231k views
2 votes
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].

User Elyana
by
4.3k points

1 Answer

6 votes

Answer:

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

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]

User Joe Antony
by
3.9k points