18.2 Definition of the Fourier Transform

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18.2 Definition of the Fourier Transform

We saw in the previous chapter that a nonsinusoidal periodic function can be represented by a Fourier series, provided that it satisfies the Dirichlet conditions. What happens if a function is not periodic? Unfortunately , there are many important nonperiodic functions—such as a unit step or an exponential function—that we cannot represent by a Fourier series. As we shall see, the Fourier transform allows a transformation from the time to the frequency domain, even if the function is not periodic.

Suppose we w ant to find the Fourier transform of a nonperiodic function p(t), shown in Fig. 18.1(a). We consider a periodic function f(t) whose shape over one period is the same as p(t), as shown in Fig. 18.1(b). If we let the period T → ∞, only a single pulse of width τ [the desired nonperiodic function in Fig. 18.1(a)] remains, because the adjacent

Effect of increasing T on the spectrum of the periodic pulse trains in Fig. 18.1(b) using the appropriately modified Eq. (17.66).

pulses have been moved to infinity. Thus, the function f(t) is no longer periodic. In other words, f(t) = p(t) as T → ∞. It is interesting to consider the spectrum of f(t) for A = 10 and τ = 0.2 (see Section 17.6). Figure 18.2 shows the effect of increasing T on the spectrum. First, we notice that the general shape of the spectrum remains the same, and the frequenc y at which the envelope first becomes zero remains the same. However, the amplitude of the spectrum and the spacing between adjacent components both decrease, while the number of harmonics increases. Thus, over a range of frequencies, the sum of the amplitudes of the harmonics remains almost constant. As the total “strength” or ener gy of the components within a band must remain unchanged, the amplitudes of the harmonics must decrease as T increases. Because f = 1∕T, as T increases, f or ω decreases, so that the discrete spectrum ultimately becomes continuous.

To further understand this connection between a nonperiodic function and its periodic counterpart, consider the e xponential form of a Fourier series in Eq. (17.58), namely,

$$f(t) = \sum_{n = -\infty}^{\infty} c_n e^{jn\omega_0 t}$$

(18.1)

where

$$c_n = \frac{1}{T} \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t} dt$$

(18.2)

The fundamental frequency is

$$\omega_0 = \frac{2\pi}{T} \tag{18.3}$$

and the spacing between adjacent harmonics is

$$\Delta \omega = (n+1)\omega_0 - n\omega_0 = \omega_0 = \frac{2\pi}{T}$$

(18.4)

Substituting Eq. (18.2) into Eq. (18.1) gives

$$f(t) = \sum_{n=-\infty}^{\infty} \left[ \frac{1}{T} \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t} dt \right] e^{jn\omega_0 t}$$

\n

$$= \sum_{n=-\infty}^{\infty} \left[ \frac{\Delta \omega}{2\pi} \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t} dt \right] e^{jn\omega_0 t}$$

\n

$$= \frac{1}{2\pi} \sum_{n=-\infty}^{\infty} \left[ \int_{-T/2}^{T/2} f(t) e^{-jn\omega_0 t} dt \right] \Delta \omega e^{jn\omega_0 t}$$

(18.5)

If we let T → ∞, the summation becomes inte gration, the incremental spacing ∆ω becomes the dif ferential separation dω, and the discrete harmonic frequency nω0 becomes a continuous frequenc y ω. Thus, as T → ∞,

$$\sum_{n=-\infty}^{\infty} \Rightarrow \int_{-\infty}^{\infty}$$

\n

$$\Delta \omega \Rightarrow d\omega \qquad (18.6)$$

\n

$$n\omega_0 \Rightarrow \omega$$

so that Eq. (18.5) becomes

$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \left[ \int_{-\infty}^{\infty} f(t) e^{-j\omega t} dt \right] e^{j\omega t} d\omega \tag{18.7}$$

Some authors use F(jω) instead of F(ω) to represent the Fourier transform.

The term in the brackets is known as the Fourier transform of f(t) and is represented by F(ω). Thus,

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

(18.8)

where is the Fourier transform operator. It is evident from Eq. (18.8) that:

The Fourier transform is an integral transformation of f (t) from the time domain to the frequency domain.

