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MODERN CONTROL
SYSTEMS ENGINEERING
COURSE : CS421
INSTRUCTOR:
DR. RICHARD H. MGAYA
Date: October 25th, 2013
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Analysis of Steady-State Error
Final Value Theorem
Examination of the asymptotic behavior of a discrete system
without the need for the evaluation of thez-transforms
associated with transfer functions
Recap: Sequences that have a limiting value for large kmust be
stable functions All poles must lie inside the unit circle
Function should be expandable into the form
Inverse transformation
Dr. Richard H. Mgaya
11)( 1 i
n
z i
i
ppz
zB
z
Az
zF
n
i
k
iipBAkf1
)(
... i
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Analysis of Steady-State Error
Final Value Theorem Cont
As kgets large the terms contributed by the summation sign
will approach zero.
Evaluation of Aby partial fraction expansion:
From eqn. i
Multiply by and letzapproaches unity
Formal mathematical theorem:
Dr. Richard H. Mgaya
z
z 1
)(1
lim 1 zFz
zA z
)(1
lim)(lim 1 zFz
zkf zk
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Analysis of Steady-State Error
Effect of Sampling on the Steady-State Error
No general conclusion about the steady-state error
Error depend on the placement of the sampler
Continuous Systems:
Steady-state error is based on the open-loop transfer function
Static error constants
Discrete Systems:
Placement of the sampler changes the open-loop transfer
function
Dr. Richard H. Mgaya
,)(lim1
1)(
0 sGe
s
step
and
ssGe
s
ramp ,)(lim
1)(
0
)(lim
1)(
2
0 sGse
s
parabola
vKpK aK
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Analysis of Steady-State Error
Effect of Sampling on the Steady-State Error Cont
Consider the following digital system:
Digital computer is represented as asamplerand azero-order hold
Block reduction techniquesDr. Richard H. Mgaya
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Analysis of Steady-State Error
Effect of Sampling on the Steady-State Error Cont
From the figure:
Final value for discrete signal
e*() is the final sampled value of e(t) or e(kT)
e*() for unity feedback system
Dr. Richard H. Mgaya
)()()()( zGzEzRzE
)(1
)()(
zG
zRzE
)(1lim)( 11* zEze z
)(1
lim)( 1*
zEz
ze z
)(1
)(1lim)( 11
*
zG
zRze z
... i i
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Analysis of Steady-State Error
Unit Step
Unit step input:
Substitution to eqn. ii
Thus,
Dr. Richard H. Mgaya
ssR 1)(
1)(
z
zzR
)(lim 1 zGK zp
)(lim1
1)(
1
*
zGe
z
pKe
1
1)(*
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Analysis of Steady-State Error
Unit Ramp
Unit ramp input:
From unit step procedureeqn. ii
Thus,
Dr. Richard H. Mgaya
2
1)(s
sR
21)(
z
TzzR
)()1(lim1
1 zGz
TK
zv
)()1(lim1
1)(
1
*
zGzT
e
z
vKe
1)(*
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Analysis of Steady-State Error
Unit Parabolic
Unit parabolic input:
Similarly
Thus,
Dr. Richard H. Mgaya
3
1)(s
sR
32
12
)1()(
z
zzTzR
)()1(lim1 2
12 zGz
TK
za
aKe
1)(*
)()1(lim1
1)(
2
12
*
zGzT
e
z
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Analysis of Steady-State Error
Conclusion:
The error constants for digital systems are similar to those ofanalog systems
Multiple pole placement at the origin of thes-plane reduces the error tozero in case of analog systems
Multiple placement of pole atz= 1 of the zplane reduces the steady-
state error to zero for the case of discrete system of the type discussed
Thus,
Stability
Continuous systems stability can be determined by theRouthHurwitzcriterion
Number of unstable poles
The number of sign changes indicates the number of unstable poles andnot their location
Dr. Richard H. Mgaya
Tsezzs under1tomaps0
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Stability of Digital Control System
Jurytest is employed to asses the stability of discrete systems
Consider the characteristic equation s a sampled-data system:
Dr. Richard H. Mgaya
0)( 011
1
azazazazQ n
nn
n
0
2
1
1
2
0
0123
3210
0321
1210
0121
1210
1210
m
m
m
m
m
mllll
llll
bbbb
bbbb
aaaaa
aaaaa
zzzzz
nnn
n
nnn
nn
nn
kn
kn
k aa
aa
b
0
kn
kn
kbb
bbc
1
10
kn
kn
kcc
ccd
2
20
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Stability of Digital Control System
Necessary and sufficient conditions for stability
No roots outside or on the unit circle
Condition1: Q(1) > 0
Condition2: (-1)nQ(-1) > 0
Condition3:
Conditionn:
Dr. Richard H. Mgaya
naa 0
10 nbb
20 ncc
20 mm
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Stability of Digital Control System
Example
Given the characteristic equation below find the value ofKto
make the system just unstable
Solution:
Condition1: Q(1) > 0
Condition2: (-1)nQ(-1) > 0
Dr. Richard H. Mgaya
0368.0368.1
)066.0092.0(1
2
zz
zK
0)066.0368.0()368.1092.0(2 KzKz
0)066.0368.0()368.1092.0(1)1( KKQ 0if0)1( KQ
0)066.0368.0()368.1092.0(1{)1()1( 2 KKQ
23.105or0026.0736.2 KK
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Stability of Digital Control System
Example Cont
Condition3:
The system is marginal stable whenK= 9.58 andK= 105.23
Dr. Richard H. Mgaya
1066.0368.0 K
naa 0
1066.0368.0 K
58.9066.0
368.01
K
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Stability of Digital Control System
Digital System Stability via s-Plane
Mapping ofjaxis points on thes-plane to unit circle points
on thez-plane, i.e., bilinear transformation
Transformation maps right-half plane points on the s-plane to point
outside the unit circle on the z-plane. It also maps the left-half points on
the s-plane to the point inside the unit circle on the z-plane.
