Question

For the signals $x(t)$ and $y(t)$ shown in the figure, $z(t) = x(t) * y(t)$ is maximum at $t = T_{1$. Then $T_{1}$ in seconds is \underline{\hspace{1cm}} (Round off to the nearest integer).} \begin{center} \includegraphics[width=0.5\textwidth]{13.jpeg} \end{center}

Hint:

When convolving a rectangular pulse with another signal, the result is simply the running integral (sliding window) of the second signal over the pulse width. Look for where this windowed average is maximum.

Answer 1

Step 1: Write the definition of convolution.
\[ z(t) = (x*y)(t) = \int_{-\infty}^{\infty} x(\tau) \, y(t-\tau) \, d\tau \] Here, $x(t)$ is a rectangular pulse: \[ x(t) = \begin{cases} 1, & 0 \leq t \leq 1
0, & \text{otherwise} \end{cases} \] $y(t)$ is a ramp that starts from $y(1)=-2$ and increases linearly to $y(5)=2$. Its slope is $1$, so: \[ y(t) = t-3, 1 \leq t \leq 5 \]

Step 2: Simplify convolution integral.
Since $x(\tau)$ is nonzero only in $[0,1]$, \[ z(t) = \int_{0}^{1} y(t-\tau) \, d\tau \] Substitute variable $u = t-\tau$, then as $\tau$ goes from $0 \to 1$, $u$ goes from $t \to t-1$. Thus, \[ z(t) = \int_{t-1}^{t} y(u) \, du \] This means $z(t)$ is the area of $y(u)$ between $u = t-1$ and $u = t$.

Step 3: Analyze interval of $y(t)$.
Since $y(t)$ exists between $1 \leq t \leq 5$, convolution overlap occurs for $t-1 \geq 1$ and $t \leq 5$. So effective interval is: \[ 2 \leq t \leq 5 \]

Step 4: Compute integral.
Within support: \[ z(t) = \int_{t-1}^{t} (u-3) \, du = \left[\tfrac{u^{2}}{2} - 3u\right]_{t-1}^{t} \] \[ z(t) = \left(\tfrac{t^{2}}{2} - 3t\right) - \left(\tfrac{(t-1)^{2}}{2} - 3(t-1)\right) \] Simplify: \[ z(t) = \tfrac{t^{2}}{2} - 3t - \Big(\tfrac{t^{2} - 2t + 1}{2} - 3t + 3\Big) \] \[ z(t) = \tfrac{t^{2}}{2} - 3t - \tfrac{t^{2}}{2} + t - \tfrac{1}{2} + 3t - 3 \] \[ z(t) = ( \tfrac{t^{2}}{2} - \tfrac{t^{2}}{2}) + (-3t + t + 3t) + (-\tfrac{1}{2}-3) \] \[ z(t) = t - 3.5 \]

Step 5: Find maximum.
$z(t)$ is linear in $t$ with slope $+1$, so maximum occurs at the largest $t$ within range $[2,5]$. Thus, \[ T_{1} = 5 \, \text{s}. \] % Final Answer \[ \boxed{T_{1} = 5 \,\text{s}} \]