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Lecture 15:The Bilateral Z-Transform

Module by: Vu Dinh Thanh, Tam Huynh-Ngoc, Anh Tuan Hoang, Truc Pham-Dinh. E-mail the authors

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

Lecture #15:

THE BILATERAL Z-TRANSFORM

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

Outline:

  • Review of last lecture
  • The bilateral Z-transform

– Definition

– Properties

  • Inventory of transform pairs
  • Conclusion

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 ] = n k = 0 K a k y h [ n + k ] = 0 by assu min g y h [ n ] = 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

  • H~(z)H~(z) size 12{ {H} cSup { size 8{ "~" } } \( z \) } {}characterizes system  compute H~(z)H~(z) size 12{ {H} cSup { size 8{ "~" } } \( z \) } {}efficiently.
  • In steady state, response to XznXzn size 12{"Xz" rSup { size 8{n} } } {} is H~(z)znH~(z)zn size 12{ {H} cSup { size 8{ "~" } } \( z \) z rSup { size 8{n} } } {} .
  • Represent arbitrary x[n] as superpositions of XznXzn size 12{"Xz" rSup { size 8{n} } } {}on z.
  • Compute response y[n] as superpositions of H~(z)XznH~(z)Xzn size 12{ {H} cSup { size 8{ "~" } } \( z \) "Xz" rSup { size 8{n} } } {} on z.

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

  • Using the Z-transform properties,
  • Determining the Z-transforms of elementary DT time functions,
  • Combining the results of the above two items.

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.

Figure 1
Figure 1 (graphics1.png)

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

graphics2.png

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

Figure 2
Figure 2 (graphics3.png)

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.

graphics4.png

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.

graphics5.png

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 } {}

Figure 3
Figure 3 (graphics6.png)

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 } {}

Figure 4
Figure 4 (graphics7.png)

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

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

Figure 5
Figure 5 (graphics8.png)

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

Figure 6
Figure 6 (graphics9.png)

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 a1z<1a1z<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 } {}

Figure 7
Figure 7 (graphics10.png)

6/ Summary of causal and anti-causal geometric sequences

  • Causal and anti-causal sequences can have the same X~(z)X~(z) size 12{ {X} cSup { size 8{ "~" } } \( z \) } {}formula but different ROCs

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 {} } } {}

  • Causal sequences have ROCs outside a circle that intersects the pole.
  • Anti-causal sequences have ROCs inside a circle that intersects the pole.
  • Bounded sequences have ROCs that include the unit circle.
  • Unbounded sequences have ROCs that do not include the unit circle.

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 { \( z - a rSup { size 8{ - 1} } \) \( z - a \) } } {} } } {}

Figure 8
Figure 8 (graphics11.png)

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

  • Z-transform properties.
  • the Z-transforms of elementary time functions.

Exercises .

Solutions of Exercises.

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