Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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can you explain to me why the convolution of these two sinc fucntions which are rect functions equal a rect function? I thought when you convolute a rectangle and a rectangle with two different lengths, it makes a trapezoid shape. The answer says it is rectangle funtion so my code is corrcect but i dont understand why this is true. I need a step by step solution and explanation.
here is the problem: Consider the signal x(t) = 2 sinc(6t + 3).
the output y (t) = (x *h )(t) when x(t) is
passed through a filter with impulse response h (t) = 4sinc(2t -1). the matlab code was just to help me visuallize
Transcribed Image Text:we/untitled.m
% Define frequency axis
f = linspace(-10*pi, 10*pi, 1000); % Frequency axis from -10pi to 10pi
% Define the Fourier Transforms of the sinc functions as rectangular pulses
FT_sinc1 = double(abs(f) <= (6*pi)); % Rectangular pulse for 2*sinc(6t), width 12pi, height 1/6
FT_sinc2 = double(abs(f) <= (2*pi)); % Rectangular pulse for 4*sinc(2t), width 4pi, height 1/2
% Result of multiplication (convolution in time domain)
result_FT = FT_sinc1 .* FT_sinc2;
% Plotting
figure;
subplot(3,1,1); % FT of 2*sinc(6t)
plot(f/(2*pi), FT_sinc1);
title('Fourier Transform of 2*sinc(6t)');
xlabel('Frequency (Hz)');
ylabel('Magnitude');
grid on;
subplot(3,1,2); % FT of 4*sinc(2t)
plot(f/(2*pi), FT_sinc2);
title('Fourier Transform of 4*sinc(2t)');
xlabel('Frequency (Hz)');
ylabel('Magnitude');
grid on;
subplot(3,1,3); % Result of convolution
plot(f/(2*pi), result_FT);
title('Result of Convolution in Frequency Domain');
xlabel('Frequency (Hz)');
ylabel('Magnitude');
grid on;
d Window.
Magnitude
Magnitude
0.5
Magnitude
0.5
Fourier Transform of 2*sinc(6t)
0
-3
-2
-1
-4
-3
-2
0
Frequency (Hz)
Fourier Transform of 4*sinc(2t)
0
Frequency (Hz)
Result of Convolution in Frequency Domain
2
3
5
2
3
4
5
-5
-3
-2
-1
0
2
3
4
5
Frequency (Hz)
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