Question

What is the transient 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) (0.5)^nu(n)

(b) 0.5(6.3)^nu(n)

(c) 6.3(0.5)^n

(d) 6.3(0.5)^nu(n)

I have been asked this question in an internship interview.

My question comes from Analysis of LTI System in Z Domain in chapter Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

The correct answer is (d) 6.3(0.5)^nu(n)

Easy explanation: 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 natural or transient response is

ynr(n)= 6.3(0.5)^nu(n)