# Convolution with two unbounded signals

So I have a convolution problem. I have two signals:

I need to find the output, y(t), through graphical convolution.

So I set the convolution integral up like this: $$\int h(\tau)*x(t-\tau)d\tau$$

I set it up this way because I believed it easier to 'flip and slide' the exponential function than the unit step function. So, then I substituted in $h(\tau)$ and $x(t-\tau)$ like so: $$\int_{t-\inf}^{0} 1*e^{t-\tau}d\tau$$ This then gave me: $$-e^{t-\tau}|_{t-\inf}^0$$ Which equals $-e^t$. This is fine, and acceptable in my opinion. However, when moving to the next section, where the exponential is partially outside of the unit step function, I get issues. The equation I used here was this: $$\int_{-\inf}^0e^{t-\tau}d\tau + \int_{t-\inf}^te^{t-\tau}d\tau$$ This includes the previous section from infinity to 0, and adds the new section, where the exponential has now moved partially outside of the step function, being bound by $t-\inf$, as before, and $t$. This then gives me this: $$-e^{t-\tau}|_{-\inf}^0 + -e^{-t-\tau}|_{t-\inf}^t$$ This is where I start to have problems. $-e^{t-\tau}|_{-\inf}^0$ evaluates to: $$-e^t + e^{t+\inf}$$ This is, of course, an issue, as $e^{t+\inf}$ is infinite, and cannot be used as it doesn't converge.

Is there a different way I can graphically convolve these two signals without running into issues with infinity?

Now, I have done this problem using $h(t-\tau)$ and $x(\tau)$. This runs into a similar issue. For the first section, where the step function is partially outside of the exponential, I get this: $$\int_{-\inf}^0e^-td\tau$$ That evaluates to this: $$-te^{-t}$$ Fine. Good. Now, when I do the next section, where the step function has passed $\tau = 0$, and now the limit of the step function is going along the curve of the exponential, I have this: $$\int_{-\inf}^0e^{-t}d\tau + \int_0^te^{-t}d\tau$$ This evaluates to $\tau e^{-t}|_{-\inf}^0+\tau e^{-t}|_0^t$, or $\inf +te^{-t}$.

So I run into a similar issue to what I had had before.

So I decided to do this convolution using Laplace transforms. Since: $$x(t) = e^{-t}$$ and $$h(t) = u(-t)$$ I found these transformations: $$H(s) = \frac1s; X(x) = \frac{1}{s+1}$$ Multiplying these, I find: $$Y(s) = \frac{1}{s^2+1}$$

Applying an inverse Laplace Transform, I get: $$y(t) = 1 - e^{-t}$$

• Are you sure the function in your first graph is supposed to be $e^{-t}$ and not $e^{-t}u(t)$? Because the second of those is much more commonly encountered in signal processing and Laplace transform problems, and it eliminates the issue you're asking about. Also your graph as drawn appears to show $e^{-t}u(t)$. Commented Oct 3, 2017 at 0:11
• Ah, that's probably the issue with my solution. The one thing I would ask is how that might eliminate the issue? I would assume it does that because $\int u(t)$ is $\delta(t)$?
• It eleminates the problem because with the $u(t)$ term (flipped and shifted) you don't end up integrating out to infinity but only out to $\tau=t$. (Some of your equations are integrating with respect to $\tau$ but don't have a $\tau$ in the expression being integrated...I think you've missed some cases where $t$ gets changed to $\tau$.) Commented Oct 3, 2017 at 0:42
• Also, $\infty$ in Mathjax is \infty. Commented Oct 3, 2017 at 0:43
• Ah I was looking for $\infty$ but I was more worried about getting my point across. As for not having a $\tau$, that was just me mistranscribing my notes. So what I see here is that I need to use t to 0, instead of $-\infty$ to 0.
The main thing I see wrong with your approach is that as drawn, your $x(t)$ is actually $x(t)= u(t)e^{-t}$, not $e^{-t}$. Making this change should remove one the infinite limits on your integration (when the exponential goes to infinity).