calculating convolution of two exponential functions

Question

I have a problem calculating and visualizing the following continuous convolution.

Let $x(t) = e^{-t}u(t+2) \tag{1}$ and $h(t) = e^{t}u(-t) \tag{2}$ find $y(t) = x(t)*h(t) \tag{3}$ Okay from the definition of step functions, $x(t)$ is simply $e^{-t}$ over the interval t = -2 to $\infty$ and zero elsewhere.

My main problem is visualizing $h(t-\tau)$. If I plug $-\tau$ into equation (2), I basically flip the function in that equation about the origin, and I get a function $$h(-\tau) = e^{-\tau}u(\tau) \tag{4}$$ that goes from $1$ to $-\infty$ over the interval $(0,\infty)$. I think that you slide this function over the interval $(-\infty,\infty)$, and the convolution integral will differ depending on whether the trailing edge of the function begins before -2 or after -2 (or the value of $t$ in $h(t-\tau)$. However, I become confused because the graph provided illustrating $h(t-\tau)$ shows an exponential function that is increasing over the relevant intervals. So for example, over the interval t = $(-2,\infty)$ what is illustrated as $h(t-\tau)$ is a curve that grows from near 0 to infinity over that interval. I do not know what I've done wrong. Please help.

Answer 1

$$y(t)=x(t)*h(t)=\int_{-\infty}^{\infty} x(t-w)h(w)dw=\int_t^0 x(t-w)h(w)dw=\int_t^0 e^{-(t-w)}e^{-w}dw=-te^{-t}$$ for $-2\leq t\leq0$ and $0$ otherwise.

Answer 2

It holds

$$ \begin{align} y(t) &= \int_{-\infty}^{\infty} x(\tau)h(t-\tau)d\tau\\ &=\int_{-\infty}^{\infty} e^{-\tau} u(\tau +2)e^{t-\tau}u(-t+\tau)d\tau \end{align} $$

Now, note that the integrand is non-zero when the arguments of both step functions are greater than zero, i.e., when it holds

$$ \tau>-2 \text{ and } \tau>t \Rightarrow \tau > \max\{-2,t\}. $$

There are two cases to consider: (a) $t<-2$ and (b) $t\geq2$.

For case (a) the integral evaluates as

$$ \begin{align} y(t) &= \int_{-2}^\infty e^{-\tau} e^{t-\tau} d\tau \\ &= \frac{1}{2}e^{t+4},t<-2. \end{align} $$

For case (b) the integral evaluate as $$ \begin{align} y(t) &= \int_{t}^\infty e^{-\tau} e^{t-\tau} d\tau \\ &= \frac{1}{2}e^{-t},t\geq-2. \end{align} $$

Summarizing, the convolution equals

$$ y(t) = \begin{cases} \frac{1}{2}e^{t+4} & ,t\geq-2 \\[2ex] \frac{1}{2}e^{-t} & ,t\geq-2 \end{cases} $$