852 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

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852 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

we select the frequency range 0 ≤ < 2π (or 0 ≤ r < 20), we obtain the Fourier series as

$$x[n] = \sin 0.1\pi n = \frac{1}{2j} (e^{j0.1\pi n} - e^{j1.9\pi n})$$

This series is equivalent to that in Eq. (9.8) because the two exponentials ej1.9π*ⁿ* and e−j0.1π*ⁿ* are equivalent. This follows from the fact that ej1.9π*ⁿ* = ej1.9π*ⁿ* ×e−j2π*ⁿ* = e−j0.1π*ⁿ*.

We could have selected the spectrum over any other range of width = 2π in Figs. 9.1b and 9.1c as a valid discrete-time Fourier series. The reader may verify this by proving that such a spectrum starting anywhere (and of width = 2π) is equivalent to the same two components on the right-hand side of Eq. (9.8).

DR ILL 9.1 DTFS Spectra on Alternate Intervals

From the spectra in Fig. 9.1, write the Fourier series corresponding to the interval −10 ≥ r > −30 (or −π ≥ > −3π). Show that this Fourier is equivalent to that in Eq. (9.8).

DR ILL 9.2 Discrete-Time Fourier Series of a Sum of Sinusoids

Find the period and the DTFS for

x[n] = 4 cos 0.2πn+6 sin 0.5πn

over the interval 0 ≤ r ≤ 19. Use Eq. (9.4) to compute Dr.

ANSWERS

N⁰ = 20 and x[n] = 2ej0.2π*ⁿ* +(3e−jπ/²)ej0.5π*ⁿ* +(3ejπ/²)ej1.5π*ⁿ* +2ej1.8π*ⁿ*

DR ILL 9.3 Fundamental Period of Discrete-Time Sinusoids

Find the fundamental periods N0, if any, for: (a) sin(301πn/4) and (b) cos 1.3n.

ANSWERS

(a) N⁰ = 8, (b) N⁰ does not exist because the sinusoid is not periodic.

Compute and plot the discrete-time Fourier series for the periodic sampled gate function shown in Fig. 9.2a.

Figure 9.2 (a) Periodic sampled gate pulse and (b) its Fourier spectrum.

In this case, N⁰ = 32 and ⁰ = 2π/32 = π/16. Therefore,

$$x[n] = \sum_{r = \langle 32 \rangle} \mathcal{D}_r e^{jr(\pi/16)n}$$

where

$$\mathcal{D}_r = \frac{1}{32} \sum_{n=(32)} x[n] e^{-jr(\pi/16)n}$$

For our convenience, we shall choose the interval −16 ≤ n ≤ 15 for this summation, although any other interval of the same width (32 points) would give the same result.†

$$\mathcal{D}_r = \frac{1}{32} \sum_{n=-16}^{15} x[n] e^{-j r (\pi/16)n}$$

^† In this example we have used the same equations as those for the DFT in Ex. 8.9, within a scaling constant. In the present example, the values of x[n] at n = 4 and −4 are taken as 1 (full value), whereas in Ex. 8.9 these values are 0.5 (half the value). This is the reason for the slight difference in spectra in Figs. 9.2b and 8.19d. Unlike continuous-time signals, discontinuity is a meaningless concept in discrete-time signals.

854 CHAPTER 9 FOURIER ANALYSIS OF DISCRETE-TIME SIGNALS

Now, x[n] = 1 for −4 ≤ n ≤ 4 and is zero for all other values of n. Therefore,

$$\mathcal{D}_r = \frac{1}{32} \sum_{n=-4}^{4} e^{-jr(\pi/16)n} \tag{9.9}$$

This is a geometric progression with a common ratio e−j(π/16)^r . Therefore (see Sec. B.8-3),†

$$\mathcal{D}_r = \frac{1}{32} \left[ \frac{e^{-j(5\pi r/16)} - e^{j(4\pi r/16)}}{e^{-j(\pi r/16)} - 1} \right]$$

\n

$$= \left(\frac{1}{32}\right) \frac{e^{-j(0.5\pi r/16)} \left[e^{-j(4.5\pi r/16)} - e^{j(4.5\pi r/16)}\right]}{e^{-j(0.5\pi r/16)} \left[e^{-j(0.5\pi r/16)} - e^{j(0.5\pi r/16)}\right]}$$

\n

$$= \left(\frac{1}{32}\right) \frac{\sin\left(\frac{4.5\pi r}{16}\right)}{\sin\left(\frac{0.5\pi r}{16}\right)}$$

\n

$$= \left(\frac{1}{32}\right) \frac{\sin(4.5r\Omega_0)}{\sin(0.5r\Omega_0)} \qquad \Omega_0 = \frac{\pi}{16}$$

(9.10)

This spectrum (with its periodic extension) is depicted in Fig. 9.2b.

DISCRETE-TIME FOURIER SERIES USING MATLAB

Let us confirm our results by using MATLAB to directly compute the DTFS according to Eq. (9.4).

>> N_0 = 32; n = (0:N_0-1); Omega_0 = 2*pi/N_0;
>> x_n = [ones(1,5) zeros(1,23) ones(1,4)];
>> for r = 0:N_0-1,
>> X_r(r+1) = sum(x_n.*exp(-j*r*Omega_0*n))/N_0;
>> end
>> r = n; stem(r,real(X_r),'k.');
>> xlabel('r'); ylabel('X_r'); axis([0 31 -.1 0.3]);

The MATLAB result, shown in Fig. 9.3, matches Fig. 9.2b. Alternatively, scaling the FFT by N⁰ produces the exact same result (Fig. 9.3).

X_r = fft(x_n)/N_0; stem(r,real(X_r),‘k.’); >> xlabel(‘r’); ylabel(‘X_r’); axis([0 31 -.1 0.3]);

$$\frac{1}{32} \sum_{n=-4}^{4} x[n] = \frac{9}{32}$$

Fortunately, the value of D0, as computed from Eq. (9.10), also happens to be 9/32. Hence, Eq. (9.10) is valid for all r.

^† Strictly speaking, the geometric progression sum formula applies only if the common ratio

^e−j(π/16)^r = 1. When ^{r =} 0, this ratio is unity. Hence, Eq. (9.10) is valid for values of ^r = 0. For the case r = 0, the sum in Eq. (9.9) is given by