Properties of the Fourier transform.

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Properties of the Fourier transform.

Property	f(t)	F(ω)
Linearity	a1 f1(t) + a2 f2(t)	a1F1(ω) + a2F2(ω)
Scaling	f(at)	_1 ω F( a ) ∣a∣
Time shift	f(t − a)	e−jωaF(ω)
Frequency shift	e jω0t f(t)	F(ω − ω0)

TABLE 18.1	(continued)
Property	f(t)	F(ω)
Modulation	cos(ω0t)f(t)	__1 [F(ω + ω0) + F(ω − ω0)] 2
Time differentiation	df __ dt	jωF(ω)
	d n f ___ dtn	n ( jω) F(ω)
Time integration	t f(t) dt ∫ −∞	F(ω) _____ jω    + πF(0)δ(ω)
Frequency differentiation	n t f(t)	d n ____ n ( j) dωn F(ω)
Reversal	f(−t)	F(−ω) or F*(ω)
Duality	F(t)	2πf(−ω)
Convolution in t	f1(t) * f2(t)	F1(ω)F2(ω)
Convolution in ω	f1(t)f2(t)	___1 F1(ω) * F2(ω) 2π

TABLE 18.2

Fourier transform pairs.

f(t)	F(ω)
δ(t)	1
1	2π δ(ω)
u(t)	π δ(ω) + ___1 jω
u(t + τ) − u(t − τ)	2 ______ sin ωτ ω
∣t∣	−2 ___ ω2
sgn(t)	___2 jω
e−at u(t)	______ 1 a + jω
eat u(−t)	______ 1 a − jω
n e−at u(t) t	__________ n! n+1 (a + jω)
e−a∣t∣	_______ 2a a2 + ω2
ejω0t	2πδ(ω − ω0)
sin ω0t	jπ[δ(ω + ω0) − δ(ω − ω0)]
cos ω0t	π[δ(ω + ω0) + δ(ω − ω0)]
e−at sin ω0tu(t)	____________ ω0 2 2 (a + jω) + ω 0
e−at cos ω0 tu(t)	a + jω ____________ 2 2 (a + jω) + ω 0

Figure 18.13 The signum function of Example 18.4.

Example 18.4 Find the Fourier transforms of the following functions: (a) signum function sgn(t), shown in Fig. 18.13, (b) the double-sided e xponential e⁻a∣t^∣ , and (c) the sinc function (sin t)∕t.

Solution:

(a) We can obtain the F ourier transform of the signum function in three ways.

■ METHOD 1 We can write the signum function in terms of the unit step function as

$$sgn(t) = f(t) = u(t) - u(-t)$$

But from Eq. (18.36),

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

Applying this and the reversal property, we obtain

$$F[\text{sgn}(t)] = U(\omega) - U(-\omega)$$ $$= \left(\pi\delta(\omega) + \frac{1}{j\omega}\right) - \left(\pi\delta(-\omega) + \frac{1}{-j\omega}\right) = \frac{2}{j\omega}$$

■ METHOD 2 Because δ(ω) = δ(−ω), another w ay of writing the signum function in terms of the unit step function is

$$f(t) = \text{sgn}(t) = -1 + 2u(t)$$

Taking the Fourier transform of each term gives

$$F(\omega) = -2\pi\delta(\omega) + 2\left(\pi\delta(\omega) + \frac{1}{j\omega}\right) = \frac{2}{j\omega}$$

■ METHOD 3 We can take the derivative of the signum function in Fig. 18.13 and obtain

$$f'(t) = 2\delta(t)$$

Taking the transform of this,

$$j\omega F(\omega) = 2$$

$\Rightarrow$ $F(\omega) = \frac{2}{j\omega}$

as obtained previously.

(b) The double-sided exponential can be expressed as

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

where y(t) = e ⁻atu(t) so that Y(ω) = 1∕(a + jω). Applying the re versal property,

$$\mathcal{F}[e^{-a|t|}] = Y(\omega) + Y(-\omega) = \left(\frac{1}{a+j\omega} + \frac{1}{a-j\omega}\right) = \frac{2a}{a^2 + \omega^2}$$

(c) From Example 18.2,

$$\mathcal{F}\left[u\left(t+\frac{\tau}{2}\right)-u\left(t-\frac{\tau}{2}\right)\right]=\tau\frac{\sin(\omega\tau/2)}{\omega\tau/2}=\tau\,\text{sinc}\,\frac{\omega\tau}{2}$$

Setting τ∕2 = 1 gives

$$\mathcal{F}[u(t+1) - u(t-1)] = 2\frac{\sin \omega}{\omega}$$

Applying the duality property yields

$$\mathcal{F}\left[2\frac{\sin t}{t}\right] = 2\pi \left[U(\omega+1) - U(\omega-1)\right]$$

or

$$\mathcal{F}\left[\frac{\sin t}{t}\right] = \pi[U(\omega + 1) - U(\omega - 1)]$$

Determine the F ourier transforms of these functions: (a) g ate function Practice Problem 18.4 g(t) = u(t) − u(t − 1), (b) f(t) = 10t −2t u(t), and (c) sawtooth pulse p(t) = 75t[u(t) − u(t − 2)].

Answer: (a)

$$(1 - e^{-j\omega}) \left[ \pi \delta(\omega) + \frac{1}{j\omega} \right]$$

, (b) $\frac{10}{(2 + j\omega)^2}$ ,
(c) $\frac{75(e^{-j2\omega} - 1)}{\omega^2} + \frac{150j}{\omega} e^{-j2\omega}$ .

