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

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 7:32
• $\tau$ is a dummy time variable. Any letter will do! – Chu Jul 20 '19 at 13:23

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.

• @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. – user287001 Jul 20 '19 at 12:01

$$\h(t-\tau)\$$, is mirror image of the impulse response $$\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.