In general, F(ω) is a complex function; its magnitude is called the amplitude spectrum, while its phase is called the phase spectrum. Thus, F(ω) is the spectrum.

Equation (18.7) can be written in terms of F(ω), and we obtain the inverse Fourier transform as

$$f(t) = \mathcal{F}^{-1}[F(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega$$

(18.9)

The function f(t) and its transform F(ω) form the Fourier transform pairs:

$$f(t) \qquad \Leftrightarrow \qquad F(\omega) \tag{18.10}$$

given that one can be derived from the other.

The F ourier transform F(ω) e xists when the F ourier inte gral in Eq. (18.8) converges. A sufficient but not necessary condition that f (t) has a Fourier transform is that it be completely integrable in the sense that

$$\int_{-\infty}^{\infty} |f(t)| \, dt < \infty \tag{18.11}$$

For example, the Fourier transform of the unit ramp function tu(t) does not exist, because the function does not satisfy the condition above.

To avoid the complex algebra that explicitly appears in the F ourier transform, it is sometimes expedient to temporarily replace jω with s and then replace s with jω at the end.

Find the F ourier transform of the follo wing functions: (a) δ(t − t0), Example 18.1 (b) e j^ω0^t , (c) cos ω⁰ t.

Solution:

(a) For the impulse function,

$$F(\omega) = \mathcal{F}[\delta(t - t_0)] = \int_{-\infty}^{\infty} \delta(t - t_0) e^{-j\omega t} dt = e^{-j\omega t_0} \qquad (18.1.1)$$

where the sifting property of the impulse function in Eq. (7.32) has been applied. For the special case t0 = 0, we obtain

$$\mathcal{F}[\delta(t)] = 1 \tag{18.1.2}$$

This shows that the magnitude of the spectrum of the impulse function is constant; that is, all frequencies are equally represented in the impulse function.

(b) We can find the Fourier transform of e j^ω0^t in two ways. If we let

$$F(\omega) = \delta(\omega - \omega_0)$$

then we can find f(t) using Eq. (18.9), writing

$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \delta(\omega - \omega_0) e^{j\omega t} d\omega$$

Using the sifting property of the impulse function gives

$$f(t) = \frac{1}{2\pi} e^{j\omega_0 t}$$

Inasmuch as F(ω) and f (t) constitute a F ourier transform pair , so too must 2πδ(ω − ω0) and e j^ω0^t ,

$$\mathcal{F}[e^{j\omega_0 t}] = 2\pi\delta(\omega - \omega_0)$$

(18.1.3)

Alternatively, from Eq. (18.1.2),

$$\delta(t) = \mathcal{F}^{-1}[1]$$

Using the inverse Fourier transform formula in Eq. (18.9),

$$\delta(t) = \mathcal{F}^{-1}[1] = \frac{1}{2\pi} \int_{-\infty}^{\infty} 1 e^{j\omega t} d\omega$$

or

$$\int_{-\infty}^{\infty} e^{j\omega t} d\omega = 2 \pi \delta(t)$$

(18.1.4)

Interchanging variables t and ω results in

$$\int_{-\infty}^{\infty} e^{j\omega t} dt = 2 \pi \delta(\omega)$$

(18.1.5)

Using this result, the Fourier transform of the given function is

$$\mathcal{F}[e^{j\omega_0 t}] = \int_{-\infty}^{\infty} e^{j\omega_0 t} e^{-j\omega t} dt = \int_{-\infty}^{\infty} e^{j(\omega_0 - \omega)} dt = 2 \pi \delta(\omega_0 - \omega)$$

Because the impulse function is an e ven function, with δ(ω⁰ − ω) = δ(ω − ω0),

$$\mathcal{F}[e^{j\omega_0 t}] = 2\pi \delta(\omega - \omega_0)$$

(18.1.6)

By simply changing the sign of ω0, we readily obtain

$$\mathcal{F}[e^{-j\omega_0 t}] = 2\pi \delta(\omega + \omega_0)$$

(18.1.7)

