Question

Consider the continuous time signal: 2. , π (sin (2t) (Sin (8t) A discrete time signal x[n] -xs(t) -x(nTs) is created by sampling x() with sampling interval, 2it 60 a) Plot the Fourier Transform of the sampled signal, i.e. Xs (jo). b) Plot the DTFT of the sampled signal, ie, X(eja) o) Repeat (a) with 7, 2π d) Repeat (b) with , 18

Answer 1

close all,
clear all,
clc,

Ts = (2*pi)/60;
Fs=1/Ts;
w = -2*pi:Ts:2*pi;
t =-1:Ts:1;

for r=1:length(t)
x_t(r) = Ts*(pi/2)*(sin(2*t(r))/(pi*t(r)))*(sin(8*t(r))/(pi*t(r)));
end
subplot(2,1,1); plot(x_t); title('Original Signal x_t = Ts*(pi/2)*(sin(2*t)/(pi*t))*(sin(8*t)/(pi*t)), Ts = (2*pi)/60');

L=length(x_t);
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(x_t,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2);

% Plot single-sided amplitude spectrum.
subplot(2,1,2); plot(f,2*abs(Y(1:NFFT/2)));
str = strcat('FFT at N = ',num2str(NFFT),' Points'); title(str);
xlabel('Frequency (Hz)'); ylabel('|Y(f)|');

figure,
Ts = (2*pi)/18;
Fs=1/Ts;
w = -2*pi:Ts:2*pi;
t =-1:Ts:1;

for r=1:length(t)
x_t(r) = Ts*(pi/2)*(sin(2*t(r))/(pi*t(r)))*(sin(8*t(r))/(pi*t(r)));
end
subplot(2,1,1); plot(x_t); title('Original Signal x_t = Ts*(pi/2)*(sin(2*t)/(pi*t))*(sin(8*t)/(pi*t)), Ts = (2*pi)/18');

L=length(x_t);
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(x_t,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2);

% Plot single-sided amplitude spectrum.
subplot(2,1,2); plot(f,2*abs(Y(1:NFFT/2)));
str = strcat('FFT at N = ',num2str(NFFT),' Points'); title(str);
xlabel('Frequency (Hz)'); ylabel('|Y(f)|');

Original Signal xt = Ts"(pi/2)"(sin(2e)/(pi*t))"(sh@*)/(рії). Ts = (7pp60 0.3 0.25 0.15 0.05 20 12 10 0.05 FFT at N 32 Points 0.12 0.08 E0.06 0.04 0.02 3.5 2.5 0.5 Frequency (Hz)

Original Signal xt = Ts"(pi/2)"(sin㎝)/(pi*t))"(sin@*)/(pnj. Ts-prpi)/18 1.2 0.2 0.2 10 18 20 FFT at N-32 Points 0.25 0.2 0.15 0.05 0.5 Frequency (Hz)