Direct Z Transform

Sources

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

Example 1 - Region of Convergence Part 1

Problem 1.a.

$$x_1[n]=\{\underset{\uparrow }{1},2,5,7,0,1\}$$

$$\begin{array}{rl}{x}_1[n]& =\delta [n]+2\delta [n-1]+5\delta [n-2]+7\delta [n-3]+\delta [n-5]\\ & \\ X_1(z)& =\sum _{n=0}^{5}x(n)z^{-n}\\ & =x[0]z^{-0}+x[1]z^{-1}+x[2]z^{-2}+x[3]z^{-3}+x[4]z^{-4}+x[5]z^{-5}\end{array}$$ $$\boxed{X_1(z)=1+2z^{-1}+5z^{-2}+7z^{-3}+z^{-5}}$$

Region of Convergence (ROC): entire z-plane

Problem 1.b.

$$x_2[n]=\{1,2,\underset{\uparrow }{5},7,0,1\}$$

$$\boxed{X_2(z)=z^2+2z+5+7z^{-1}+z^{-3}}$$

Problem 1.c.

$$x_3[n]=\{\underset{\uparrow }{0},0,1,2,5,7,0,1\}$$

$$\boxed{X_3(z)=z^{-2}+2z^{-3}+5z^{-4}+7z^{-5}+z^{-7}}$$

ROC: entire z-plane, except $z=0$

Problem 1.d.

$$\begin{array}{r}{x}_4[n]=\delta [n]\end{array}$$

$$\begin{array}{rl}{X}_4(z)& =\sum _{n=-\infty }^{\infty }\delta [n]z^{-n}\\ & =\delta [0]z^{-0}\\ & =1(1)\end{array}$$ $$\boxed{X_4(z)=1}$$

ROC: entire z-plane

Problem 1.e.

$$\begin{array}{rlr}{x}_5[n]=\delta [n-k],& & k>0\end{array}$$

$$\begin{array}{rl}{X}_5(z)& =\sum _{-\infty }^{\infty }\delta [n-k]z^{-n}\\ & =\delta [0]z^{-k}\end{array}$$ $$\boxed{X_5(z)=z^{-k},k>0}$$

Problem 1.f.

$$\begin{array}{rlr}{x}_6[n]=\delta [n+k],& & k>0\end{array}$$

$$\boxed{X_6(z)=z^k,k>0}$$

Example 2 - Region of Convergence Part 2

Problem 2.a.

$$\begin{array}{r}x[n]={\alpha }^{n}u[n]=\{\begin{array}{ll}{a}^n& n\ge 0\\ 0& n<0\end{array}\end{array}$$

$$\begin{array}{rlr}X(z)& =\sum _{n=0}^{\infty }{\alpha }^n{z}^{-n}& \\ & =\sum _{n=0}^{\infty }(\alpha z^{-1}{)}^n& \\ & =\frac{1}{1-\alpha z^{-1}}& |\alpha z^{-1}|<1\\ & =\frac{1}{1-\alpha z^{-1}}& |z|>|\alpha |\end{array}$$

Tip 2.a.1. Summation of Exponentials $(0\le n<\infty )$

$$\begin{array}{rlr}\sum _{n=0}^{\infty }{A}^n& =1+A+A^2+A^3+\dots & \\ & =\frac{1}{1-A}& |A|<1\end{array}$$

Tip 2.a.2. $|z|>|\alpha |$

$$\begin{array}{rlr}|\alpha z^{-1}|& <1& \text{negative exponent means inverse/reciprocate}\\ |\alpha \left(\frac{1}{z}\right)|& <1& \text{multiply both sides by }z\\ |\alpha |& <|z|& \\ |z|& >|a|& \end{array}$$

Transform pairs:

$$\boxed{a^{n}u[n]\stackrel{\text{z}}{\leftrightarrow }\frac{1}{1-a{z}^{-1}}\text{, }|z|>|\alpha |}$$

let $\alpha =1$:

$$\boxed{u[n]\stackrel{\text{z}}{\leftrightarrow }\frac{1}{1-z^{-1}}\text{, }|z|>1}$$

Problem 2.b.

