# How to convolve an arbitrary signal with a causal decaying exponential?

I need to simplify the following convolution

$$x(t)\star [e^{-2t} u(t)]$$

where $u(t)$ is the unit step function. I'm very confused with this. Using the definition of convolution of continuous-time signals, I obtained

$$\int\limits_{-\infty}^t {x(\tau)e^{-2(t-\tau)}u(t-\tau)}{d\tau}$$

However, I have no idea what to do at this step. I'm stuck. Is there any property / theorem / trick that can help me simplify this?

Indeed you have reached what can be reached, may be the following additional line can be obtained by moving the $t$ function $e^{-2t}$ out of the integral and replacing the $u(t-\tau)$ by $1$ as:

$$y(t) = e^{-2t} \int\limits_{-\infty}^t {x(\tau)e^{2\tau}{}d\tau}$$

In the general case you cannot simplify it any further...

• How can you replace $u(t-\tau)$ by 1? Is it always 1 from - infinity to t?
– anon
Dec 6, 2017 at 20:30
• Indeed the better question is how did you determined the upper limit of the intgral as $t$ in the first place? i.e., $$\int_{-\infty}^{\infty} f(\tau)u(t-\tau) d\tau = \int_{-\infty}^{t} f(\tau) u(t-\tau) d\tau$$. If you reach this line then you implicitly say that $$u(t-\tau) = \begin{cases} 1 &, \text{ for } -\infty < \tau < t \\ 0 &, \text{ for } ~~~~~ t < \tau < \infty \\ \end{cases}$$ I think now it's clear... Dec 6, 2017 at 20:37
• Yes sorry my bad. Should have thought of that. Thank you for the explanation. Unfortunately it seems that there is no way x(t) can get out of the integration right?
– anon
Dec 6, 2017 at 20:43
• you welcome. In the general case you cannot get $x(\tau)$ out of integration... Dec 6, 2017 at 20:46