Output for h = u(t)−u(t−3), x = u(t+4)−u(t)

The output is the convolution $y(t) = \int_{-\infty}^{\infty} h(\tau)\,x(t-\tau)\,d\tau$.

Given:

• $h(t) = u(t) - u(t-3)$: unit pulse on $0 \le t < 3$ (width 3).
• $x(t) = u(t+4) - u(t)$: unit pulse on $-4 \le t < 0$ (width 4).

The output lasts from $0 + (-4) = -4$ to $3 + 0 = 3$, and convolving two unequal rectangles gives a trapezoid with flat top height = smaller width = 3.

Limits: $x(t-\tau) = 1$ when $-4 \le t-\tau < 0$, i.e. $t < \tau \le t+4$. Combine with $0 \le \tau < 3$ from $h$.

Case 1: $t < -4$ ($t+4 < 0$): no overlap, $y(t) = 0$.

Case 2: $-4 \le t < -1$ (partial overlap, $0 \le \tau \le t+4$):

$$y(t) = \int_0^{t+4} d\tau = t + 4$$

Case 3: $-1 \le t < 0$ (window $(t,\ t+4)$ covers all of $[0,3]$):

$$y(t) = \int_0^{3} d\tau = 3$$

Case 4: $0 \le t < 3$ (partial overlap, $t \le \tau \le 3$):

$$y(t) = \int_t^{3} d\tau = 3 - t$$

Case 5: $t \ge 3$: no overlap, $y(t) = 0$.

Answer:

$$y(t) = \begin{cases} 0, & t < -4 \\ t + 4, & -4 \le t < -1 \\ 3, & -1 \le t < 0 \\ 3 - t, & 0 \le t < 3 \\ 0, & t \ge 3 \end{cases}$$

Equivalently $y(t) = r(t+4) - r(t+1) - r(t) + r(t-3)$, where $r(t) = t\,u(t)$ is the ramp.

Check: area of $y$ = (area of $h$) × (area of $x$) = $3 \times 4 = 12$; trapezoid area $= \frac{1}{2}(7 + 1)\times 3 = 12$.

 y(t)
   3 |        +----+
     |       /|    |\
     |      / |    | \
     |     /  |    |  \
   0 +----+---+----+---+----> t
         -4  -1    0   3