Question 2 of 6: Graphical Convolution of a Pulse with a Triangle

Question 2: Graphical Convolution of a Pulse with a Triangle (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Read from Figure 1 of the examination paper:

[Figure not reproduced: Figure 1(a) redrawn: the input pulse x(t), height 1 on -2. See the official exam paper.]

[Figure not reproduced: Figure 1(b) redrawn: the impulse response h(t), a triangle of peak 2 at t = 3 supported on 1. See the official exam paper.]

Find. The output $y(t)=x(t)*h(t)$ in closed piecewise form, together with its sketch, its support and its peak value.

Approach. Because $x(t)$ is a constant-height pulse, the convolution integral collapses to a running area of $h$: sliding the reversed pulse across $h$ simply integrates $h$ over a window of fixed width 3. Define the running integral $H(s)=\int_{-\infty}^{s}h(\lambda)\,d\lambda$ once, then read every region off it.

1. Set up the sliding-window integral. With $y(t)=\int_{-\infty}^{\infty}x(\tau)h(t-\tau)\,d\tau$ and $x=1$ only on $-2\le\tau\le 1$, substitute $s=t-\tau$: $$y(t)=\int_{-2}^{1}h(t-\tau)\,d\tau=\int_{t-1}^{t+2}h(s)\,ds=H(t+2)-H(t-1).$$ The window has fixed width 3 and slides to the right as $t$ increases.
2. Build the running integral of the triangle. On the rising flank $h(s)=s-1$ and on the falling flank $h(s)=5-s$, so $$H(s)=\begin{cases}0,&s\lt 1\\[2pt] \tfrac{(s-1)^{2}}{2},&1\le s\le 3\\[2pt] 4-\tfrac{(5-s)^{2}}{2},&3\le s\le 5\\[2pt] 4,&s\gt 5.\end{cases}$$ The total area $H(\infty)=4$ agrees with $\tfrac12\times 4\times 2$.
3. Locate the breakpoints. The leading edge $t+2$ crosses $s=1,3,5$ at $t=-1,1,3$; the trailing edge $t-1$ crosses the same values at $t=2,4,6$. The output therefore has five regions and support $-1\le t\le 6$ — the sum of the two supports, as convolution always requires.
4. Evaluate region by region. Substituting the two window edges into $H$ gives $$y(t)=\begin{cases}\tfrac{(t+1)^{2}}{2}, & -1\le t\le 1\\[3pt]4-\tfrac{(3-t)^{2}}{2}, & 1\le t\le 2\\[3pt]4-\tfrac{(3-t)^{2}}{2}-\tfrac{(t-2)^{2}}{2}, & 2\le t\le 3\\[3pt]4-\tfrac{(t-2)^{2}}{2}, & 3\le t\le 4\\[3pt]\tfrac{(6-t)^{2}}{2}, & 4\le t\le 6\\[3pt]0,&\text{otherwise.}\end{cases}$$
5. Find the peak. Only the middle region can hold the maximum: differentiating, $y^{\prime}(t)=(3-t)-(t-2)=5-2t$, which vanishes at $t=2.5$. Substituting, $y(2.5)=4-\tfrac{(0.5)^{2}}{2}-\tfrac{(0.5)^{2}}{2}$, i.e. $$\boxed{\,y_{\max}=3.75\ \text{at}\ t=2.5\,}$$ which is below the ceiling of 4 because the 3-wide window is narrower than the 4-wide triangle and can never enclose all of it.
6. Check the total area. Convolution multiplies areas: $\int y\,dt$ must equal $\left(\int x\right)\left(\int h\right)=3\times 4=12$. Integrating the piecewise result region by region returns exactly 12, confirming both the breakpoints and the algebra.