Question

Determine the output of LTI with input x[n] =2δ(n)-δ(n-1) and h[n] given as below.

The question was posed to me in an interview for internship.

Asked question is from Convolution : Impulse Response Representation for LTI Systems topic in division Time Domain Representation for LTI Systems of Signals and Systems

Answer 1

To determine the output \( y[n] \) of a Linear Time-Invariant (LTI) system with input \( x[n] = 2\delta[n] - \delta[n-1] \) and impulse response \( h[n] \), we use the convolution sum:

\[

y[n] = x[n] * h[n] = \sum_{k=-\infty}^{\infty} x[k] \cdot h[n - k]

\]

Given \( x[n] = 2\delta[n] - \delta[n-1] \), we can break down the convolution as follows:

1. **Convolve** \( 2\delta[n] \) with \( h[n] \):

   - Since \( 2\delta[n] \) is scaled by 2, it scales the response to \( h[n] \) by 2.

   - \( 2\delta[n] * h[n] = 2h[n] \).

2. **Convolve** \( -\delta[n-1] \) with \( h[n] \):

   - Shifting \( \delta[n] \) by 1 shifts the impulse response \( h[n] \) by 1.

   - \( -\delta[n-1] * h[n] = -h[n-1] \).

Combining these, we get:

\[

y[n] = 2h[n] - h[n-1]

\]

Thus, the output \( y[n] \) is \( y[n] = 2h[n] - h[n-1] \). This result depends directly on the values of \( h[n] \) provided in the problem.