform of the time-shifted rectangular pulse x(t) illustrated in Fig. 7.24a.

The pulse x(t) is the gate pulse rect(t/τ ) in Fig. 7.10a delayed by 3τ/4 seconds. Hence, according to Eq. (7.29), its Fourier transform is the Fourier transform of rect(t/τ ) multiplied by e−jω(3τ /4) . Therefore,

$$X(\omega) = \tau \operatorname{sinc}\left(\frac{\omega \tau}{2}\right) e^{-j\omega(3\tau/4)}$$

The amplitude spectrum |X(ω)| (depicted in Fig. 7.24b) of this pulse is the same as that indicated in Fig. 7.10c. But the phase spectrum has an added linear term −3ωτ/4. Hence, the phase spectrum of x(t) (Fig. 7.24a) is identical to that in Fig. 7.10d plus a linear term −3ωτ/4, as shown in Fig. 7.24c.

PHASE SPECTRUM USING PRINCIPAL VALUES

There is an alternate way of spectral representation of X(ω). The phase angle computed on a calculator or by using a computer subroutine is generally the principal value (modulo 2π value) of the phase angle, which always lies in the range −π to π. For instance, the principal value of angle 3π/2 is −π/2, and so on. The principal value differs from the actual value by ±2π radians (and its integer multiples) in a way that ensures that the principal value remains within −π to π. Thus, the principal value will show jump discontinuities of ±2π whenever the actual

710 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM

phase crosses ±π. The phase plot in Fig. 7.24c is redrawn in Fig. 7.24d using the principal value for the phase. This phase pattern, which contains phase discontinuities of magnitudes 2π and π, becomes repetitive at intervals of ω = 8π/τ .

DR ILL 7.5 Fourier Transform Time-Shifting Property

Use pair 18 of Table 7.1 and the time-shifting property to show that the Fourier transform of sinc [ω0(t − T)] is (π/ω0)rect(ω/2ω0)e−jω*^T* . Sketch the amplitude and phase spectra of the Fourier transform.

THE FREQUENCY-SHIFTING PROPERTY If

$$x(t) \Longleftrightarrow X(\omega)$$

then

$$x(t)e^{j\omega_0 t} \Longleftrightarrow X(\omega - \omega_0) \tag{7.30}$$

Proof. By definition,

$$\mathcal{F}[x(t)e^{j\omega_0 t}] = \int_{-\infty}^{\infty} x(t)e^{j\omega_0 t}e^{-j\omega t} dt = \int_{-\infty}^{\infty} x(t)e^{-j(\omega - \omega_0)t} dt = X(\omega - \omega_0)$$

According to this property, the multiplication of a signal by a factor ej^ω0*^t* shifts the spectrum of that signal by ω = ω0. Note the duality between the time-shifting and the frequency-shifting properties.

Changing ω⁰ to −ω⁰ in Eq. (7.30) yields

$$x(t)e^{-j\omega_0 t} \Longleftrightarrow X(\omega + \omega_0) \tag{7.31}$$

Because ej^ω0*^t* is not a real function that can be generated, frequency shifting in practice is achieved by multiplying x(t) by a sinusoid. Observe that

$$x(t)\cos \omega_0 t = \frac{1}{2} [x(t)e^{j\omega_0 t} + x(t)e^{-j\omega_0 t}]$$

From Eqs. (7.30) and (7.31), it follows that

$$x(t)\cos\omega_0 t \Longleftrightarrow \frac{1}{2}[X(\omega - \omega_0) + X(\omega + \omega_0)]\tag{7.32}$$

This result shows that the multiplication of a signal x(t) by a sinusoid of frequency ω⁰ shifts the spectrum X(ω) by ±ω0, as depicted in Fig. 7.25.

Multiplication of a sinusoid cosω0t by x(t) amounts to modulating the sinusoid amplitude. This type of modulation is known as amplitude modulation. The sinusoid cosω0t is called the carrier, the signal x(t) is the modulating signal, and the signal x(t) cosω0t is the modulated signal. Further discussion of modulation and demodulation appears in Sec. 7.7.

