Example of Transformation of the independent variable:

Example: Provided that x(t) = e^-5t u(t) is sampled at 50 Hz, Search an expression for x(n). Plot x(t), x(n) and x(2n). Sketch the spectrum of x(n).

Solution: The sampling time is T = 0.02 sec. Reproducing t with nT we get x(nT) = e^-5nt u(nT ) , or 

We illustrate below three plots: (1) The continuous-time signal x(t), (2) The sampled (at 50 Hz) version x(n), and (3) x(2n), the 2-fold down-sampled version of x(n); that is similar to sampling x(t) at 25 Hz.

t = 0 : 1/512: 1; xt = exp (-5*t); %x(t) evaluated at 512 points

subplot(3, 1, 1), plot(t, xt); legend ('x(t) = exp(-5t)');

xlabel ('time, sec.'), ylabel('x(t)'); grid; title ('x(t) - Continuous-time')

%

t1 = 0 : 0.02: 1; xn = exp (-5*t1); %Sampled at 50 Hz.

subplot(3, 1, 2), stem(t1, xn); legend ('x(n) at 50 Hz');

xlabel ('time, sec.'), ylabel('x(n)'); grid; title ('x(nT) at T = 0.02 sec')

%

t2 = 0 : 0.04: 1; xt2 = exp (-5*t2); %Sampled at 25 Hz

subplot(3, 1, 3), stem(t2, xt2); legend ('2-fold down-sampled');

xlabel ('time, sec.'), ylabel('x(2n)'); grid; title ('x(nT) at T = 0.04 sec.')

Note that X(s) = L (e^-5t u(t)) = 1 (s + 5) . Defines below is the MATLAB plot of the magnitude spectrum |X(jΩ)| of the continuous-time signal x(t) taking the function plot. Omega is a vector, consequently we need "./" instead of "/" etc. The main point to be prepared here is that X(jΩ) expands asymptotically to ∞, so, strictly speaking, x(t) is not band-limited. Consequently, the spectrum X(ω) of the sampled signal x(n) has some inbuilt aliasing.

t = 0 : 1/512: 1; xt = exp (-5*t); %x(t) evaluated at 512 points

subplot(3, 1, 1), plot(t, xt); legend ('x(t) = exp(-5t)');

xlabel ('time'), ylabel('x(t)'); grid; title ('x(t) - Continuous-time')

%

Omega = -6*pi: pi/256: 6*pi; X = 1./(5.+ j .*Omega);

subplot(3, 1, 2), plot(Omega, abs(X), 'k'); legend ('Spectrum of x(t)');

xlabel ('Omega, rad/sec'), ylabel('|X(Omega)|'); grid; title ('Magnitude')%

subplot(3, 1, 3), plot(Omega, angle(X), 'k'); legend ('Spectrum of x(t)');

xlabel ('Omega, rad/sec'), ylabel('Phase of X(Omega)'); grid; title ('Phase')

Coming to the discrete-time signal, the spectrum of x(n) = aⁿ u(n) = (0.905)ⁿ u(n) is its DTFT

The MATLAB segment is

t1 = 0 : 0.02: 1; xn = exp (-5*t1); %Sampled at 50 Hz.

subplot(3, 1, 1), stem(t1, xn); legend ('x(n) at 50 Hz');

xlabel ('time, sec.'), ylabel('x(n)'); grid; title ('x(nT) at T = 0.02 sec')

%

b = [1]; %Numerator coefficient

a = [1, -0.905]; %Denominator coefficients

w = -6*pi: pi/256: 6*pi; [Xw] = freqz(b, a, w);

subplot(3, 1, 2), plot(w, abs(Xw)); legend ('Spectrum of x(n)');

xlabel('Frequency \omega, rad/sample'), ylabel('Magnitude of X(\omega)'); grid

subplot(3, 1, 3), plot(w, angle(Xw)); legend ('Spectrum of x(n)');

xlabel('Frequency \omega, rad/sample'), ylabel('Phase of X(\omega)'); grid

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