y(t) for h = e^(−t)u(t−3), x = u(t−2)

For an LTI system, $y(t) = x(t) * h(t) = \displaystyle\int_{-\infty}^{\infty} x(\tau)\,h(t - \tau)\,d\tau$.

Given $x(t) = u(t-2)$ and $h(t) = e^{-t}u(t-3)$.

Step 1: Write both functions in $\tau$

$$x(\tau) = u(\tau - 2), \qquad h(t-\tau) = e^{-(t-\tau)}\,u(t - \tau - 3)$$

• $x(\tau) \neq 0$ when $\tau > 2$.
• $h(t-\tau) \neq 0$ when $t - \tau > 3$, i.e. $\tau < t - 3$.

 x(tau)=1 for tau>2      h(t-tau) for tau<t-3
          ___________   ___________
         |              decaying   |
 --------+-------->   <------------+------ tau
         2                        t-3
 overlap: 2 < tau < t-3  (exists only if t-3 > 2)

Step 2: Range of $t$

The overlap $2 < \tau < t - 3$ exists only when $t - 3 > 2$, i.e. $t > 5$.

• For $t < 5$: no overlap, $y(t) = 0$.

Step 3: Evaluate for $t > 5$

$$\begin{aligned} y(t) &= \int_{2}^{t-3} 1\cdot e^{-(t-\tau)}\,d\tau = e^{-t}\int_{2}^{t-3} e^{\tau}\,d\tau \\ &= e^{-t}\left[e^{\tau}\right]_{2}^{t-3} = e^{-t}\left(e^{t-3} - e^{2}\right) \\ &= e^{-3} - e^{-(t-2)} \end{aligned}$$

Result

$$y(t) = \left(e^{-3} - e^{-(t-2)}\right)u(t-5)$$

Answer: $y(t) = 0$ for $t < 5$, and $y(t) = e^{-3} - e^{-(t-2)}$ for $t \ge 5$.

Checks:

• At $t = 5$: $e^{-3} - e^{-3} = 0$ (continuous start).
• As $t \to \infty$: $y \to e^{-3} \approx 0.0498$.
• Shift view: $u(t)*e^{-t}u(t) = (1 - e^{-t})u(t)$; here $h(t) = e^{-3}\,e^{-(t-3)}u(t-3)$, so total shift is $2 + 3 = 5$: $y(t) = e^{-3}(1 - e^{-(t-5)})u(t-5)$, which is the same expression.