# Convultion with Delta Function

• Caspian

#### Caspian

I can remember from Differential Equations that any function convolved with a delta function results in a copy of the function located at the impulese.

That is, $$x(t) * \delta(t-5) = x(t-5)$$

However, I can't remember why. This is really irritating me since I need to use this concept for my courses, yet I can't remember why this is true. This makes sense... but I get stuck when trying to evaluate the following integral:

$$\int_0^t \delta(t - \tau) d \tau$$

Any help would be appreciated.

Thanks!

Last edited:
How is the convolution defined? Also, remember $\delta (t-5) = (\delta \circ \epsilon )(t)$ where $\epsilon : t \mapsto t-5$. I haven't worked with convolutions, but looking it up the definition on Mathworld, the equation above seems obvious:

$$x * (\delta \circ \epsilon ) = \int _{-\infty} ^{\infty} x(\tau )(\delta \circ \epsilon )(t - \tau )d\tau$$

$$= \int _{-\infty} ^{\infty} x(\tau )\delta ((t - \tau) - 5)d\tau$$

$$= \int _{-\infty} ^{\infty} x(\tau )\delta ((t - 5) - \tau)d\tau$$

$$= x(t - 5)$$

where the last line follows by the definition of the delta function.

Thanks, AKG!

I don't know why that waws stumping me -- I really appreciate your help.