Lecture #15:

THE BILATERAL Z-TRANSFORM

Motivation: Method for representing DT signals as superpositions of complex geometric (exponential) functions

Outline:

– Definition

– Properties

Review of last lecture

Solve linear difference equation for a causal exponential input

∑ k = 0 K a k y [ n + k ] = ∑ l = 0 L b l x [ n + l ] for x [ n ] = Xz n u [ n ] ∑ k = 0 K a k y [ n + k ] = ∑ l = 0 L b l x [ n + l ] for x [ n ] = Xz n u [ n ] size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y \[ n+k \] ={}} Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } x \[ n+l \] } matrix { {} # {} } ital "for" matrix { {} # {} } x \[ n \] = ital "Xz" rSup { size 8{n} } u \[ n \] } {}

Solve homogeneous equation for n > 0

∑ k = 0 K a k y h [ n + k ] = 0 by assu min g y h [ n ] = Aλ n ∑ k = 0 K a k y h [ n + k ] = 0 by assu min g y h [ n ] = Aλ n size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y rSub { size 8{h} } \[ n+k \] ={}} 0 matrix { {} # {} } ital "by" matrix { {} # {} } ital "assu""min"g matrix { {} # {} } y rSub { size 8{h} } \[ n \] =Aλ rSup { size 8{n} } } {}

Solve characteristic polynomial for λ.

∑ k = 0 K a k λ n + k = 0 ∑ k = 0 K a k λ n + k = 0 size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } λ rSup { size 8{n+k} } ={}} 0} {}

Solve for a particular solution for n > 0

∑ k = 0 K a k y p [ n + k ] = ∑ l = 0 L b l x [ n + l ] for x [ n ] = Xz n u [ n ] ∑ k = 0 K a k y p [ n + k ] = ∑ l = 0 L b l x [ n + l ] for x [ n ] = Xz n u [ n ] size 12{ Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } y rSub { size 8{p} } \[ n+k \] ={}} Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } x \[ n+l \] } matrix { {} # {} } ital "for" matrix { {} # {} } x \[ n \] = ital "Xz" rSup { size 8{n} } u \[ n \] } {}

Assuming yp[n]= Yznyp[n]= Yzn size 12{y rSub { size 8{p} } \[ n \] =" Yz" rSup { size 8{n} } } {} and solving for Y yields

Y = H ~ ( z ) X = ∑ l = 0 L b l z l ∑ k = 0 K a k z k X Y = H ~ ( z ) X = ∑ l = 0 L b l z l ∑ k = 0 K a k z k X size 12{Y= {H} cSup { size 8{ "~" } } \( z \) X= { { Sum cSub { size 8{l=0} } cSup { size 8{L} } {b rSub { size 8{l} } } z rSup { size 8{l} } } over { Sum cSub { size 8{k=0} } cSup { size 8{K} } {a rSub { size 8{k} } } z rSup { size 8{k} } } } X} {}

Logic for an analysis method for DT LTI systems

I. THE BILATERAL Z-TRANSFORM

1/ Definition

The bilateral Z-transform is defined by the analysis formula

X ~ ( z ) = ∑ n = − ∞ ∞ x [ n ] z − n X ~ ( z ) = ∑ n = − ∞ ∞ x [ n ] z − n size 12{ {X} cSup { size 8{ "~" } } \( z \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x \[ n \] z rSup { size 8{ - n} } } } {}

X~(z)X~(z) size 12{ {X} cSup { size 8{ "~" } } \( z \) } {} is defined for a region in z — called the region of convergence — for which the sum exists.