Also, by setting ω0 = 0,

$$\mathcal{F}[1] = 2\pi \delta(\omega) \tag{18.1.8}$$

(c) By using the result in Eqs. (18.1.6) and (18.1.7), we get

$$\mathcal{F}[\cos \omega_0 t] = \mathcal{F} \left[ \frac{e^{j\omega_0 t} + e^{-j\omega_0 t}}{2} \right]$$

= $\frac{1}{2} \mathcal{F} [e^{j\omega_0 t}] + \frac{1}{2} \mathcal{F} [e^{-j\omega_0 t}]$ (18.1.9)
= $\pi \delta(\omega - \omega_0) + \pi \delta(\omega + \omega_0)$

The Fourier transform of the cosine signal is shown in Fig. 18.3.

Fourier transform of f(t) = cos ω0t.

Practice Problem 18.1 Determine the Fourier transforms of the follo wing functions: (a) g ate function g(t) = 10u(t − 1) − 10u(t − 2), (b) 12δ(t − 2), (c) 15 sin ω0t.

Answer: (a) 10(e⁻j^ω − e⁻j2*^ω)∕jω, (b) 12ej*^2ω, (c) j15π[δ(ω + ω0) − δ(ω − ω0)].

Derive the Fourier transform of a single rectangular pulse of width τ and Example 18.2 height A, shown in Fig. 18.4.

Solution:

$$F(\omega) = \int_{-\tau/2}^{\tau/2} Ae^{-j\omega t} dt = -\frac{A}{j\omega} e^{-j\omega t} \Big|_{-\tau/2}^{\tau/2}$$ $$= \frac{2A}{\omega} \left( \frac{e^{j\omega \tau/2} - e^{-j\omega \tau/2}}{2j} \right)$$ $$= A\tau \frac{\sin \omega \tau/2}{\omega \tau/2} = A\tau \text{ sinc } \frac{\omega \tau}{2}$$

If we make A = 10 and τ = 2 as in Fig. 17.27 (like in Section 17.6), then

F(ω) = 20 sinc ω

whose amplitude spectrum is shown in Fig. 18.5. Comparing Fig. 18.4 with the frequency spectrum of the rectangular pulses in Fig. 17.28, we notice that the spectrum in Fig. 17.28 is discrete and its envelope has the same shape as the Fourier transform of a single rectangular pulse.

Figure 18.4

Figure 18.5

Amplitude spectrum of the rectangular pulse in Fig. 18.4: for Example 18.2.

0 t 25 ‒1 f(t) 1 Obtain the Fourier transform of the function in Fig. 18.6. Practice Problem 18.2 Answer: 50(cos ω− 1) ____________ ^jω.

Figure 18.6 For Practice Prob. 18.2.

‒25

Obtain the Fourier transform of the “switched-on” e xponential function Example 18.3 shown in Fig. 18.7.

0,

t > 0 t < 0

Solution:

From Fig. 18.7,

Hence,

$$F(\omega) = \int_{-\infty}^{\infty} f(t)e^{-j\omega t} dt = \int_{0}^{\infty} e^{-at} e^{-j\omega t} dt = \int_{0}^{\infty} e^{-(a+j\omega)t} dt$$ $$= \frac{-1}{a+j\omega} e^{-(a+j\omega)t} \Big|_{0}^{\infty} = \frac{1}{a+j\omega}$$

^f(t) = e⁻atu(t) = {e⁻at,

For Example 18.3.

Practice Problem 18.3 Determine the Fourier transform of the “switched-off” exponential function in Fig. 18.8.

Answer:

$$\frac{37.5}{a - j\omega}$$

.

18.3 Properties of the Fourier Transform

We now develop some properties of the Fourier transform that are useful in finding the transforms of complicated functions from the transforms of simple functions. F or each property, we will first state and derive it, and then illustrate it with some examples.

