how to get convolution for two signals without using conv command

so i have two signals the first is h = e^-3t and the another is g = e^-5t

i already done the convoloution of these two using the conv command but i need another method without the command.

any help?

Answer 1

[Jan]

If you read the documentation, you find the mathematical definition of conv:

doc conv

The code is easy to implement using loops. Try it. If you have specific problems, post your code here and ask for details.

If you spend the time for searching in the net (meta-message...), you find this also:

• https://www.mathworks.com/matlabcentral/fileexchange/44466-beating-matlab-s-convolution-function-conv-m-for-long-real-sequences
• https://www.mathworks.com/matlabcentral/answers/388344-how-can-i-implement-a-convolution-function-in-matlab-and-perform-it-on-the-following-signals-and-plo

[mohammad yaman habra]

thank you

Answer 2

[Matt J]

You're probably meant to use the convolution property of FFTs.

[mohammad yaman habra]

no i have already tried this method as following :

>> t=1:0.1:10;

>> h= exp(-3*t);

>> g = exp(-5*t);

>> y = 0.1*conv(h,g);

>> plot(y)

and i need to do the same pocess but without the conv command

[Matt J]

There are no FFTs in the code you've shown.

Answer 3

[John D'Errico]

Several ways you could do this. If your signal is represented as a discrete sequence, then just use a double loop, accumulating the result into a vector of appropriate length.

If you were intended to work on the signals in symbolic form, then a z-transform can also be used for convolution.

syms t
ztrans(exp(-3*t))

ans = 

I showed that much to explain how to write an exponential. Note that the exponential you wrote will use the function exp in MATLAB.

Or, can you solve a convolution problem using the Laplace tranform? If this is a convolution INTEGRAL, as opposed to the sum implied by a discretized signal, then Laplace transforms would be appropriate.

https://planetmath.org/laplacetransformofconvolution

laplace(exp(-3*t))

ans = 

You should see that both will be capable of computing the convolution of two signals, represented as functions. The Laplace transform applies to a convolution integral, a z transform to a discrete signal. My guess is, if you already did it using conv, then you would be intended to use a z-transform. (Hint.)