Cl i l t lClassical control - UPC Universitat Politècnica...
Transcript of Cl i l t lClassical control - UPC Universitat Politècnica...
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Control & Guidance 2011
Enginyeria Tècnica d'AeronàuticaA ióesp. en Aeronavegació
Escola d'Enginyeria de Telecomunicació i Aeroespacial de Castelldefels
Adeline de Villardi de Montlaur
Cl i l t lCl i l t lClassical controlClassical control
Control and guidance
Slide 1
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Cl i l t lCl i l t lClassical control1 P t i ti ti
Classical control1 P t i ti ti1. Parametric estimation
2 Steady state error
1. Parametric estimation
2 Steady state error2. Steady state error
3. Root locus
2. Steady state error
3. Root locus
4. Controllers4. Controllers
5. Frequency response5. Frequency response
6. Bode diagrams6. Bode diagrams
Control and guidance
Slide 2
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Study properties of the response of the system:
desired angle of attack αrefd
actual angle of attack αdisplacement
Study properties of the response of the system:
LONGITUDINAL CONTROL LONGITUDINAL
DYNAMICS
speed urefg
speed udisplacement of elevator δe
CONTROL SYSTEM DYNAMICS
t l l f tt k
SENSORS:
actual angle of attack αspeed u
SENSORS: INS,
Anemometer
1- Parametric estimation1- Parametric estimationControl and guidance
Slide 3
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1- Parametric estimation1- Parametric estimation
Temporal methods:Temporal methods:
a. Firsta. First--order systemsorder systems
b. Secondb. Second--order systemsorder systems
c. Higherc. Higher--order systemsorder systems
Control and guidance
Slide 4
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Fi tFi t d td t
1- Parametric estimation1- Parametric estimation
A first-order system is defined by a first-order
a. Firsta. First--order systemsorder systems
A first order system is defined by a first order
differential equation:
s1K
)s(R)s(Y)s(G)t(Kr)t(y)t(y L
s1)s(R
τ: system time constantτ: system time constant
K: gain
Electrical/mechanical examples
Control and guidance
Slide 5
examples
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Fi tFi t d td t
1- Parametric estimation1- Parametric estimation
Impulse response
a. Firsta. First--order systemsorder systems
Impulse response
1)s(R)]t([L 1K)s(Y )()]([ s1
using the inverse Laplace transform, the impulse
response is:
0teK)t(yt
0te)t(y
Control and guidance
Slide 6
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a Firsta First order systemsorder systems
1- Parametric estimation1- Parametric estimation
Impulse response 1
a. Firsta. First--order systemsorder systems
0teK)t(yt
0.8
0.9
y(t))(y
0.6
0.7
)Tangent slope in 0:
y( )
0 3
0.4
0.5y(t)K)t(dy
g p
0.1
0.2
0.320tdt
r(t)
0 1 2 3 4 5 6 70
t
Control and guidance
Slide 7
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Fi tFi t d td t
1- Parametric estimation1- Parametric estimation
Step response: response to a unit step function
a. Firsta. First--order systemsorder systems
p p p p
1KK
)1(K)s(Y
1)s(R)]t(u[L s1ss)s1(
)(s
)()]([
i th i L l t f th tusing the inverse Laplace transform, the step response
or indicial response is:
0te1K)t(yt
0te1K)t(y
Control and guidance
Slide 8
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Fi tFi t d td t
1- Parametric estimation1- Parametric estimation
St 9
10
a. Firsta. First--order systemsorder systems
y(t)Step response:
t
8
9y( )
te1K)t(y
6
7
)Tangent slope in 0: 4
5y(t)
2
3
K
dt)t(dy r(t)
0 1 2 3 4 5 6 70
1
t
dt 0t
Example
Control and guidance
Slide 9
t
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b S db S d d td t
1- Parametric estimation1- Parametric estimation
A second-order system is defined by a second-order
b. Secondb. Second--order systemsorder systems
A second order system is defined by a second order
differential equation:
)t(ra)t(yb)t(yb)t(yb 0012
0a)s(Y)(G01
22
0
bsbsb)s(R)()s(G
012)(
Electrical/Mechanical examples
Control and guidance
Slide 10
ect ca / ec a ca e a p es
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
It can be factorized to emphasize particular
b. Secondb. Second--order systemsorder systems
parameters:2KK)s(Y
2nn
2n
2 s2sK
1s2s
K)s(R)s(Y)s(G
nn
12
with K: system gain (corresponds to final value for a unit step function)
ωn: undamped natural frequency
ζ: damping factor (ζ>0)Control and guidance
Slide 11
ζ: damping factor (ζ 0)
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
Step response:2nK)s(Y
b. Secondb. Second--order systemsorder systems
p p
Response depends on the poles of the transfer function
2nn
2 s2s)s(R Response depends on the poles of the transfer function
0s2s 2nn
2
2442
s2n
2n
2n
1444let2
22n
2n
2n
2 di i i t’ i d d ζ l→ discriminant’s sign depends on ζ value
→ poles and response's properties depend on ζ valueControl and guidance
Slide 12
poles and response s properties depend on ζ value
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
ζ>1 Over-damped movement (non-oscillatory modes)
b. Secondb. Second--order systemsorder systems
Real and negative poles: 1s 2n21
2n
2n K1K1)s(Y
n2,1
212nn
2 ssssss2ss)s(Y
111
Development in simple fractions:
22112n
sss1
sss1
12s1K)s(Y
Control and guidance
Slide 13
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
ζ>1 Over-damped movement (non-oscillatory modes)1
Step Response
b. Secondb. Second--order systemsorder systems
tsts
Inverse Laplace transform:0.8
1
y(t)y(t)
