Answer:)
a.) We see that the fundamental period is such that it is the
least common multiple of 10, 20 and 30, which is 60, and we can
also see that the three signals give the same value for
at n and n + 60. Then, we have that
The code for generating the signal is as
follows:
N = 60;
n = 0:N-1;
x_2 = sin(0.2*pi*n) + sin(0.1*pi*n) + sin(0.2*pi*n/3);
b.)The code for generating is as
follows:
h_2 = (0.5).^n;
c.) We see the convolution as follows:
y_2 = conv(x_2, h_2);
One can see the result using y_2, which gives the full output of y_2.
d.)The FFT of
is found as :
X_2 = fft(x_2);
And similarly the FFT is found
as:
H_2 = fft(h_2);
e.) The product of these two, i.e the DTFS of
is given as :
Y_2 = N * X_2 .* H_2;
f.) Using ifft(), we get that the inverse is :
y_inv = ifft(Y_2)
g.) We see that the convolution in time domain, is similar to
using fft() and then ifft() on the DTFS. This corroborates the fact
that FFT converts convolution in time domain to multiplication of
the respective DTFS in frequency domain. However, the magnitudes
are different (which is due to our multiplication of the frequency
domain components by N), and also, the output of y_inv is
repetitive and matches y_2 exactly within
, and repeats outside that interval.
The plots are attached for reference:


Solution required in MATLAB 1. Convolution and Discrete-Time Fourier Series (DTFS) (a) Generate a periodic signal r2[n]...
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