Convolution integral; e^(−t)u(t) * e^(−2t)u(t−3)

Derivation of the convolution integral

The convolution integral gives the output $y(t)$ of a continuous-time LTI system for any input $x(t)$, once the impulse response $h(t)$ is known.

Step 1 – Represent the input by impulses (sifting property):

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

So $x(t)$ is a continuous sum of shifted impulses $\delta(t-\tau)$, each weighted by $x(\tau)\,d\tau$.

Step 2 – Response to one impulse: by definition, $\delta(t) \rightarrow h(t)$.

Step 3 – Time invariance: $\delta(t-\tau) \rightarrow h(t-\tau)$.

Step 4 – Linearity (homogeneity): $x(\tau)\,\delta(t-\tau)\,d\tau \rightarrow x(\tau)\,h(t-\tau)\,d\tau$.

Step 5 – Linearity (additivity): adding (integrating) the responses to all the weighted impulses,

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

This is the convolution integral. By changing the variable $\lambda = t-\tau$ it can also be written $y(t)=\int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$.

Graphical steps: (1) change $t$ to $\tau$; (2) fold $h(\tau)$ to get $h(-\tau)$; (3) shift by $t$ to get $h(t-\tau)$; (4) multiply by $x(\tau)$; (5) integrate over the overlap; (6) repeat for all $t$.

Convolution of $x(t)=e^{-t}u(t)$ and $y(t)=e^{-2t}u(t-3)$

Let $z(t) = x(t) * y(t)$:

$$z(t) = \int_{-\infty}^{\infty} e^{-\tau}u(\tau)\; e^{-2(t-\tau)}u(t-\tau-3)\,d\tau$$

Limits:

• $u(\tau) = 1$ only for $\tau \ge 0$.
• $u(t-\tau-3) = 1$ only for $\tau \le t-3$.

So the product is non-zero only for $0 \le \tau \le t-3$, which needs $t \ge 3$.

Case 1: $t < 3$ – no overlap, so $z(t) = 0$.

Case 2: $t \ge 3$:

$$\begin{aligned} z(t) &= \int_{0}^{t-3} e^{-\tau}\,e^{-2t}\,e^{2\tau}\,d\tau \\ &= e^{-2t}\int_{0}^{t-3} e^{\tau}\,d\tau \\ &= e^{-2t}\left[e^{\tau}\right]_0^{t-3} \\ &= e^{-2t}\left(e^{t-3} - 1\right) \\ &= e^{-(t+3)} - e^{-2t} \end{aligned}$$

Check: at $t = 3$, $z(3) = e^{-6} - e^{-6} = 0$, so the output starts smoothly from zero. At $t = 4$, $z(4) = e^{-7} - e^{-8} \approx 0.000576$.

Answer:

$$z(t) = x(t)*y(t) = \left(e^{-(t+3)} - e^{-2t}\right)u(t-3)$$

Shortcut check: $e^{-2t}u(t-3) = e^{-6}\,e^{-2(t-3)}u(t-3)$, so it is $e^{-6}$ times $e^{-2t}u(t)$ delayed by 3. Since $e^{-t}u(t) * e^{-2t}u(t) = (e^{-t}-e^{-2t})u(t)$, the result is $e^{-6}\left(e^{-(t-3)} - e^{-2(t-3)}\right)u(t-3) = \left(e^{-(t+3)} - e^{-2t}\right)u(t-3)$, the same answer.