To sketch a signal x(t) cos ω0t, we observe that

$$x(t)\cos\omega_0 t = \begin{cases} x(t) & \text{when } \cos\omega_0 t = 1\\ -x(t) & \text{when } \cos\omega_0 t = -1 \end{cases}$$

Therefore, x(t) cos ω0t touches x(t) when the sinusoid cos ω0t is at its positive peaks and touches −x(t) when cos ω0t is at its negative peaks. This means that x(t) and −x(t) act as envelopes for the signal x(t) cos ω0t (see Fig. 7.25). The signal −x(t) is a mirror image of x(t) about the horizontal axis. Figure 7.25 shows the signals x(t) and x(t) cos ω0t and their spectra.

Figure 7.25 Amplitude modulation of a signal causes spectral shifting.

EXAMPLE 7.15 Spectral Shifting by Amplitude Modulation

Find and sketch the Fourier transform of the modulated signal x(t) cos 10t in which x(t) is a gate pulse rect(t/4), as illustrated in Fig. 7.26a.

From pair 17 of Table 7.1, we find rect(t/4) ⇐⇒ 4 sinc (2ω), which is depicted in Fig. 7.26b. From Eq. (7.32) it follows that

$$x(t)\cos 10t \Longleftrightarrow \frac{1}{2}[X(\omega+10) + X(\omega-10)]$$

In this case, X(ω) = 4 sinc (2ω). Therefore,

x(t) cos 10t ⇐⇒ 2 sinc [2(ω +10)] +2 sinc [2(ω −10)]

The spectrum (Fig. 7.26c) of x(t) cos 10t is obtained by shifting X(ω) in Fig. 7.26b to the left by 10 and also to the right by 10, and then multiplying it by 0.5, as depicted in Fig. 7.26d.

DR ILL 7.6 Fourier Transform of an Amplitude-Modulated Signal

Sketch signal e−|t^| cos 10t. Find the Fourier transform of this signal and sketch its spectrum. Answer: ^X(ω) ^{= 1} (ω−10)2+^{1 + 1} (ω+10)2+¹ . See Fig. 7.21b for the spectrum of ^e−a|t^| .

DR ILL 7.7 Amplitude Modulation Using a Phase-Shifted Carrier

Show that

x(t) cos(ω0t +θ ) ⇐⇒ 1
                     2

                       X(ω −ω0)ejθ +X(ω +ω0)e−jθ

APPLICATIONS OF MODULATION

Modulation is used to shift signal spectra. Some of the situations that call for spectrum shifting are presented next.

1. If several signals, all occupying the same frequency band, are transmitted simultaneously over the same transmission medium, they will all interfere; it will be impossible to separate or retrieve them at a receiver. For example, if all radio stations decide to broadcast audio signals simultaneously, a receiver will not be able to separate them. This problem is solved

Old is gold, but sometimes it is fool’s gold.

by using modulation, whereby each radio station is assigned a distinct carrier frequency. Each station transmits a modulated signal. This procedure shifts the signal spectrum to its allocated band, which is not occupied by any other station. A radio receiver can pick up any station by tuning to the band of the desired station. The receiver must now demodulate the received signal (undo the effect of modulation). Demodulation therefore consists of another spectral shift required to restore the signal to its original band. Note that both modulation and demodulation implement spectral shifting; consequently, demodulation operation is similar to modulation (see Sec. 7.7).

This method of transmitting several signals simultaneously over a channel by sharing its frequency band is known as frequency-division multiplexing (FDM).

2. For effective radiation of power over a radio link, the antenna size must be of the order of the wavelength of the signal to be radiated. Audio signal frequencies are so low (wavelengths are so large) that impracticably large antennas would be required for radiation. Here, shifting the spectrum to a higher frequency (a smaller wavelength) by modulation solves the problem.

