Question

Consider a 10 Hz sine wave sampled at 35 Hz. If the samples are plotted and displayed using linear interpolation, you will see that the reconstructed waveform will not resemble the original continuous-time sine wave. Show how sinc interpolation can be used to reconstruct the original continuous-time signal. (Hint: In this question you are being asked to resample the sine wave to a higher sampling rate of, say, 105Hz, by sinc interpolation – i.e., convolution with a sinc function)

Please show the MATLAB codes and figure results.

Answer 1

MATLAB Code with result for sampling reconstruction with sinc interpolation function;

clc;
clear all;
close all;
a1=20*pi;
a2=210*pi;
Ts=1/550;
t=0:Ts:0.2;
x1=cos(a1.*t);
x2=cos(a2.*t);
subplot(2,2,1);   
plot(t,x1);
title('Information signal');
xlabel('t');
ylabel('x(t)');
subplot(2,2,2);
n=0:0.01:20;
plot(t,x2);
title('Carrier sinusoidal signal');
xlabel('t');
ylabel('x(t)');
subplot(2,2,3);
n=0:0.01:20;
stem(t,x2);
title('Sampled signal');
xlabel('t');
ylabel('x(t)');
n=-10:30;
x=cos(pi.*n./4);
T=1/1000;
for l=1:2000
t=(l-1)*T/100;
h=sinc((t-n.*T)./T);
xr(l)=x*h.';
end
subplot(2,2,4)
plot(n,x);

Simulation Result;

The fourth graph shows the reconstructed signal using inc interpolation – i.e., convolution with a sinc function. This shows that with sinc interpolation function we can easily reconstruct our original continuous-time signal.