Convolution Proof of time scaling property

[woohs1216]

Hello

I don't quiet understand how the integration in the picture works...
I must have forgotten something...

Can anyone explain what is used?

 

[davenn]

Hi
welcome to PF

looks like a maths problem , not EE have asked for it to be moved

take care to post questions in the correct section :)

Cheers
Dave

 

[HallsofIvy]

I see no place, in what you have posted, where "c(at)" is defined! In any case, the part you have enclosed in red, on the left, is identical to the calculation on the right. Do you not have a problem with that? In any case, it is impossible to say why that integral is equal to c(at) without knowing how the fuction, c, is defined.

 

[Lord Crc]

The interesting part happens between the very first integral and the one you've outlined in red. That's when the change in variable happens and when the scaling factor comes into the mix.

I'm pretty sure c(t) has the form of the outlined integral, as that is the standard convolution integral, and as such is just a convenience.

Of course I might be wrong :-)

 

[slider142]

I agree with Lord Crc. They have not actually integrated anything. They have just defined a function c by the equation
$$c(at) = \int_{-\infty}^{\infty} x(m)g(at - m)\, dm$$
Since the variable m is integrated over, the value of the integral is indeed only a function of at. They probably chose the letter c to mean "convolution integral".