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# Lecture 16:The Z Transform - Method of Solution

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 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

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 z1z1 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.

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 + αp 1 p 2 1 1 pz 1 V ~ o ( z ) = αp 1 p 2 1 1 p 1 z 1 + αp 1 p 2 1 1 pz 1 size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } $$z$$ = - { {αp} over {1 - p rSup { size 8{2} } } } left [ { {1} over {1 - p rSup { size 8{ - 1} } z rSup { size 8{ - 1} } } } right ]+ { {αp} over {1 - p rSup { size 8{2} } } } left [ { {1} over {1 - ital "pz" rSup { size 8{ - 1} } } } right ]} {}

Therefore,

v 0 [ n ] = αp 1 p 2 p n u [ n 1 ] + αp 1 p 2 p n u [ n ] v 0 [ n ] = αp 1 p 2 p n u [ n 1 ] + αp 1 p 2 p n u [ n ] size 12{v rSub { size 8{0} } $n$ = { {αp} over {1 - p rSup { size 8{2} } } } p rSup { size 8{ - n} } u $- n - 1$ + { {αp} over {1 - p rSup { size 8{2} } } } p rSup { size 8{n} } u $n$ } {}

which can be written compactly as

v o [ n ] = αp 1 p 2 p n v o [ n ] = αp 1 p 2 p n size 12{v rSub { size 8{o} } $n$ = left [ { {αp} over {1 - p rSup { size 8{2} } } } right ]p rSup { size 8{ \lline n \lline } } } {}

As the quantity α = rg increases, the spatial distribution of voltage gets narrower and narrower. Recall that r is the series resistance and g is the shunt conductance of the ladder.

II. DISCRETIZED CT SYSTEM

1/ Differential equation — RC Circuit

In a previous lecture, we considered the CT lowpass filter shown below.

The equilibrium equation is

dv o ( t ) dt = 1 RC v o ( t ) + 1 RC v i ( t ) dv o ( t ) dt = 1 RC v o ( t ) + 1 RC v i ( t ) size 12{ { { ital "dv" rSub { size 8{o} } $$t$$ } over { ital "dt"} } = - { {1} over { ital "RC"} } v rSub { size 8{o} } $$t$$ + { {1} over { ital "RC"} } v rSub { size 8{i} } $$t$$ } {}

To solve this equation numerically in a computer, the CT signals are discretized and the derivative is approximated.

2/ Forward Euler algorithm

a/ Discretization

Define

v i [ n ] = v i ( t ) t = nT = v i ( nT ) and v o [ n ] = v o ( t ) t = nT = v o ( nT ) v i [ n ] = v i ( t ) t = nT = v i ( nT ) and v o [ n ] = v o ( t ) t = nT = v o ( nT ) size 12{v rSub { size 8{i} } $n$ =v rSub { size 8{i} } $$t$$ \lline rSub { size 8{t= ital "nT"} } =v rSub { size 8{i} } $$ital "nT"$$ matrix { {} # {} } ital "and" matrix { {} # {} } v rSub { size 8{o} } $n$ =v rSub { size 8{o} } $$t$$ \lline rSub { size 8{t= ital "nT"} } =v rSub { size 8{o} } $$ital "nT"$$ } {}

and approximate the derivative by the forward Euler algorithm

dv o ( t ) dt t = nT v o ( ( n + 1 ) T ) v o ( nT ) T dv o ( t ) dt t = nT v o ( ( n + 1 ) T ) v o ( nT ) T size 12{ { { ital "dv" rSub { size 8{o} } $$t$$ } over { ital "dt"} } \rline rSub { size 8{t= ital "nT"} } approx { {v rSub { size 8{o} } $$\( n+1$$ T \) - v rSub { size 8{o} } $$ital "nT"$$ } over {T} } } {}

b/ Difference equation

Substituting the approximation for the derivative into the differential equation, we obtain

v o ( ( n + 1 ) T ) v o ( nT ) T = 1 RC v o ( nT ) + 1 RC v i ( nT ) v o ( ( n + 1 ) T ) v o ( nT ) T = 1 RC v o ( nT ) + 1 RC v i ( nT ) size 12{ { {v rSub { size 8{o} } $$\( n+1$$ T \) - v rSub { size 8{o} } $$ital "nT"$$ } over {T} } = - { {1} over { ital "RC"} } v rSub { size 8{o} } $$ital "nT"$$ + { {1} over { ital "RC"} } v rSub { size 8{i} } $$ital "nT"$$ } {}