Linearity

If F1(ω) and F2(ω) are the F ourier transforms of f1(t) and f2(t), respectively, then

$$\mathcal{F}[a_1 f_1(t) + a_2 f_2(t)] = a_1 F_1(\omega) + a_2 F_2(\omega)$$

(18.12)

where a1 and a2 are constants. This property simply states that the Fourier transform of a linear combination of functions is the same as the linear combination of the transforms of the indi vidual functions. The proof of the linearity property in Eq. (18.12) is straightforward. By definition,

$$\mathcal{F}[a_1 f_1(t) + a_2 f_2(t)] = \int_{-\infty}^{\infty} [a_1 f_1(t) + a_2 f_2(t)] e^{-j\omega t} dt$$

=

$$\int_{-\infty}^{\infty} a_1 f_1(t) e^{-j\omega t} dt + \int_{-\infty}^{\infty} a_2 f_2(t) e^{-j\omega t} dt$$

=

$$a_1 F_1(\omega) + a_2 F_2(\omega)$$

(18.13)

For e xample, sin ω0t = __1 2j (e^jω0*^t* − e⁻jω0*^t* ). Using the linearity property,

$$F[\sin \omega_0 t] = \frac{1}{2j} [\mathcal{F}(e^{j\omega_0 t}) - \mathcal{F}(e^{-j\omega_0 t})]$$ $$= \frac{\pi}{j} [\delta(\omega - \omega_0) - \delta(\omega + \omega_0)]$$

(18.14)

$$= j\pi [\delta(\omega + \omega_0) - \delta(\omega - \omega_0)$$

Time Scaling

If F(ω) = [ f(t)], then

$$\mathcal{F}[f(at)] = \frac{1}{|a|} F(\frac{\omega}{a})$$

(18.15)

where a is a constant. Equation (18.15) sho ws that time e xpansion (∣a∣ > 1) corresponds to frequency compression, or conversely, time compression (∣a∣ < 1) implies frequenc y expansion. The proof of the time-scaling property proceeds as follows.

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

(18.16)

If we let x = at, so that dx = a dt, then

$$\mathcal{F}[f(at)] = \int_{-\infty}^{\infty} f(x)e^{-j\omega x/a} \frac{dx}{a} = \frac{1}{a}F\left(\frac{\omega}{a}\right)$$

(18.17)

For example, for the rectangular pulse p(t) in Example 18.2,

$$\mathcal{F}[p(t)] = A\tau \operatorname{sinc} \frac{\omega \tau}{2}$$

(18.18a)

Using Eq. (18.15),

$$\mathcal{F}[p(2t)] = \frac{A\tau}{2}\operatorname{sinc}\frac{\omega\tau}{4}$$

(18.18b)

It may be helpful to plot p(t) and p(2t) and their F ourier transforms. Because

$$p(t) = \begin{cases} A, & \frac{\tau}{2} < t < \frac{\tau}{2} \\ 0, & \text{otherwise} \end{cases} \tag{18.19a}$$

then replacing every t with 2t gives

$$p(2t) = \begin{cases} A, & -\frac{\tau}{2} < 2t < \frac{\tau}{2} \\ 0, & \text{otherwise} \end{cases} = \begin{cases} A, & -\frac{\tau}{4} < t < \frac{\tau}{4} \\ 0, & \text{otherwise} \end{cases}$$

(18.19b)

showing that p(2t) is time compressed, as shown in Fig. 18.9(b). To plot both Fourier transforms in Eq. (18.18), we recall that the sinc function has zeros when its argument is nπ, where n is an integer. Hence, for the transform of p(t) in Eq. (18.18a), ωτ∕2 = 2πfτ∕2 = nπ → f = n∕τ, and for the transform of p(2t) in Eq. (18.18b), ωτ∕4 = 2π f τ∕4 = nπ → f = 2n∕τ. The plots of the Fourier transforms are shown in Fig. 18.9, which shows that time compression corresponds with frequency expansion. We should expect this intuitively, because when the signal is squashed in time, we expect it to change more rapidly, thereby causing higher-frequency components to exist.

Time Shifting

If F(ω) = [ f(t)], then

$$\mathcal{F}[f(t-t_0)] = e^{-j\omega t_0} F(\omega)$$

(18.20)

that is, a delay in the time domain corresponds to a phase shift in the frequency domain. To derive the time shifting property, we note that

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

(18.21)

Figure 18.9

The effect of time scaling: (a) transform of the pulse, (b) time compression of the pulse causes frequency expansion.