The inverse transform is defined by the synthesis formula

x [ n ] = 1 2πj ∫ C X ~ ( z ) z n − 1 dz x [ n ] = 1 2πj ∫ C X ~ ( z ) z n − 1 dz size 12{x \[ n \] = { {1} over {2πj} } Int rSub { size 8{C} } { {X} cSup { size 8{ "~" } } \( z \) } z rSup { size 8{n - 1} } ital "dz"} {}

Since z is a complex quantity, X~(z)X~(z) size 12{ {X} cSup { size 8{ "~" } } \( z \) } {}is a complex function of a complex variable. Hence, the synthesis formula involves integration in the complex z domain. We shall not perform this integration in this subject. The synthesis formula will be used only to prove theorems and not to compute time functions directly.

a/ Approach

An inventory of time functions and their Z-transforms will be developed by

b/ Notation

We shall use two useful notations — Z{x[n]} signifies the Z-transform of x[n] and a Z-transform pair is indicated by

x [ n ] ⇔ Z X ~ ( z ) x [ n ] ⇔ Z X ~ ( z ) size 12{x \[ n \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) } {}

2/ Properties

a/ Linearity

ax 1 [ n ] + bx 2 [ n ] ⇔ Z a X ~ 1 ( z ) + b X ~ 2 ( z ) ax 1 [ n ] + bx 2 [ n ] ⇔ Z a X ~ 1 ( z ) + b X ~ 2 ( z ) size 12{ ital "ax" rSub { size 8{1} } \[ n \] + ital "bx" rSub { size 8{2} } \[ n \] { dlrarrow } cSup { size 8{Z} } a {X} cSup { size 8{ "~" } } rSub { size 8{1} } \( z \) +b {X} cSup { size 8{ "~" } } rSub { size 8{2} } \( z \) } {}

The proof follows from the definition of the Z-transform as a sum.

X ~ ( z ) = ∑ n = − ∞ ∞ ( ax 1 [ n ] + bx 2 [ n ] ) z − n X ~ ( z ) = a ∑ n = − ∞ ∞ x 1 [ n ] z − n + b ∑ n = − ∞ ∞ x 2 [ n ] z − n X ~ ( z ) = a X 1 ~ ( z ) + b X 2 ~ ( z ) X ~ ( z ) = ∑ n = − ∞ ∞ ( ax 1 [ n ] + bx 2 [ n ] ) z − n X ~ ( z ) = a ∑ n = − ∞ ∞ x 1 [ n ] z − n + b ∑ n = − ∞ ∞ x 2 [ n ] z − n X ~ ( z ) = a X 1 ~ ( z ) + b X 2 ~ ( z ) alignl { stack { size 12{ {X} cSup { size 8{ "~" } } \( z \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { \( ital "ax" rSub { size 8{1} } \[ n \] + ital "bx" rSub { size 8{2} } \[ n \] \) z rSup { size 8{ - n} } } } {} # {X} cSup { size 8{ "~" } } \( z \) =a Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x rSub { size 8{1} } \[ n \] z rSup { size 8{ - n} } } +b Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {x rSub { size 8{2} } \[ n \] z rSup { size 8{ - n} } } {} # {X} cSup { size 8{ "~" } } \( z \) =a {X rSub { size 8{1} } } cSup { size 8{ "~" } } \( z \) +b {X rSub { size 8{2} } } cSup { size 8{ "~" } } \( z \) {} } } {} {}

b/ Delay by k

x [ n − k ] ⇔ Z z − k X ~ ( z ) x [ n − k ] ⇔ Z z − k X ~ ( z ) size 12{x \[ n - k \] { dlrarrow } cSup { size 8{Z} } z rSup { size 8{ - k} } {X} cSup { size 8{ "~" } } \( z \) } {}

This result can be seen using the synthesis formula,

x [ n − k ] = 1 2πj ∫ X ~ ( z ) z n − k − 1 dz x [ n − k ] = 1 2πj ∫ z − k X ~ ( z ) z n − 1 dz x [ n − k ] = 1 2πj ∫ X ~ ( z ) z n − k − 1 dz x [ n − k ] = 1 2πj ∫ z − k X ~ ( z ) z n − 1 dz alignl { stack { size 12{x \[ n - k \] = { {1} over {2πj} } Int { {X} cSup { size 8{ "~" } } \( z \) z rSup { size 8{n - k - 1} } ital "dz"} } {} # x \[ n - k \] = { {1} over {2πj} } Int {z rSup { size 8{ - k} } {X} cSup { size 8{ "~" } } \( z \) z rSup { size 8{n - 1} } ital "dz"} {} } } {}

c/ Multiply by n

nx [ n ] ⇔ Z − z d X ~ ( x ) dz nx [ n ] ⇔ Z − z d X ~ ( x ) dz size 12{ ital "nx" \[ n \] { dlrarrow } cSup { size 8{Z} } - z { {d {X} cSup { size 8{ "~" } } \( x \) } over { ital "dz"} } } {}

