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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

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  • \$\begingroup\$ \ge = \$\ge\$. Also use $$ to center your formulae. \$\endgroup\$ – Eugene Sh. May 22 '18 at 18:33
  • \$\begingroup\$ Shouldn't this be a better question for Math SE? \$\endgroup\$ – Helena Wells Jul 23 '20 at 6:27
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The functions to be convolved are \$h(t) \$ and \$u(t) \$, as shown in the top two diagrams in the figure below:

Convolution example 1

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:

Convolution 5

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:

$$ y(t) = e^{-(t -1)}-2e^{-t} = e^{1 -t}-2e^{-t} $$

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  • \$\begingroup\$ 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 ? \$\endgroup\$ – user215805 Mar 28 at 19:37
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    \$\begingroup\$ @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. \$\endgroup\$ – Ed V Mar 28 at 19:46
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    \$\begingroup\$ 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! \$\endgroup\$ – Weird Mar 28 at 23:25
  • \$\begingroup\$ Glad to be of help! And many thanks for the green checkmark! \$\endgroup\$ – Ed V Mar 28 at 23:55
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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.

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