Question 9.4: z transforms of a causal exponential and a causal exponentia......

Using

\alpha^{n}\,\mathbf{u}[n]\overset{\mathcal Z}{\longleftrightarrow } {\frac{z}{z-\alpha}}={\frac{1}{1-\alpha z^{-1}}},\;\;|z|\gt |\alpha|

we get

e^{-n/40}\operatorname{u}[n]\overset{\mathcal Z}{\longleftrightarrow } {\frac{z}{z-e^{-1/40}}},~|z|\gt |e^{-1/40}|.

Therefore

\mathrm{X}(z)={\frac{z}{z-e^{-1/40}}},\;\;|z|\gt \left|e^{-1/40}\right|.

We can rewrite \mathrm{x}_{m}[n] as

x_{m}[n]=e^{-n/40}{\frac{e^{j2\pi n/8}-e^{-j2\pi n/8}}{j2}}\mathbf{u}[n]

or

{\bf x}_{m}[n]=-\frac{j}{2}[e^{-n/40}e^{j2\pi n/8}-e^{-n/40}e^{-j2\pi n/8}]{\bf u}[n].

Then, starting with

e^{-n/40}\ \mathrm{u}[n]{\overset{\mathcal Z}{\longleftrightarrow } {\frac{z}{z-e^{-1/40}}}},\ \ |z|\gt \left|e^{-1/40}\right|

and, using the change-of-scale property \alpha^{n}\,{{g}}[n]\overset{\mathcal Z}{\longleftrightarrow } {G}(z/\alpha), we get

e^{j2\pi n/8}e^{-n/40}\,\mathrm{u}[n]\overset{\mathcal Z}{\longleftrightarrow } {\frac{z e^{-j2\pi/8}}{z e^{-j2\pi/8}-e^{-1/40}}},\ \ |z|\gt \left|e^{-1/40}\right|

and

e^{-j2\pi n/8}e^{-n/40}\mathbf{u}[n]\overset{\mathcal Z}{\longleftrightarrow } {\frac{z e^{j2\pi/8}}{z e^{j2\pi/8}-e^{-1/40}}},\ \ |z|\gt |e^{-1/40}|

and

-{\frac{j}{2}}\Bigl[e^{-n/40}e^{j2\pi n/8}-e^{-n/40}e^{-j2\pi n/8}\Bigr]u[n]  \overset{\mathcal Z}{\longleftrightarrow } -\frac{j}{2}\biggl[\frac{z e^{-j2\pi/8}}{z e^{-j2\pi/8}-e^{-1/40}}-\frac{z e^{j2\pi/8}}{z e^{j2\pi/8}-e^{-1/40}}\biggr],~\vert z\vert\gt |e^{-1/40}|

or

{\bf X}_{m}(z)=-\frac{j}{2}\biggl[\frac{z e^{-j2\pi/8}}{z e^{-j2\pi/8}-e^{-1/40}}-\frac{z e^{j2\pi/8}}{z e^{j2\pi/8}-e^{-1/40}}\biggr]={\frac{z e^{-1/40}\sin(2\pi/8)}{z^{2}-2z e^{-1/40}\cos(2\pi/8)+e^{-1/20}}}\ ,\ \left|z\right|\gt \left|e^{-1/40}\right|

or

\mathrm{X}_{m}(z)={\frac{0.6896z}{z^{2}-1.3793z+0.9512}}

={\frac{0.6896z}{(z-0.6896-j0.6896)(z-0.6896+j0.6896)}},\ \ |z|\gt| e^{-1/40}|

(Figure 9.13).