This result can be seen using the analysis formula.

d X ~ ( z ) dz = ∑ n = − ∞ ∞ − nx [ n ] z − n − 1 − z d X ~ ( z ) dz = ∑ n = − ∞ ∞ nx [ n ] z − n d X ~ ( z ) dz = ∑ n = − ∞ ∞ − nx [ n ] z − n − 1 − z d X ~ ( z ) dz = ∑ n = − ∞ ∞ nx [ n ] z − n alignl { stack { size 12{ { {d {X} cSup { size 8{ "~" } } \( z \) } over { ital "dz"} } = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { - ital "nx" \[ n \] z rSup { size 8{ - n - 1} } } } {} # - z { {d {X} cSup { size 8{ "~" } } \( z \) } over { ital "dz"} } = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } { ital "nx" \[ n \] z rSup { size 8{ - n} } } {} } } {}

Most proofs of Z-transform properties are simple. Some of the important properties are summarized here.

R, R1, and R2 are the ROCs of X~(z)X~(z) size 12{ {X} cSup { size 8{ "~" } } \( z \) } {}, X1~(z)X1~(z) size 12{ {X rSub { size 8{1} } } cSup { size 8{ "~" } } \( z \) } {}, and X2~(z)X2~(z) size 12{ {X rSub { size 8{2} } } cSup { size 8{ "~" } } \( z \) } {}, respectively. * Exceptions may occur at z = 0 and z = ∞.

II. Z-TRANSFORMS OF SIMPLE TIME FUNCTIONS

1/ Unit sample function

δ [ n ] = { 1 if n = 0 0 otherwise δ [ n ] = { 1 if n = 0 0 otherwise size 12{δ \[ n \] = left lbrace matrix { 1 matrix { {} # {} } ital "if" matrix { {} # {} } n=0 {} ## 0 matrix { {} # {} } ital "otherwise"{} } right none } {}

The Z-transform of the unit sample is

Z { δ [ n ] } = ∑ n = − ∞ ∞ δ [ n ] z − n = z 0 = 1 Z { δ [ n ] } = ∑ n = − ∞ ∞ δ [ n ] z − n = z 0 = 1 size 12{Z lbrace δ \[ n \] rbrace = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \[ n \] z rSup { size 8{ - n} } =z rSup { size 8{0} } =1} } {}

for all values of z, i.e., the ROC is the entire z plane.

2/ Unit step function

u [ n ] = { 1 for n ≥ 0 0 for n < 0 u [ n ] = { 1 for n ≥ 0 0 for n < 0 size 12{u \[ n \] = left lbrace matrix { 1 matrix { {} # {} } ital "for" matrix { {} # {} } n >= 0 {} ## 0 matrix { {} # {} } ital "for" matrix { {} # {} } n<0{} } right none } {}

The unit step and unit sample functions are simply related.

δ [ n ] = u [ n ] - u [ n - 1 ] δ [ n ] = u [ n ] - u [ n - 1 ] size 12{δ \[ n \] " ="u \[ n \] "- "u \[ n "- "1 \] } {}

i.e., the unit sample is the first difference of the unit step function.

u [ n ] = ∑ n = − ∞ ∞ δ [ m ] u [ n ] = ∑ n = − ∞ ∞ δ [ m ] size 12{u \[ n \] = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {δ \[ m \] } } {}

i.e., the unit step is a running sum of the unit sample.

