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Solution required in MATLAB1. Convolution and Discrete-Time Fourier Series (DTFS) (a) Generate a periodic signal r2[n] with the fundamental period N ral

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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 x_2[n] at n and n + 60. Then, we have that

60 30

The code for generating the signal x_2[n] 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 h_2[n] 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 X_2[k] of x_2[n] is found as :

X_2 = fft(x_2);

And similarly the FFT H_2[k] is found as:

H_2 = fft(h_2);

e.) The product of these two, i.e the DTFS of x_2[n] \text{ and } h_2[n] 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 0\leq n \leq N-1 , and repeats outside that interval.

The plots are attached for reference:

Plot of ifft[fftsx) - fft(h))) 200 100 200 10 20 30 40 50 500 23-45 5432101

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