Question 9.7: z transform of an anti causal signal Find the z transform of......

Using

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

Identify {\alpha}  as -0.3^{-1} .Then

-(-0.3^{-1})^{n}\,\mathrm{u}[-n-1]\overset{\mathcal Z}{\longleftrightarrow }\frac{z}{z+0.3^{-1}},\;|z|\lt \left|-0.3^{-1}\right|

-(-10/3)^{n}\,\mathrm{u}[-n-1]\overset{\mathcal Z}{\longleftrightarrow }\frac{z}{z+10/3},\ |z|\lt |10/3|

Use the time shifting property.

-(-10/3)^{n-1}\,\mathbf{u}[-(n-1)-1]\overset{\mathcal Z}{\longleftrightarrow }z^{-1}{\frac{z}{z+10/3}}={\frac{1}{z+10/3}},\ |z|\lt |10/3|

-(-3/10)(-10/3)^{n}\,\mathrm{u}[-n]\overset{\mathcal Z}{\longleftrightarrow }\frac{z}{z+10/3},\,\ \vert z\vert\lt \vert10/3\vert

(3/10)(-10/3)^{n}\,\mathrm{u}[-n]\overset{\mathcal Z}{\longleftrightarrow }\frac{1}{z+10/3},\,\left|z\right|\lt \left|10/3\right|

Using the linearity property, multiply both sides by 4/(3/10) or 40/3.

4(-0.3)^{-n}\,\mathrm{{u}}[-n]\overset{\mathcal Z}{\longleftrightarrow }{\frac{40/3}{z+10/3}}={\frac{40}{3z+10}},\ \left|z\right|\lt |10/3|