The input-output relationship of an LTI system is given below.

\( \text{For an input } x[n] \text{ shown below,} \)

\( \text{the peak value of the output when } x[n] \text{ passes through } h \text{ is:} \)
Step 1: Understand the given system.
In an LTI system, the output \( y[n] \) is the convolution of the input \( x[n] \) with the system's impulse response \( h[n] \):
\[
y[n] = x[n] h[n].
\]
The system is linear and time-invariant, meaning the output is a weighted sum of the shifted impulse response, depending on the values of the input.
Step 2: Convolution of \( x[n] \) and \( h[n] \).
Looking at the given input \( x[n] \) and the impulse response \( h[n] \), we calculate the convolution \( y[n] = x[n] h[n] \). This convolution will give us the output signal, and we are interested in the peak value of the output.
Step 3: Analyze the peak value of the output.
Upon calculating the convolution, we find that the peak value of the output is 5, which occurs at the appropriate time index.
Final Answer: 5.
Given \( y(t) = e^{-3t}u(t) u(t + 3) \), where denotes convolution operation. The value of \( y(t) \) as \( t \to \infty \) is \(\underline{\hspace{2cm}}\) (rounded off to two decimal places).
$x[n]$ is convolved with $h[n]$ to give $y[n]$. If $y[2] = 1$ and $y[3] = 0$, then $h[0] = \underline{\hspace{1cm}}.$ (Graphs are not uniformly scaled)



