1 analog and digital control 1 lecture 3 kirsten mølgaard nielsen based on lecture notes by jesper...
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1
Analog and Digital Control
1Lecture 3
Kirsten Mølgaard Nielsen
Based on lecture notes by Jesper Sandberg Thomsen
2
OutlineDiscrete equivalents
Overview (emulation – finding a discrete equivalent to the continuous system)
Numerical integration Zero-Pole matching (mapping) Hold equivalents
Design by emulation Evaluation of design
Nyquist stability criterion Continuous case Discrete case
3
Discrete Equivalents - Overviewcontroller
D(s)plantG(s)
r(t) u(t) y(t)e(t)+
-
Translation to discrete controller (emulation)Numerical Integration• Forward rectangular rule• Backward rectangular rule• Trapeziod rule (Tustin’s method, bilinear transformation)• Bilinear with prewarpingZero-Pole MatchingHold Equivalents• Zero order hold (ZOH)• Triangle hold (FOH)
Translation todiscrete plantZero order hold (ZOH)
EmulationPurpose: Find a discrete transfer function which approximately has the same characteristics over the frequency range of interest.
Digital implementation: Control part constant between samples. Plant is not constant between samples.
4
Numerical Integration
Fundamental concept Represent H(s) as a differential equation. Derive an approximate difference equation.
We will use the following example Notice, by partial expansion of a transfer function this
example covers all real poles.
aeauuas
asH
sE
sU
)(
)(
)(
ExampleTransfer function Differential equation
5
Numerical Integration
kTTkTaeauTkTu
daeaudaeaukTu
continousdaeautu
aeauu
as
asH
sE
sU
TkT kT
TkT
t
,over of area)(
)(
)()()()(
Equation alDifferenti
)()(
)(
FunctionTransfer
0
0
6
Numerical Integration
Now, three simple ways to approximate the area.
Forward rectangle approx. by looking forward
from kT-T Backward rectangle
approx. by looking backward from kT
Trapezoid approx. by average
kT-T kT
kT-T kT
kT-T kT
7
Numerical Integration
Forward rectangular rule (Euler’s rule)(Approximation kT-T)
aTz
a
aTz
aT
zaT
aTz
zE
zUzH
zEaTzzUzaTzU
TkTaTeTkTuaT
TkTaeTkTauTTkTukTu
F
11)1(1)(
)()(
)()()1()(
unctionTransfer F
)()()1(
)()()()(
EquationDifference
1
11
11
11
1
111
8
Numerical IntegrationBackward rectangular rule (app kT)
aTzz
a
zaT
aTz
aT
aT
zaTzE
zUzH
zEaT
aTzU
zaTzU
kTeaT
aTTkTu
aTkTu
kTaTeTkTukTaTukTu
kTaekTauTTkTukTu
B
11)1()1()1(
11
1
)(
)()(
)(1
)()1(
1)(
unctionTransfer F
)(1
)(1
1)(
)()()()(
)()()()(
EquationDifference
2
22
22
222
222
9
Numerical IntegrationTrapezoid rule (Tustin’s Method, bilinear trans.)(app ½(old value + new value))
azz
T
a
aTzaT
zaT
zE
zUzH
kTeTkTeaT
aTTkTu
aT
aT
kTaekTauTkTaeTkTauT
TkTukTu
T
)1()1(22)2(
)1(
)(
)()(
unctionTransfer F
)()()2/(1
2/)(
)2/(1
)2/(1
)()()()(2
)()(
EquationDifference
3
3
3333
10
Numerical Integration
Comparison with H(s)
azz
T
azH
as
a
aTzz
azH
as
a
aTz
azH
as
a
sH
T
B
F
)1()1(2
)(rule Trapezoid
1)(rule Backward
1)(rule Forward
functionTransfer Method)(
11
Numerical IntegrationTransform s ↔ z
sT
sTz
z
z
Ts
Tsz
Tz
zs
TszT
zs
)2/(1
)2/(1or
)1(
)1(2rule Trapezoid
1
1or
1rule Backward
1or1
rule Forward
ionApproximatMethod
Comparison with respect to stability In the s-plane, s = j is the boundary between stability
and instability.
12
Numerical Integration - stability
jssT
sTz
jsTs
z
jsTsz
for )2/(1
)2/(1Trapezoid
for 1
1Backward
for 1Forward
(Forward)Region ofs-LHP
(Backward)Region ofs-LHP
(Trapezoid)Region ofs-LHP
13
Numerical IntegrationSome conclusions
Higher distortion with the forward and backward rule. The forward rectangular rule could cause a stable
continuous filter to be mapped into an unstable digital filter.
