Inverse Z-Transform

Sources

1. InverseZTransform (Lecture Slides & Annotations)
2. Lecture Notes

Example 1 - Inverse z-transform Part 1

Problem 1.a.

Find the inverse z-transform of

$$X(z)=\frac{1}{1-1.5z^{-1}+0.5z^{-2}}$$

a.) $\text{ROC: }|z|>1$

b.) $\text{ROC: }|z|<0.5$

c.) $\text{ROC: }0.5<|z|<1$

Positive power of $z$

$$\begin{array}{rl}X(z)& =\frac{1}{1-1.5z^{-1}+0.5z^{-2}}\cdot \frac{z^2}{z^2}\\ & =\frac{z^2}{z^2-1.5z+0.5}\end{array}$$

Set a side one $z$

$$\begin{array}{rl}\frac{X(z)}{z}& =\frac{z}{z^2-1.5z+0.5}\\ & =\frac{z}{(z-0.5)(z-1)}\\ \frac{X(z)}{z}& =\frac{A_1}{z-0.5}+\frac{A_2}{z-1}\end{array}$$ $$\boxed{A_k=(z-P_k)\frac{X(z)}{z}{|}_{z=P_k}}$$ $$\begin{array}{rl}{A}_1& =(z-0.5)\frac{X(z)}{z}{|}_{z=0.5}\\ & =(z-0.5)\frac{1}{(z-0.5)(z-1)}{|}_{z=0.5}\\ & =\frac{z}{z-1}{|}_{z=0.5}\\ & =-1\\ A_2& =(z-1)\frac{X(z)}{z}{|}_{z=1}\\ & =\frac{z}{z-0.5}{|}_{z=1}\\ & =2\end{array}$$ $$\begin{array}{rl}\frac{X(z)}{z}& =\frac{-1}{z-0.5}+\frac{2}{z-1}\\ X(z)& =\frac{-z}{z-0.5}+\frac{2z}{z-1}\\ X(z)& =\frac{-1}{1-0.5z^{-1}}+\frac{2}{1-z^{-1}}\end{array}$$

a.) ROC: |z| > 1 (Causal)

$$\boxed{\begin{array}{rl}x[n]& =-(0.5{)}^{n}u[n]+2(1{)}^{n}u[n]\\ & =-(0.5{)}^{n}u[n]+2u[n]\end{array}}$$

b.) ROC: |z| < 0.5 (Anticausal)

$$\boxed{x[n]=(0.5{)}^{n}u[-n-1]-2u[-n-1]}$$

c.) ROC: 0.5 < |z| < 1 (Noncausal)

$$\boxed{x[n]=\underset{|z|>0.5}{\underbrace{-(0.5{)}^{n}u[n]}}-\underset{|z|<1}{\underbrace{2u[-n-1]}}}$$

