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Sampling and Reconstruction of Analog Signals 85
which is a real-valued function since x a(t) isareal and even signal. To evaluate
X a(jΩ) numerically, we have to first approximate x a(t) byafinite-duration
grid sequence x G(m). Using the approximation e −5 ≈ 0, we note that x a(t)
can be approximated by a finite-duration signal over −0.005 ≤ t ≤ 0.005 (or
equivalently, over [−5, 5] msec). Similarly from (3.36), X a(jΩ) ≈ 0 for Ω ≥
2π (2000). Hence choosing
∆t =5× 10 −5 ≪ 1 =25× 10−5
2(2000)
we can obtain x G(m) and then implement (3.35) in MATLAB.
% Analog Signal
>> Dt = 0.00005; t = -0.005:Dt:0.005; xa = exp(-1000*abs(t));
% Continuous-time Fourier Transform
>>Wmax = 2*pi*2000; K = 500; k = 0:1:K; W = k*Wmax/K;
>>Xa = xa * exp(-j*t’*W) * Dt; Xa = real(Xa);
>>W = [-fliplr(W), W(2:501)]; % Omega from -Wmax to Wmax
>>Xa = [fliplr(Xa), Xa(2:501)]; % Xa over -Wmax to Wmax interval
>>subplot(2,1,1);plot(t*1000,xa);
>>xlabel(’t in msec.’); ylabel(’xa(t)’)
>>title(’Analog Signal’)
>>subplot(2,1,2);plot(W/(2*pi*1000),Xa*1000);
>>xlabel(’Frequency in KHz’); ylabel(’Xa(jW)*1000’)
>>title(’Continuous-time Fourier Transform’)
Figure 3.11 shows the plots of x a(t) and X a(jΩ). Note that to reduce the number
of computations, we computed X a(jΩ) over [0, 4000π] rad/sec (or equivalently,
over [0, 2] KHz) and then duplicated it over [−4000π, 0] for plotting purposes.
The displayed plot of X a(jΩ) agrees with (3.36).
□
□ EXAMPLE 3.19 To study the effect of sampling on the frequency-domain quantities, we will
sample x a(t) inExample 3.18 at 2 different sampling frequencies.
a. Sample x a(t) atF s = 5000 sam/sec to obtain x 1(n). Determine and plot
X 1(e jω ).
b. Sample x a(t) atF s = 1000 sam/sec to obtain x 2(n). Determine and plot
X 2(e jω ).
Solution
a. Since the bandwidth of x a(t) is2KHz, the Nyquist rate is 4000 sam/sec,
which is less than the given F s. Therefore aliasing will be (almost) nonexistent.
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