The Z-transform of the unit step is

Z { u [ n ] } = ∑ n = − ∞ ∞ u [ n ] z − n = ∑ n = 0 ∞ z − n = ∑ n = 0 ∞ z − 1 n Z { u [ n ] } = ∑ n = − ∞ ∞ u [ n ] z − n = ∑ n = 0 ∞ z − n = ∑ n = 0 ∞ z − 1 n size 12{Z lbrace u \[ n \] rbrace = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {u \[ n \] z rSup { size 8{ - n} } ={}} Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {z rSup { size 8{ - n} } ={}} Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { left [z rSup { size 8{ - 1} } right ] rSup { size 8{n} } } } {}

This is a sum of a geometric series of the form

∑ n = 0 ∞ α n = 1 1 − α for ∣ α ∣ < 1 ∑ n = 0 ∞ α n = 1 1 − α for ∣ α ∣ < 1 size 12{ Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } {α rSup { size 8{n} } } = { {1} over {1 - α} } matrix { {} # {} } ital "for" matrix { {} # {} } \lline α \lline <1} {}

Therefore, the geometric series for Z{u[n]} converges provided that |z−1| < 1 or |z| > 1. Therefore,

Z { u [ n ] } = 1 1 − z − 1 = z z − 1 for ∣ z > 1 ∣ Z { u [ n ] } = 1 1 − z − 1 = z z − 1 for ∣ z > 1 ∣ size 12{Z lbrace u \[ n \] rbrace = { {1} over {1 - z rSup { size 8{ - 1} } } } = { {z} over {z - 1} } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z>1 \lline } {}

The region of convergence is outside a circle of radius one called the unit circle.

On the unit circle, z = ejΩz = ejΩ size 12{"z "=" e" rSup { size 8{j %OMEGA } } } {}. As we shall see, the unit circle in the z-plane plays a role analogous to the jω-axis in the s-plane.

3/ Causal exponential DT time function

If x[n]= anu[n]x[n]= anu[n] size 12{x \[ n \] =" a" rSup { size 8{n} } u \[ n \] } {} then the Z-transform is

X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ > ∣ a ∣ X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ > ∣ a ∣ size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline > \lline a \lline } {}

The sum converges provided |az−1| < 1 or |z| > |a| so that

X ~ ( z ) = ∑ n = − ∞ ∞ a n z − n u [ n ] = ∑ n = 0 ∞ az − 1 n = 1 1 − az − 1 X ~ ( z ) = ∑ n = − ∞ ∞ a n z − n u [ n ] = ∑ n = 0 ∞ az − 1 n = 1 1 − az − 1 size 12{ {X} cSup { size 8{ "~" } } \( z \) = Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {a rSup { size 8{n} } z rSup { size 8{ - n} } u \[ n \] } = Sum cSub { size 8{n=0} } cSup { size 8{ infinity } } { left [ ital "az" rSup { size 8{ - 1} } right ] rSup { size 8{n} } } = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } } {}

The region of convergence (ROC) of X~(z)X~(z) size 12{ {X} cSup { size 8{ "~" } } \( z \) } {} is |z| > |a|, i.e., outside a circle of radius a.

Thus we have

x [ n ] = a n u [ n ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ > ∣ a ∣ x [ n ] = a n u [ n ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ > ∣ a ∣ size 12{x \[ n \] =a rSup { size 8{n} } u \[ n \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline > \lline a \lline } {}

What happens if a < 0?

x [ n ] = a n u [ n ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ > ∣ a ∣ x [ n ] = a n u [ n ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ > ∣ a ∣ size 12{x \[ n \] =a rSup { size 8{n} } u \[ n \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline > \lline a \lline } {}

4/ Relation between causal DT time functions and pole-zero diagrams

Demo of relation of pole-zero diagram to time function.