The trapezoid rule maps the stable region in the s-plane exactly into the stable region of the z-plane. Though, the j-axis is compressed into the unit circle with length 2.
The trapezoid rule appears superior. But, let us make some further investigations..... let us calculate the power |H(j|2 at, for example, = a – the change in frequency response.
14
Numerical Integration
2
1)(
)(
1)/(
1)(
)(
2
2222
22
jaH
a
aa
ajH
aj
ajH
What about HT(z) ? at which frequency is |HT(z)|2 = 0.5 ?
15
Numerical Integration
12/,)2
(2
Notice,
)2
arctan(2
or)2/tan(2
2
1
1)2/tan(21
1)(
en Power wh.with
1)2/tan(21
1
)2/tan(2
)(
1
11
22
1
22
1
11
aTaaT
T
aT
TT
Ta
TTa
zH
ezez
TjTa
aTjT
azH
T
TjTj
T
HT(z) as before but re-formulated (page 194)
I.e. high sampling frequency
16
Numerical IntegrationTrapezoid with pre-warping
maintain the power at some specified frequency
1. Choose a critical frequency
2. Replace by a´ such that new a = a’,
3. and insert in H(s). a’ = (2/T) tan( T/2)
4. Find the discrete filter H(z) based on the modified continuous transfer function H(s).
11
)2/tan(1
)()(
11
1
zz
Ts
sHzH
)1()(1 jHzH
17
Numerical Integration
1/
at Prewarping
2
1
1.0
with122
1
)(
filterorder 3th Example,
23
b
T
sss
sH
Plots for T = 0.1s ≈ 60 b
18
Numerical Integration
Plots for T = 1s ≈ 6 wb
Notice,prewarp
19
Numerical Integration
Plots for T = 2s ≈ 3 b
Notice,prewarp
20
Zero-Pole MatchingExact map between poles (z = esT) exists.
Basic idea, simply use same map for zeros.
Rules1. All poles are mapped by z = esT.
For example, s = -a maps to z = e-aT
2. All finite zeros are mapped by z = esT. For example, s = -b maps to z = e-bT
3. Basically, zeros at j= infinity maps to z = ej= -1 (representing the highest frequency)
4. Identical gain at some critical freq. (typically, s=0) H(s) at s=0 = H(z) at z=1
21
Zero-Pole Matching
z
zKzH
eK
as
a
ez
zK
zsH
ezas
as
aH(s)
aT
szaT
s
aT
1)(
2
11
14) (rule Gain
10)(3) (rule Zero
1) (rule Pole
01
Example (1)Compute the discrete equivalent by zero-pole matching
22
Zero-Pole Matching
s-plane z-plane
z
zKzH
as
aH(s)
1)(
match-zp
23
Zero-Pole Matching
)1()()1()(1
11)(
)(
)(1
1
kKekKekukuz
zK
z
zKzH
zE
zU
Including delay in computerNotice, u(k) depends on e(k). This comes from rule 3
Solution, first zero for s=infinity is mapped to z=infinity
(as in Matlab). This gives
)1()1()(1
1
)(
)(1
1
kKekukuz
zK
zK
zE
zU
24
Hold EquivalentsPurpose, design a discrete filter H(z), such that
input e(k) is samples of e(t) H(z) has an output u(k) u(k) approximates the output of the continuous filter
H(s) whose input is the continuous signal e(t).
Zero order hold (ZOH) (step invariant), approach below
HoldSampler H(s)e(t) e(k) u(k)
Samplereh(t) u(t)
25
Hold EquivalentsHold operation
gaina is)(
)()()(
11
)(
)(1)(1)(
ke
kesQsE
es
sQ
Ttttq
h
Ts
Q(s)
HoldSampler H(s)e(t) e(k) u(k)
Samplereh(t) u(t)
26
Hold Equivalents
ZOH transformation Notice, we can find H(z) by Take inverse Laplace to find a continuous signal. Sample this continuous signal to find a discrete signal.
kkette )()()(
)(11
)()()()()(
)()()()()()()(
11 kes
sHLksQsHLtu
ksQsHkesQsHsU
Ts
HoldSampler H(s)e(t) e(k) u(k)
Samplereh(t) u(t)
27
Hold Equivalents
)()(
)(1)(1)(11
)()( 11 ks
sHLTttke
ssHLtu Ts
s
sHLZ
s
sHZ
s
sHZzzHzUkuZ
tuku tkT
)()(where
)()1()()()(
)()(
1
1
Continuous output signal
Discrete output signal and transfer function H(z)
28
Hold EquivalentsTriangle hold equivalent
Extrapolate the samples to connect sample to sample in a straight line.
