Question

What is the z-transform of the signal x(n)=a^n(sinω0n)u(n)?

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

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

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

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

This question was posed to me in an interview for job.

This is a very interesting question from Properties of Z Transform in section Z Transform and its Application – Analysis of the LTI Systems of Digital Signal Processing

Answer 1

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

To explain I would say: we know that by the linearity property of z-transform of sin(ω0n)u(n) is X(z)=\(\frac{z^{-1} sin\omega_0}{1-2z^{-1} cos\omega_0+z^{-2}}\)

Now by the scaling in the z-domain property, we have z-transform of an (sin(ω0n))u(n) as

X(az)=\(\frac{az^{-1} sin\omega_0}{1-2az^{-1} cosω_0+a^2 z^{-2}}\)