Cascade of two CT LTI systems

For two LTI systems in cascade, the equivalent impulse response is the convolution of the individual impulse responses (associative property):

$$h(t) = h_1(t) * h_2(t) = \int_{-\infty}^{\infty} h_1(\tau)\,h_2(t-\tau)\,d\tau$$

x(t) -->[ h1(t) ]-->[ h2(t) ]--> y(t)
   ==  x(t) -->[ h1(t)*h2(t) ]--> y(t)

Given: $h_1(t) = e^{-t/2}u(t)$; $h_2(t) = u(t) - u(t-5)$, a unit pulse on $0 \le t < 5$.

$h_2(t-\tau) = 1$ when $0 \le t-\tau < 5$, i.e. $t-5 < \tau \le t$. Also $h_1(\tau) \ne 0$ only for $\tau \ge 0$.

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

Case 1: $t < 0$ – no overlap: $h(t) = 0$.

Case 2: $0 \le t < 5$ – limits $0$ to $t$:

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

Case 3: $t \ge 5$ – limits $t-5$ to $t$:

$$\begin{aligned} h(t) &= 2\left(e^{-(t-5)/2} - e^{-t/2}\right) \\ &= 2\left(e^{2.5} - 1\right)e^{-t/2} \approx 22.365\,e^{-t/2} \end{aligned}$$

Answer:

$$h(t) = \begin{cases} 0, & t < 0 \\ 2\left(1 - e^{-t/2}\right), & 0 \le t < 5 \\ 2\left(e^{2.5} - 1\right)e^{-t/2}, & t \ge 5 \end{cases}$$

Compactly: $h(t) = 2\left(1 - e^{-t/2}\right)u(t) - 2\left(1 - e^{-(t-5)/2}\right)u(t-5)$.

Values for the sketch:

Peak: $h(5) = 2(1 - e^{-2.5}) = 1.836$, which is continuous from both sides.

Sketch

 h(t)
1.84|            *
    |         .-' \
    |      .-'     \
    |    .'         `.
    |  .'             `-._
    | /                   `--.__
  0 +-------------+---------------> t
    0             5     10

It rises like a charging capacitor ($2(1-e^{-t/2})$) from 0 to 1.836 over $0 \le t \le 5$, then decays exponentially to zero with time constant 2 s for $t > 5$.