$$\begin{array}{r}x[n]=-{\alpha }^{n}u[-n-1]=\{\begin{array}{ll}0& n\ge 0\\ -{\alpha }^n& n\le -1\end{array}\end{array}$$

Tip 2.b.1. What to do with $u[-n-1]$?

$$\begin{array}{rlr}u[n]& \to 0\le n<\infty & \text{fold and shift}\\ u[-n-1]& \to -\infty <n\le -1& \end{array}$$

$$\begin{array}{rlr}X(z)& =\sum _{n=-\infty }^{-1}-{\alpha }^n{z}^{-n}& \\ & =-\sum _{n=-\infty }^{-1}{\alpha }^n{z}^{-n}& \\ & =-\sum _{n=1}^{\infty }{\alpha }^{-n}{z}^n& \\ & =-\frac{{\alpha }^{-1}z}{1-{\alpha }^{-1}z}& |a^{-1}z|<1\\ & =-\frac{1}{\alpha z^{-1}-1}& |z|<|\alpha |\\ & =\frac{1}{1-\alpha z^{-1}}& |z|<|\alpha |\end{array}$$

Tip 2.b.2. Summation of Exponentials $(1\le n<\infty )$

$$\begin{array}{rlr}\sum _{n=1}^{\infty }{A}^n& =A+A^2+A^3+\dots & \\ & & \\ & =A(1+A+A^2+\dots )& \\ & =A(\frac{1}{1-A})& |A|<1\end{array}$$

Transform pairs

$$\boxed{-a^{n}u[-n-1]\stackrel{\text{z}}{\leftrightarrow }\frac{1}{1-a{z}^{-1}}\text{, }|z|<|\alpha |}$$

Problem 2.c.

$$x[n]={\alpha }^{n}u[n]+b^{n}u[-n-1]$$

$$\boxed{\begin{array}{rl}X(z)=\underset{|z|>|\alpha |}{\underbrace{\frac{1}{1-\alpha z^{-1}}}}& -\underset{|z|<|b|}{\underbrace{\frac{1}{1-b{z}^{-1}}}}\end{array}}$$

Case $|\alpha |>|b|$:

ROC:

There is no overlapping region, hence, the z-transform does not exist

Case $|\alpha |<|b|$:

ROC :

$$\begin{array}{rlr}X(z)=\frac{1}{1-\alpha z^{-1}}& -\frac{1}{1-b{z}^{-1}}& |\alpha |<|z|<|b|\end{array}$$

Example 3 - Linearity Part 1

Problem

$$x[n]=\{3(2{)}^n-4(3{)}^n\}u[n]$$

$$x[n]=a_1{x}_1[n]+a_2{x}_2[n]$$

1st term:

$$\begin{array}{rlr}{a}_1& =3& \\ x_1[n]& =2^n& \\ X_1(z)& =\frac{1}{1-2z^{-1}}& |z|>2\end{array}$$

2nd term:

$$\begin{array}{rlr}{a}_2& =-4& \\ x_2[n]& =3^n& \\ X_2(z)& =\frac{1}{1-3z^{-1}}& |z|>3\end{array}$$

Combined:

$$\begin{array}{rl}X(z)& =a_1{x}_1(z)+a_2{x}_2(z)\\ & =\frac{3}{1-2z^{-1}}-\frac{4}{1-3z^{-1}}\end{array}$$

ROC: $|z|>3$

Tip 3. ROC

Causal:

ROC: $|z|>max\{P_1,P_2,\dots ,P_n\}$

Non-Causal:

ROC: $|z|<min\{P_1,P_2,\dots ,P_n\}$

Example 4 - Linearity Part 2

Problem 4.a.

$$x[n]=\cos ({\omega }_{0}n)u[n]$$

TIP 4.a.1. cos, sin, and jsin

$$\begin{array}{rl}\cos ({\omega }_{0}n)& =\frac{e^{j{\omega }_{0}n}+e^{-j{\omega }_{0}n}}{2}\\ \sin ({\omega }_{0}n)& =\frac{e^{j{\omega }_{0}n}-e^{-j{\omega }_{0}n}}{j2}\\ j\sin ({\omega }_{0}n)& =\frac{e^{j{\omega }_{0}n}-e^{-j{\omega }_{0}n}}{2}\end{array}$$