This equation can be written as a difference equation

v o [ n + 1 ] = 1 T RC v o [ n ] + T RC v i [ n ] v o [ n + 1 ] = 1 T RC v o [ n ] + T RC v i [ n ] size 12{v rSub { size 8{o} } $n+1$ = left [1 - { {T} over { ital "RC"} } right ]v rSub { size 8{o} } $n$ + { {T} over { ital "RC"} } v rSub { size 8{i} } $n$ } {}

This equation was solved iteratively for vi[n]=u[n]vi[n]=u[n] size 12{v rSub { size 8{i} } $n$ =u $n$ } {} and vo[0]=0vo[0]=0 size 12{v rSub { size 8{o} } $0$ =0} {} to yield the solution

v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] size 12{v rSub { size 8{o} } $n$ = $$1 - \( 1 - α$$ rSup { size 8{n} } \) u $n$ } {}

where α = T/RC. We now solve this difference equation using the Z-transform method.

c/ System function

The Z-transform of the difference equation

v o [ n + 1 ] = ( 1 α ) v o [ n ] + αv i [ n ] v o [ n + 1 ] = ( 1 α ) v o [ n ] + αv i [ n ] size 12{v rSub { size 8{o} } $n+1$ = $$1 - α$$ v rSub { size 8{o} } $n$ +αv rSub { size 8{i} } $n$ } {}

is

z V ~ o ( z ) = ( 1 α ) V ~ o ( z ) + α V ~ i ( z ) z V ~ o ( z ) = ( 1 α ) V ~ o ( z ) + α V ~ i ( z ) size 12{z {V} cSup { size 8{ "~" } } rSub { size 8{o} } $$z$$ = $$1 - α$$ {V} cSup { size 8{ "~" } } rSub { size 8{o} } $$z$$ +α {V} cSup { size 8{ "~" } } rSub { size 8{i} } $$z$$ } {}

so that the system function is

H ~ ( z ) = V o ~ ( z ) V i ~ ( z ) = α z ( 1 α ) = αz 1 1 ( 1 α ) z 1 H ~ ( z ) = V o ~ ( z ) V i ~ ( z ) = α z ( 1 α ) = αz 1 1 ( 1 α ) z 1 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{ "~" } } $$z$$ } } = { {α} over {z - $$1 - α$$ } } = { {αz rSup { size 8{ - 1} } } over {1 - $$1 - α$$ z rSup { size 8{ - 1} } } } } {}

The region of convergence of H~(z)H~(z) size 12{ {H} cSup { size 8{ "~" } } $$z$$ } {} needs to be specified from the system description. Because we are simulating a CT system consisting of an RC circuit, we are dealing with a causal system, one that does not respond before it is stimulated. Then we know that the ROC is |z| > |1 − α|.

d/ Input and output Z-transforms

The input is vi[n]= u[n]vi[n]= u[n] size 12{v rSub { size 8{i} } $n$ =" u" $n$ } {}. Therefore,

V ~ i ( z ) = 1 1 z 1 for z > 1 V ~ i ( z ) = 1 1 z 1 for z > 1 size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{i} } $$z$$ = { {1} over {1 - z rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline >1} {}

Therefore,

V ~ o ( z ) = 1 1 z 1 αz 1 1 ( 1 α ) z 1 for z > 1 z > 1 α V ~ o ( z ) = 1 1 z 1 αz 1 1 ( 1 α ) z 1 for z > 1 z > 1 α size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } $$z$$ = left [ { {1} over {1 - z rSup { size 8{ - 1} } } } right ] left [ { {αz rSup { size 8{ - 1} } } over {1 - $$1 - α$$ z rSup { size 8{ - 1} } } } right ] matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline >1& \lline z \lline > \lline 1 - α \lline } {}

This is a proper rational function and we expand it in a partial fraction expansion as follows

