Value of the convolution of two rectangular pulses at t = 2
BlinkNBuild practice problem · GATE standard, authored and verified in-house
A continuous-time LTI system has impulse response $h(t) = u(t) - u(t-3)$ and is driven by $x(t) = u(t) - u(t-2)$, where $u(t)$ is the unit step. The output is $y(t) = x(t) * h(t)$.
The value of $y(2)$ is ________.
Show the step-by-step derivation
Answer
2 unitless (accept 1.98 to 2.02) unitless (accept 1.98 to 2.02)
Step-by-step derivation
- Identify the two signals as rectangular pulses of unit height: $x(t) = 1$ for $0 \le t < 2$ and zero elsewhere; $h(t) = 1$ for $0 \le t < 3$ and zero elsewhere.
- Write the convolution integral: $$y(t) = \int_{-\infty}^{\infty} x(\tau)\, h(t-\tau)\, d\tau.$$
- $x(\tau)$ is $1$ on the interval $[0,\,2]$. The reflected-and-shifted $h(t-\tau)$ is $1$ when $0 \le t - \tau < 3$, i.e. for $\tau \in (t-3,\; t]$.
- Because both signals have unit height, the integrand is $1$ wherever the two intervals overlap and $0$ elsewhere. So $y(t)$ is simply the length of the overlap of $[0,\,2]$ and $[t-3,\;t]$.
- At $t = 2$ the second interval is $[2-3,\;2] = [-1,\;2]$. Intersecting with $[0,\,2]$ gives $[0,\,2]$, whose length is $2$.
- Therefore $y(2) = \mathbf{2}$.
- Shape check. Convolving pulses of width $2$ and $3$ gives a trapezoid: it rises linearly on $0 \le t \le 2$, is flat at the value $\min(2,3) = 2$ for $2 \le t \le 3$, and falls linearly to zero at $t = 2 + 3 = 5$. The point $t = 2$ is the start of the flat top, so the peak value $2$ is exactly what we expect. ✓
The trap this question is built around
Assuming the peak equals the product of the heights. For unit-height pulses the peak of the convolution is the narrower width, not the height product - here $\min(2,3) = 2$, not $1 \times 1 = 1$. And the total duration is the sum of the widths, $2 + 3 = 5$.