Question

Find the Z-transform of cos⁡ωn u(n).

(a) \(\frac{z(z+cos⁡ω)}{z^2-2z cos⁡ω+1}\)

(b) \(\frac{z(z-cos⁡ω)}{z^2-2z cos⁡ω+1}\)

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

(d) \(\frac{z(z+cos⁡ω)}{z^2+2z cos⁡ω+1}\)

I have been asked this question during an online exam.

My question comes from The Z-Transform topic in portion Z-Transform and Digital Filtering of Signals and Systems

Answer 1

The correct answer is (b) \(\frac{z(z-cos⁡ω)}{z^2-2z cos⁡ω+1}\)

The best explanation: Given x(n) = cos⁡ωn u(n)

We know that \(

u(n)=\begin{cases}

1 &\text{\(n≥0\)} \\

0 &\text{\(n<0\)} \\

\end{cases}\)

Z[cos⁡ωn u(n)] = \(Z\Big[\frac{e^jωn+e^{-jωn}}{2} u(n)\Big] = \frac{1}{2} Z[e^{jωn} u(n)] + \frac{1}{2} Z[e^{-jωn} u(n)]\)

\(= \frac{1}{2} \left(\frac{z}{z-e^{jω}} + \frac{z}{z-e^{-jω}}\right) = \frac{1}{2} \Big[\frac{z(z-e^{-jω}) + z(z-e^{jω})}{(z-e^{jω})(z-e^{-jω})}\Big]\)

\(= \frac{1}{2} \Big\{\frac{z[2z-(e^{jω}+e^{-jω})]}{z^2-z(e^{jω}+e^{-jω})+1}\Big\} = \frac{z(z-cos⁡ω)}{z^2-2z cos⁡ω+1}\).