Tip 4.a.2 Trig Identity

$$\cos ({\omega }_{0}n)+j\sin ({\omega }_{0}n)=e^{j{\omega }_{0}n}$$

$$\begin{array}{rlr}x[n]& =\cos (w_{0}n)u[n]& \\ X(z)& =\frac{1}{2}\frac{1}{1-e^{j{\omega }_0}{z}^{-1}}+\frac{1}{2}\frac{1}{1-e^{-j{\omega }_0}{z}^{-1}},& |z|>|e^{\pm j{\omega }_0}|,|z|>1\\ & =\frac{1}{2}\frac{1-e^{-j{\omega }_0}{z}^{-1}+1-e^{j{\omega }_0}{z}^{-1}}{(1-e^{j{\omega }_0}{z}^{-1})(1-e^{-j{\omega }_0}{z}^{-1})}& \\ & =\frac{1}{2}\frac{2-2\left(\frac{e^{j{\omega }_0}+e^{-j{\omega }_0}}{2}\right)z^{-1}}{1-2\left(\frac{e^{j{\omega }_0}+e^{-j{\omega }_0}}{2}\right)z^{-1}+z^{-2}}& \end{array}$$

WARNING

I get the feeling that the reference is incomplete for this one, so I’m guessing that this is not the final answer yet

Example 5 - Time Shifting

Problem 5

$$x[n]=\{\begin{array}{ll}1& 0\le n\le N-1\\ 0& \text{elsewhere}\end{array}$$

$$x[n]=\{\underset{0\text{ to }N-1}{\underbrace{\underset{\uparrow }{1},1,1,\dots ,1}}\}$$ $$\begin{array}{rlr}x[n]& =u[n]-u[n-N]& \\ X(z)& =Z\{u[n]\}-Z\{u[n-N]\}& \\ & =Z\{u[n]\}-z^{-N}Z\{u[n]\}& \\ & =Z\{u[n]\}(1-z^{-N}),& \frac{1}{1-z^{-1}}\end{array}$$ $$\boxed{\begin{array}{rlr}X(z)& =\frac{1-z^{-N}}{1-z^{-1}},& |z|>1\end{array}}$$

Example 6 - Scaling in the Z-Domain

Problem 6.a.

$$x[n]=a^n\cos ({\omega }_{0}n)u[n]$$

$$\begin{array}{rlr}\text{let }{x}_1[n]& =\cos ({\omega }_{0}n)u[n]& \\ X_1(z)& =\frac{1-z^{-1}\cos ({\omega }_0)}{1-2z^{-1}\cos ({\omega }_0)+z^{-2}},& |z|>1\end{array}$$ $$\begin{array}{rl}X(z)& =X_1(a^{-1}z)\\ & =\frac{1-(a^{-1}z{)}^{-1}\cos ({\omega }_0)}{1-2(a^{-1}z)\cos ({\omega }_0)+({\alpha }^{-1}z{)}^{-2}}\end{array}$$ $$\boxed{X(z)=\frac{1-a{z}^{-1}\cos ({\omega }_0)}{1-2a{z}^{-1}\cos ({\omega }_0)+a^2{z}^{-2}}\text{, }|z|>|a|}$$

Problem 6.b.

$$x[n]=a^n\sin ({\omega }_{0}n)u[n]$$

$$\begin{array}{rl}\text{let }{x}_1[n]& =\sin ({\omega }_{0}n)u[n]\\ X_1(z)& =\frac{z^{-1}\sin ({\omega }_0)}{1-2z^{-1}\cos ({\omega }_0)+z^{-2}}\text{, }|z|>1\end{array}$$ $$\begin{array}{rl}X(z)& =X(a^{-1}z)\\ & =\frac{(a^{-1}z{)}^{-1}\sin ({\omega }_0)}{1-2(a^{-1}z{)}^{-1}\cos ({\omega }_0)+(a^{-1}z{)}^{-2}}\end{array}$$ $$\boxed{X(z)=\frac{a{z}^{-1}\sin ({\omega }_0)}{1-2a{z}^{-1}\cos ({\omega }_0)+a^2{z}^{-2}}\text{, }|z|>|a|}$$