Transformation of the denominator of the pulse transfer
functionD(z) to the denominator of a continuous transfer
function G(s)
Routh-Hurwitz stability criterion can then be applied to the
transformed system
Dr. Richard H. Mgaya
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Stability of Digital Control System
Digital System Stability via s-Plane
Bilinear transformation:
Provides tools for applyings-plane design and analysis to
digital system
If
and the inverse
Expansion of the logarithmic function
Bilinear transforms
Dr. Richard H. Mgaya1
12
z
z
Ts
Ts
ez
zT
s ln1
3
1
1
3
1
1
12ln
z
z
z
zz
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Stability of Digital Control System
Digital System Stability via s-Plane
Inverse
Substitutes = +j
Thus,
Dr. Richard H. Mgaya
1
1
s
sz
j
jz
)1(
)1(
22
22
)1(
)1(
z
0 when1
0 when1
0 when1
z
z
z
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Stability of Digital Control System
Digital System Stability via s-Plane
Example: Given T(z) =N(z)/D(z), where
Use Routh-Hurwitz criterion to find the number of z-pane poles
of T(z) inside, outside, on the unit circle. Is the system stable?
Substitute:Inverse of bilinear transform intoD(z)= 0
Routh-Hurwitz table:
Dr. Richard H. Mgaya
1.02.0)( 23 zzzzD
0174519 23 sss
017-
045.89-
17-19-
45-1
0
1
2
3
s
s
s
s
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Stability of Digital Control System
Digital System Stability via s-Plane
Example: Cont
Routh-Hurwitz table shows one root on the right half-plane
and two roots on the left half-plane.
Thus, T(z) has one pole outside the unit circle, two poles in the
unit circle and no pole on the unit circle
Conclusion: System is unstablepoles outside the unit circle
Dr. Richard H. Mgaya
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Root locus Analysis in z-plane
Root Locus Construction Rules
1. Starting points (K= 0): The root loci starts at open-loop poles
2. Terminating points (K=): The root loci terminate at open-loop zeros
when they exists, otherwise at infinity.
3. Number of distinct root loci: This is equal to the order of the characteristicpolynomial
4. Symmetry of root loci: The root loci are symmetrical about the real axis
5. Root locus location on real axis: A point on the real axis is part of the loci
if the sum of the open-loop poles and zeros to the right of the point is zero
6. Break away points: The points where a locus breaks away from real axis
are the root of the equation
Dr. Richard H. Mgaya
0)( zGHdz
d
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Root locus Analysis in z-plane
Root Locus Construction Rules
7. Unit circle crossover: Can be obtained by determining the
value ofKfor marginal stability using theJurytest
Example: For a given transfer function G(s), determine the
breakaway point, the value ofKfor marginal stability and the unitcircle cross-over for T= 0.5 seconds
z-Transform:
Dr. Richard H. Mgaya
368.0368.1
066.0092.0)(
2 zz
zKzG
)2(
11)(
sss
eKsG
Ts
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Root locus Analysis in z-plane
Root Locus Construction Rules
Characteristic equation:
Breakaway points:
Dr. Richard H. Mgaya
0)}({ zGHdz
d
0368.0368.1
)066.0092.0(1
2
zz
zK
0)066.0368.0()368.1092.0(2 KzKz
0)368.12)(066.0092.0()368.0368.1( 2 zzKzz
01239.0132.0092.0 2 zz
084.2
647.0
2
1
z
z
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Root locus Analysis in z-plane
Root Locus Construction Rules
K for marginal stability:Jury test
Unit circle crossover:K into the characteristic equation
Roots
Dr. Richard H. Mgaya
01487.02 zz
23.105
58.9
2
1
K
K
97.0244.01 jz
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Root locus Analysis in z-plane
Root Locus Construction Rules
Root locus analysis of a discrete system is the plot of the roots
of the characteristic equation 1+G(z)=0 inz-plane as a function
ofK.
Dr Richard H Mgaya