# integral of two added function can't be implemented

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Faisal Al-Wazir on 28 Oct 2022
Commented: Paul on 5 Nov 2022
im trying to answer q5 here
clc
clear
u = @(t) double(t>=0);
h = @(t) exp(-t).*(u(t)-u(t-1));
h1= @(t) u(t)-2.*u(t-1)+u(t-2)
h1 = function_handle with value:
@(t)u(t)-2.*u(t-1)+u(t-2)
h2= @(t) h(t) + h1(t)
h2 = function_handle with value:
@(t)h(t)+h1(t)
fplot(h2,[-3,3])
grid on
ylim([-1 2])
clc
clear
t=@(t) t
t = function_handle with value:
@(t)t
u = @(t) double(t>=0);
h = @(t) exp(-t).*(u(t)-u(t-1));
h1= @(t) u(t)-2.*u(t-1)+u(t-2)
h1 = function_handle with value:
@(t)u(t)-2.*u(t-1)+u(t-2)
h2= @(t) h(t) + h1(t)
h2 = function_handle with value:
@(t)h(t)+h1(t)
i=integral(h2,-inf,t)
Error using integral
Limits of integration must be double or single scalars.
fplot(i,[-3,3])
grid on
ylim([-1 2])

Paul on 28 Oct 2022
Hi Faisal,
If the plot for the answer to Q4 is corret, then does the integration really need to start from -inf?
However, I suggest revisiting Q4, that solution does not look correct. Why would the output of the system be the sum of the impulse response and the input?
Paul on 5 Nov 2022
Yes, as I stated previously, I was showing an example that would have to be modified for the specific problem at hand.
Your current code changes the definition of x(t) from the original Q4. For Q5, consider extending tval to a longer time.

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