If we let x = t − t0 so that dx = dt and t = x + t0, then

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

= $e^{-j\omega t_0} \int_{-\infty}^{\infty} f(x)e^{-j\omega x} dx = e^{-j\omega t_0} F(\omega)$ (18.22)

Similarly, [ f(t + t0)] = e jωt⁰ F(ω). For example, from Example 18.3,

$$\mathcal{F}[e^{-at}u(t)] = \frac{1}{a + j\omega} \tag{18.23}$$

The transform of f(t) = e⁻(t−2)u(t − 2) is

$$F(\omega) = \mathcal{F}[e^{-(t-2)} u(t-2)] = \frac{e^{-j2\omega}}{1+j\omega}$$

(18.24)

Frequency Shifting (or Amplitude Modulation)

This property states that if F(ω) = [ f(t)], then

$$\mathcal{F}[f(t)e^{j\omega_0 t}] = F(\omega - \omega_0)$$

(18.25)

meaning, a frequency shift in the frequency domain adds a phase shift to the time function. By definition,

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

=

$$\int_{-\infty}^{\infty} f(t)e^{-j(\omega - \omega_0)t} dt = F(\omega - \omega_0)$$

(18.26)

For e xample, cos ω0t = _1 2 (e j^ω0^t + e⁻jω0*^t* ). Using the property in Eq. (18.25),

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

(18.27)

This is an important result in modulation where frequenc y components of a signal are shifted. If, for e xample, the amplitude spectrum of f(t) is as shown in Fig. 18.10(a), then the amplitude spectrum of f(t)cosω0t will be as shown in Fig. 18.10(b). We will elaborate on amplitude modulation in Section 18.7.1.

Amplitude spectra of: (a) signal f(t), (b) modulated signal f(t)cos ω0t.

Time Differentiation

Given that F(ω) = [ f(t)], then

$$\mathcal{F}[f'(t)] = j\omega F(\omega)$$

(18.28)

In other words, the transform of the derivative of f(t) is obtained by multiplying the transform of f(t) by jω. By definition,

$$f(t) = \mathcal{F}^{-1}[F(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t} d\omega$$

(18.29)

Taking the derivative of both sides with respect to t gives

$$f'(t) = \frac{j\omega}{2\pi} \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega = j\omega \mathcal{F}^{-1}[F(\omega)]$$

or

$$F[f'(t)] = j\omega F(\omega)$$

(18.30)

Repeated applications of Eq. (18.30) give

$$\mathcal{F}[f^{(n)}(t)] = (j\omega)^n F(\omega)$$

(18.31)

For example, if f(t) = e⁻atu(t), then

$$f'(t) = -ae^{-at} u(t) + e^{-at} \delta(t) = -af(t) + e^{-at} \delta(t)$$

(18.32)

Taking the Fourier transforms of the first and last terms, we obtain

$$j\omega F(\omega) = -aF(\omega) + 1$$

$\Rightarrow$ $F(\omega) = \frac{1}{a + j\omega}$ (18.33)

which agrees with the result in Example 18.3.

Time Integration

Given that F(ω) = [ f(t)], then

$$\mathcal{F}\left[\int_{-\infty}^{t} f(\tau) d\tau\right] = \frac{F(\omega)}{j\omega} + \pi F(0)\delta(\omega)$$

(18.34)

that is, the transform of the inte gral of f(t) is obtained by di viding the transform of f(t) by jω and adding the result to the impulse term that reflects the dc component F(0). Someone might ask, “How do we know that when we take the Fourier transform for time integration, we should integrate over the interval [−∞, t] and not [−∞, ∞]?” When we integrate over [−∞, ∞], the result does not depend on time an ymore, and the Fourier transform of a constant is what we will eventually get. But when we integrate over [−∞, t], we get the inte gral of the function from the past to time t, so that the result depends on t and we can take the Fourier transform of that.

If ω is replaced by 0 in Eq. (18.8),

$$F(0) = \int_{-\infty}^{\infty} f(t) dt$$

(18.35)

indicating that the dc component is zero when the integral of f(t) over all time vanishes. The proof of the time inte gration in Eq. (18.34) will be given later when we consider the convolution property.

