# Find the Z-transform of cos⁡ωn u(n).

+1 vote
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

## 1 Answer

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by (42.1k points)
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)] = $Z\Big[\frac{e^jωn+e^{-jωn}}{2} u(n)\Big] = \frac{1}{2} Z[e^{jωn} u(n)] + \frac{1}{2} Z[e^{-jωn} u(n)]$

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

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

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