Question 8.3: Consider the two sinusoids x1(t) = cos(Ω0t), x2(t) = cos((Ω0......
Consider the two sinusoids
x_1(t) = \cos(Ω_0t),\quad x_2(t) = \cos((Ω_0 + Ω_1)t)\qquad −∞ ≤ t ≤ ∞
where Ω_1 > 2Ω_0. Show that if we sample these signals using T_s = 2π/Ω_1, we cannot differentiate the sampled signals, i.e. x_1(nT_s) = x_2(nT_s). Let Ω_0 = 1\text{ and }Ω_1 = 7 rad/s, use MATLAB to show the above graphically. Explain the significance of this.
Script File
Sampling the two signals using T_s = 2π/Ω_1 we have
x_{1}(n T_{s})=\cos(\Omega_{0}n T_{s}),\quad x_{2}(n T_{s})=\cos((\Omega_{0}+\Omega_{1})n T_{s})\qquad-\infty\leq n\leq\infty,
but since Ω_1T_s = 2π the sinusoid x_2(nT_s) can be written
\begin{array}{l l l}{{x_{2}(n T_{s})}}&{{=}}&{{\cos((\Omega_{0}T_{s}+2\pi)n)}}\\ {{}}&{{=}}&{{\cos(\Omega_{0}T_{s}n)=x_{1}(n T_{s}),}}\end{array}
making it impossible to differentiate the two sampled signals. Since the sampling frequency Ω_1 is smaller than twice the maximum frequency of x_2(t)\text{ or }2(Ω_0 + Ω_1) the sampled version of this signal is aliased. See Fig. 8.6 for Ω_0 = 1\text{ and }Ω_0 + Ω_1 = 8.
The following script shows the aliasing effect when Ω_0 = 1\text{ and }Ω_1 = 7 rad/s. Notice that x_1(t) is sampled satisfying the Nyquits sampling rate condition (Ω_s = Ω_1 = 7 > 2Ω_0 = 2 (rad/s)), while x_2(t)\text{ is not }(Ω_s = Ω_1 = 7 < 2(Ω_0 + Ω_1) = 16 rad/s).
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% Example 8.3
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clear all; clf
omega_0=1;omega_s=7;
T=2*pi/omega_0; t=0:0.001:T; % a period of x_1
x1=cos(omega_0*t); x2=cos((omega_0+omega_s)*t);
N=length(t); Ts=2*pi/omega_s; % sampling period
M=fix(Ts/0.001); imp=zeros(1,N);
for k=1:M:N-1.
imp(k)=1;
end
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
legend(’x_1(t)’,’x_2(t)’,’x_1(nTs)’); xlabel(’t’);
ylabel(’x_1(t), x_2(t), x_1(nTs)’)
The result in the frequency domain is shown in Fig. 8.7: the spectra of the two sinusoids are different but the spectra of the sampled signals are identical.