V ~ o ( z ) = 1 1 z 1 1 1 ( 1 α ) z 1 for z > 1 z > 1 α V ~ o ( z ) = 1 1 z 1 1 1 ( 1 α ) z 1 for z > 1 z > 1 α size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } $$z$$ = { {1} over {1 - z rSup { size 8{ - 1} } } } - { {1} over {1 - $$1 - α$$ z rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline >1& \lline z \lline > \lline 1 - α \lline } {}

Because the ROC is outside a circle enclosing all the poles, the solution is

v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] size 12{v rSub { size 8{o} } $n$ = $$1 - \( 1 - α$$ rSup { size 8{n} } \) u $n$ } {}

e/ Step response

CT and DT system step responses for α = T/RC, RC = 1,

v o ( t ) = ( 1 e t ) u ( t ) and v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] v o ( t ) = ( 1 e t ) u ( t ) and v o [ n ] = ( 1 ( 1 α ) n ) u [ n ] size 12{v rSub { size 8{o} } $$t$$ = $$1 - e rSup { size 8{ - t} }$$ u $$t$$ matrix { {} # {} } ital "and" matrix { {} # {} } v rSub { size 8{o} } $n$ = $$1 - \( 1 - α$$ rSup { size 8{n} } \) u $n$ } {}

The DT solution is increasingly accurate as T is made smaller. Note that when

T > 2, the DT solution diverges. Why?

f/ Natural frequencies of the CT and DT systems

The natural frequency of the CT system is s = −1/(RC) and the natural frequency of the DT system is z = 1− α = 1− T/(RC) .

g/ Relation of natural frequencies in s- and z-planes

The CT system has a natural frequency of s = −1/RC and the DT system for the forward Euler algorithm has the natural frequency z = 1− (T/RC) = 1+sT . Thus, the forward Euler algorithm maps the s-plane into the z-plane by the mapping z = 1+sT . The mapping is shown below.

s z 0 1 2 T 1 1 + jωT s z 0 1 2 T 1 1 + jωT alignl { stack { size 12{s matrix { {} # {} } rightarrow matrix { {} # {} } z} {} # size 12{0 matrix { {} # {} } rightarrow matrix { {} # {} } 1} {} # size 12{ - { {2} over {T} } matrix { {} # {} } rightarrow matrix { {} # {} } - 1} {} # size 12{jω matrix { {} # {} } rightarrow matrix { {} # {} } 1+jωT} {} } } {}

The conclusion is that for s <2Ts <2T size 12{"s "< { { - 2} over {T} } } {} , z < −1 and the DT approximation to the CT system is unstable. Therefore, to extend the range of stability, decrease T.

3/ Backward Euler algorithm

The forward Euler algorithm is one of many possible approximations to the derivative. Another is called the backward Euler algorithm.

a/ Discretization

Define

v i [ n ] = v i ( t ) t = nT = v i ( nT ) and v o [ n ] = v o ( t ) t = nT = v o ( nT ) v i [ n ] = v i ( t ) t = nT = v i ( nT ) and v o [ n ] = v o ( t ) t = nT = v o ( nT ) size 12{v rSub { size 8{i} } $n$ =v rSub { size 8{i} } $$t$$ \lline rSub { size 8{t= ital "nT"} } =v rSub { size 8{i} } $$ital "nT"$$ matrix { {} # {} } ital "and" matrix { {} # {} } v rSub { size 8{o} } $n$ =v rSub { size 8{o} } $$t$$ \lline rSub { size 8{t= ital "nT"} } =v rSub { size 8{o} } $$ital "nT"$$ } {}

and approximate the derivative by the backward Euler algorithm

dv o ( t ) dt t = nT v o ( nT ) v o ( ( n 1 ) T ) T dv o ( t ) dt t = nT v o ( nT ) v o ( ( n 1 ) T ) T size 12{ { { ital "dv" rSub { size 8{o} } $$t$$ } over { ital "dt"} } \rline rSub { size 8{t= ital "nT"} } approx { {v rSub { size 8{o} } $$ital "nT"$$ - v rSub { size 8{o} } $$\( n - 1$$ T \) } over {T} } } {}

