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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
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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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