When we pad an aperiodic signal with  zeros, we are improving its frequency resolution, i.e., the more zeros we  attach to the original signal the better the frequency resolution, as we  obtain the frequency representation at a larger number of frequencies  around the unit circle.

(a) Consider an aperiodic signal x[n] = u[n] − u[n − 10], compute its  DFT by means of the fft function padding it with 10 and then 100 zeros. Plot the magnitude response using stem. Comment on the  frequency resolution of the two DFTs.

(b) When the signal is periodic, one cannot pad a period with zeros.  When computing the FFT in theory we generate a periodic signal  of period L equal or larger than the length of the signal when the  signal is aperiodic, but if the signal is periodic we must let L be  the signal fundamental period or a multiple of it. Adding zeros to  the period makes the signal be different from the periodic signal.  Consider x[n] = cos(πn/5), − ∞ < n < ∞ be a periodic signal, and  do the following:

• Consider exactly one period of x[n], and compute the FFT of this sequence.

• Consider 10 periods of x[n]and compute the FFT of this sequence.

• Consider attaching 10 zeros to one period and compute the FFT of the resulting sequence.

If we consider the first of these cases giving the correct DFT of x[n],  how many harmonic frequencies does it show? What happens when we consider the 10 periods? Are the harmonic frequencies  the same as before? What are the values of the DFT in frequencies  in between the harmonic frequencies? What happened to the mag-nitude at the original frequencies? Finally, does the last FFT relate  at all to the first FFT?