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:
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