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104<br />

Signals and Spectra Chap. 2<br />

We approximate this integral by using a finite series, where t = k ¢t, f 0 = 1/T, dt =¢t, and<br />

¢t = T/N. Then<br />

c n L 1 N-1<br />

T a w(k ¢t)e -j(2p/N)nk ¢t<br />

k=0<br />

(2–185)<br />

Using Eq. (2–176), we find that the Fourier series coefficient is related to the DFT by<br />

c n L 1 N X(n)<br />

(2–186)<br />

The DFT returns X(n) values for n = 0, 1, Á , N - 1. Consequently, Eq. (2–186) must be<br />

modified to give c n values for negative n. For positive n, we use<br />

c n = 1 N X(n), 0 … n 6 N 2<br />

(2–187a)<br />

and for negative n we use<br />

c n = 1 N X(N + n), - N 2 6 n 6 0<br />

(2–187b)<br />

Example 2–21 USE THE FFT TO COMPUTE THE SPECTRUM OF A SINUSOID<br />

Let<br />

w(t) = 3 sin (v 0 t + 20°)<br />

where v 0 = 2pf 0 and f 0 = 10 Hz.<br />

Because w(t) is periodic, Eq. (2–109) is used to obtain the spectrum<br />

(2–188)<br />

W(f) =© c n d(f - nf 0 )<br />

where {c n } are the complex Fourier series coefficients for w(t). Furthermore, because<br />

sin(x) = (e jx - e -jx )/(2j),<br />

3 sin(v 0 t + 20°) = a 3 2j ej20 be jv 0t + a -3<br />

2j e-j20 be -jv 0t<br />

Consequently, the FS coefficients are known to be<br />

c 1 = a 3 2j ej20 b = 1.5l -70°<br />

(2–189a)<br />

c 1 = a -3<br />

(2–189b)<br />

2j e-j20 b = 1.5l +70°<br />

and the other c n ’s are zero. Now see if this known correct answer can be computed by using the<br />

DFT. See Example2_21.m for the solution. Referring to Fig. 2–22, we observe that MATLAB computes<br />

the FFT and plots the spectrum. The computed result checks with the known analytical result.

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