Finding $z$-transform of a sequence

Question

I would like to know how to find the $z$-transform of the sequence: $$x[n] = \exp(0.1 \cdot n)\cos(0.25 \cdot \pi \cdot n)u[n]$$ I don't know if I have to express the $\cos$ function in exponential.

Answer 1

First re-wirte $$\exp(0.1n)=e^{0.1n}=(e^{0.1})^n=a^n$$

Consider$$x[n]=a^n\cos(\omega_0 n)u[n],\, |a|>0$$ The z-transform is $$\begin{align} X(z)&=\sum_{n=-\infty}^{\infty}a^n\cos(\omega_0 n)u[n]z^{-n}\\ &=\sum_{n=0}^{\infty}a^n\frac{e^{j\omega_0 n}+e^{-j\omega_0 n}}{2}z^{-n}\\ &=\frac{1}{2}\left(\sum_{n=0}^{\infty}(ae^{j\omega_0}z^{-1})^ n+(ae^{-j\omega_0}z^{-1})^ n\right)\\ &=\frac{1}{2}\left(\frac{1}{1-ae^{j\omega_0}z^{-1}}+\frac{1}{1-ae^{-j\omega_0}z^{-1}}\right)\\ &=\frac{1}{2}\left(\frac{2-a(e^{j\omega_0}+e^{-j\omega_0})z^{-1}}{1-ae^{j\omega_0}z^{-1}-ae^{-j\omega_0}z^{-1}+a^2z^{-2}}\right)\\ &=\frac{1-az^{-1}\cos(\omega_0)}{1-2az^{-1}\cos(\omega_0)+a^2z^{-2}}\\ \end{align}$$ where the ROC is $|az^{-1}|<1\Rightarrow |z|>a$