Conclusions on relation between causal DT time functions and pole-zero diagrams

5/ Anti-causal exponential DT time function

If x[n]= -anu[-n- 1]x[n]= -anu[-n- 1] size 12{x \[ n \] =" -a" rSup { size 8{n} } u \[ "-n" "- "1 \] } {} then the Z-transform is

X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ < ∣ a ∣ X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ < ∣ a ∣ size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline < \lline a \lline } {}

The sum converges provided ∣a−1z∣<1∣a−1z∣<1 size 12{ \lline a rSup { size 8{ - 1} } z \lline <1} {} or |z| < |a| so that

X ~ ( z ) = − ∑ n = − ∞ ∞ a n u [ − n − 1 ] z − n = − ∑ n = − ∞ − 1 az − 1 n = − ∑ n = 1 ∞ a − 1 z n = − a − 1 z 1 − a − 1 z = 1 1 − a − 1 z X ~ ( z ) = − ∑ n = − ∞ ∞ a n u [ − n − 1 ] z − n = − ∑ n = − ∞ − 1 az − 1 n = − ∑ n = 1 ∞ a − 1 z n = − a − 1 z 1 − a − 1 z = 1 1 − a − 1 z alignl { stack { size 12{ {X} cSup { size 8{ "~" } } \( z \) = - Sum cSub { size 8{n= - infinity } } cSup { size 8{ infinity } } {a rSup { size 8{n} } u \[ - n - 1 \] z rSup { size 8{ - n} } } = - Sum cSub { size 8{n= - infinity } } cSup { size 8{ - 1} } { left [ ital "az" rSup { size 8{ - 1} } right ] rSup { size 8{n} } } } {} # matrix { {} # {} } = - Sum cSub { size 8{n=1} } cSup { size 8{ infinity } } { left [a rSup { size 8{ - 1} } z right ] rSup { size 8{n} } } = - { {a rSup { size 8{ - 1} } z} over {1 - a rSup { size 8{ - 1} } z} } = { {1} over {1 - a rSup { size 8{ - 1} } z} } {} } } {}

The region of convergence (ROC) of X~(z)X~(z) size 12{ {X} cSup { size 8{ "~" } } \( z \) } {}is |z| < |a|, i.e., inside a circle of radius a.

Thus we have

x [ n ] = − a n u [ − n − 1 ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ < ∣ a ∣ x [ n ] = − a n u [ − n − 1 ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ < ∣ a ∣ size 12{x \[ n \] = - a rSup { size 8{n} } u \[ - n - 1 \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline < \lline a \lline } {}

6/ Summary of causal and anti-causal geometric sequences

x [ n ] = a n u [ n ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ > ∣ a ∣ x [ n ] = − a n u [ − n − 1 ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ < ∣ a ∣ x [ n ] = a n u [ n ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ > ∣ a ∣ x [ n ] = − a n u [ − n − 1 ] ⇔ Z X ~ ( z ) = 1 1 − az − 1 for ∣ z ∣ < ∣ a ∣ alignl { stack { size 12{x \[ n \] =a rSup { size 8{n} } u \[ n \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline > \lline a \lline } {} # x \[ n \] = - a rSup { size 8{n} } u \[ - n - 1 \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline < \lline a \lline {} } } {}

7/ Two-sided DT time function

Next we consider the two-sided sequence

x [ n ] = a ∣ n ∣ = a − n u [ − n − 1 ] + a n u [ n ] x [ n ] = a ∣ n ∣ = a − n u [ − n − 1 ] + a n u [ n ] size 12{x \[ n \] =a rSup { size 8{ \lline n \lline } } =a rSup { size 8{ - n} } u \[ - n - 1 \] +a rSup { size 8{n} } u \[ n \] } {}

where we assume 0 < a < 1. We use previous results to obtain the Z-transform

X ~ ( z ) = 1 1 − a − 1 z − 1 + 1 1 − az − 1 X ~ ( z ) = 1 1 − a − 1 z − 1 + 1 1 − az − 1 size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - a rSup { size 8{ - 1} } z rSup { size 8{ - 1} } } } + { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } } {}