b/ Difference equation and system function

Substituting the approximation for the derivative into the differential equation as before, we now obtain

v o ( nT ) v o ( ( n 1 ) T ) T = 1 RC v o ( nT ) + 1 RC v i ( nT ) v o ( nT ) v o ( ( n 1 ) T ) T = 1 RC v o ( nT ) + 1 RC v i ( nT ) size 12{ { {v rSub { size 8{o} } $$ital "nT"$$ - v rSub { size 8{o} } $$\( n - 1$$ T \) } over {T} } = - { {1} over { ital "RC"} } v rSub { size 8{o} } $$ital "nT"$$ + { {1} over { ital "RC"} } v rSub { size 8{i} } $$ital "nT"$$ } {}

This equation can be written as a difference equation

v o [ n + 1 ] = 1 1 + α v o [ n ] + α 1 + α v i [ n + 1 ] v o [ n + 1 ] = 1 1 + α v o [ n ] + α 1 + α v i [ n + 1 ] size 12{v rSub { size 8{o} } $n+1$ = left [ { {1} over {1+α} } right ]v rSub { size 8{o} } $n$ + left [ { {α} over {1+α} } right ]v rSub { size 8{i} } $n+1$ } {}

where α = T/RC. We shall analyze a slightly modified difference equation that gives somewhat more accurate results

v o [ n + 1 ] = 1 1 + α v o [ n ] + α 1 + α v i [ n ] v o [ n + 1 ] = 1 1 + α v o [ n ] + α 1 + α v i [ n ] size 12{v rSub { size 8{o} } $n+1$ = left [ { {1} over {1+α} } right ]v rSub { size 8{o} } $n$ + left [ { {α} over {1+α} } right ]v rSub { size 8{i} } $n$ } {}

The system function for this difference equation is

H ~ ( z ) = V ~ o ( z ) V ~ i ( z ) = α 1 + α z 1 1 + α = α 1 + α z 1 1 1 1 + α z 1 H ~ ( z ) = V ~ o ( z ) V ~ i ( z ) = α 1 + α z 1 1 + α = α 1 + α z 1 1 1 1 + α z 1 size 12{ {H} cSup { size 8{ "~" } } $$z$$ = { { {V} cSup { size 8{ "~" } } rSub { size 8{o} } $$z$$ } over { {V} cSup { size 8{ "~" } } rSub { size 8{i} } $$z$$ } } = { { { {α} over {1+α} } } over {z - { {1} over {1+α} } } } = { { { {α} over {1+α} } z rSup { size 8{ - 1} } } over {1 - { {1} over {1+α} } z rSup { size 8{ - 1} } } } } {}

The region of convergence of H~(z)H~(z) size 12{ {H} cSup { size 8{ "~" } } $$z$$ } {}is z>11+αz>11+α size 12{ \lline z \lline > lline { {1} over {1+α} } rline } {}

c/ Input and output Z-transforms

The input is vi[n]= u[n]vi[n]= u[n] size 12{v rSub { size 8{i} } $n$ =" u" $n$ } {}. Therefore,

V ~ i ( z ) = 1 1 z 1 for z > 1 V ~ i ( z ) = 1 1 z 1 for z > 1 size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{i} } $$z$$ = { {1} over {1 - z rSup { size 8{ - 1} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline >1} {}

Therefore,

V ~ o ( z ) = 1 1 z 1 α 1 + α z 1 1 z 1 1 + α for z > 1 z > 1 1 + α V ~ o ( z ) = 1 1 z 1 α 1 + α z 1 1 z 1 1 + α for z > 1 z > 1 1 + α size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } $$z$$ = left [ { {1} over {1 - z rSup { size 8{ - 1} } } } right ] left [ { { { {α} over {1+α} } z rSup { size 8{ - 1} } } over {1 - { {z rSup { size 8{ - 1} } } over {1+α} } } } right ] matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline >1& \lline z \lline > lline { {1} over {1+α} } rline } {}

This is a proper rational function and we expand it in a partial fraction expansion as follows

