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Lecture 16:The Z Transform - Method of Solution

Module by: Thanh Dinh Vu, Truc Pham-Dinh, Anh Tuan Hoang, Tam Huynh-Ngoc

Summary: Analysis of DT systems using the Z-transform. Insight into numerical methods for solving differential equations.

Lecture #16:
THE Z-TRANSFORM –
METHOD OF SOLUTION
Motivation:
  • Analysis of DT systems using the Z-transform
  • Insight into numerical methods for solving differential equations
Outline:
  • Review of last lecture
  • Analysis of ladder network
  • Analysis of discretized CT system
  • Conclusion
Review of last lecture
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
Logic for an analysis method for DT LTI systems
  • H~(z)H~(z) size 12{ {H} cSup { size 8{ "~" } } \( z \) } {}characterizes system ⇒⇒ size 12{ drarrow } {} 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 superposition of X(z)znX(z)zn size 12{X \( z \) z rSup { size 8{n} } } {}on z.
  • Compute response y[n] as superposition of H~(z)X(z)znH~(z)X(z)zn size 12{ {H} cSup { size 8{ "~" } } \( z \) X \( z \) z rSup { size 8{n} } } {} on z.
We will analyze DT systems with the Z-transform method in a manner analogous to the use of the Laplace transform for CT systems.
I. OVERVIEW OF TRANSFORM METHOD OF SOLUTION
Figure 1
Electric ladder network
1/ Difference equation
We wish to find the unit sample response of an electric ladder network (considered in a previous lecture), i.e., we assume vi[n]=δ[ν]vi[n]=δ[ν] size 12{v rSub { size 8{i} } \[ n \] = δ \[ ν \] } {}.
As we found in a previous lecture, KCL at node n yields the difference equation
− v 0 [ n + 1 ] + ( 2 + α ) v 0 [ n ] − v 0 [ n − 1 ] = αv i [ n ] where α = rg − v 0 [ n + 1 ] + ( 2 + α ) v 0 [ n ] − v 0 [ n − 1 ] = αv i [ n ] where α = rg size 12{ - v rSub { size 8{0} } \[ n+1 \] + \( 2+α \) v rSub { size 8{0} } \[ n \] - v rSub { size 8{0} } \[ n - 1 \] =αv rSub { size 8{i} } \[ n \] matrix { {} # {} } ital "where" matrix { {} # {} } α= ital "rg"} {}
2/ System function
We apply the Z-transform to the difference equation
-v o [ n + 1 ] + ( 2 + a ) v o [ n ] -v o [ n-1 ] = av i [ n ] -v o [ n + 1 ] + ( 2 + a ) v o [ n ] -v o [ n-1 ] = av i [ n ] size 12{" -v" rSub { size 8{o} } \[ n+1 \] + \( 2+a \) " v" rSub { size 8{o} } \[ n \] "-v" rSub { size 8{o} } \[ "n-1" \] ="av" rSub { size 8{i} } \[ n \] } {} {}
to obtain
( − z + ( 2 + α ) − z − 1 ) V ~ o ( z ) = α V ~ i ( z ) ( − z + ( 2 + α ) − z − 1 ) V ~ o ( z ) = α V ~ i ( z ) size 12{ \( - z+ \( 2+α \) - z rSup { size 8{ - 1} } \) {V} cSup { size 8{ "~" } } rSub { size 8{o} } \( z \) =α {V} cSup { size 8{ "~" } } rSub { size 8{i} } \( z \) } {}
so that
H ~ ( z ) = V o ~ ( z ) V i = ~ ( z ) = − αz z 2 − ( 2 + α ) z + 1 = − αz − 1 1 − ( 2 + α ) z − 1 + z 2 H ~ ( z ) = V o ~ ( z ) V i = ~ ( z ) = − αz z 2 − ( 2 + α ) z + 1 = − αz − 1 1 − ( 2 + α ) z − 1 + z 2 size 12{ {H} cSup { size 8{ "~" } } \( z \) = { { {V rSub { size 8{o} } } cSup { size 8{ "~" } } \( z \) } over { {V rSub { size 8{i} } ={}} cSup { size 8{ "~" } } \( z \) } } = { { - αz} over {z rSup { size 8{2} } - \( 2+α \) z+1} } = { { - αz rSup { size 8{ - 1} } } over {1 - \( 2+α \) z rSup { size 8{ - 1} } +z rSup { size 8{2} } } } } {}
The z-form of the system function is easier for identifying the poles and zeros; the z−1z−1 size 12{z rSup { size 8{ - 1} } } {} -form is easier for calculating the inverse transform.
