Question

What is the steady-state response of the system described by the difference equation y(n)=0.5y(n-1)+x(n) when the input signal is x(n)= 10cos(πn/4)u(n) and the system is initially at rest?

(a) 13.56cos(πn/4 -28.7^o)

(b) 13.56cos(πn/4 +28.7^o)u(n)

(c) 13.56cos(πn/4 -28.7^o)u(n)

(d) None of the mentioned

I had been asked this question in a job interview.

My query is from Analysis of LTI System in Z Domain in section Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

The correct option is (c) 13.56cos(πn/4 -28.7^o)u(n)

The explanation is: The system function for the system is

H(z)=\(\frac{1}{1-0.5z^{-1}}\)

and therefore the system has a pole at z=0.5. The z-transform of the input signal is

Y(z)=H(z)X(z)

=\(\frac{10(1-(\frac{1}{\sqrt{2}}) z^{-1})}{(1-0.5z^{-1})(1-e^{jπ/4} z^{-1})(1-e^{-jπ}/4 z^{-1})}\)  

=\(\frac{6.3}{1-0.5z^{-1}} + \frac{6.78e^{-j28.7}}{(1-e^{jπ/4} z^{-1})} + \frac{6.78e^{j28.7}}{(1-e^{-jπ/4} z^{-1})}\)

The forced state response  or steady-state response is

yfr(n)=13.56cos(πn/4 -28.7^0)u(n)