Question

What is the impulse response of the system described by the second order difference equation y(n)-3y(n-1)-4y(n-2)=x(n)+2x(n-1)?

(a) [-\(\frac{1}{5}\) (-1)^n–\(\frac{6}{5}\) (4)^n]u(n)

(b) [\(\frac{1}{5}\) (-1)^n–\(\frac{6}{5}\) (4)^n]u(n)

(c) [\(\frac{1}{5}\) (-1)^n+\(\frac{6}{5}\) (4)^n]u(n)

(d) [-\(\frac{1}{5}\) (-1)^n+\(\frac{6}{5}\) (4)^n]u(n)

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

This key question is from Discrete Time Systems Described by Difference Equations in section Discrete Time Signals and Systems of Digital Signal Processing

Answer 1

Correct option is (d) [-\(\frac{1}{5}\) (-1)^n+\(\frac{6}{5}\) (4)^n]u(n)

Easy explanation: The homogenous solution of the given equation is yh(n)=C1(-1)^n+C2(4)^n—-(1)

To find the impulse response, x(n)=δ(n)

now, for n=0 and n=1 we get

y(0)=1 and

y(1)=3+2=5

From equation (1) we get

y(0)=C1+C2 and

y(1)=-C1+4C2

On solving the above two set of equations we get

C1=-\(\frac{1}{5}\) and C2=\(\frac{6}{5}\)

=>h(n)= [-\(\frac{1}{5}\) (-1)^n + \(\frac{6}{5}\) (4)^n]u(n).