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

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

The Fourier series of a periodic signal x(t) with period T⁰ is given by

$$x(t) = \sum_{n = -\infty}^{\infty} D_n e^{jn\omega_0 t} \qquad \omega_0 = \frac{2\pi}{T_0}$$

Taking the Fourier transform of both sides, we obtain†

$$X(\omega) = 2\pi \sum_{n=-\infty}^{\infty} D_n \delta(\omega - n\omega_0)$$

(7.22)

Figure 7.14 (a) The uniform impulse train and (b) its Fourier transform.

As shown in Eq. (6.24) from Ex. 6.9, the Fourier coefficients Dn for δ*^T0 (t) are constant Dn = 1/T0. From Eq. (7.22), the Fourier transform of δ^T*⁰ (t) is therefore

$$X(\omega) = \frac{2\pi}{T_0} \sum_{n=-\infty}^{\infty} \delta(\omega - n\omega_0) = \omega_0 \delta_{\omega_0}(\omega), \quad \text{where } \omega_0 = \frac{2\pi}{T_0}$$

The corresponding spectrum is shown in Fig. 7.14b.

^† We assume here that the linearity property can be extended to an infinite sum.

EXAMPLE 7.9 Fourier Transform of the Unit Step Function

Find the Fourier transform of the unit step function u(t).

Trying to find the Fourier transform of u(t) by direct integration leads to an indeterminate result because

$$U(\omega) = \int_{-\infty}^{\infty} u(t)e^{-j\omega t}dt = \int_{0}^{\infty} e^{-j\omega t}dt = \left. \frac{-1}{j\omega}e^{-j\omega t} \right|_{0}^{\infty}$$

The upper limit of e−jω*^t* as t→∞ yields an indeterminate answer. So we approach this problem by considering u(t) to be a decaying exponential e−atu(t) in the limit as a → 0 (Fig. 7.15a). Thus,

$$u(t) = \lim_{a \to 0} e^{-at} u(t)$$

and

$$U(\omega) = \lim_{a \to 0} \mathcal{F}\lbrace e^{-at} u(t) \rbrace = \lim_{a \to 0} \frac{1}{a + j\omega}$$

Expressing the right-hand side in terms of its real and imaginary parts yields

$$U(\omega) = \lim_{a \to 0} \left[ \frac{a}{a^2 + \omega^2} - j \frac{\omega}{a^2 + \omega^2} \right] = \lim_{a \to 0} \left[ \frac{a}{a^2 + \omega^2} \right] + \frac{1}{j\omega}$$

Figure 7.15 Derivation of the Fourier transform of the step function.

The function a/(a² + ω²) has interesting properties. First, the area under this function (Fig. 7.15b) is π regardless of the value of a:

$$\int_{-\infty}^{\infty} \frac{a}{a^2 + \omega^2} \, d\omega = \tan^{-1} \frac{\omega}{a} \bigg|_{-\infty}^{\infty} = \pi$$

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

Second, when a → 0, this function approaches zero for all ω = 0, and all its area (π) is concentrated at a single point ω = 0. Clearly, as a → 0, this function approaches an impulse of strength π. Thus,

$$U(\omega) = \pi \delta(\omega) + \frac{1}{j\omega} \tag{7.23}$$

Note that u(t) is not a “true” dc signal because it is not constant over the interval −∞ to ∞. To synthesize “true” dc, we require only one everlasting exponential with ω = 0 (impulse at ω = 0). The signal u(t) has a jump discontinuity at t = 0. It is impossible to synthesize such a signal with a single everlasting exponential ejω*^t* . To synthesize this signal from everlasting exponentials, we need, in addition to an impulse at ω = 0, all the frequency components, as indicated by the term 1/jω in Eq. (7.23).

EXAMPLE 7.10 Fourier Transform of the Sign Function

Find the Fourier transform of the sign function sgn(t) [pronounced signum (t)], depicted in Fig. 7.16.

t sgn (t) 1 0 -¹ Figure 7.16 The signum function sgn(t). Observe that sgn(t)+1 = 2u(t) ⇒ sgn(t) = 2u(t)−1 Using Eqs. (7.20) and (7.23) and the linearity property, we obtain sgn(t) ⇐⇒ 2 jω

Table 7.1 provides many common Fourier transform pairs.

No.	x(t)	X(ω)
1	e−atu(t)	1 a+jω	a > 0
2	eatu(−t)	1 a−jω	a > 0
3	e−a t	2a a2 +ω2	a > 0
4	te−atu(t)	1 (a+jω)2	a > 0
5	ne−atu(t) t	n! (a+jω)n+1	a > 0
6	δ(t)	1
7	1	2πδ(ω)
8	ejω0t	2πδ(ω −ω0)
9	cos ω0t	π[δ(ω −ω0) +δ(ω +ω0)]
10	sin ω0t	jπ[δ(ω +ω0)−δ(ω −ω0)]
11	u(t)	1 πδ(ω)+ jω
12	sgnt	2 jω
13	cos ω0t u(t)	π jω 2 [δ(ω −ω0) +δ(ω +ω0)] + ω2 0 −ω2
14	sin ω0t u(t)	π ω0 [δ(ω −ω0)−δ(ω +ω0)] + ω2 2j 0 −ω2
15	e−atsin ω0t u(t)	ω0 (a+jω)2 +ω2 0	a > 0
16	e−at cos ω0t u(t)	a+jω (a+jω)2 +ω2 0	a > 0
17	rect t τ	τ sincωτ 2
18	W π sinc(Wt)	rect ω 2W
19	t τ	τ ωτ sinc2 2 4
20	Wt W 2π sinc2 2	ω 2W
21	“∞ δ(t −nT)	“∞ ω0 δ(ω −nω0)	2π ω0 = T
22	n=−∞ 2/2σ2 e−t	n=−∞ √2πe−σ2ω2/2 σ

TABLE 7.1 Select Fourier Transform Pairs

DR ILL 7.2 Inverse Fourier Transform of a Rectangular Pulse

Show that the inverse Fourier transform of X(ω) illustrated in Fig. 7.17 is x(t) = (ω0/π )sinc (ω0t). Sketch x(t).

DR ILL 7.3 Fourier Transform of a General Sinusoid

Show that cos(ω0t +θ ) ⇐⇒ π[δ(ω +ω0)e−j^θ +δ(ω −ω0)ej^θ ].