# 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 $$ to center your formulae. – Eugene Sh. May 22 '18 at 18:33 ## 1 Answer Writing down convolution product results in$$ (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.