Convolution of e^(0.5t) window with a pulse

Use the convolution integral, sliding the pulse $h$ across the fixed input $x$:

$$y(t) = \int_{-\infty}^{\infty} x(\tau)\,h(t-\tau)\,d\tau$$

Given:

• $x(\tau) = e^{0.5\tau}$ for $0 < \tau < 5$.
• $h(t-\tau) = 1$ when $1 < t-\tau < 3$, i.e. $t-3 < \tau < t-1$.

So the shifted window is $(t-3,\ t-1)$, of width 2, and it must overlap $(0,\ 5)$.

x(tau):        |0=========5|
window:  (t-3 ----- t-1)  slides right as t increases

Useful integral: $\int_{a}^{b} e^{0.5\tau}d\tau = 2\left(e^{0.5b} - e^{0.5a}\right)$.

Case 1: $t < 1$ ($t-1 < 0$): no overlap, $y(t) = 0$.

Case 2: $1 < t < 3$ (partial overlap, window enters): limits $0$ to $t-1$.

$$y(t) = \int_0^{t-1} e^{0.5\tau}d\tau = 2\left(e^{0.5(t-1)} - 1\right)$$

Case 3: $3 < t < 6$ (window fully inside): limits $t-3$ to $t-1$.

$$\begin{aligned} y(t) &= 2\left(e^{0.5(t-1)} - e^{0.5(t-3)}\right) \\ &= 2\left(1 - e^{-1}\right)e^{0.5(t-1)} \approx 1.2642\,e^{0.5(t-1)} \end{aligned}$$

Case 4: $6 < t < 8$ (window leaving): limits $t-3$ to $5$.

$$y(t) = 2\left(e^{2.5} - e^{0.5(t-3)}\right)$$

Case 5: $t > 8$ ($t-3 > 5$): no overlap, $y(t) = 0$.

Answer:

$$y(t) = \begin{cases} 0, & t < 1 \\ 2\left(e^{0.5(t-1)} - 1\right), & 1 \le t < 3 \\ 2\left(e^{0.5(t-1)} - e^{0.5(t-3)}\right), & 3 \le t < 6 \\ 2\left(e^{2.5} - e^{0.5(t-3)}\right), & 6 \le t < 8 \\ 0, & t \ge 8 \end{cases}$$

Key values (continuity check):

The pieces meet at $t = 3$ and $t = 6$. The output rises from 0 at $t=1$, peaks at about 15.40 at $t = 6$, and falls to 0 at $t = 8$. The total duration is $5 + 2 = 7$ (from $0+1$ to $5+3$), as expected for convolution.

 y(t)
15.4|                 *
    |              .-' \
    |          .--'     \
 3.4|      .--*          \
    |   .-'               \
  0 +--*-----+-------------*--> t
       1     3        6    8