Find the Fourier transform of the function in Fig. 18.14. Example 18.5

Solution:

The Fourier transform can be found directly using Eq. (18.8), b ut it is much easier to find it using the derivative property. We can express the function as

$$f(t) = \begin{cases} 1+t, & -1 < t < 0 \\ 1-t, & 0 < t < 1 \end{cases}$$

Its first derivative is shown in Fig. 18.15(a) and is given by

Figure 18.15 First and second derivatives of f(t) in Fig. 18.14; for Example 18.5.

f(t) ‒1 0 1 t 1 Figure 18.14

For Example 18.5.

Its second derivative is in Fig. 18.15(b) and is given by

$$f''(t) = \delta(t+1) - 2\delta(t) + \delta(t-1)$$

Taking the Fourier transform of both sides,

$$(j\omega)^2 F(\omega) = e^{j\omega} - 2 + e^{-j\omega} = -2 + 2 \cos \omega$$

or

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

Example 18.6 Obtain the inverse Fourier transform of:

Obtain the inverse Fourier transform of:
\n(a)

$$F(\omega) = \frac{10j\omega + 4}{(j\omega)^2 + 6j\omega + 8}$$

(b) $G(\omega) = \frac{\omega^2 + 21}{\omega^2 + 9}$

Solution:

(a) To avoid complex algebra, we can replace jω with s for the moment. Using partial fraction expansion,

$$F(s) = \frac{10s + 4}{s^2 + 6s + 8} = \frac{10s + 4}{(s + 4)(s + 2)} = \frac{A}{s + 4} + \frac{B}{s + 2}$$

where

$$A = (s + 4)F(s)|_{s=-4} = \frac{10s + 4}{(s + 2)}|_{s=-4} = \frac{-36}{-2} = 18$$ $$B = (s + 2)F(s)|_{s=-2} = \frac{10s + 4}{(s + 4)}|_{s=-2} = \frac{-16}{2} = -8$$

Substituting A = 18 and B = −8 in F(s) and s with jω gives

$$F(j\omega) = \frac{18}{j\omega + 4} + \frac{-8}{j\omega + 2}$$

With the aid of Table 18.2, we obtain the inverse transform as

$$f(t) = (18e^{-4t} - 8e^{-2t})u(t)$$

(b) We simplify G(ω) as

$$G(\omega) = \frac{\omega^2 + 21}{\omega^2 + 9} = 1 + \frac{12}{\omega^2 + 9}$$

With the aid of Table 18.2, the inverse transform is obtained as

$$g(t) = \delta(t) + 2e^{-3|t|}$$

Find the inverse Fourier transform of:
\n(a)

$$H(\omega) = \frac{6(3 + j2\omega)}{(1 + j\omega)(4 + j\omega)(2 + j\omega)}$$

\n(b) $Y(\omega) = \pi\delta(\omega) + \frac{1}{j\omega} + \frac{2(1 + j\omega)}{(1 + j\omega)^2 + 16}$
\nAnswer: (a) $h(t) = (2e^{-t} + 3e^{-2t} - 5e^{-4t}) u(t)$ ,
\n(b) $y(t) = (1 + 2e^{-t} \cos 4t)u(t)$ .

18.4 Circuit Applications

The Fourier transform generalizes the phasor technique to nonperiodic functions. Therefore, we apply F ourier transforms to circuits with nonsinusoidal excitations in exactly the same way we apply phasor techniques to circuits with sinusoidal excitations. Thus, Ohm’s law is still valid:

$$V(\omega) = Z(\omega)I(\omega) \tag{18.52}$$

where V(ω) and I(ω) are the F ourier transforms of the v oltage and current and Z(ω) is the impedance. We get the same e xpressions for the impedances of resistors, inductors, and capacitors as in phasor analysis, namely,

$$\begin{array}{ccc}\nR & \Rightarrow & R \\ L & \Rightarrow & j\omega L \\ C & \Rightarrow & \frac{1}{j\omega C}\n\end{array}$$

\n(18.53)

Once we transform the functions for the circuit elements into the fre quency domain and tak e the F ourier transforms of the e xcitations, we can use circuit techniques such as v oltage division, source transforma tion, mesh analysis, node analysis, or Thevenin’s theorem, to find the unknown response (current or v oltage). Finally , we tak e the in verse Fourier transform to obtain the response in the time domain.

Although the F ourier transform method produces a response that exists for −∞ < t < ∞, F ourier analysis cannot handle circuits with initial conditions.

The transfer function is ag ain defined as the ratio of the output response Y(ω) to the input excitation X(ω); that is,

$$H(\omega) = \frac{Y(\omega)}{X(\omega)}\tag{18.54}$$

Find the inverse Fourier transform of: Practice Problem 18.6

$$Y(\omega) = H(\omega)X(\omega) \tag{18.55}$$

X(ω) H(ω) Y(ω)

Figure 18.17

The frequenc y domain input-output relationship is portrayed in Fig. 18.17. Equation (18.55) sho ws that if we kno w the transfer func tion and the input, we can readily find the output. The relationship in Eq. (18.54) is the principal reason for using the F ourier transform in circuit analysis. Notice that H(ω) is identical to H(s) with s = jω. Also, if the input is an impulse function [i.e., x(t) = δ(t)], then X(ω) = 1, so that the response is

$$Y(\omega) = H(\omega) = \mathcal{F}[h(t)] \tag{18.56}$$

indicating that H(ω) is the Fourier transform of the impulse response h(t).

