Output of e^(−2t)u(t) to a rectangular input

LTI system

An LTI system is a system that is both linear (obeys superposition) and time-invariant (a shift in input gives the same shift in output). It is fully described by its impulse response $h(t)$, and the output for any input is the convolution $y(t) = x(t) * h(t)$.

Output for $h(t) = e^{-2t}u(t)$, $x(t) = 1$ for $|t| < 5$

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

• $x(\tau) = 1$ for $-5 < \tau < 5$.
• $h(t - \tau) = e^{-2(t-\tau)}$ for $t - \tau > 0$, i.e. $\tau < t$.

So the integrand is non-zero for $-5 < \tau < \min(t, 5)$.

   x(tau)                  h(t-tau)
    ___________          .-'|
   |           |     .-'    |
 --+-----+-----+--.--------+---> tau
  -5     0     5           t

Case 1: $t < -5$. No overlap.

$$y(t) = 0$$

Case 2: $-5 \le t < 5$. Overlap from $-5$ to $t$.

$$\begin{aligned} y(t) &= \int_{-5}^{t} e^{-2(t-\tau)}\,d\tau = e^{-2t}\left[\frac{e^{2\tau}}{2}\right]_{-5}^{t} \\ &= \frac{e^{-2t}}{2}\left(e^{2t} - e^{-10}\right) = \frac{1}{2}\left(1 - e^{-2(t+5)}\right) \end{aligned}$$

Case 3: $t \ge 5$. Overlap from $-5$ to $5$.

$$\begin{aligned} y(t) &= \int_{-5}^{5} e^{-2(t-\tau)}\,d\tau = \frac{e^{-2t}}{2}\left(e^{10} - e^{-10}\right) \\ &= \frac{1}{2}\left(e^{-2(t-5)} - e^{-2(t+5)}\right) \end{aligned}$$

Result

$$y(t) = \begin{cases} 0, & t < -5 \\[2pt] \dfrac12\left(1 - e^{-2(t+5)}\right), & -5 \le t < 5 \\[6pt] \dfrac12\left(e^{-2(t-5)} - e^{-2(t+5)}\right), & t \ge 5 \end{cases}$$

 y(t)
 0.5 |      .------------.
     |    /               \
     |   /                  `.
  ---+--/--------+-----------+-`---___--> t
       -5        0           5

Checks: $y(-5) = 0$; at $t = 5$ both pieces give $\tfrac12(1 - e^{-20}) \approx 0.5$ (continuous); $y \to 0$ as $t \to \infty$. Physically, the output rises toward 0.5 with time constant 0.5 s while the input is on, then decays after $t = 5$.