# convolution of $e^{-t}$ and 1-t

I cannot solve the convolution based on $h= e^{-t}$ for $t\ge0$ and $u(t) = 1-t$ when $0 \le t \le1$.
Every time I try I keep getting a factor with $te^{-t}$ whilst the answer shows:
$y(t) = 0$ if $t<0$
$y(t) = 2-t-2e^{-t}$ if $0\le t\le 1$
$y(t) = e^{1-t}-2e^{-t}$ if $t>1$

And I get:
$(1-e^{-t})+te^{-t}+(e^{-t}-1)$
After solving the integration by parts with boundaries of 0>t

• \ge = $\ge$. Also use  to center your formulae. – Eugene Sh. May 22 '18 at 18:33 • Shouldn't this be a better question for Math SE? – Helena Wells Jul 23 '20 at 6:27 ## 2 Answers The functions to be convolved are $$\h(t) \$$ and $$\u(t) \$$, as shown in the top two diagrams in the figure below: Figure source: I drew it. The third diagram shows the folded $$\u(t) \$$, i.e., $$\u(-t) \$$ and the fourth diagram shows the folded and shifted $$\u(t) \$$, i.e., $$\u(-t-\tau) \$$. Note that $$\\tau \$$ is the shift variable. There will be two non-zero overlap integral scenarios: for Overlap A and Overlap B. These are schematically shown in the fifth and sixth diagrams in the figure. The convolution is given by the following overlap integral: $$h(t) * u(t) = \int_{-\infty}^{+\infty}h(t) u(-t-\tau) dt$$ For the Overlap A scenario, the convolution integral reduces to: $$h(t) * u(t) = \int_{0}^{\tau}h(t) u(-t-\tau) dt$$ while for the Overlap B scenario, the convolution integral reduces to: $$h(t) * u(t) = \int_{\tau -1}^{\tau}h(t) u(-t-\tau) dt$$ Evaluation of convolution integral for Overlap A \begin{align} h(t)*u(t) &= \int_0^\tau h(t) u(-t-\tau) dt \\ &= \int_0^\tau e^{-t}(1+t-\tau)dt \\ &= (1-\tau) \int_0^\tau e^{-t}dt + \int_0^\tau te^{-t}dt \\ \end{align} The two definite integrals are just special cases of well known indefinite integrals: $$\int e^{ax}dx = e^{ax}/a$$ and $$\int xe^{ax}dx = e^{ax}(ax - 1)/a^2$$ With $$\a = -1\$$ and $$\x \$$ replaced by $$\t \$$ in both integrals, we have $$\int e^{-t}dt = -e^{-t}$$ and $$\int te^{-t}dt = e^{-t}(-t - 1) = -e^{-t}(t + 1)$$ Thus, continuing the evaluation for Overlap A: \begin{align} h(t)*u(t) &= (1-\tau) \left[-e^{-t} \right]_0^\tau + \left[e^{-t}(-t-1) \right]_0^\tau \\ &= (\tau -1)(e^{-\tau}-1)-[e^{-\tau}(\tau + 1) - 1] \\ &= \tau e^{-\tau}-e^{-\tau}-\tau+1-\tau e^{-\tau}-e^{-\tau}+1 \\ &= 2-\tau -2e^{-\tau} \end{align} Replacing $$\\tau \$$ by $$\t \$$ then gives the desired result: $$y(t) = 2-t -2e^{-t}$$ Evaluation of convolution integral for Overlap B \begin{align} h(t)*u(t) &= \int_{\tau -1}^\tau h(t) u(-t-\tau) dt \\ &= \int_{\tau -1}^\tau e^{-t}(1+t-\tau)dt \\ &= (1-\tau) \int_{\tau -1}^\tau e^{-t}dt + \int_{\tau -1}^\tau te^{-t}dt \\ &= (1-\tau) \left[-e^{-t} \right]_{\tau -1}^\tau + \left[e^{-t}(-t-1) \right]_{\tau -1}^\tau \\ &= (\tau -1)[e^{-\tau}-e^{-(\tau -1)}]-(\tau +1)e^{-\tau}+\tau e^{-(\tau -1)} \\ &= \tau e^{-\tau}-e^{-\tau}-\tau e^{-(\tau -1)}+e^{-(\tau -1)}-\tau e^{-\tau}-e^{-\tau}+\tau e^{-(\tau -1)} \\ &= e^{-(\tau -1)}-2e^{-\tau} \end{align} Replacing $$\\tau \$$ by $$\t \$$ then gives the desired result: $$y(t) = e^{-(t -1)}-2e^{-t} = e^{1 -t}-2e^{-t}$$ Summary: $$\ y(t) = 0 \$$ if $$\t<0\$$ $$\ y(t) = 2-t-2e^{-t} \$$ if $$\0\le t\le 1\$$ $$\ y(t) = e^{1-t}-2e^{-t} \$$ if $$\t>1\$$ Check: Area under $$\h(t) = \int_0^\infty h(t) dt = 1 \$$. Area under $$\u(t) = \int_0^1 u(t) dt = 1/2 \$$. Area under $$\y(t) \$$ is $$\int_0^\infty y(t) dt = \int_0^1 (2-t -2e^{-t}) dt + \int_1^\infty (e^{1 -t}-2e^{-t}) dt = (\frac{2}{e} -\frac{1}{2})+(1-\frac{2}{e}) = \frac{1}{2}$$ •••••••••••••••••••••••••••••••••••••••••••• The alternative convolution option Suppose it is desired to fold and shift $$\h(t) \$$ instead of $$\u(t) \$$. Then the first figure is replaced by this figure: Figure source: I drew it. The functions to be convolved are $$\h(t) \$$ and $$\u(t) \$$, as shown in the top two diagrams in