2 Ω vi (t) 1 F vo(t) + ‒ + ‒

Figure 18.18 For Example 18.7.

Solution:

The Fourier transform of the input voltage is

$$V_i(\omega) = \frac{2}{3 + j\omega}$$

and the transfer function obtained by voltage division is

$$H(\omega) = \frac{V_o(\omega)}{V_i(\omega)} = \frac{1/j\omega}{2 + 1/j\omega} = \frac{1}{1 + j2\omega}$$

Hence,

$$V_o(\omega) = V_i(\omega)H(\omega) = \frac{2}{(3 + j\omega)(1 + j2\omega)}$$

or

$$V_o(\omega) = \frac{1}{(3 + j\omega)(0.5 + j\omega)}$$

By partial fractions,

$$V_o(\omega) = \frac{-0.4}{3 + j\omega} + \frac{0.4}{0.5 + j\omega}$$

Taking the inverse Fourier transform yields

$$v_o(t) = 0.4(e^{-0.5t} - e^{-3t})u(t)$$

1 H vi (t) 4 Ω vo(t) + ‒ + ‒ Figure 18.19 For Practice Prob. 18.7.

Practice Problem 18.7 Determine

$$v_o(t)$$

in Fig. 18.19 if $v_i(t) = 5\text{sgn}(t) = (-5 + 10u(t))$ V.

Answer: −5 + 10(1 − e⁻4*^t* )u(t) V.

Using the F ourier transform method, find io(t) in Fig. 18.20 when Example 18.8 is(t) = 10 sin 2t A.

Solution:

By current division,

$$H(\omega) = \frac{I_o(\omega)}{I_s(\omega)} = \frac{2}{2 + 4 + 2/j\omega} = \frac{j\omega}{1 + j\omega^2}$$

If is(t) = 10 sin 2t, then

$$I_s(\omega) = j\pi 10[\delta(\omega + 2) - \delta(\omega - 2)]$$

Hence,

$$I_o(\omega) = H(\omega)I_s(\omega) = \frac{10\pi\omega[\delta(\omega - 2) - \delta(\omega + 2)]}{1 + j\omega^2}$$

The inverse Fourier transform of Io(ω) cannot be found using Table 18.2. We resort to the inverse Fourier transform formula in Eq. (18.9) and write

respect to the inverse Fourier transform formula in Eq. (18.9) and
\n

$$i_o(t) = \mathcal{F}^{-1}[I_o(\omega)] = \frac{1}{2\pi} \int_{-\infty}^{\infty} \frac{10\pi\omega[\delta(\omega - 2) - \delta(\omega + 2)]}{1 + j\omega^2} e^{j\omega t} d\omega$$

We apply the sifting property of the impulse function, namely,

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

or

$$\int_{-\infty}^{\infty} \delta(\omega - \omega_0) f(\omega) \, d\omega = f(\omega_0)$$

and obtain

$$i_o(t) = \frac{10\pi}{2\pi} \left[ \frac{2}{1+j6} e^{j2t} - \frac{-2}{1-j6} e^{-j2t} \right]$$

= $10 \left[ \frac{e^{j2t}}{6.082e^{j80.54^\circ}} + \frac{e^{-j2t}}{6.082e^{-j80.54^\circ}} \right]$
= $1.644[e^{j(2t-80.54^\circ)} + e^{-j(2t-80.54^\circ)}]$
= $3.288 \cos(2t - 80.54^\circ)$ A

Find the current io(t) in the circuit in Fig. 18.21, gi ven that is(t) = Practice Problem 18.8 50 cos 4t A.

Answer: 27.95 cos(4t + 26.57°) A.

Figure 18.21 For Practice Prob. 18.8.

18.5 Parseval’s Theorem

Parseval’s theorem demonstrates one practical use of the F ourier transform. It relates the ener gy carried by a signal to the F ourier transform of the signal. If p(t) is the power associated with the signal, the energy carried by the signal is

$$W = \int_{-\infty}^{\infty} p(t) dt$$

(18.57)

To be able to compare the energy content of current and voltage signals, it is convenient to use a 1-Ω resistor as the base for energy calculation. For a 1-Ω resistor, p(t) = v² (t) = i 2 (t) = f ² (t), where f(t) stands for either voltage or current. The energy delivered to the 1-Ω resistor is

$$W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) \, dt \tag{18.58}$$

Parseval’s theorem states that this same ener gy can be calculated in the frequency domain as

$$W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega$$

(18.59)

Parseval’s theorem states that the total energy delivered to a 1-Ω resistor equals the total area under the square of ^f (t) or 1∕2 π times the total area under the square of the magnitude of the Fourier transform of f (t).

Parseval’s theorem relates energy associated with a signal to its Fourier transform. It pro vides the ph ysical significance of F(ω), namely, that ∣F(ω)∣ 2 is a measure of the ener gy density (in joules per hertz) corre sponding to f (t).