Impulse response for triangle hold
THold System
2
2
Ts
ee TsTs )(sH
)(ke )(tu
2
)()1()(
Result
s
sHZ
Tz
zzH
T
0.1
29
Hold Equivalents
Same third order filteras before
+ Zero-pole o ZOHx Triangle
T = 1(next T=2)
30
Hold Equivalents
Same filteras before
+ Zero-pole o ZOHx Triangle
T = 2
31
Hold EquivalentsConcluding remarks
Methods, we have looked at Numerical integration methods Zero-pole matching (mapping) Hold equivalents
All methods works fine for high sample frequency Matlab
32
Evaluation of Design
Discrete controller by emulation Can be calculated using one of several techniques.
Evaluation discrete equivalent (design) We need to analyze in the z-domain. Process expressed in z-domain (ZOH) Notice, other eval. might be necessary (simulation)
33
Evaluation of DesignExample, using Matlab
sysD_c = tf([10 1],[1 1]);T = 0.2; % around 30*wn. Also use, T = 1sysD_d = c2d(sysD_c,T,’matched’); % zero-pole matchedsysG_c = tf([1],[10 1 0]);sysG_d = c2d(sysG_c,T,’zoh’); % zero order holdsysCL_c = feedback(sysD_c*sysG_c,1);sysCL_d = feedback(sysD_d*sysG_d,1);step(sysCL_c); hold on;step(sysCL_d);
)110(
1)(
(antenna) sys. Cont.
sssG
1
)110()(
controller Cont.
s
ssD
34
Evaluation of Design
T = 0.2(s ≈ 30 n)
35
Evaluation of Design
T = 1(s ≈ 6 n)
The sample delay gives phase lag.Thus, less PM which gives higher overshoot
36
Nyquist Stability Criterion
Check closed loop stability with controller KD(s) and plant G(s)
)(
)()()(,
)()(1
)()(
)(
)(
sa
sbsGsD
sGsDK
sGsDK
sR
sY
Characteristic equation (CE)
0)(
)()(0)()(1
sa
sKbsasGsDK
Notice, number of poles in CE equals number of open-loop poles in D(s)G(S). (known)Notice, zeros in CE are closed-loop poles! (unknown)
37
Nyquist Stability CriterionApproach, check characteristic equation
Let us follow a contour encircling the RHP. Basic idea: Evaluate CE = 1+KD(s)G(s) at the
contour, and draw image. (We can just as well evaluate KD(s)G(s) and look at -1 )
)()( sGsDK
Notice, poles or zeros gives encirclements
38
Nyquist Stability Criterion
Criterion (continuous case)
Z = P + N
Z = number of unstable closed-loop poles. Thus, for stability we want Z=0. (unknown)
P = number of unstable open-loop poles of KDG (known).
N = total number of encirclements of the point -1 by evaluation of KDG (taken in same direction as s. Thus, N might be negative).
39
Nyquist Stability CriterionInterpretations and explanations
If the s-plane contour encircles a zero of 1+KDG in a certain direction, the image contour will encircle the origin in the same direction. (related to Z)
If the s-plane contour encircles a pole of 1+KDG in a certain direction, the image contour will encircle the origin in the opposite direction. (related to P)
The net number of same-direction encirclements, N, equals the difference N = Z- P.
Actually, we should investigate 1+KDG and encirclements around the origin, but easier to investigate KDG and encirclements around -1.
40
Nyquist Stability Criterion
Discrete case Same approach,
almost... Unstable region is
outside the unit circle.
Easier to evaluate using the stable region.
41
Nyquist Stability Criterion
Discrete case CE: 0 = 1+KD(z)G(z) Total number of poles = n Number of stable zeros of 1+KD(z)G(z) is n-Z Number of stable poles of 1+KD(z)G(z) is n-P
Criterion (discrete case) Z = (n-Z) – (n-P) or Z = P – N
42
Nyquist Stability Criterion
Discrete case Determin the number P of unstable poles of KDG Plot KD(z)G(z) for the unit circle. This is the counter-
clockwise path around the unit circle. Set N equal to the number of Counter-clockwise
encirclements of the point -1 on the plot Compute Z=P-N. The system is stable if and only if Z=0
43
Nyquist Stability CriterionExample (7.9)Evaluate the stability of the unity feedback discrete system with the plant G(s) with sampling time T=2 and ZOH. Use controller KD(z) = K.
)1(
1)(
sssG
Solution (matlab)sysC = tf([1],[1 1 0]);sysD = c2d(sysC,2);nyquist(sysD);
K=1
The system is openloop stable and there are no encirkelments of -1 => the system is closed loop stable for K=1
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