CONVOLUTION

The time-convolution property and its dual, the frequency-convolution property, state that if

$$x_1(t) \Longleftrightarrow X_1(\omega) \quad \text{and} \quad x_2(t) \Longleftrightarrow X_2(\omega)$$

then

$$x_1(t) * x_2(t) \Longleftrightarrow X_1(\omega) X_2(\omega) \quad \text{(time convolution)} \tag{7.33}$$

7.3 Some Properties of the Fourier Transform 715

and

$$x_1(t)x_2(t) \Longleftrightarrow \frac{1}{2\pi}X_1(\omega) * X_2(\omega) \quad \text{(frequency convolution)} \tag{7.34}$$

Proof. By definition,

$$\mathcal{F}|x_1(t) * x_2(t)| = \int_{-\infty}^{\infty} e^{-j\omega t} \left[ \int_{-\infty}^{\infty} x_1(\tau) x_2(t-\tau) d\tau \right] dt$$ $$= \int_{-\infty}^{\infty} x_1(\tau) \left[ \int_{-\infty}^{\infty} e^{-j\omega t} x_2(t-\tau) d\tau \right] d\tau$$

The inner integral is the Fourier transform of x2(t − τ ), given by [time-shifting property in Eq. (7.29)] X2(ω)e−jωτ . Hence,

$$\mathcal{F}[x_1(t) * x_2(t)] = \int_{-\infty}^{\infty} x_1(\tau) e^{-j\omega \tau} X_2(\omega) d\tau = X_2(\omega) \int_{-\infty}^{\infty} x_1(\tau) e^{-j\omega \tau} d\tau = X_1(\omega) X_2(\omega)$$

Let H(ω) be the Fourier transform of the unit impulse response h(t), that is,

h(t) ⇐⇒ H(ω)

Application of the time-convolution property to y(t) = x(t) ∗ h(t) yields [assuming that both x(t) and h(t) are Fourier transformable]

$$Y(\omega) = X(\omega)H(\omega) \tag{7.35}$$

The frequency-convolution property of Eq. (7.34) can be proved in exactly the same way by reversing the roles of x(t) and X(ω).

EXAMPLE 7.16 Time-Convolution Property to Show the Time-Integration Property

Use the time-convolution property to show that if

$$x(t) \Longleftrightarrow X(\omega)$$

then

$$\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(\omega)}{j\omega} + \pi X(0)\delta(\omega)$$

Because

$$u(t-\tau) = \begin{cases} 1 & \tau \leq t \\ 0 & \tau > t \end{cases}$$

it follows that

$$x(t) * u(t) = \int_{-\infty}^{\infty} x(\tau)u(t-\tau) d\tau = \int_{-\infty}^{t} x(\tau) d\tau$$

716 CHAPTER 7 CONTINUOUS-TIME SIGNAL ANALYSIS: THE FOURIER TRANSFORM

Now, from the time-convolution property [Eq. (7.33)], it follows that

$$x(t) * u(t) = \int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow X(\omega) \left[ \frac{1}{j\omega} + \pi \delta(\omega) \right]$$ $$= \frac{X(\omega)}{j\omega} + \pi X(0) \delta(\omega)$$

In deriving the last result, we used Eq. (1.10).

DR ILL 7.8 Fourier Transform Time-Convolution Property

Use the time-convolution property to show that:

$$(a) x(t) * \delta(t) = x(t)$$

(b) ^e−atu(t)∗e−btu(t) ^{= 1 b}−^a [e−at ⁻e−bt]u(t)

TIME DIFFERENTIATION AND TIME INTEGRATION If

$$x(t) \Longleftrightarrow X(\omega)$$

then†

$$\frac{dx(t)}{dt} \Longleftrightarrow j\omega X(\omega) \quad \text{(time differentiation)} \tag{7.36}$$

and

$$\int_{-\infty}^{t} x(\tau) d\tau \Longleftrightarrow \frac{X(\omega)}{j\omega} + \pi X(0)\delta(\omega) \quad \text{(time integration)} \tag{7.37}$$

Proof. Differentiation of both sides of Eq. (7.10) yields

$$\frac{dx(t)}{dt} = \frac{1}{2\pi} \int_{-\infty}^{\infty} j\omega X(\omega) e^{j\omega t} d\omega$$ $$\int_{-\infty}^{\infty} \left| \frac{dx(t)}{dt} \right| dt < \infty$$

^† Valid only if the transform of dx/dt exists. In other words, dx/dt must satisfy the Dirichlet conditions. The first Dirichlet condition implies

We also require that x(t) → 0 as t → ±∞. Otherwise, x(t) has a dc component, which gets lost in differentiation, and there is no one-to-one relationship between x(t) and dx/dt.