To deri ve Eq. (18.59), we be gin with Eq. (18.58) and substitute Eq. (18.9) for one of the f(t)‘s. We obtain

$$W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) dt = \int_{-\infty}^{\infty} f(t) \left[ \frac{1}{2\pi} \int_{-\infty}^{\infty} F(\omega) e^{j\omega t} d\omega \right] dt \qquad (18.60)$$

The function f(t) can be mo ved inside the inte gral within the brack ets, since the integral does not involve time:

$$W_{1\Omega} = \frac{1}{2\pi} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(t) F(\omega) e^{j\omega t} d\omega dt$$

(18.61)

Reversing the order of integration,

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

(18.62)

In fact, ∣F(ω)∣ 2 is sometimes known as the energy spectral density of signal f (t). But if z = x + jy, zz* = (x + jy)(x − jy) = x² + y² = ∣z∣ 2 . Hence,

$$W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega$$

(18.63)

as expected. Equation (18.63) indicates that the energy carried by a signal can be found by integrating either the square of f(t) in the time domain or 1 ∕2π times the square of F(ω) in the frequency domain.

Because ∣F(ω)∣ 2 is an even function, we may integrate from 0 to ∞ and double the result; that is,

$$W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) \, dt = \frac{1}{\pi} \int_{0}^{\infty} |F(\omega)|^2 \, d\omega \tag{18.64}$$

We may also calculate the energy in any frequency band ω¹ < ω< ω2 as

$$W_{1\Omega} = \frac{1}{\pi} \int_{\omega_1}^{\omega_2} |F(\omega)|^2 \, d\omega \tag{18.65}$$

Notice that Pa rseval’s theorem as stated here applies to nonpe riodic functions. Pa rseval’s theorem for periodic functions w as pre sented in Sections 17.5 and 17.6. As evident in Eq. (18.63), Parseval’s theorem shows that the ener gy associated with a nonperiodic signal is spread ove r the entire frequenc y spectrum, whereas the ener gy of a periodic signal is concentrated at the frequencies of its harmonic components.

The voltage across a 10- ^Ω resistor is v(t) = 5*^e* Example 18.9 ⁻3*^t u*(t) V. Find the total energy dissipated in the resistor.

Solution:

• 1. Define. The problem is well defined and clearly stated.
• 2. Present. We are given the voltage across the resistor for all time and are asked to find the energy dissipated by the resistor. We note that the voltage is zero for all time less than zero. Thus, we only need to consider the time from zero to infinity.
• 3. Alternative. There are basically two ways to find this answer. The first would be to find the answer in the time domain. We will use the second approach to find the answer using Fourier analysis.
• 4. Attempt. In the time domain,

$$W_{10\Omega} = 0.1 \int_{-\infty}^{\infty} f^2(t) dt = 0.1 \int_{0}^{\infty} 25e^{-6t} dt$$ $$= 2.5 \frac{e^{-6t}}{-6} \Big|_{0}^{\infty} = \frac{2.5}{6} = 416.7 \text{ mJ}$$

5. Evaluate. In the frequency domain,

$$F(\omega) = V(\omega) = \frac{5}{3 + j\omega}$$

so that

$$|F(\omega)|^2 = F(\omega)F(\omega)^* = \frac{25}{9 + \omega^2}$$

Hence, the energy dissipated is

$$W_{10\Omega} = \frac{0.1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega = \frac{0.1}{\pi} \int_{0}^{\infty} \frac{25}{9 + \omega^2} d\omega$$

= $\frac{2.5}{\pi} \left( \frac{1}{3} \tan^{-1} \frac{\omega}{3} \right) \Big|_{0}^{\infty} = \frac{2.5}{\pi} \left( \frac{1}{3} \right) \left( \frac{\pi}{2} \right) = \frac{2.5}{6} = 416.7 \text{ mJ}$

6. Satisfactory? We have satisf actorily solv ed the problem and can present the results as a solution to the problem.

Practice Problem 18.9 (a) Calculate the total ener gy absorbed by a 1- ^Ω resistor with i(t) = 10e⁻2∣t^∣ A in the time domain. (b) Repeat (a) in the frequency domain.

Answer: (a) 50 J, (b) 50 J.

Example 18.10 Calculate the fraction of the total ener gy dissipated by a 1-Ω resistor in the frequency band −10 < ω< 10 rad/s when the v oltage across it is v(t) = e⁻2*^t u*(t).

Solution:

Given that f(t) = v(t) = e⁻2*^t u*(t), then

$$F(\omega) = \frac{1}{2 + j\omega} \qquad \Rightarrow \qquad |F(\omega)|^2 = \frac{1}{4 + \omega^2}$$

The total energy dissipated by the resistor is

$$W_{1\Omega} = \frac{1}{\pi} \int_0^\infty |F(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^\infty \frac{d\omega}{4 + \omega^2}$$ $$= \frac{1}{\pi} \left( \frac{1}{2} \tan^{-1} \frac{\omega}{2} \Big|_0^\infty \right) = \frac{1}{\pi} \left( \frac{1}{2} \right) \frac{\pi}{2} = 0.25 \text{ J}$$

The energy in the frequencies −10 < ω< 10 rad/s is

$$W = \frac{1}{\pi} \int_0^{10} |F(\omega)|^2 d\omega = \frac{1}{\pi} \int_0^{10} \frac{d\omega}{4 + \omega^2} = \frac{1}{\pi} \left( \frac{1}{2} \tan^1 \frac{\omega}{2} \Big|_0^{10} \right)$$ $$= \frac{1}{2\pi} \tan^{-1} 5 = \frac{1}{2\pi} \left( \frac{78.69^\circ}{180^\circ} \pi \right) = 0.218 \text{ J}$$

Its percentage of the total energy is

$$\frac{W}{W_{1\Omega}} = \frac{0.218}{0.25} = 87.4$$

percent

A 2-Ω resistor has i(t) = 2e Practice Problem 18.10 ^−t u(t) A. What percentage of the total energy is in the frequency band −4 < ω< 4 rad/s?

