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
Review of last lecture
Solve linear difference equation for a causal exponential input
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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 {
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} ital "for" matrix {
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} x \[ n \] = ital "Xz" rSup { size 8{n} } u \[ n \] } {}
Solve homogeneous equation for n > 0
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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 {
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} y rSub { size 8{h} } \[ n \] =Aλ rSup { size 8{n} } } {}
Solve characteristic polynomial for λ.
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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
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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 {
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} ital "for" matrix {
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} 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
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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
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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
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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
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size 12{x \[ n \] { dlrarrow } cSup { size 8{Z} } {X} cSup { size 8{ "~" } } \( z \) } {}
2/ Properties
a/ Linearity
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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.
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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
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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,
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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
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