Example 2 - Inverse z-transform Part 2

Problem 2

Find the inverse z-transform of

$$X(z)=\frac{1}{(1+z^{-1})(1-z^{-1}{)}^2}$$

Positive powers of $z$

$$\begin{array}{rl}X(z)& =\frac{1}{(1+z^{-1})(1-z^{-1})}\cdot \frac{z^3}{z^3}\\ X(z)& =\frac{z^3}{(z+1)(z-1{)}^2}\\ \frac{X(z)}{z}& =\frac{z^2}{(z+1)(z-1{)}^2}\\ \frac{X(z)}{z}& =\frac{A_1}{z+1}+\frac{A_2}{z-1}+\frac{A_3}{(z-1{)}^2}\end{array}$$ $$\begin{array}{rl}{A}_1& =(z+1)\frac{X(z)}{z}{|}_{z=-1}\\ & =\cancel{(z+1)}\frac{z^2}{\cancel{(z+1)}(z-1)}{|}_{z=-1}\\ & =\frac{z^2}{(z-1{)}^2}{|}_{z=-1}\\ A_1& =\frac{1}{4}\end{array}$$ $$\begin{array}{rl}{A}_2& =\frac{d}{dz}\left\{(z-1{)}^2\frac{X(z)}{z}\right\}{|}_{z=1}\\ & =\frac{d}{dz}\left\{\cancel{(z-1{)}^2}\frac{z^2}{(z+1)(z-1{)}^2}\right\}{|}_{z=1}\\ & =\frac{d}{dz}\left\{\frac{z^2}{z+1}\right\}{|}_{z=1}\\ & =\frac{(z+1)(2z)-z^2(1)}{(z+1{)}^2}{|}_{z=1}\\ A_2& =\frac{3}{4}\end{array}$$ $$\begin{array}{rl}{A}_3& =\frac{z^2}{z+1}{|}_{z=1}\\ A_3& =\frac{1}{2}\end{array}$$ $$\begin{array}{rl}\frac{X(z)}{z}& =\frac{1}{4}\frac{1}{z+1}+\frac{3}{4}\frac{1}{z-1}+\frac{1}{2}\frac{1}{(z-1{)}^2}\\ X(z)& =\frac{1}{4}\frac{z}{z+1}+\frac{3}{4}\frac{z}{z-1}+\frac{1}{2}\frac{z}{(z-1{)}^2}\\ & =\frac{1}{4}\frac{1}{1+z^{-1}}+\frac{3}{4}\frac{1}{1-z^{-1}}+\frac{1}{2}\frac{z}{{\left(\frac{z}{z}-\frac{1}{z}\right)}^2}\end{array}$$

Consider:

$$\frac{z}{z^2{\left(\frac{z}{z}-\frac{1}{z}\right)}^2}=\frac{z}{z^2{\left(1-z^{-1}\right)}^2}=\frac{z^{-1}}{(1-z^{-1}{)}^2}$$

Tip 2. $n{a}^{n}u[n]$

$$n{a}^{n}u[n]⟷\frac{a{z}^{-1}}{(1-a{z}^{-1}{)}^2}\text{, }|z|>|a|$$

$$z^{-}1\left\{X(z)=\frac{1}{4}\frac{1}{1+z^{-1}}+\frac{3}{4}\frac{1}{1-z^{-1}}+\frac{1}{2}\frac{z^{-1}}{{\left(1-z^{-1}\right)}^2}\right\}{z}^{-}1$$ $$\boxed{x(n)=\frac{1}{4}(-1{)}^{n}u[n]+\frac{3}{4}u[n]+\frac{1}{2}nu[n]}$$

Example 3 - Inverse z-transform Part 2

Problem 3

Find the inverse z-transform of

$$X(z)=\frac{1+z^{-1}}{1-z^{-1}+0.5z^{-2}}$$

$$\begin{array}{c}M=1,N=2,N>M\end{array}$$

This is already a proper fraction:

Positive powers of $z$:

$$\begin{array}{rl}X(z)& =\frac{z^2+z}{z^2-z+0.5}\\ \frac{X(z)}{z}& =\frac{z+1}{z^2-z+0.5}\\ & =\frac{A_1}{z-P_1}+\frac{A_2}{z-P_2}\end{array}$$ $$\begin{array}{rl}{P}_1& =\frac{1}{2}+\frac{j}{2}\\ P_2& =P_1^{\ast }=\frac{1}{2}-\frac{j}{2}\end{array}$$ $$\begin{array}{rl}\frac{X(z)}{z}& =\frac{z+1}{(z-P_1)(z-P_2)}\\ & =\frac{A_1}{z-P_1}+\frac{A_2}{z-P_2}\end{array}$$ $$\boxed{\begin{array}{rl}{A}_2& =A_1^{\ast }\\ P_2& =P_1^{\ast }\end{array}}$$ $$\begin{array}{rl}{A}_1& =(z-P_1)\frac{X(z)}{z}{|}_{z=P_1}\\ & =\frac{z+1}{z-P_2}{|}_{z=P_1}\\ & =\frac{P_1+1}{P_1-P_2}\end{array}$$ $$\begin{array}{c}{A}_1=\frac{1}{2}-j\frac{3}{2},A_2=A_1^{\ast }=\frac{1}{2}+j\frac{3}{2}\end{array}$$ $$\begin{array}{rl}\frac{X(z)}{z}& =\frac{A_1}{z-P_1}+\frac{A_1^{\ast }}{z-P_1^{\ast }}\\ X(z)& =\frac{A_1}{1-P_1{z}^{-1}}+\frac{A_1^{\ast }}{1-P_1^{\ast }{z}^{-1}}\end{array}$$ $$\underline{x[n]=A_1(P_1{)}^{n}u[n]+A_1^{\ast }(P_1^{\ast }{)}^{n}u[n]}$$