Answer: 84.4 percent.

18.6 Comparing the Fourier and Laplace Transforms

It is worthwhile to take some moments to compare the Laplace and Fourier transforms. The following similarities and differences should be noted:

• 1. The Laplace transform defined in Chapter 15 is one-sided in that the integral is over 0 < t < ∞, making it only useful for positi ve-time functions, f(t), t > 0. The Fourier transform is applicable to func tions defined for all time.
• 2. For a function f(t) that is nonzero for positive time only (i.e.,

$$f(t) = 0, t < 0) \text{ and } \int_0^{\infty} |f(t)| \, dt < \infty \text{, the two transforms are related by}$$

\n

$$F(\omega) = F(s)|_{s=j\omega} \tag{18.66}$$

This equation also shows that the Fourier transform can be regarded as a special case of the Laplace transform with s = jω. Recall that s = σ + jω. Therefore, Eq. (18.66) sho ws that the Laplace trans form is related to the entire s plane, whereas the Fourier transform is restricted to the jω axis. See Fig. 15.1.

• 3. The Laplace transform is applicable to a wider range of functions than the F ourier transform. F or e xample, the function tu(t) has a Laplace transform b ut no F ourier transform. But F ourier trans forms exist for signals that are not physically realizable and have no Laplace transforms.
• 4. The Laplace transform is better suited for the analysis of transient problems involving initial conditions, because it permits the inclu sion of the initial conditions, whereas the F ourier transform does not. The Fourier transform is especially useful for problems in the steady state.
• 5. The Fourier transform pro vides greater insight into the frequenc y characteristics of signals than does the Laplace transform.

Some of the similarities and dif ferences can be observed by comparing Tables 15.1 and 15.2 with Tables 18.1 and 18.2.

In other words, if all the poles of F(s) lie in the left-hand side of the s plane, then one can obtain the Fourier transform F(ω) from the corresponding Laplace transform F(s) by merely replacing s by jω. Note that this is not the case, for example, for u(t) or cos atu(t).

18.7 Applications

Besides its usefulness for circuit analysis, the F ourier transform is used extensively in a variety of fields such as optics, spectroscopy, acoustics, computer science, and electrical engineering. In electrical engineering, it is applied in communications systems and signal processing, where fre quency response and frequency spectra are vital. Here we consider tw o simple applications: amplitude modulation (AM) and sampling.

18.7.1 Amplitude Modulation

Electromagnetic radiation or transmission of information through space has become an indispensable part of a modern technological society . However, transmission through space is only efficient and economical at high frequencies (above 20 kHz). To transmit intelligent signals such as for speech and music—contained in the lo w-frequency range of 50 Hz to 20 kHz is e xpensive; it requires a huge amount of po wer and large antennas. A common method of transm itting low-frequency audio information is to transmit a high-frequency signal, called a carrier, which is controlled in some w ay to correspond to the audio informa tion. Three characteristics (amplitude, frequency, or phase) of a carrier can be controlled so as to allow it to carry the intelligent signal, called the modulating signal. Here we will only consider the control of the carrier’s amplitude. This is known as amplitude modulation.

Amplitude modulation (AM) is a process whereby the amplitude of the carrier is controlled by the modulating signal.

AM is used in ordinary commercial radio bands and the video portion of commercial television.

Suppose the audio information, such as voice or music (or the modulating signal in general) to be transmitted is m(t) = Vm cos ωmt, while the high-frequency carrier is c(t) = Vc cos ωct, where ωc ≫ ωm. Then an AM signal f(t) is given by

$$f(t) = V_c \left[1 + m(t)\right] \cos \omega_c t \tag{18.67}$$

Figure 18.22 illustrates the modulating signal m(t), the carrier c(t), and the AM signal f(t). We can use the result in Eq. (18.27) together with the Fourier transform of the cosine function (see Example 18.1 or Table 18.1) to determine the spectrum of the AM signal:

$$F(\omega) = \mathcal{F}[V_c \cos \omega_c t] + \mathcal{F}[V_c m(t) \cos \omega_c t]$$

= $V_c \pi [\delta(\omega - \omega_c) + \delta(\omega + \omega_c)]$

• $\frac{V_c}{2} [M(\omega - \omega_c) + M(\omega + \omega_c)]$ (18.68)

where M(ω) is the F ourier transform of the modulating signal m(t). Shown in Fig. 18.23 is the frequenc y spectrum of the AM signal. Fig ure 18.23 indicates that the AM signal consists of the carrier and tw o other sinusoids. The sinusoid with frequenc y ωc − ωm is kno wn as the lower sideband, while the one with frequenc y ωc + ωm is known as the upper sideband.

Frequency spectrum of AM signal.

Notice that we ha ve assumed that the modulating signal is sinu soidal to mak e the analysis easy . In real life, m(t) is a nonsinusoidal, band-limited signal—its frequency spectrum is within the range between 0 and ωu = 2π fu (i.e., the signal has an upper frequency limit). Typically, fu = 5kHz for AM radio. If the frequenc y spectrum of the modulating signal is as sho wn in Fig. 18.24(a), then the frequenc y spectrum of the AM signal is sho wn in Fig. 18.24(b). Thus, to avoid any interference, carriers for AM radio stations are spaced 10 kHz apart.

At the recei ving end of the transmission, the audio informa tion is recovered from the modulated carrier by a process kno wn as demodulation.

