Question

11.43* A rectangular pulse extends from – Xoto +x, and has a height of 1.0. Sketch its autocorrelation, Crf(X). How wide is eff(X)? Is it an even or odd function? Where does it start (become nonzero) and where does it end?

please solve this problem as possible as detail.

Answer 1

We have a rectangular pulse that extends from – Io to +To and it has a height of 1.0. The width of the autocorrelation is twice the width of the function, since the width of the function is 2.00 then the width of the autocorrelation is 4.30), and the it is an even function that starts at -2.70 and ends at +2.00. To make a plot of the autocorrelation function I used python, where the code and the plot are shown in the following figure.

from scipy import signal import matplotlib.pyplot as plt import numpy as np sig = np.repeat ([0.,1., 0.], 100) win = np.repeat ([0.,1., 0.], 100) filtered = signal.correlate (sig, win, mode='same') / sum (win) fig, (ax_orig, ax_win, ax_filt) = plt.subplots (3, 1, sharex=True) ax_orig.plot(sig) ax_orig.set_title('$f(x) $') ax_orig.margins (0, 0.1) ax_win.plot(win) ax_win.set_title('$f(x) $') ax_win.margins (0, 0.1) ax_filt.plot(filtered) ax_filt.set_title('$f(x)* f(x) $') ax_filt.margins (0, 0.1) fig.tight_layout() fig.show()

f(x) 1.0 0.5 0.0 -Хо f(x) хо 1.0 0.5 0.0 — Хо f(x) = f(x) хо 1.0 0.5 0.0 - —2xo — Хо Хо 2xo

The width of the function is 2.0, then the width of the autocorrelation is 4.10, and the it is an even function that starts at -2.00 and ends at +2.30

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