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# Lecture 15:The Bilateral Z-Transform

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.

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 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 } {}

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 { $$z - a rSup { size 8{ - 1} }$$ $$z - a$$ } } {} } } {}

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 .

## Content actions

PDF | EPUB (?)

### What is an EPUB file?

EPUB is an electronic book format that can be read on a variety of mobile devices.

My Favorites (?)

'My Favorites' is a special kind of lens which you can use to bookmark modules and collections. 'My Favorites' can only be seen by you, and collections saved in 'My Favorites' can remember the last module you were on. You need an account to use 'My Favorites'.

| A lens I own (?)

#### Definition of a lens

##### Lenses

A lens is a custom view of the content in the repository. You can think of it as a fancy kind of list that will let you see content through the eyes of organizations and people you trust.

##### What is in a lens?

Lens makers point to materials (modules and collections), creating a guide that includes their own comments and descriptive tags about the content.

##### Who can create a lens?

Any individual member, a community, or a respected organization.

##### What are tags?

Tags are descriptors added by lens makers to help label content, attaching a vocabulary that is meaningful in the context of the lens.

| External bookmarks