MATLAB Convolution: Finding the Convolution of Two Functions with Step Inputs

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The discussion focuses on using MATLAB to compute the convolution of two step functions, f(t) and g(t). The initial attempt resulted in a convolution output that reached a maximum of 10, despite the individual functions having maximum values of 1. It was clarified that multiplying the convolution result by the step size (0.1) is necessary to accurately approximate the integral, as MATLAB's convolution function follows a linear algebra definition. This adjustment aligns the output with the expected values based on the rectangular area interpretation of convolution. Understanding the relationship between discrete convolution and integral approximation is crucial for accurate results in MATLAB.
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Homework Statement


Use MATLAB to find the convolution between
a) ##f(t) = u(t) -u(t -3)## and ##g(t) = u(t) -u(t -1)##

Homework Equations



The Attempt at a Solution


t = -10: 0.1: 10;
f = heavisde(t) -heaviside(t -3);
g = heaviside(t) -heaviside(t -1);
t = -20: 0.1: 20;
c = conv(f, g);
plot(t, c)

The graph of the convolution has values from 0 to 10. I don't get how the convolution can get to 10 if the functions being convolved only have maximum values of 1. I changed the line to
c = 0.1*conv(f, g);
and this produced the right output. Why do I need to multiply by the step size?
 
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Convolution is a sum over element-wise products. To turn this into the equivalent approximation for the integral, you have to multiply by the step size. Think about the rectangular area. Matlab applies the linear algebra definition of convolution.
 
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Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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