Question

What is the z-transform of the signal x(n)=sin(jω0n)u(n)?

(a) \(\frac{z^{-1} sin\omega_0}{1+2z^{-1} cos\omega_0+z^{-2}}\)

(b) \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0-z^{-2}}\)

(c) \(\frac{z^{-1} cos\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)

(d) \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)

I have been asked this question in final exam.

This key question is from Properties of Z Transform topic in portion Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

The correct choice is (d) \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)

Best explanation: By Euler’s identity, the given signal x(n) can be written as

x(n) = sin(jω0n)u(n)=\(\frac{1}{2j}[e^{jω_0n} u(n)-e^{-jω_0n} u(n)]\)

Thus X(z)=\(\frac{1}{2j}[\frac{1}{1-e^{jω_0} z^{-1}}-\frac{1}{1-e^{-jω_0} z^{-1}}]\)

On simplification, we obtain

=> \(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\).