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) ispassed through a filter with impulse response h (t) = 4sinc(2t -1). the matlab code was just to help me visuallize we/untitled.m% Define frequency axisf = linspace(-10*pi, 10*pi, 1000); % Frequency axis from -10pi to 10pi% Define the Fourier Transforms of the sinc functions as rectangular pulsesFT_sinc1 = double(abs(f) <= (6*pi)); % Rectangular pulse for 2*sinc(6t), width 12pi, height 1/6FT_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;% Plottingfigure;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 convolutionplot(f/(2*pi), result_FT);title('Result of Convolution in Frequency Domain');xlabel('Frequency (Hz)');ylabel('Magnitude');grid on;d Window.MagnitudeMagnitude0.5Magnitude0.5Fourier Transform of 2*sinc(6t)0-3-2-1-4-3-20Frequency (Hz)Fourier Transform of 4*sinc(2t)0Frequency (Hz)Result of Convolution in Frequency Domain2352345-5-3-2-102345Frequency (Hz)

The input signal is given by⇒x(t)=2sinc(6t+3).The impulse response of a filter is given…

Answered: Consider the signal x(t) = 2 sinc(6t +…

Introductory Circuit Analysis (13th Edition)

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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

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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