$A_K$ and $P_K$

$$\begin{array}{r}{A}_K=|A_K|e^{j{\alpha }_K}\\ P_K=r_K{e}^{j\beta K}\end{array}$$

$$x[n]=2|A_K|r_k^n\cos (\underset{45^{\circ }n\text{ or}\frac{\pi }{4}n}{\underbrace{{\beta }_{K}n}}+{\alpha }_K)$$

if $k=1$

$$\begin{array}{rl}{A}_1& =\frac{1}{2}-j\frac{3}{2}=\frac{\sqrt{10}}{2}{e}^{-j71.57^{\circ }}\\ P_1& =\frac{1}{2}+j\frac{1}{2}=\frac{1}{\sqrt{2}}{e}^{j\frac{\pi }{4}}\end{array}$$ $$\begin{array}{rl}x[n]& =2\left(\frac{\sqrt{10}}{2}\right){\left(\frac{1}{\sqrt{2}}\right)}^n\cos \left(\frac{\pi }{4}n-71.57^{\circ }\right)u[n]\end{array}$$ $$\boxed{x[n]=\sqrt{10}{\left(\frac{1}{\sqrt{2}}\right)}^n\cos \left(\frac{\pi }{4}n-71.57^{\circ }\right)u[n]}$$

Example 4 - Time Shifting Property

Problem 4

$$x_1[n]=x[n-2]$$

$$\begin{array}{r}{X}_1^{+}(z)=z^{-2}{X}_1^{+}(z)+x_1[-1]z^{-1}+x[-2]\end{array}$$

Tip 4. $x[n]$ and $x^{+}(z)$

$$\begin{array}{rl}x[n]& =a^n\\ X^{+}(z)& =\frac{1}{1-a{z}^{-1}}\end{array}$$

$$\boxed{X^{+}(z)=\frac{z^{-2}}{1-a{z}^{-1}}+a^{-1}{z}^{-1}+a^{-2}}$$

Example 5 - Using Z-Transform to Solve an LCCDE

Problem

$$y[n]=\frac{3}{5}y[n-1]-\frac{2}{25}y[n-2]+x[n-1]+\frac{1}{2}x[n-2]$$

Find:

1. The impulse response

2. The zero-state step response

3. The step response if $y[-1]=1$ and $y[n-2]=2$.

Take the one-sided $z$-transform of both sides

$$\begin{array}{rl}{Y}^{+}(z)& =\frac{3}{5}\left\{z^{-1}{Y}^{+}(z)+y[-1]\right\}-\frac{2}{25}\left\{z^{-2}{Y}^{+}(z)+y[-1]z^{-1}+y[-2]\right\}+\{z^{-1}{X}^{+}(z)+x[-1]\}+\frac{1}{2}\{z^{-2}{X}^{+}(z)+x[-1]z^{-1}+x[-2]\}\\ \left(1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}\right)Y^{+}(z)& =\frac{3}{5}y[-1]-\frac{2}{25}y[-1]z^{-1}-\frac{2}{25}y[-2]+\left(z^{-1}+\frac{1}{2}{z}^{-2}\right)x^{+}(z)+x[-1]+\frac{1}{2}x[-1]z^{-1}+\frac{1}{2}x[-2]\end{array}$$