Operation	x(t)	X(ω)
Scalar multiplication	kx(t)	kX(ω)
Addition	x1(t)+x2(t)	X1(ω) +X2(ω)
Conjugation	x∗(t)	X∗(−ω)
Duality	X(t)	2πx(−ω)
Scaling (a real)	x(at)	1 ω X a a
Time shifting	x(t −t0)	X(ω)e−jωt0
Frequency shifting (ω0 real)	x(t)ejω0t	X(ω −ω0)
Time convolution	x1(t)∗x2(t)	X1(ω)X2(ω)
Frequency convolution	x1(t)x2(t)	1 2π X1(ω)∗X2(ω)
Time differentiation	dnx(t) dtn	(jω)nX(ω)
Time integration	# t x(u)du −∞	X(ω) +πX(0)δ(ω) jω

TABLE 7.2 Fourier Transform Properties

This result shows that

$$\frac{dx(t)}{dt} \Longleftrightarrow j\omega X(\omega)$$

Repeated application of this property yields

$$\frac{d^n x(t)}{dt^n} \Longleftrightarrow (j\omega)^n X(\omega)$$

The time-integration property [Eq. (7.37)] has already been proved in Ex. 7.16. Table 7.2 summarizes the most important properties of the Fourier transform.

EXAMPLE 7.17 Fourier Transform Time-Differentiation Property

Use the time-differentiation property to find the Fourier transform of the triangle pulse (t/τ ) illustrated in Fig. 7.27a. Verify the correctness of the spectrum by using it to synthesize a periodic replication of the original time-domain signal with τ = 1.

Figure 7.27 Finding the Fourier transform of a piecewise-linear signal using the time-differentiation property.

To find the Fourier transform of this pulse, we differentiate the pulse successively, as illustrated in Fig. 7.27b and 7.27c. Because dx/dt is constant everywhere, its derivative, d²x/dt2, is zero everywhere. But dx/dt has jump discontinuities with a positive jump of 2/τ at t = ±τ/2, and a negative jump of 4/τ at t = 0. Recall that the derivative of a signal at a jump discontinuity is an impulse at that point of strength equal to the amount of jump. Hence, d²x/dt2, the derivative of dx/dt, consists of a sequence of impulses, as depicted in Fig. 7.27c; that is,

$$\frac{d^2x(t)}{dt^2} = \frac{2}{\tau} \left[ \delta \left( t + \frac{\tau}{2} \right) - 2\delta(t) + \delta \left( t - \frac{\tau}{2} \right) \right]$$

From the time-differentiation property [Eq. (7.36)],

$$\frac{d^2x(t)}{dt^2} \Longleftrightarrow (j\omega)^2 X(\omega) = -\omega^2 X(\omega)$$

Also, from the time-shifting property [Eq. (7.29)],

$$\delta(t-t_0) \Longleftrightarrow e^{-j\omega t_0}$$

Combining these results, we obtain

$$-\omega^2 X(\omega) = \frac{2}{\tau} \left[ e^{j(\omega \tau/2)} - 2 + e^{-j(\omega \tau/2)} \right] = \frac{4}{\tau} \left( \cos \frac{\omega \tau}{2} - 1 \right) = -\frac{8}{\tau} \sin^2 \left( \frac{\omega \tau}{4} \right)$$

and

$$X(\omega) = \frac{8}{\omega^2 \tau} \sin^2\left(\frac{\omega \tau}{4}\right) = \frac{\tau}{2} \left[ \frac{\sin\left(\frac{\omega \tau}{4}\right)}{\frac{\omega \tau}{4}} \right]^2 = \frac{\tau}{2} \text{sinc}^2\left(\frac{\omega \tau}{4}\right)$$

The spectrum X(ω) is depicted in Fig. 7.27d. This procedure of finding the Fourier transform can be applied to any function x(t) made up of straight-line segments with x(t) → 0 as |t|→∞. The second derivative of such a signal yields a sequence of impulses whose Fourier transform can be found by inspection. This example suggests a numerical method of finding the Fourier transform of an arbitrary signal x(t) by approximating the signal by straight-line segments.

