1 Modern Control Theory (Digital Control) Lecture 2.
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1
Modern Control Theory
(Digital Control)
Lecture 2
2
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
3
Signal analysis – discrete signals Analysis
look at different characteristic signals z-transform, poles and zeros signals
unit pulse unit step exponential general sinusoid
4
Signal analysis – discrete signals The z transform
k
kk
k
zezE
keZzE
eeee
)(
))(()(
as transform-z thedefine we
........,.....,,
valuesdiskrete Given the
210
5
Signal analysis – discrete signals The Unit Pulse
k
kk
k
k
k
zzzE
Z
k
k
ke
1)(
0,0
0,1
)(
01
1
6
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
7
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
8
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)
9
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
10
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
11
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
k=0
2
1
345
12
Signal analysis – discrete signals Samples per oscillation (cycle), cont.
cyclesamples845
360
have we45For
N
13
Signal analysis – discrete signals
14
Signal analysis – discrete signalsPoleplacements
15
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
16
Correspondence with cont. signals
sTez
Pole map
10,0,
1,0,
1
:pole plane-s General
rrzs
rrzs
zjs
js
17
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
18
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
19
Correspondence with cont. signals Fixed n,
varying Fixed ,varying n
20
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.
21
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)
22
Signal analysis – step response
Discrete stepresponsesfor = 18°
Overshootincreases withthe zero Z2
23
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
25
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
26
Signal analysis – step response
Mainly effect on rise time
Rise time expressedas number of samples.
The rise time increaseswith the pole
27
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)
28
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
29
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
31
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
32
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.
33
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
34
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)+
-
35
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
36
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).
37
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
38
Frequency response
Primary effect,Additional lag
Approx.phase lagT/2
Half samplefrequency
39
Frequency response
Approx.phase lagT/2
Accurate upto T = /2
40
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
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