5.1. Impulse Response

$h[n]$:

$$\begin{array}{rl}\text{Input}& :x[n]=\delta [n]\\ \text{Initial conditions}& :y[-1]=y[-2]=0\end{array}$$ $$\begin{array}{rl}\left(1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}\right)Y^{+}(z)& =\left(z^{-1}+\frac{1}{2}{z}^{-2}\right)X^{+}(z)\\ H(z)& =\frac{Y^{+}(z)}{X^{+}(z)}=\frac{z^{-1}+\frac{1}{2}{z}^{-2}}{1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}}\end{array}$$

Perform the following long division:

$$\frac{2}{25}{z}^{-2}-\frac{3}{5}{z}^{-1}+1\overline{)\frac{1}{2}{z}^{-2}+z^{-1}}$$

Resulting in $\frac{19}{4}{z}^{-1}-\frac{25}{4}$

$$H(z)=\frac{25}{4}+\frac{\frac{19}{4}{z}^{-1}-\frac{25}{4}}{1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}}$$

Consider:

$$\begin{array}{rl}{H}_1(z)& =\frac{\frac{19}{4}{z}^{-1}-\frac{25}{4}}{1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}}\cdot \frac{z^2}{z^2}\\ \frac{H_1(z)}{z}& =\frac{\frac{19}{4}-\frac{25}{4}z}{\left(z-\frac{1}{5}\right)\left(z-\frac{2}{5}\right)}=\frac{A_1}{z-\frac{1}{5}}+\frac{A_2}{z-\frac{2}{5}}\\ A_1& =\frac{\frac{19}{4}-\frac{25}{4}z}{z-\frac{2}{5}}{|}_{z=\frac{1}{5}}=-\frac{35}{2}\\ A_2& =\frac{\frac{19}{4}-\frac{25}{4}z}{z-\frac{1}{5}}{|}_{z=\frac{2}{5}}=\frac{45}{4}\\ H_1(z)& =-\frac{35}{2}\frac{1}{1-\frac{1}{5}{z}^{{}_{-1}}}+\frac{45}{4}\frac{1}{1-\frac{2}{5}{z}^{-1}}\\ H(z)& =\frac{25}{4}-\frac{35}{2}\frac{1}{1-\frac{1}{5}{z}^{-1}}+\frac{45}{4}\frac{1}{1-\frac{2}{5}{z}^{-1}}\end{array}$$ $$\boxed{h[n]=\frac{25}{4}\delta [n]-\frac{35}{2}{\left(\frac{1}{5}\right)}^{n}u[n]+\frac{45}{4}{\left(\frac{2}{5}\right)}^{n}u[n]}$$

5.2. Zero-State Step Response

$$\begin{array}{rl}x[n]& =u[n]\\ \text{zero state: }y[-1]& =y[-2]=0\end{array}$$ $$\begin{array}{rl}{x}^{+}(z)& =z^{+}\{u[n]\}\\ & =\frac{1}{1-z^{-1}}\end{array}$$ $$\begin{array}{rl}{Y}^{+}(z)\left(1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}\right)& =\left(\frac{1}{2}{z}^{-2}+z^{-1}\right)X^{+}(z)\\ Y^{+}(z)\left(1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}\right)& =\left(\frac{1}{2}{z}^{-2}+z^{-1}\right)\left(\frac{1}{1-z^{-1}}\right)\\ Y^{+}(z)& =\frac{\frac{1}{2}{z}^{-2}+z^{-1}}{\left(1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}\right)(1-z^{-1})}\cdot \frac{z^3}{z^3}\\ & =\frac{\frac{1}{2}z+z^2}{\left(z^2-\frac{3}{5}z+\frac{2}{25}\right)(z-1)}\\ \frac{Y^{+}(z)}{z}& =\frac{z+\frac{1}{2}}{\left(z-\frac{1}{5}\right)\left(z-\frac{2}{5}\right)(z-1)}\\ & =\frac{A_1}{z-\frac{1}{5}}+\frac{A_2}{z-\frac{2}{5}}+\frac{A_3}{z-1}\end{array}$$ $$\begin{array}{rl}{A}_1& =\frac{z+\frac{1}{2}}{\left(z-\frac{2}{5}\right)(z-1)}{|}_{z=\frac{1}{5}}=\frac{35}{8}\\ A_2& =\frac{z+\frac{1}{2}}{\left(z-\frac{1}{5}\right)(z-1)}{|}_{z=\frac{2}{5}}=-\frac{15}{2}\\ A_3& =\frac{z+\frac{1}{2}}{\left(z-\frac{1}{5}\right)\left(z-\frac{2}{5}\right)}{|}_{z=1}=\frac{25}{8}\end{array}$$ $$\begin{array}{rl}{Y}^{+}(z)& =\frac{35}{8}\frac{1}{1-\frac{1}{5}{z}^{-1}}-\frac{15}{2}\frac{1}{1-\frac{2}{5}{z}^{-1}}+\frac{25}{8}\frac{1}{1-z^{-1}}\end{array}$$ $$\boxed{y[n]=\frac{35}{8}{\left(\frac{1}{5}\right)}^{n}u[n]-\frac{15}{2}{\left(\frac{2}{5}\right)}^{n}u[n]+\frac{25}{8}u[n]}$$