Example 18.11 A music signal has frequency components from 15 Hz to 30 kHz. If this signal could be used to amplitude modulate a 1.2-MHz carrier , find the range of frequencies for the lower and upper sidebands.

Solution:

The lo wer sideband is the dif ference of the carrier and modulating frequencies. It will include the frequencies from

$$1,200,000 - 30,000 \text{ Hz} = 1,170,000 \text{ Hz}$$

to

$$1,200,000 - 15$$

Hz = 1,199,985 Hz

The upper sideband is the sum of the carrier and modulating frequencies. It will include the frequencies from

1,200,000 + 15 Hz = 1,200,015 Hz

to

$$1,200,000 + 30,000 \text{ Hz} = 1,230,000 \text{ Hz}$$

Practice Problem 18.11 If a 2-MHz carrier is modulated by a 4-kHz intelligent signal, determine the frequencies of the three components of the AM signal that results.

Answer: 2,004,000 Hz, 2,000,000 Hz, 1,996,000 Hz.

Figure 18.25

(a) Continuous (analog) signal to be sampled, (b) train of impulses, (c) sampled (digital) signal.

(c)

18.7.2 Sampling

In analog systems, signals are processed in their entirety . However, in modern digital systems, only samples of signals are required for pro cessing. This is possible as a result of the sampling theorem gi ven in Section 17.8.1. The sampling can be done by using a train of pulses or impulses. We will use impulse sampling here.

Consider the continuous signal g(t) shown in Fig. 18.25(a). This can be multiplied by a train of impulses δ(t − nTs) shown in Fig. 18.25(b), where Ts is the sampling interval and fs = 1∕Ts is the sampling frequency or the sampling rate. The sampled signal gs(t) is therefore

$$g_s(t) = g(t) \sum_{n=-\infty}^{\infty} \delta(t - nT_s) = \sum_{n=-\infty}^{\infty} g(nT_s) \delta(t - nT_s)$$

(18.69)

The Fourier transform of this is

$$G_{s}(\omega) = \sum_{n=-\infty}^{\infty} g(nT_{s}) \mathcal{F}[\delta(t - nT_{s})] = \sum_{n=-\infty}^{\infty} g(nT_{s})e^{-jn\omega T_{s}}$$

(18.70)

It can be shown that

$$\sum_{n=-\infty}^{\infty} g(nT_s)e^{-jn\omega T_s} = \frac{1}{T_s}\sum_{n=-\infty}^{\infty} G(\omega + n\omega_s)$$

(18.71)

where ωs = 2π∕Ts. Thus, Eq. (18.70) becomes

$$G_s(\omega) = \frac{1}{T_s} \sum_{n=-\infty}^{\infty} G(\omega + n\omega_s)$$

(18.72)

This shows that the F ourier transform Gs(ω) of the sampled signal is a sum of translates of the Fourier transform of the original signal at a rate of 1∕Ts.

To ensure optimum recovery of the original signal, what must be the sampling interval? This fundamental question in sampling is answered by an equivalent part of the sampling theorem:

A band-limited signal, with no frequency component higher than W hertz, may be completely recovered from its samples taken at a frequency at least twice as high as 2W samples per second.

In other words, for a signal with bandwidth W hertz, there is no loss of information or overlapping if the sampling frequency is at least twice the highest frequency in the modulating signal. Thus,

$$\frac{1}{T_s} = f_s \ge 2W\tag{18.73}$$

The sampling frequency fs = 2W is known as the Nyquist frequency or rate, and 1∕fs is the Nyquist interval.

A telephone signal with a cutoff frequency of 5 kHz is sampled at a rate Example 18.12 60 percent higher than the minimum allowed rate. Find the sampling rate.

Solution:

The minimum sample rate is the Nyquist rate = 2W = 2 × 5 = 10 kHz. Hence,

$$f_s = 1.60 \times 2W = 16 \text{ kHz}$$

An audio signal that is band-limited to 12.5 kHz is digitized into 8-bit Practice Problem 18.12 samples. What is the maximum sampling interv al that must be used to ensure complete recovery?

Answer: 40 μs.

18.8 Summary

1. The Fourier transform con verts a nonperiodic function f(t) into a transform F(ω), where

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

2. The inverse Fourier transform of F(ω) is

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

• 3. Important Fourier transform properties and pairs are summarized in Tables 18.1 and 18.2, respectively.
• 4. Using the F ourier transform method to analyze a circuit in volves finding the Fourier transform of the e xcitation, transforming the circuit element into the frequency domain, solving for the unkno wn response, and transforming the response to the time domain using the inverse Fourier transform.
• 5. If H(ω) is the transfer function of a network, then H(ω) is the Fourier transform of the network’s impulse response; that is,

$$H(\omega) = \mathcal{F}[h(t)]$$

The output Vo(ω) of the netw ork can be obtained from the input Vi(ω) using

$$V_o(\omega) = H(\omega)V_i(\omega)$$

6. Parseval’s theorem gives the energy relationship between a function f(t) and its Fourier transform F(ω). The 1-Ω energy is

$$W_{1\Omega} = \int_{-\infty}^{\infty} f^2(t) dt = \frac{1}{2\pi} \int_{-\infty}^{\infty} |F(\omega)|^2 d\omega$$

The theorem is useful in calculating energy carried by a signal either in the time domain or in the frequency domain.

7. Typical applications of the Fourier transform are found in amplitude modulation (AM) and sampling. F or AM application, a w ay of determining the sidebands in an amplitude-modulated wave is derived from the modulation property of the F ourier transform. F or sampling application, we found that no information is lost in sampling (required for digital transmission) if the sampling frequency is equal to at least twice the Nyquist rate.

