A continuous time signal is given below: x(t) = 10 + 3 sin (20 pi t + pi/3) + 5

Question

A continuous time signal is given below: x(t) = 10 + 3 sin (20 pi t + pi/3) + 5 cos(40 pi t) a. This is sampled at t = 0.01 n to get a the discrete-time signal x[n], which is then applied to an ideal DAC to obtain a reconstructed continuous time signal y_r(t). i. Determine x[n] and graph its samples, using Matlab, along with the signal x(t) in one plot, plot a few cycles of x(t). ii. Determine the reconstructed signal, y_r(t), as a sinusoidal signal and compare it to x(t). b. Repeat part a for sampling times of t = 0.05n and t = 0.1n, what conclusion can you draw from these three cases?

Explanation / Answer

clc;
close all;
clear all;

T = 0.001; %t = nT = 0.001n => T = 0.001

t = 0:0.001:1;
xt = 10 + 3*sin(20*pi*t + (pi/3)) + 5*cos(40*pi*t);

n = t./0.001;
xn = 10 + 3*sin(20*pi*0.001*n + (pi/3)) + 5*cos(40*pi*0.001*n);

figure
subplot(2,1,1);
plot(t,xt);

subplot(2,1,2);
stem(n,xn);

for l = 1:100000
    t = (l-1)*T/100;
    h=sinc((t-n.*T)./T);
    xr(l) = xn*h.';
end

figure
plot(xr);

T = 0.05; %t = nT = 0.001n => T = 0.001

t = 0:0.001:1;
xt = 10 + 3*sin(20*pi*t + (pi/3)) + 5*cos(40*pi*t);

n = t./0.001;
xn = 10 + 3*sin(20*pi*0.001*n + (pi/3)) + 5*cos(40*pi*0.001*n);

figure
subplot(2,1,1);
plot(t,xt);

subplot(2,1,2);
stem(n,xn);

for l = 1:100000
    t = (l-1)*T/100;
    h=sinc((t-n.*T)./T);
    xr(l) = xn*h.';
end

figure
plot(xr);

T = 0.1; %t = nT = 0.001n => T = 0.001

t = 0:0.001:1;
xt = 10 + 3*sin(20*pi*t + (pi/3)) + 5*cos(40*pi*t);

n = t./0.001;
xn = 10 + 3*sin(20*pi*0.001*n + (pi/3)) + 5*cos(40*pi*0.001*n);

figure
subplot(2,1,1);
plot(t,xt);

subplot(2,1,2);
stem(n,xn);

for l = 1:100000
    t = (l-1)*T/100;
    h=sinc((t-n.*T)./T);
    xr(l) = xn*h.';
end

figure
plot(xr);

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.