Therefore, we have

X ~ ( z ) = 1 1 − a − 1 z − 1 + 1 1 − az − 1 for a < ∣ z ∣ < 1 / a X ~ ( z ) = 1 1 − a − 1 z − 1 + 1 1 − az − 1 for a < ∣ z ∣ < 1 / a size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {1} over {1 - a rSup { size 8{ - 1} } z rSup { size 8{ - 1} } } } + { {1} over {1 - ital "az" rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } a< \lline z \lline <1/a} {}

which can be written as

X ~ ( z ) = z − 1 ( a − a − 1 ) ( 1 − a − 1 z − 1 ) ( 1 − az − 1 ) = z ( a − a − 1 ) ( z − a − 1 ) ( z − a ) x [ n ] = a ∣ n ∣ ⇔ Z X ~ ( z ) = z ( a − a − 1 ) ( z − a − 1 ) ( z − a ) X ~ ( z ) = z − 1 ( a − a − 1 ) ( 1 − a − 1 z − 1 ) ( 1 − az − 1 ) = z ( a − a − 1 ) ( z − a − 1 ) ( z − a ) x [ n ] = a ∣ n ∣ ⇔ Z X ~ ( z ) = z ( a − a − 1 ) ( z − a − 1 ) ( z − a ) alignl { stack { size 12{ {X} cSup { size 8{ "~" } } \( z \) = { {z rSup { size 8{ - 1} } \( a - a rSup { size 8{ - 1} } \) } over { \( 1 - a rSup { size 8{ - 1} } z rSup { size 8{ - 1} } \) \( 1 - ital "az" rSup { size 8{ - 1} } \) } } = { {z \( a - a rSup { size 8{ - 1} } \) } over { \( z - a rSup { size 8{ - 1} } \) \( z - a \) } } } {} # x \[ n \] =a rSup { size 8{ \lline n \lline } } { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) = { {z \( a - a rSup { size 8{ - 1} } \) } over { \( z - a rSup { size 8{ - 1} } \) \( z - a \) } } {} } } {}

8/ Conclusions on relation between DT time functions and pole-zero diagrams

More rapidly changing exponential time functions, both growing and decaying, have poles that lie further from the point z = 1 in the z-plane.

Two-minute miniquiz problem

Problem 11-1

Find the Z-transform including the ROC for

x [ n ] = ( 0 . 5 ) n − 4 u [ n − 4 ] x [ n ] = ( 0 . 5 ) n − 4 u [ n − 4 ] size 12{x \[ n \] = \( 0 "." 5 \) rSup { size 8{n - 4} } u \[ n - 4 \] } {}

Solution

We use the Z-transform of the causal exponential time function and time delay property to solve this problem.

( 0 . 5 ) n u [ n ] ⇔ Z 1 1 − 0 . 5z − 1 for ∣ z ∣ > 0 . 5 ( 0 . 5 ) n − 4 u [ n − 4 ] ⇔ Z z − 4 1 − 0 . 5z − 1 for ∣ z ∣ > 0 . 5 ( 0 . 5 ) n u [ n ] ⇔ Z 1 1 − 0 . 5z − 1 for ∣ z ∣ > 0 . 5 ( 0 . 5 ) n − 4 u [ n − 4 ] ⇔ Z z − 4 1 − 0 . 5z − 1 for ∣ z ∣ > 0 . 5 alignl { stack { size 12{ \( 0 "." 5 \) rSup { size 8{n} } u \[ n \] { matrix { {} # {} } { dlrarrow } cSup { size 8{Z} } } cSup {} matrix { {} # {} } { {1} over {1 - 0 "." 5z rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline >0 "." 5} {} # \( 0 "." 5 \) rSup { size 8{n - 4} } u \[ n - 4 \] { matrix { {} # {} } { dlrarrow } cSup { size 8{Z} } } cSup {} matrix { {} # {} } { {z rSup { size 8{ - 4} } } over {1 - 0 "." 5z rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline >0 "." 5 {} } } {}

Note that, the delayed sequence has the same ROC as the original sequence.

III. CONCLUSIONS

The Z-transform is capable of representing a rich class of DT time functions. Z-transform pairs can be obtained by combining

Exercises .

Solutions of Exercises.

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This work is licensed by Truc Pham-Dinh, Tam Huynh-Ngoc, Anh Tuan Hoang, and Thanh Dinh Vu under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.

Last edited by vocw on Jul 7, 2009 4:03 am -0500.