the figure. The third diagram shows the folded $$\h(t) \$$, i.e., $$\h(-t) \$$ and the fourth diagram shows the folded and shifted $$\h(t) \$$, i.e., $$\h(-t-\tau) \$$. Note that $$\\tau \$$ is the shift variable. There will be two non-zero overlap integral scenarios: for Overlap A and Overlap B. These are schematically shown in the fifth and sixth diagrams in the figure. The convolution is given by the following overlap integral: $$h(t) * u(t) = \int_{-\infty}^{+\infty}h(-t-\tau) u(t) dt$$ For the alternative Overlap A scenario, the convolution integral reduces to: $$h(t) * u(t) = \int_{0}^{\tau}h(-t-\tau) u(t) dt$$ while for the alternative Overlap B scenario, the convolution integral reduces to: $$h(t) * u(t) = \int_0^1 h(-t-\tau) u(t) dt$$ Evaluation of convolution integral for Overlap A \begin{align} h(t)*u(t) &= \int_0^\tau h(-t-\tau) u(t) dt \\ &= \int_0^\tau e^{t-\tau} (1-t)dt \\ &= e^{-\tau} \left[\int_0^{\tau}e^t dt - \int_0^{\tau}te^t dt \right] \\ &= e^{-\tau} \left[[e^t]_0^{\tau} - [e^t (t-1)]_0^{\tau} \right] \\ &= e^{-\tau} \left[e^{\tau} -1 -\tau e^{\tau} + e^{\tau} -1 \right] \\ &= e^{-\tau} \left[2e^{\tau} -\tau e^{\tau} -2 \right] \\ &= 2-\tau -2e^{-\tau} \\ \end{align} Replacing $$\\tau \$$ by $$\t \$$ then gives the desired result: $$y(t) = 2-t -2e^{-t}$$ Evaluation of convolution integral for Overlap B \begin{align} h(t)*u(t) &= \int_0^1 h(-t-\tau) u(t) dt \\ &= \int_0^1 e^{t-\tau} (1-t)dt \\ &= e^{-(\tau -1)}-2e^{-\tau} \\ \end{align} Replacing $$\\tau \$$ by $$\t \$$ then gives the desired result: $$y(t) = e^{-(t -1)}-2e^{-t} = e^{1 -t}-2e^{-t}$$ • Hi,+1 very detailed explanation ,is it always necessary to draw graphs and then figure out different overlap regions ? Can we directly integrate using formula ignoring graphs ? – user215805 Mar 28 at 19:37 • @user215805 Thanks for the upvote! It is definitely not necessary (or always necessary) to draw the graphs, but I find it very useful to keep from screwing up! I have done more complicated convolutions by hand and keeping the notation and integration limits sorted out is really the main thing. – Ed V Mar 28 at 19:46 • I appreciate the very detailed answer, Since then I had much more practise and now I am able to do these better. But for anyone googling a similar question, this will help them out greatly! – Weird Mar 28 at 23:25 • Glad to be of help! And many thanks for the green checkmark! – Ed V Mar 28 at 23:55 Writing down convolution product results in (h * u)(t) = \int_{-\infty}^{+\infty}h(t - v)\cdot u(v) dv $$The parts of the domain we're interested in are:$$t-v\ge 0 \Rightarrow v \leq t$$for $h(t-v)$$$0 \leq v \leq 1$$for $u(v)$ We can then find that for $0 \leq t \leq 1$:$$\begin{align} (h*u)(t) &= \int_0^th(t - v)\cdot u(v) dv \\ &= \int_0^t e^{-(t-v)}(1-v)dv \\ &= e^{-t} \left( \int_0^te^vdv - \int_0^tve^vdv \right) \end{align}$$Solving the first integral is rather easy:$$\int_0^te^vdv = \left[e^v\right]_0^t = e^t - 1$$Solving the second integral requires integration by parts:$$\begin{align} \int_0^tve^vdv &= \int_0^tvd(e^v) \\ &= \left[v\cdot e^v\right]_0^t - \int_0^tve^vdv \end{align}$$As you get the same integration as before, you can write:$$\begin{align} I &= \left[ v\cdot e^v\right]_0^t - I \\ \Rightarrow 2I &= \left[ v\cdot e^v\right]_0^t \\ \Rightarrow I &= \frac{[v\cdot e^v]_0^t}{2}=\frac{te^t}{2} \end{align}$$Plugging it all in yields:$$ \begin{align} y(t) &= e^{-t} \left( e^t - 1 - \frac{te^t}{2} \right) \\ &= 1 - e^{-t} - \frac{t}{2} \end{align}$$This answer is off by a factor 2, but it resembles the given answer. I don't see where I was wrong. For $t \geq 1$, you can write$$ \begin{align} y(t) &= \int_0^1h(t-v)u(v)dv \\ &= e^{-t} \left( e - 1 - \frac{e}{2} \right) \\ &= \frac{e^{1-t}}{2} - e^{-t} \end{align}

This is again off by the same factor of 2 compared to your reference.