Question

H is a discrete-time LTI system with impulse response h[n] = δ[n] − 2δ[n − 1] + 3δ[n − 2]. Find the output signal y[n] in terms of the input signal x[n] (use the convolution equation and the properties of convolution with deltas). Hint: since the input signal x[n] is unknown, your answer should express the output signal y[n] as a linear combination of the shifts of x[n].

Answer 1

Answer:

The output is given as $y(n)=x[n]-2x[n-1]+3x[n-2]$

Step-by-step explanation:

As the output is given as y(n) and the input is given as x(n) thus the

equation of the output is given as

$y(n)=h(n)*x(n)$

Here the impulse response h(n) is given as

$h(n)=\delta[n]-2\delta[n-1]+3\delta[n-2]$

So the output is given as

$y(n)=h(n)*x(n)\\y(n)=(\delta[n]-2\delta[n-1]+3\delta[n-2])*x(n)$

By distributive property,$(ax1[n] + bx2[n]) * h[n] = ax1[n] * h[n] + bx2[n] * h[n]$

$y(n)=h(n)*x(n)\\y(n)=\delta[n]*x(n)-2\delta[n-1]*x(n)+3\delta[n-2]*x(n)$

Now by the convolution properties with delta $\delta[n- k] *x[n] = x[n-k]$

so

$y(n)=\delta[n]*x(n)-2\delta[n-1]*x(n)+3\delta[n-2]*x(n)\\y(n)=\delta[n-0]*x(n)-2\delta[n-1]*x(n)+3\delta[n-2]*x(n)\\y(n)=x[n]-2x[n-1]+3x[n-2]$

So the output is given as $y(n)=x[n]-2x[n-1]+3x[n-2]$