Modern Control Theory (Digital Control)
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Modern Control Theory
(Digital Control)
Lecture 2
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Outline Signal analysis and dynamic response
Discrete signals Discrete time – discrete signal plot z-Transform – poles and zeros in the z-plane
Correspondence with continuous signals Step response
Effect of additional zeros Effect of additional poles
s-Plane specifications z-Plane specifications Frequency response
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Signal analysis – discrete signals Analysis
look at different characteristic signals z-transform, poles and zeros signals
unit pulse unit step exponential general sinusoid
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Signal analysis – discrete signals The z transform
k
kk
k
zezE
keZzE
eeee
)(
))(()(
as transform-z thedefine we
........,.....,,
valuesdiskrete Given the
210
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Signal analysis – discrete signals The Unit Pulse
k
kk
k
k
k
zzzE
Z
k
k
ke
1)(
0,0
0,1
)(
01
1
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Signal analysis – discrete signals The unit Step
11
1
)(1)(
0,0)(1
0,1)(1
)(1)(
1
02
2
z
z
z
zzkzE
Z
kk
kk
kke
k
k
k
k
Zeros : z=0Poles : z=1
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Signal analysis – discrete signals Exponential
rz
z
rz
rzzrzE
Z
k
krke
k
k
k
kk
k
1
0
1
03
3
1
1
)()(
0,0
0,)(
Zeros : z=0Poles : z=r
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Signal analysis – discrete signals General Sinusoid
jj
k
kj
k
kjkk
jkk
jkjkkk
rez
z
zre
zrezerzE
Z
kerke
keerkkrke
1
0
1
05
5
4
1
1
)()(
)(1)(
)(12
1)(1)cos()(
(let us look at the terms, one by one, and use linearity)
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Signal analysis – discrete signals
22
4
5
)cos(2
))cos((
))((
)()(
2
1
2
1)(
)(
rzrz
rzz
rezrez
rezzrezz
rez
z
rez
zzE
rez
zzE
jj
jj
jj
j
Zeros : z=0, z=r cosPoles : z=r exp(j) , z=r exp(-j)
Plots shown for
45,7.0 r
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Signal analysis – discrete signals
Transients r > 1, growing signal (unstable) r = 1, constant amplitude signal r < 1, decreasing signal (the closer r is to 0 the shorter the
settling time. In fact, we can compute settling time in terms of samples N.)
)(1)cos()(4 kkrke k
Conclusions General sinusoid
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Signal analysis – discrete signals Samples per oscillation (cycle)
number of samples in a cycle is determined by or, N = samples/cycle depends on pole placements depend on
cycle
samples2
))(cos()cos(
N
Nkk
We have
dependence of
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Signal analysis – discrete signals Samples per oscillation (cycle), cont.
cyclesamples845
360
have we45For
N
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Signal analysis – discrete signals
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The duration of a time signal is related to the radius of the pole locations
the closer r is to 0 the shorter setteling time The number of the samples per cycle N is related to the
angle
cycle
samples360
N
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Signal analysis – discrete signalsPoleplacements
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Correspondence with cont. signals Continuous signal
))(()(
)(1)cos()(
jbasjbas
assY
kbtety Lat
Discrete signal
))((
))cos(()(
)(1)cos()()(
jbTaTjbTaT
ZkaT
eezeez
bTrzzzY
kbTkeky
Poles: s = -a + jb,s = -a - jb
Poles: z = exp(-aT - jbT)z = exp(-aT + jbT)
sTez
Pole map
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Correspondence with cont. signals
sTez
Pole map
10,0,
1,0,
1
:pole plane-s General
rrzs
rrzs
zjs
js
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Correspondence with cont. signals Recall, poles in the s-plane
22
22
2))(()(
nn
n
dd
n
ssjsjssH
222222
222
1
part, Imag.
part, Real
frequency, natural Undamped
)sin(ratio, Damping
nnnnd
dn
n
n
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Correspondence with cont. signals
sTez
Pole map
2
2
1,and
,where
1:pole plane-s General
nn
aT
j
nnd
ba
bTer
rezjbas
jjs
Fixed ,varying n
Fixed ,varying n
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Correspondence with cont. signals Fixed n,
varying Fixed ,varying n
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Correspondence with cont. signals Notice, in the vicinity of z = 1, the map of and n looks like the s-plane in the vicinity of s = 0.
