According to the Time Shifting Property of the z-transform, if X(z) is the z-transform of x(n), then what is the z-transform of x(n-k)? a) z^kX(z) b) z^(-k)X(z) c) X(z-k) d) X(z+k)
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Solve the difference equation $x_{k+1}-2 x_{k}=0$ given the initial condition that $x_{0}=3$ Taking the z-transform of each term gives: $$ \left.Z\left\{x_{k+1}\right\}-2 Z\left\{x_{k}\right\}=Z \mid 0\right\} $$ Since from equation (6), $Z\left\{x_{k+m}\right\}=z^{m} F(z)-\left[z^{m} x_{0}+z^{m-1} x_{1}+\ldots+z x_{m-1}\right]$ then $\quad\left(z^{1} Z\{k\}-\left[z^{1}(3)\right]\right)-2 Z\left\{x_{k}\right\}=0$ i.e. $\left.\quad z Z\left\{x_{k}\right\}-3 z-2 Z \mid x_{k}\right\}=0$ i.e. $\quad(z-2) Z\left\{x_{k}\right\}=3 z$ and $\quad Z\left\{x_{k}\right\}=\frac{3 z}{z-2}$ Taking the inverse $z$-transform gives: $$ \left\{x_{k}\right\}=Z^{-1}\left(\frac{3 z}{z-2}\right)=3 Z^{-1}\left(\frac{z}{z-2}\right) $$ i.e. $\quad\left\{x_{k}\right\}=3\left(2^{k}\right)$ from 6 of Table $79.1$
Determine the inverse z-transform of $F(z)=\frac{z}{z^{2}-3 z+2}$ Since $\mathrm{F}(\mathrm{z})=\frac{z}{z^{2}-3 z+2}$ then $\frac{F(z)}{z}=\frac{1}{z^{2}-3 z+2}$ Using partial fractions, let $$ \begin{aligned} \frac{1}{z^{2}-3 z+2}=\frac{1}{(z-1)(z-2)} &=\frac{A}{(z-1)}+\frac{B}{(z-2)} \\ &=\frac{A(z-2)+B(z-1)}{(z-1)(z-2)} \end{aligned} $$ from which, $1=\mathrm{A}(\mathrm{z}-2)+\mathrm{B}(\mathrm{z}-1)$ Letting $z=1$ gives: $1=-A$ i.e. $A=-1$ Letting $z=2$ gives: $1=B$ Hence, $$ \frac{F(z)}{z}=\frac{-1}{(z-1)}+\frac{1}{(z-2)}=\frac{1}{(z-2)}-\frac{1}{(z-1)} $$ and $\quad F(z)=\frac{z}{(z-2)}-\frac{z}{(z-1)}$ Thus, $$ \begin{aligned} Z^{-1} \mathrm{~F}(\mathrm{z}) &=Z^{-1}\left\{\frac{z}{(z-2)}-\frac{z}{(z-1)}\right\} \\ &=Z^{-1}\left\{\frac{z}{(z-2)}\right\}-Z^{-1}\left\{\frac{z}{(z-1)}\right\} \end{aligned} $$ From 6 in Table 79.1, $\boldsymbol{Z}^{-1} \mathbf{F}(\mathbf{z})=(2)^{k}-(1)^{k}=(\mathbf{2})^{k}-\mathbf{1}$
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