Change of variable from t to \$\tau\$ during convolution

Question

I am not able to understand that in the standard convolution formula how we can change the variable from t to \$\tau\$.

$$\int{x(\tau)\cdot h(t-\tau)d\tau }$$

Isn't this incorrect mathematically?

\$\tau\$ = TAU is used to improve signals with this known repeating time constant. — Tony Stewart EE75, Jul 20 '19 at 07:32

score 2 · Answer 1 · edited Jul 20 '19 at 11:32

2

In addition there should be somewhere mentioned the integration period, it's from time=0 to current time t.

$$\int_0^t{x(\tau)\cdot h(t-\tau)d\tau }$$

Tau is the formal variable for integration process, t is the running physical time. The integral should be calculated completely from blank board for every t. That's quite a job for a signal processor. Fortunately there are accelerating algorithms developed.

edited Jul 20 '19 at 11:32

Dave Tweed

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answered Jul 20 '19 at 07:57

@Dave Tweed Thanks for typesetting the integral. I must add that in formal writings the integration period is from minus infinity to +infinity to cover also non-causal filterings and signals which are non-zero before t=0. – Jul 20 '19 at 12:01

score 0 · Answer 2 · answered Jul 20 '19 at 13:54

\$h(t-\tau)\$, is mirror image of the impulse response \$h(t)\$, reflected about the vertical axis, and shifted (delayed) by \$\tau\$. Integrating between 0 and \$t\$ effectively slides the reflected impulse past the input signal, multiplying and adding (i.e. integrating) as it goes, thus mimicking what happens in practice.

Google: 'convolution - fold slide multiply add', for a gif

https://en.wikipedia.org/wiki/Convolution#/media/File:Convolution_of_spiky_function_with_box2.gif

Change of variable from t to \$\tau\$ during convolution

2 Answers2