Review Questions

18.1 Which of these functions does not have a Fourier transform?

(a) et	(b) te−3t
u(−t)	u(t)
(c) 1∕t	(d) ∣t∣u(t)

18.2 The Fourier transform of e^j2^t is:

(a)

$$\frac{1}{2 + j\omega}$$

\n(b) $\frac{1}{-2 + j\omega}$
\n(c) $2\pi\delta(\omega - 2)$
\n(d) $2\pi\delta(\omega + 2)$

18.3 The inverse Fourier transform of e⁻j^ω ______ ² + ^jω is (a) e⁻2*^t* (b) e⁻2*^t u*(t − 1) (c) e⁻2(t−1) (d) e⁻2(t−1)u(t − 1)

18.4 The inverse Fourier transform of δ(ω) is:

(a) δ(t) (b) u(t) (c) 1 (d) 1∕2π

18.5 The inverse Fourier transform of jω is:

(a) δʹ(t) (b) uʹ(t) (c) 1∕t (d) undefined

18.6 Evaluating the integral ∫ −∞ ∞ 10δ(ω) ______ 4 + ω² dω results in:

(a) 0 (b) 2 (c) 2.5 (d) ∞

18.7 The integral ∫ −∞ ∞ 10δ(ω− 1) __________ 4 + ω² dω gives: (a) 0 (b) 2 (c) 2.5 (d) ∞ 18.8 The current through an initially uncharged 1-F capacitor is δ(t) A. The voltage across the capacitor is:

(a) u(t) V	(b) −1∕2 + u(t) V
(c) e−t u(t) V	(d) δ(t) V

18.9 A unit step current is applied through a 1-H inductor. The voltage across the inductor is:

(a) u(t) V	(b) sgn(t) V
(c) e−t u(t) V	(d) δ(t) V

Problems

† Sections 18.2 and 18.3 Fourier Transform and its Properties

18.1 Obtain the Fourier transform of the function in Fig. 18.26.

Figure 18.26

For Prob. 18.1.

18.2 Using Fig. 18.27, design a problem to help other students better understand the Fourier transform given a wave shape.

Figure 18.27

For Prob. 18.2.

Figure 18.28 For Prob. 18.3.

18.10 Parseval’s theorem is only for nonperiodic functions.

(a) True (b) False

Answers: 18.1c, 18.2c, 18.3d, 18.4d, 18.5a, 18.6c, 18.7b, 18.8a, 18.9d, 18.10b

18.4 Find the Fourier transform of the waveform shown in Fig. 18.29.

† We have marked (with the MATLAB icon) the problems where we are asking the student to find the Fourier transform of a wave shape. We do this because you can use MATLAB to plot the results as a check.

18.7 Find the Fourier transforms of the signals in Fig. 18.32.

18.8 Obtain the Fourier transforms of the signals shown in Fig. 18.33.

Figure 18.33 For Prob. 18.8.

Figure 18.34 For Prob. 18.9.

Figure 18.35 For Prob. 18.10.

18.11 Find the Fourier transform of the “sine-wave pulse” shown in Fig. 18.36.

Figure 18.36 For Prob. 18.11.

18.12 Find the Fourier transform of the following signals.

(a) f1(t) = e⁻3*^t* sin(10t)u(t) (b) f2(t) = e⁻4*^t* cos(10t)u(t)

18.13 Find the Fourier transform of the following signals:

(a)

$$f(t) = \cos(at - \pi/3)$$

, $-\infty < t < \infty$
\n(b) $g(t) = u(t + 1) \sin \pi t$ , $-\infty < t < \infty$
\n(c) $h(t) = (1 + A \sin at) \cos bt$ , $-\infty < t < \infty$ ,
\nwhere A, a, and b are constants
\n(d) $i(t) = 1 - t$ , $0 < t < 4$

• 18.14 Design a problem to help other students better
  • understand finding the Fourier transform of a variety of time varying functions (do at least three).
• 18.15 Find the Fourier transforms of the following functions:

(a)

$$f(t) = \delta(t + 3) - \delta(t - 3)$$

\n(b) $f(t) = \int_{-\infty}^{\infty} 2\delta(t - 1) dt$
\n(c) $f(t) = \delta(3t) - \delta'(2t)$

18.16 Determine the Fourier transforms of these functions: *

)

(a)

$$f(t) = 8/t^2$$

(b) $g(t) = 4/(4 + t^2)$

18.17 Find the Fourier transforms of:

(a) 2 cos 2tu(t)

• (b) 0.5 sin 10tu(t)
• 18.18 Given that F(ω) = [ f(t)], prove the following results, using the definition of Fourier transform:

(a)

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

\n(b) $\mathcal{F}\left[\frac{df(t)}{dt}\right] = j\omega F(\omega)$
\n(c) $\mathcal{F}[f(-t)] = F(-\omega)$
\n(d) $\mathcal{F}[tf(t)] = j\frac{d}{d\omega} F(\omega)$

18.19 Find the Fourier transform of

$$f(t) = 2 \cos 2\pi t [u(t) - u(t-1)]$$

18.20 (a) Show that a periodic signal with exponential Fourier series

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

has the Fourier transform

$$F(\omega) = \sum_{n = -\infty}^{\infty} c_n \delta(\omega - n\omega_0)$$

where $\omega_0 = 2\pi/T$ .