5.3. Zero-State Step Response

$$\begin{array}{c}x[n]=u[n]y[-1]=1y[-2]=2\end{array}$$ $$\begin{array}{rl}{Y}^{+}(z)\left(1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}\right)& =\frac{3}{5}(1)-\frac{2}{25}(1)z^{-1}-\frac{2}{25}(2)+\left(\frac{1}{2}{z}^{-2}+z^{-}1\right)\left(\frac{1}{1-z^{-1}}\right)\\ & =\frac{\left(\frac{11}{25}-\frac{2}{25}{z}^{-1}\right)(1-z^{-1})+\frac{1}{2}{z}^{-2}+z^{-1}}{(1-z^{-1})}\\ Y^{+}(z)& =\frac{\frac{11}{25}+\frac{34}{25}{z}^{-1}+\frac{29}{50}{z}^{-2}}{\left(1-\frac{3}{5}{z}^{-1}+\frac{2}{25}{z}^{-2}\right)(1-z^{-1})}\cdot \frac{z^3}{z^3}\\ & =\frac{\frac{11}{25}{z}^3+\frac{34}{25}{z}^2+\frac{29}{50}z}{\left(z-\frac{1}{5}\right)\left(z-\frac{2}{5}\right)(z-1)}\\ \frac{Y^{+}(z)}{z}& =\frac{A_1}{z-\frac{1}{5}}+\frac{A_2}{z-\frac{2}{5}}+\frac{A_3}{z-1}\end{array}$$ $$\begin{array}{c}{A}_1=\frac{1087}{200}{A}_2=-\frac{1493}{150}{A}_3=\frac{119}{24}\end{array}$$ $$\boxed{y[n]=\left[\frac{1087}{200}{\left(\frac{1}{5}\right)}^n-\frac{1493}{150}{\left(\frac{2}{5}\right)}^n+\frac{119}{24}\right]u[n]}$$

Example not in the slides - Time Shifting Property

Problem

$$x[n-2]=x^{+}(z)+\sum _{n=1}^{2}x[-n]z^n$$

$$\begin{array}{rl}{z}^{+}\left\{x[n-2]\right\}& =z^{-2}\left\{X^{+}(z)+\sum _{n=1}^{2}x[-n]z^n\right\}\\ & =z^{-2}\left\{X^{+}(z)+x[-1]z+x[-2]z^2\right\}\\ & =z^{-2}{X}^{+}(z)+X[-1]z^{-1}+x[-2]\end{array}$$ $$\begin{array}{rl}{z}^{+}x[n-2]& =z^{-2}{x}^{+}(z)+x[-1]+x[-2]\\ z^{+}x[n-1]& =z^{-1}{x}^{+}(z)+x[-1]\\ z^{+}x[n-3]& =z^{-3}{x}^{+}(z)+x[-1]z^{-2}+x[-2]z^{-1}+x[-3]\end{array}$$