For example, we know that [δ(t)] = 1 and that integrating the impulse function gives the unit step function [see Eq. (7.39a)]. By applying the property in Eq. (18.34), we obtain the F ourier transform of the unit step function as

$$\mathcal{F}[u(t)] = \mathcal{F}\left[\int_{-\infty}^{t} \delta(\tau) d\tau\right] = \frac{1}{j\omega} + \pi \delta(\omega)$$

(18.36)

Reversal

If F(ω) = [ f(t)], then

$$\mathcal{F}[f(-t)] = F(-\omega) = F^*(\omega)$$

(18.37)

where the asterisk denotes the comple x conjugate. This property states that reversing f(t) about the time axis reverses F(ω) about the frequency axis. This may be regarded as a special case of time scaling for which a = −1 in Eq. (18.15).

For example, 1 = u(t) + u(−t). Hence,

$$F[1] = F[u(t)] + F[u(-t)]$$ $$= \frac{1}{j\omega} + \pi\delta(\omega)$$ $$- \frac{1}{j\omega} + \pi\delta(-\omega)$$ $$= 2\pi\delta(\omega)$$

Duality

This property states that if F(ω) is the Fourier transform of f(t), then the Fourier transform of F(t) is 2πf(−ω); we write

$$\mathcal{F}[f(t)] = F(\omega) \qquad \Rightarrow \qquad \mathcal{F}[F(t)] = 2\pi f(-\omega) \qquad (18.38)$$

This expresses the symmetry property of the Fourier transform. To derive this property, we recall that

$$f(t) = \mathcal{F}^{-1}[F(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega$$

or

$$2\pi f(t) = \int_{-\infty}^{\infty} F(\omega)e^{j\omega t} d\omega$$

(18.39)

Replacing t by −t gives

$$2\pi f(-t) = \int_{-\infty}^{\infty} F(\omega)e^{-j\omega t} d\omega$$

If we interchange t and ω, we obtain

$$2\pi f(-\omega) = \int_{-\infty}^{\infty} F(t)e^{-j\omega t} dt = \mathcal{F}[F(t)]$$

(18.40)

as expected.

For example, if f(t) = e −∣t^∣ , then

$$F(\omega) = \frac{2}{\omega^2 + 1}$$

(18.41)

By the duality property, the Fourier transform of F(t) = 2∕(t 2 + 1) is

$$2\pi f(\omega) = 2\pi e^{-|\omega|} \tag{18.42}$$

Figure 18.11 sho ws another e xample of the duality property . It illus trates the f act that if f(t) = δ(t) so that F(ω) = 1, as in Fig. 18.11(a), then the Fourier transform of F(t) = 1 is 2πf(ω) = 2πδ(ω) as shown in Fig. 18.11(b).

Convolution

Recall from Chapter 15 that if x(t) is the input excitation to a circuit with an impulse function of h(t), then the output response y(t) is given by the convolution integral

$$y(t) = h(t) * x(t) = \int_{-\infty}^{\infty} h(\lambda) x(t - \lambda) d\lambda$$

(18.43)

Because f(t) is the sum of the signals in Figs. 18.7 and 18.8, F(ω) is the sum of the results in Example 18.3 and Practice Prob. 18.3.

A typical illustration of the duality property of the Fourier transform: (a) transform of impulse, (b) transform of unit dc level.

If X(ω), H(ω), and Y(ω) are the Fourier transforms of x(t), h(t), and y(t), respectively, then

$$Y(\omega) = \mathcal{F}[h(t) * x(t)] = H(\omega)X(\omega)$$

(18.44)

which indicates that con volution in the time domain corresponds with multiplication in the frequency domain.