Figure 18.37

For Prob. 18.20(b).

18.21 Show that

$$\int_{-\infty}^{\infty} \left( \frac{\sin a\omega}{a\omega} \right)^2 d\omega = \frac{\pi}{a}$$

Hint: Use the fact that

$$\mathcal{F}[u(t+a) - u(t-a)] = 2a\left(\frac{\sin a\omega}{a\omega}\right).$$

18.22 Prove that if F(ω) is the Fourier transform of f(t),

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

18.23 If the Fourier transform of f(t) is

transform of

$$f(t)$$

is
\n

$$F(\omega) = \frac{10}{(2 + j\omega)(5 + j\omega)}$$

determine the transforms of the following:

(a)

$$f(-3t)$$

(b) $f(2t - 1)$ (c) $f(t) \cos 2t$
(d) $\frac{d}{dt}f(t)$ (e) $\int_{-\infty}^{t} f(t) dt$

−∞ 18.24 Given that [ f(t)t] = ( j∕ω)(e⁻j^ω − 1), find the Fourier transforms of:

(a)

$$x(t) = f(t) + 3$$

\n(b) $y(t) = f(t - 2)$
\n(c) $h(t) = f'(t)$
\n(d) $g(t) = 4f(\frac{2}{3}t) + 10f(\frac{5}{3}t)$

18.25 Obtain the inverse Fourier transform of the following signals.

(a)

$$G(\omega) = \frac{5}{j\omega - 2}$$

\n(b) $H(\omega) = \frac{12}{\omega^2 + 4}$
\n(c) $X(\omega) = \frac{10}{(j\omega - 1)(j\omega - 2)}$

18.26 Determine the inverse Fourier transforms of the following:

(a)

$$F(\omega) = \frac{e^{-j2\omega}}{1 + j\omega}$$

\n(b) $H(\omega) = \frac{1}{(j\omega + 4)^2}$
\n(c) $G(\omega) = 2u(\omega + 1) - 2u(\omega - 1)$

^* An asterisk indicates a challenging problem. (c) G(ω) = 2u(ω + 1) − 2u(ω− 1)

18.27 Find the inverse Fourier transforms of the following functions:

(a)

$$F(\omega) = \frac{100}{j\omega(j\omega + 10)}$$

\n(b) $G(\omega) = \frac{10 j\omega}{(-j\omega + 2)(j\omega + 3)}$
\n(c) $H(\omega) = \frac{60}{-\omega^2 + j40\omega + 1300}$
\n(d) $Y(\omega) = \frac{\delta(\omega)}{(j\omega + 1)(j\omega + 2)}$

18.28 Find the inverse Fourier transforms of:

Find the inverse Fourier t
\n(a)

$$\frac{n\delta(\omega)}{(5+j\omega)(2+j\omega)}$$

\n(b) $\frac{10\delta(\omega+2)}{j\omega(j\omega+1)}$
\n(c) $\frac{20\delta(\omega-1)}{(2+j\omega)(3+j\omega)}$
\n(d) $\frac{5n\delta(\omega)}{5+j\omega} + \frac{5}{j\omega(5+j\omega)}$

• 18.29 Determine the inverse Fourier transforms of: *
  • (a) F(ω) = 4δ(ω + 3) + δ(ω) + 4δ(ω− 3)
  • (b) G(ω) = 4u(ω + 2) − 4u(ω− 2)
  • (c) H(ω) = 6 cos 2ω
• 18.30 For a linear system with input x(t) and output y(t), find the impulse response for the following cases:

(a)

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

, $y(t) = u(t) - u(-t)$
\n(b) $x(t) = e^{-t}u(t)$ , $y(t) = e^{-2t}u(t)$
\n(c) $x(t) = \delta(t)$ , $y(t) = e^{-at} \sin btu(t)$

18.31 Given a linear system with output y(t) and impulse response h(t), find the corresponding input x(t) for the following cases:

(a)

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

, $h(t) = e^{-at}u(t)$
\n(b) $y(t) = u(t+1) - u(t-1)$ , $h(t) = \delta(t)$
\n(c) $y(t) = e^{-at}u(t)$ , $h(t) = \text{sgn}(t)$

18.32 Determine the functions corresponding to the following Fourier transforms: *

(a)

$$F_1(\omega) = \frac{e^{j\omega}}{-j\omega + 1}$$

\n(b) $F_2(\omega) = 2e^{|\omega|}$
\n(c) $F_3(\omega) = \frac{1}{(1 + \omega^2)^2}$
\n(d) $F_4(\omega) = \frac{\delta(\omega)}{1 + j2\omega}$

18.33 Find f(t) if: *

(a)

$$F(\omega) = 2 \sin \pi \omega [u(\omega + 1) - u(\omega - 1)]$$

(b) $F(\omega) = \frac{1}{\omega} (\sin 2\omega - \sin \omega) + \frac{j}{\omega} (\cos 2\omega - \cos \omega)$

18.34 Determine the signal f(t) whose Fourier transform is shown in Fig. 18.38. (Hint: Use the duality property.)

Figure 18.38

For Prob. 18.34.

18.35 A signal f(t) has Fourier transform

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

Determine the Fourier transform of the following signals:

(a)

$$x(t) = f(3t - 1)$$

\n(b) $y(t) = f(t) \cos 5t$
\n(c) $z(t) = \frac{d}{dt}f(t)$
\n(d) $h(t) = f(t) * f(t)$
\n(e) $i(t) = tf(t)$