Signals and Systems - Electrical Engineering

7.2 Uniform Sampling 431

Show that if we sample these signals usingTs= 2 π/s, we cannot differentiate the sampled signals

(i.e.,x 1 (nTs)=x 2 (nTs)). Use MATLAB to show the above graphically when 0 =1 ands=7.

Explain the significance of this.

Solution

Sampling the two signals usingTs= 2 π/s, we have

x 1 (nTs)=cos( 0 nTs) −∞≤n≤∞

x 2 (nTs)=cos(( 0 +s)nTs) −∞≤n≤∞

but sincesTs= 2 π, the sinusoidx 2 (nTs)can be written as

x 2 (nTs)=cos(( 0 Ts+ 2 π)n)

=cos( 0 Tsn)=x 1 (nTs)

The following script shows the aliasing effect when 0 =1 ands=7 rad/sec. Notice thatx 1 (t)is

sampled satisfying the Nyquist sampling rate condition (s= 7 > 2  0 =2 rad/sec), whilex 2 (t)is

not (s= 7 < 2 ( 0 +s)=16 rad/sec).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

% Example 7.3 ---Two sinusoids of different frequencies being sampled

% with same sampling period -- aliasing for signal with higher frequency

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

clear all; clf

% sinusoids

omega0 = 1;omegas = 7;

T = 2∗pi/omega0; t = 0:0.001:T; % a period of x1

x1 = cos(omega 0 ∗t); x2 = cos((omega0 + omegas)∗t);

N = length(t); Ts = 2∗pi/omegas; % sampling period

M = fix(Ts/0.001); imp = zeros(1,N);

for k = 1:M:N−1.

imp(k) = 1; % sequence of impulses

end

xs = imp.∗x1; % sampled signal

plot(t,x1,’b’,t,x2,’k’); hold on

stem(t,imp.∗x1,’r’,’filled’);axis([0 max(t)−1.1 1.1]); xlabel(’t’); grid

Figure 7.5 shows the two sinusoids and the sampled signal that coincides for the two signals. The
result in the frequency domain is shown in Figure 7.6: The spectra of the two sinusoids are different
but the spectra of the sampled signals are identical. n