Question

$x[n]$ 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) 

• A. 1.85
• B. -2.50
• C. -1.90
• D. 2.38

Hint:

In convolution, identify the values of \(x[n]\) and \(h[n]\) and use the sum to calculate the desired output at specific points.

Answer 1

The Correct Option is D

The convolution of \(x[n]\) and \(h[n]\) gives \(y[n]\), and we are given that \(y[2] = 1\) and \(y[3] = 0\). To find \(h[0]\), we need to use the properties of the convolution sum.
The convolution formula is: \[ y[n] = \sum_{k=-\infty}^{\infty} x[k] \cdot h[n-k] \] We are provided with the graphs of \(x[n]\) and \(h[n]\). By inspecting the graphs and using the convolution sum, we can determine the values of \(y[2]\) and \(y[3]\).

Step 1: Identify the values of \(x[n]\) and \(h[n]\) for convolution. By looking at the graphs of \(x[n]\) and \(h[n]\), we find the contributions to \(y[2]\) and \(y[3]\). Through calculations and trial, we determine that: \[ h[0] = 2.38 \] Thus, the value of \(h[0]\) is 2.38, which corresponds to option (D).