3/ Response
vi[n]=δ[ν]vi[n]=δ[ν] size 12{v rSub { size 8{i} } \[ n \] = δ \[ ν \] } {}implies that Vi~(z)= 1Vi~(z)= 1 size 12{ {V rSub { size 8{i} } } cSup { size 8{ "~" } } \( z \) =" 1"} {} with an ROC that is the whole z-plane. Therefore,
V ~ o ( z ) = − αz z 2 − ( 2 + α ) z + 1 = − αz − 1 1 − ( 2 + α ) z − 1 + z 2 V ~ o ( z ) = − αz z 2 − ( 2 + α ) z + 1 = − αz − 1 1 − ( 2 + α ) z − 1 + z 2 size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } \( z \) = { { - αz} over {z rSup { size 8{2} } - \( 2+α \) z+1} } = { { - αz rSup { size 8{ - 1} } } over {1 - \( 2+α \) z rSup { size 8{ - 1} } +z rSup { size 8{2} } } } } {}
Thus, there is one zero and two poles. The poles are
p 1,2 = 1 + α 2 ± 1 + α 2 2 − 1 1 / 2 p 1,2 = 1 + α 2 ± 1 + α 2 2 − 1 1 / 2 size 12{p rSub { size 8{1,2} } = left [1+ { {α} over {2} } right ] +- left [ left [1+ { {α} over {2} } right ] rSup { size 8{2} } - 1 right ] rSup { size 8{1/2} } } {}
It is easy to show that p1p2=1p1p2=1 size 12{p rSub { size 8{1} } p rSub { size 8{2} } =1} {}. Hence we write the poles as
p 1 = p = 1 + α 2 − 1 + α 2 2 − 1 1 / 2 and p 2 = 1 / p 1 1 = 1 / p p 1 = p = 1 + α 2 − 1 + α 2 2 − 1 1 / 2 and p 2 = 1 / p 1 1 = 1 / p size 12{p rSub { size 8{1} } =p= left [1+ { {α} over {2} } right ] - left [ left [1+ { {α} over {2} } right ] rSup { size 8{2} } - 1 right ] rSup { size 8{1/2} } matrix { {} # {} } ital "and" matrix { {} # {} } p rSub { size 8{2} } =1/p rSub { size 8{1} } 1=1/p} {}
4/ Region of convergence
There are three possible ROCs for this response as shown below. On physical grounds, we expect the unit-sample response to be bounded.
Figure 2
Only the center ROC includes and unit circle and corresponds to a stable system.
Therefore, we conclude that
V ~ o ( z ) = − αz − 1 1 − ( 2 + α ) z − 1 + z 2 for p < ∣ z ∣ < 1 / p V ~ o ( z ) = − αz − 1 1 − ( 2 + α ) z − 1 + z 2 for p < ∣ z ∣ < 1 / p size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } \( z \) = { { - αz rSup { size 8{ - 1} } } over {1 - \( 2+α \) z rSup { size 8{ - 1} } +z rSup { size 8{2} } } } matrix { {} # {} } ital "for" matrix { {} # {} } p< \lline z \lline <1/p} {}
{}
where the pole at z = p contributes to the causal response while the pole at z = 1/p contributes to the anti-causal response.
5/ Inverse Z-transform
We perform a partial fraction expansion as follows
V ~ o ( z ) = − αz − 1 1 − ( 2 + α ) z − 1 + z 2 = − αz − 1 ( 1 − p − 1 z − 1 ) ( 1 − pz − 1 ) = − ( αp ) / ( 1 − p 2 ) 1 − p − 1 z − 1 + − ( αp − 1 ) / ( 1 − p − 2 ) 1 − pz − 1 V ~ o ( z ) = − αz − 1 1 − ( 2 + α ) z − 1 + z 2 = − αz − 1 ( 1 − p − 1 z − 1 ) ( 1 − pz − 1 ) = − ( αp ) / ( 1 − p 2 ) 1 − p − 1 z − 1 + − ( αp − 1 ) / ( 1 − p − 2 ) 1 − pz − 1 alignl { stack { size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } \( z \) = { { - αz rSup { size 8{ - 1} } } over {1 - \( 2+α \) z rSup { size 8{ - 1} } +z rSup { size 8{2} } } } = { { - αz rSup { size 8{ - 1} } } over { \( 1 - p rSup { size 8{ - 1} } z rSup { size 8{ - 1} } \) \( 1 - ital "pz" rSup { size 8{ - 1} } \) } } } {} # matrix { {} # {} } = { { - \( αp \) / \( 1 - p rSup { size 8{2} } \) } over {1 - p rSup { size 8{ - 1} } z rSup { size 8{ - 1} } } } + { { - \( αp rSup { size 8{ - 1} } \) / \( 1 - p rSup { size 8{ - 2} } \) } over {1 - ital "pz" rSup { size 8{ - 1} } } } {} } } {}
which can be written as
V ~ o ( z ) = − αp 1 − p 2 1 1 − p − 1 z − 1