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Signal analysis – step response
Investigate effect of zeros fix z1 = p1, and explore effect of z2
a (delayed) second order sys is obtained = {0.5, 0.707} (by adjusting a1 and a2)
= {18°,45°,72°} (by adj. a1 and a2) a unit step U(z) = z/(z-1) is applied to the
system (pole, z=1, and zero, z=0)
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Signal analysis – step response
Discrete stepresponsesfor = 18°
Overshootincreases withthe zero Z2
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Signal analysis – step response
The zero has little infuence on the negative axis, large influence near +1
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Signal analysis – step response
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Signal analysis – step response
Investigate effect of extra pole fix z1 = z2 = -1, and explore effect of moving
singularity p1 (from -1 to 1) = 0.5 = {18°,45°,72°} a unit step is applied to the system
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Signal analysis – step response
Mainly effect on rise time
Rise time expressedas number of samples.
The rise time increaseswith the pole
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Signal analysis – step response Conclusions
Addition of a pole or a zero between -1 and 0 Only small effect
Addition of a zero between 0 and +1 Increasing overshoot when the zero is moving towards +1
Addition of a pole between 0 and +1 Increasing rise time when the pole is moving towards +1 (the
pole dominates)
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s-Plane specifications Spec. on transients of dominant modes
dominant first order time constant (related to 3 dB bandwidth)
dominant second order rise time tr (related to natural frequency n ≈ 1.8/tr )
settling time ts (related to real part = 4.6 ts )
overshoot Mp, or damping ratio . Spec. on reference tracking
typically step or ramp input specification i.e. specifications on Kp and Kv , ess = r0 /Kv
ess is the steady state error for a ramp input of slope r0
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s-Plane specifications
ExampleWe have system with dominant 2. order mode
Specifications:
)line vertical(
lines) angel(
43
21%
CCt
CCM
s
p
Notice, spec. on n not shown
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z-Plane specifications Discrete system
similar specifications in addition, sample
time T2C
)(
)(
)(
5
43
21%
green
black
blue
Cer
CCt
CCM
T
s
p
Example (continued)Notice, sample time T must be chosen.
5Cr If fixed n
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z-Plane specifications
Example (7.2 and 7.5)A system is given by
Specifications are1) Overshoot Mp less than 16%
2) Settling time ts (1%) less than 10 sec.
3) Chose sample time T such that)110(
1)(
sssG
)3(sec2.025.020
1
Hz25.0where20
T
fff bbs
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z-Plane specifications
)2(912.0,
46.010,6.4
)1(5.0%16)1(6.0
46.0
rerer
tt
MM
TT
ss
pp
1) Overshoot Mp less than 16%
2) Settling time ts (1%) less than 10 sec.
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z-Plane specifications
Cn
Also, we might have an additional specification on rise time tr
Damping, radius
r
n
Possibleregion
912.0
5.0
r
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z-Plane specifications Steady-state errors
ZOH of plant transfer function, i.e. G(s) to G(z) Transfer function from R(z) to E(z), for investigating
the error.
)()()(1
1)( zR
zGzDzE
controllerD(z)
plantG(z)
R(z) U(z) Y(z)E(z)+
-
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z-Plane specifications Now, if r(kT) is a step, then
zero. iserror theThus,
then,1at polea has If
1
1
)1()1(1
1
1)()(1
1)1(lim)(
),( valueFinal
1)()(1
1)(
1
DGzDG
KGD
z
z
zGzDze
e
z
z
zGzDzE
p
z
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Frequency response Frequency response methods
Gain and phase can easily be plotted. Freq. response can be measured directly on a
physical plant. Nyquist's stability criterion can be applied. Error constants can be seen on gain plot. Corrections to gain a phase by additional poles and
zeros. Effect can easily be observed – in terms of cross over frequency, gain margin, phase margin.
Frequency response methods can also be applied for discrete systems (example).
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Frequency responseDiscrete Bode Plot, Example (7.8)Plot the discrete frequency response corresponding to
)1(
1)(
sssG
Transform to z-domain by ZOH, with sample time T = 0.2, 1 and 2.Solution. Use Matlab c2d(sys,T).
Matlabsysc = tf([1],[1 1 0]);
sysd1 = c2d(sysc,0.2);sysd2 = c2d(sysc,1);sysd3 = c2d(sysc,2);
bode(sysc,'-',sysd1,'-.', sysd2,':', sysd3,'-',)
)135.0)(1(
523.0135.1)(
)368.0)(1(
718.0368.0)(
)8187.0)(1(
9355.00187.0)(
3
2
1
zz
zzG
zz
zzG
zz
zzG
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Frequency response
Primary effect,Additional lag
Approx.phase lagT/2
Half samplefrequency
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Discrete Equivalents - Overview
controllerD(s)
plantG(s)
r(t) u(t) y(t)e(t)+
-
Translation to discrete controller (emulation)Numerical Integration• Forward rectangular rule• Trapeziod rule (Tustin’s method, bilinear transformation)• Bilinear with prewarpingZero-Pole MatchingHold Equivalents• Zero order hold (ZOH)• Triangle hold
Translation todiscrete plantZero order hold (ZOH)
Lecture 3