V ~ o ( z ) = 1 1 z 1 1 1 z 1 1 + α for z > 1 z > 1 1 + α V ~ o ( z ) = 1 1 z 1 1 1 z 1 1 + α for z > 1 z > 1 1 + α size 12{ {V} cSup { size 8{ "~" } } rSub { size 8{o} } $$z$$ = { {1} over {1 - z rSup { size 8{ - 1} } } } - { {1} over {1 - { {z rSup { size 8{ - 1} } } over {1+α} } } } matrix { {} # {} } ital "for" matrix { {} # {} } \lline z \lline >1 matrix { {} # {} } & matrix { {} # {} } \lline z \lline > lline { {1} over {1+α} } rline } {}

Because the ROC is outside a circle enclosing all the poles, the sequence is causal

v o [ n ] = 1 1 1 + α n n [ n ] v o [ n ] = 1 1 1 + α n n [ n ] size 12{v rSub { size 8{o} } $n$ = left [1 - left [ { {1} over {1+α} } right ] rSup { size 8{n} } right ]n $n$ } {}

d/ Step response

CT and DT system step responses for α = T/RC, RC = 1,

v o ( t ) = ( 1 e t ) u ( t ) and v o [ n ] = 1 1 1 + α n u [ n ] v o ( t ) = ( 1 e t ) u ( t ) and v o [ n ] = 1 1 1 + α n u [ n ] size 12{v rSub { size 8{o} } $$t$$ = $$1 - e rSup { size 8{ - t} }$$ u $$t$$ matrix { {} # {} } ital "and" matrix { {} # {} } v rSub { size 8{o} } $n$ = left [1 - left [ { {1} over {1+α} } right ] rSup { size 8{n} } right ]u $n$ } {}

The DT solution is bounded for all values of T. Why?

e/ Natural frequencies of the CT and DT systems

The natural frequency of the CT system is s = −1/(RC) and the natural frequency of the DT system is

z = 1 1 + α = 1 1 + T / ( RC ) z = 1 1 + α = 1 1 + T / ( RC ) size 12{z= { {1} over {1+α} } = { {1} over {1+T/ $$ital "RC"$$ } } } {}

f/ Relation of natural frequencies — backward Euler algorithm

The CT system has a natural frequency of s = −1/RC and the DT system for the backward Euler algorithm has the natural frequency.

z = 1 1 + T / ( RC ) = 1 1 sT z = 1 1 + T / ( RC ) = 1 1 sT size 12{z= { {1} over {1+T/ $$ital "RC"$$ } } = { {1} over {1 - ital "sT"} } } {}

Thus, the forward Euler algorithm maps the s-plane into the z-plane by the mapping

z = 1 1 sT z = 1 1 sT size 12{z= { {1} over {1 - ital "sT"} } } {}

The mapping is shown below.

s z 0 1 0 1 1 jωT s z 0 1 0 1 1 jωT alignl { stack { size 12{s matrix { {} # {} } rightarrow matrix { {} # {} } z} {} # size 12{0 matrix { {} # {} } rightarrow matrix { {} # {} } 1} {} # size 12{ - infinity matrix { {} # {} } rightarrow matrix { {} # {} } 0} {} # size 12{jω matrix { {} # {} } rightarrow matrix { {} # {} } { {1} over {1 - jωT} } } {} } } {}

The conclusion is that the backward Euler maps the entire left half of the s-plane into a circle of radius 1/2 centered at z = 1/2. Thus, the solution is bounded for all values of T.

g/ Conclusion

One might conclude that the backward Euler algorithm should be used for numerical analysis. However, such a conclusion would be based on analysis of an RC circuit only. Let us see what happens with an LC circuit.

III. DIFFERENTIAL EQUATION — LC CIRCUIT

1/ Differential equation

Consider the LC network shown below and the homogeneous differential equations that must be satisfied.

