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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 ] 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 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 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 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 = e j Ω 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 ] = a n u [ 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 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 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 ) 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 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 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 ] = -a n u [ -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 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 1 z < 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 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 ) 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 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 ) 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 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 ] 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 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 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 ) 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 ] 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 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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Source:  OpenStax, Signals and systems. OpenStax CNX. Jul 29, 2009 Download for free at http://cnx.org/content/col10803/1.1
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