To derive the convolution property, we take the Fourier transform of both sides of Eq. (18.43) to get

$$Y(\omega) = \int_{-\infty}^{\infty} \left[ \int_{-\infty}^{\infty} h(\lambda) x(t - \lambda) \, d\lambda \right] e^{-j\omega t} \, dt \tag{18.45}$$

Exchanging the order of inte gration and factoring h(λ), which does not depend on t, we have

$$Y(\omega) = \int_{-\infty}^{\infty} h(\lambda) \left[ \int_{-\infty}^{\infty} x(t - \lambda) e^{-j\omega t} dt \right] d\lambda$$

For the integral within the brack ets, let τ = t − λ so that t = τ + λ and dt = dτ. Then,

$$Y(\omega) = \int_{-\infty}^{\infty} h(\lambda) \left[ \int_{-\infty}^{\infty} x(\tau) e^{-j\omega(\tau+\lambda)} d\tau \right] d\lambda$$

=

$$\int_{-\infty}^{\infty} h(\lambda) e^{-j\omega\lambda} d\lambda \int_{-\infty}^{\infty} x(\tau) e^{-j\omega\tau} d\tau = H(\omega)X(\omega)$$

(18.46)

as expected. This result e xpands the phasor method be yond what w as done with the Fourier series in the previous chapter.

To illustrate the con volution property , suppose both h(t) and x(t) are identical rectangular pulses, as sho wn in Fig. 18.12(a) and 18.12(b). We recall from Example 18.2 and Fig. 18.5 that the Fourier transforms of the rectangular pulses are sinc functions, as sho wn in Fig. 18.12(c) and 18.12(d). According to the con volution property, the product of the sinc functions should gi ve us the con volution of the rectangular pulses in the time domain. Thus, the con volution of the pulses in Fig. 18.12(e) and the product of the sinc functions in Fig. 18.12(f) form a Fourier pair.

In view of the duality property, we expect that if convolution in the time domain corresponds with multiplication in the frequenc y domain,

The important relationship in Eq. (18.46) is the key reason for using the Fourier transform in the analysis of linear systems.

then multiplication in the time domain should ha ve a correspondence in the frequency domain. This happens to be the case. If f(t) = f1(t) f2(t), then

$$F(\omega) = \mathcal{F}[f_1(t)f_2(t)] = \frac{1}{2\pi}F_1(\omega) * F_2(\omega)$$

(18.47)

or

$$F(\omega) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F_1(\lambda) F_2(\omega - \lambda) d\lambda$$

(18.48)

which is convolution in the frequency domain. The proof of Eq. (18.48) readily follows from the duality property in Eq. (18.38).

Let us no w deri ve the time inte gration property in Eq. (18.34). If we replace x(t) with the unit step function u(t) and h(t) with f(t) in Eq. (18.43), then

$$\int_{-\infty}^{\infty} f(\lambda)u(t-\lambda) \, d\lambda = f(t) * u(t) \tag{18.49}$$

But by the definition of the unit step function,

∫

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

We can write this as

$$u(t - \lambda) = \begin{cases} 1, & \lambda < t \\ 0, & \lambda > t \end{cases}$$

Substituting this into Eq. (18.49) makes the interval of integration change from [−∞, ∞] to [−∞, t], and thus Eq. (18.49) becomes

$$\int_{-\infty}^{t} f(\lambda) \, d\lambda = u(t) * f(t)$$

Taking the Fourier transform of both sides yields

$$\mathcal{F}\left[\int_{-\infty}^{t} f(\lambda) d\lambda\right] = U(\omega)F(\omega)$$

\n(18.50)

But from Eq. (18.36), the Fourier transform of the unit step function is

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

Substituting this into Eq. (18.50) gives

$$\mathcal{F}\left[\int_{-\infty}^{t} f(\lambda) d\lambda\right] = \left(\frac{1}{j\omega} + \pi \delta(\omega)\right) F(\omega)$$

\n

$$= \frac{F(\omega)}{j\omega} + \pi F(0) \delta(\omega)$$

\n(18.51)

which is the time inte gration property of Eq. (18.34). Note that in Eq. (18.51), F(ω)δ(ω) = F(0)δ(ω), since δ(ω) is only nonzero at ω = 0.

Table 18.1 lists these properties of the Fourier transform. Table 18.2 presents the transform pairs of some common functions. Note the simi larities between these tables and Tables 15.1 and 15.2.

TABLE 18.1