The equilibrium equations are

dv ( t ) dt + i ( t ) C = 0 di ( t ) dt v ( t ) L = 0 dv ( t ) dt + i ( t ) C = 0 di ( t ) dt v ( t ) L = 0 alignl { stack { size 12{ { { ital "dv" $$t$$ } over { ital "dt"} } + { {i $$t$$ } over {C} } =0} {} # size 12{ { { ital "di" $$t$$ } over { ital "dt"} } - { {v $$t$$ } over {L} } =0} {} } } {}

2/ Natural frequencies

The Laplace transform of these differential equations, in matrix form, yields

s 1 C 1 L s V ( s ) I ( s ) = 0 s 1 C 1 L s V ( s ) I ( s ) = 0 size 12{ left [ matrix { s {} # { {1} over {C} } {} ## - { {1} over {L} } {} # s{} } right ] left [ matrix { V $$s$$ {} ## I $$s$$ } right ]=0} {}

Hence, a solution is possible if the determinant is zero, which yields the characteristic polynomial

s 2 + 1 LC = 0 s 2 + 1 LC = 0 size 12{s rSup { size 8{2} } + { {1} over { ital "LC"} } =0} {}

so that the natural frequencies are

s 1,2 = ± j 1 ( LC ) 1 / 2 s 1,2 = ± j 1 ( LC ) 1 / 2 size 12{s rSub { size 8{1,2} } = +- j { {1} over { $$ital "LC"$$ rSup { size 8{1/2} } } } } {}

3/ Mapping of natural frequencies for Euler algorithms

It can be shown that the mapping of natural frequencies for the LC circuit is the same as for the RC circuit.

Two-minute miniquiz problem

Problem 12-1

Describe (in words) the homogeneous solution for

• the CT LC system;
• the DT approximation to the CT system using the forward Euler algorithm;
• the DT approximation to the CT system using the backward Euler algorithm.

Solution

• The CT LC system has imaginary natural frequencies so that the homogeneous solution is a sinusoid.
• The DT approximation to the CT system using the forward Euler algorithm has natural frequencies outside the unit circle so that the homogeneous solution is a geometrically (exponentially) growing sinusoid.
• the DT approximation to the CT system using the backward Euler algorithm has natural frequencies inside the unit circle so that the homogeneous solution is a geometrically (exponentially) decaying sinusoid.

4/ Conclusions on Euler algorithms

So neither the forward nor the backward Euler algorithms give accurate numerical solutions for the LC circuit. Since one response grows and the other decays, a method that somehow

averages the two algorithms might work. This is the trapezoidal algorithm which we consider briefly.

5/ Trapezoidal algorithm

Summary of all 3 algorithms

Consider the homogeneous differential equation

dy ( t ) dt = λy ( t ) dy ( t ) dt = λy ( t ) size 12{ { { ital "dy" $$t$$ } over { ital "dt"} } =λy $$t$$ } {}

where λ is the natural frequency. [Note, this differential equation can be regarded as a scalar equation or a vector equation.] The forward Euler (FE), backward Euler (BE), and trapezoidal (T) algorithms are summarized below.

The trapezoidal algorithm mapping is shown below.

s z 0 1 1 1 + j ωT 2 1 j ωT 2 s z 0 1 1 1 + j ωT 2 1 j ωT 2 alignl { stack { size 12{s matrix { {} # {} } rightarrow matrix { {} # {} } z} {} # size 12{0 matrix { {} # {} } rightarrow matrix { {} # {} } 1} {} # size 12{ - infinity matrix { {} # {} } rightarrow matrix { {} # {} } - 1} {} # size 12{jω matrix { {} # {} } rightarrow matrix { {} # {} } { {1+j { {ωT} over {2} } } over {1 - j { {ωT} over {2} } } } } {} } } {}

The important point is that the trapezoidal algorithm maps the entire left half of the s-plane inside the unit circle of the z-plane and the jω axis of the s-plane onto the unit circle of the z-plane. Thus, both natural frequencies in s are preserved in z, and the trapezoidal algorithm gives bounded responses for the RC circuit and oscillatory responses for the LC circuit.

IV. CONCLUSIONS

The basic method for use of the Z-transform for the solution to DT systems has been demonstrated in a context to give insight into two DT systems.

• Ladder network as an example of a system that is neither causal nor anti-causal. The natural frequencies of this system define the spatial distribution of the dependent variable.
• DT approximations to CT systems were examples of causal systems. The Z-transform gave insight into the properties of numerical approximations to the solution of differential equations.

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