SYNTHESIZING A PERIODIC REPLICATION TO VERIFY SPECTRUM CORRECTNESS

While a signal’s spectrum X(ω) provides useful insight into signal character, it can be difficult to look at X(ω) and know that it is correct for a particular signal x(t). Is it obvious, for example, that ^X(ω) ^{= τ 2} sinc² (ωτ/4) is really the spectrum of a τ -duration rectangle function? Or is it possible that a mathematical error was made in the determination of X(ω)? It is difficult to be certain by simple inspection of the spectrum.

The same uncertainties exist when we are looking at a periodic signal’s Fourier series spectrum. In the Fourier series case, we can verify the correctness of a signal’s spectrum by synthesizing x(t) with a truncated Fourier series; the synthesized signal will match the original only if the computed spectrum is correct. This is exactly the approach that was taken in Ex. 6.11. And since a truncated Fourier series involves a simple sum, tools like MATLAB make waveform synthesis relatively simple, at least in the case of the Fourier series.

In the case of the Fourier transform, however, synthesis of x(t) using Eq. (7.10) requires integration, a task not well suited to numerical packages such as MATLAB. All is not lost, however. Consider Eq. (7.5). By scaling and sampling the spectrum X(ω) of an aperiodic signal x(t), we obtain the Fourier series coefficient of a signal that is the periodic replication of x(t). Similar to Ex. 6.11, we can then synthesize a periodic replication of x(t) with a truncated Fourier series to verify spectrum correctness. Let us demonstrate the idea for the current example with τ = 1.

To begin, we represent ^X(ω) ^{= τ 2} sinc² (ωτ/4) using an anonymous function in MATLAB. Since MATLAB computes sinc(x) as (sin(πx))/πx, we must scale the input by 1/π to match the notation of sinc in this book.

tau = 1; X = @(omega) tau/2*(sinc(omega*tau/(4*pi))).^2;

For our periodic replication, let us pick T⁰ = 2, which is comfortably wide enough to accommodate our (τ = 1)-width function without overlap. We use Eq. (7.5) to define the needed Fourier series coefficients Dn.

TO = 2; omega0 =

$$2*pi/TO$$

; D = $\mathcal{Q}(n)$ X(n*omega0)/TO;

Let us use 25 harmonics to synthesize the periodic replication x25(t) of our triangular signal x(t). To begin waveform synthesis, we set the dc portion of the signal.

t = (-T0:.001:T0); x25 = D(0)*ones(size(t));

To add the desired 25 harmonics, we enter a loop for 1 ≤ n ≤ 25 and add in the Dn and D−ⁿ terms. Although the result should be real, small round-off errors cause the reconstruction to be complex. These small imaginary parts are removed by using the real command.

for n = 1:25, >> x25 = x25+real(D(n)*exp(1j*omega0*n*t)+D(-n)*exp(-1j*omega0*n*t)); >> end

Lastly, we plot the resulting truncated Fourier series synthesis of x(t).

plot(t,x25,‘k’); xlabel(‘t’); ylabel(‘x_{25}(t)’);

Since the synthesized waveform shown in Fig. 7.28 closely matches a 2-periodic replication of the triangle wave in Fig. 7.27a, we have high confidence that both the computed Dn and, by extension, the Fourier spectrum X(ω) are correct.

Figure 7.28 Synthesizing a 2-periodic replication of x(t) using a truncated Fourier series.

DR ILL 7.9 Fourier Transform Time-Differentiation Property

Use the time-differentiation property to find the Fourier transform of rect(t/τ ).