Topic: Signals and Systems
$x(t) = \{1, 0\le t\le T; 0, \text{otherwise}\}$ and $h(t) = \{t, 0\le t\le 2T; 0, \text{otherwise}\}$. Calculate $y(t)=x(t)*h(t)$, where * denotes convolution for interval $T\le t\le 2T$
- A. 0
- B. $0.5t^2$
- C. $Tt-0.5T^2$
- D. $-0.5t^2+Tt+2.5T^2$
Answer: C. $Tt-0.5T^2$
Explanation: The convolution integral is $y(t) = \int_{0}^{T} x(\tau)h(t-\tau)d\tau$. Since $x(\tau)=1$ for $0\le \tau\le T$, $y(t) = \int_{0}^{T} h(t-\tau)d\tau$. Let $\lambda = t - \tau$, then $d\lambda = -d\tau$. The integral becomes $\int_{t-T}^{t} h(\lambda)d\lambda$. For the interval $T \le t \le 2T$, the limits $t-T$ and $t$ fully reside within the definition of $h(t)=t$. Thus, $y(t) = \int_{t-T}^{t} \lambda d\lambda = [\frac{\lambda^2}{2}]_{t-T}^{t} = \frac{t^2 - (t-T)^2}{2} = \frac{t^2 - (t^2 - 2tT + T^2)}{2} = Tt - 0.5T^2$.