Example 8 - Differentiation in the Z-Domain

Problem 8

$$x[n]=n{a}^{n}u[n]$$

$$\begin{array}{rlr}{x}_1[n]& =a^{n}u[n]& \\ X_1(z)& =\frac{1}{1-a{z}^{-1}},& |z|>|a|\end{array}$$ $$\begin{array}{rlr}X(z)& =(-z)\frac{d}{dz}\{X_1(z)\}& \\ & =(-z)\frac{d}{dz}\left\{\frac{1}{1-a{z}^{-1}}\right\}& \\ & =-z\frac{0-(1)(0+a{z}^{-2})}{(1-a{z}^{-1}{)}^2}& \\ & =-z\frac{-a{z}^{-2}}{(1-a{z}^{-1}{)}^2},& |z|>|a|\end{array}$$ $$\boxed{X(z)=\frac{a{z}^{-1}}{(1-a{z}^{-1}{)}^2}\text{, }|z|>|a|}$$

Example not in the slides 1 - Time Shifting

Problem

$$x[n]={\left(\frac{1}{2}\right)}^{n}u[n-2]$$

$$\text{let }{x}_1[n]={\left(\frac{1}{2}\right)}^{n}u[n]\to X_1(z)=\frac{1}{1-\frac{1}{2}{z}^{-1}}$$ $$\begin{array}{rl}X(z)& =Z\left\{{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^{n-2}u[n-2]\right\}\\ & =Z\left\{\frac{1}{4}{\left(\frac{1}{2}\right)}^{n-2}u[n-2]\right\}\\ & =\frac{1}{4}{z}^{-2}{X}_1(z)\end{array}$$ $$\boxed{\begin{array}{r}\frac{1}{4}\frac{z^{-2}}{1-\frac{1}{2}{z}^{-1}},|z|>\frac{1}{2}\end{array}}$$

Example not in the slides 2 - Time Reversal

Problem a

$$x[n]=u[-n]$$

$$\begin{array}{rlr}{x}_1[n]& =u[n]& \\ X_1(z)& =\frac{1}{1-z^{-1}},& |z|>1\end{array}$$ $$\begin{array}{rl}X(z)& =X_1(z^{-1})\\ & =\frac{1}{1-(z^{-1}{)}^{-1}}\end{array}$$ $$\boxed{X(z)=\frac{1}{1-z}\text{, }|z|<1}$$

Problem b

$$x[n]={\left(\frac{1}{2}\right)}^{n}u[-n]$$

$$\begin{array}{rlr}{x}_1[n]& ={\left(\frac{1}{2}\right)}^{n}u[-n]& \\ & =2^{n}u[n]& \\ X_1(z)& =\frac{1}{1-2z^{-1}},& |z|>2\\ X(z)& =X_1(z^{-1})& \end{array}$$ $$\boxed{X(z)=\frac{1}{1-2z}\text{, }|z|<\frac{1}{2}}$$

Example not in the slides 3 - Convolution of Two Sequences

Problem

$$\begin{array}{rl}{x}_1[n]& =\{\underset{\uparrow }{1},2,1\}\\ x_2[n]& =\{\underset{\uparrow }{1},1,1,1,1,1\}\end{array}$$

$$\begin{array}{rl}{X}_1(z)& =1+2z^{-1}+z^{-2}\\ X_2(z)& =1+z^{-1}+z^{-2}+z^{-3}+z^{-4}+z^{-5}\\ X(z)& =X_1(z)X_2(z)\\ & =(1+2z^{-1}+z^{-2})(1+z^{-1}+z^{-2}+z^{-3}+z^{-4}+z^{-5})\end{array}$$

Perform convolution using any preferred method

$$\boxed{\begin{array}{rl}X(z)& =1+3z^{-1}+4z^{-2}+4z^{-3}+4z^{-4}+4z^{-5}+3z^{-6}+z^{-7}\\ & \\ x[n]& =\{\underset{\uparrow }{1},3,4,4,4,4,3,1\}\end{array}}$$