Question

Consider the signal x[n] passing through the LSI system h[n]. What would be the output?

x[n]=δ[n]−δ[n+1]
h[n]=δ[n−1]+δ[n+1]
a. −δ[n+2]+δ[n+1]−δ[n]+δ[n−1]
b. δ[n+2]−{[n+1]
c. δ[n]−δ[n−1]
d. δ[n+1]−δ[n]+2.δ[n−1]

Answer 1

The output of the given signals passing through the LSI system is 0 at all time steps.

Step-by-step explanation:

To find the output of the signal x[n] passing through the LSI system h[n], we need to convolve the two signals. The convolution operation involves shifting and multiplying the two signals at each time step, and then summing the results. Let's calculate the convolution:

1. At n = 0, x[0] = 1 and h[0] = 0, so the output at n = 0 is 1 * 0 = 0.
2. At n = 1, x[1] = -1 and h[1] = 0, so the output at n = 1 is -1 * 0 = 0.
3. At n = 2, x[2] = 0 and h[2] = 1, so the output at n = 2 is 0 * 1 = 0.
4. At n = 3, x[3] = 0 and h[3] = 0, so the output at n = 3 is 0 * 0 = 0.

Therefore, the output of the given signals x[n] = δ[n] - δ[n+1] and h[n] = δ[n-1] + δ[n+1] through the LSI system is 0 at all time steps. Answer: Option d. δ[n]−δ[n−1].