2
ts
1
ts
2n
se
se
121K)t(y
21
0.6
e
y(t)
0.4
Am
plit
ude
Tangent slope in 0:
)t(dy0.2 =4
=1.5
0dt)t(dy
0t
0 5 10 15 20 25 300
Time (sec )
Control and guidance
Slide 14
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
ζ=1 Critically damped movement (non-oscillatory modes)
b. Secondb. Second--order systemsorder systems
Double real negative poles: n2,1s
22n
sK
s1)s(Y
nss
Development in simple fractions:
n 11K)(Y
Development in simple fractions:
n
2n
n
sssK)s(Y
Control and guidance
Slide 15
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
ζ=1 Critically damped movement (non-oscillatory modes)
b. Secondb. Second--order systemsorder systems
1Step Response
Inverse Laplace transform:y(t)
tnnet11K)t(y
0 6
0.8y( )
Tangent slope in 0:
)t(dy 0.4
0.6
Am
plit
ude
0dt)t(dy
0t
0.2
0 2 4 6 80
Time (sec )
Control and guidance
Slide 16
Time (sec )
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
ζ<1 Under-damped movement (oscillatory modes)
b. Secondb. Second--order systemsorder systems
Conjugated complex poles: 21js Conjugated complex poles:
2K1
n2,1 1js
22n
2n
n
1ssK
s1)s(Y
Development in simple fractions…p p
Control and guidance
Slide 17
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
ξ<1 Under-damped movement: Inverse Laplace transform:
b. Secondb. Second--order systemsorder systems
t1sin1
t1cose1K)t(y 2n2
2n
tn
S R 1
1 4
1.6
1.8Step Response
=0.7=0.1
Tangent slope in 0:1
1.2
1.4
litud
e
Tangent slope in 0:
0)t(dy 0 4
0.6
0.8Am
p
0dt)(y
0t
0 10 20 30 400
0.2
0.4
Control and guidance
Slide 18
Time (sec )
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
Characteristic parametersvaluefinaloutput
b. Secondb. Second--order systemsorder systems
K: gain
M: maximum overshoot : represents the value of thevaluefinalinputvaluefinaloutput
M: maximum overshoot : represents the value of the highest peak of the system response measured with
t t th f l (fi l l )respect to the reference value (final value)
tp: peak time: time needed for the response to arrive at its p p pfirst peak
T: period
t : settling timeControl and guidance
Slide 19
ts: settling time
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b Secondb Second order systemsorder systems
1- Parametric estimation1- Parametric estimation
Characteristic parameters: for second-order systems
b. Secondb. Second--order systemsorder systems
K 2
21tan eeM
natural(dampedζ1ωω 2
nd
2p1
t
frequency)
( p
2n
p1
22 2
nd 122T
Control and guidance
Slide 20
n
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b Secondb Second--order systemsorder systems1- Parametric estimation1- Parametric estimation
Obtain:b. Secondb. Second order systemsorder systems3,5
Step ResponseObtain:KM
2,5
3 r(t)Mtpt 5%
2
itude
ts5%
T
1
1,5Am
pl
Deduce:ζ
0,5
1
y(t)
ζωn
0 1 2 3 4 5 6 7 8 9 10 11 120
Time (sec)
y( )Calculate: G(s)=Y(s)/R(s)
Control and guidance
Slide 21
Time (sec)( ) ( ) ( )
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Hi hHi h d td t
1- Parametric estimation1- Parametric estimation
c. Higherc. Higher--order systemsorder systems→ characterize the transitory state of any-order systems
generally y(t)= linear combination of elementary time
functions defined by the nature (real or complex) of the
characteristic equation roots: system modes:characteristic equation roots: system modes:
• real poles: non- oscillatory modes, exponential term in
the response
• complex poles: oscillatory modes, exponential term
multiplied by sine or cosineControl and guidance
Slide 22
p y
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Hi hHi h d td t
1- Parametric estimation1- Parametric estimation
High order systems can be simplified using:
c. Higherc. Higher--order systemsorder systems
High order systems can be simplified using:
dominant poles
poles further from the imaginary axis have a
weaker contribution
1 pole near 1 zero
if there is a zero near a pole this pole contributionif there is a zero near a pole, this pole contribution
will be weak
Control and guidance
Slide 23
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Study error of the response of the system:
desired angle of attack αrefspeed u actual angle of attack αdisplacement
Study error of the response of the system:
LONGITUDINAL CONTROL LONGITUDINAL
DYNAMICS
speed urefg
speed udisplacement of elevator δe
CONTROL SYSTEM DYNAMICS
t l l f tt k
SENSORS:
actual angle of attack αspeed u
SENSORS: INS,
Anemometer
2- Steady state error2- Steady state errorControl and guidance
Slide 24
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2- Steady state error2- Steady state error
G(s)Y(s)+
-R(s) Ess
Steady State error:
ess= difference between the entry signal and the exit signal
e = “what we want minus what we get”ess = what we want minus what we get
Control and guidance
Slide 25
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2- Steady state error2- Steady state error
System’s type:y yp
Given the transfer function:2
)ss1) (s1)(s1(s)...dscs1)...(bs1)(as1(K)s(G 2N
2
)...ss1)...(s1)(s1(s
with K: system gain,
and N: number of poles in the origin
→ N = system’s typeControl and guidance
Slide 26
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2- Steady state error2- Steady state error
Definition of Steady State error:
)t(y)t(rlimetss
t
e > 0 : exit signal has not reached the entry referenceess > 0 : exit signal has not reached the entry reference
ess < 0 : exit signal is higher than the entryss g g y
Control and guidance
Slide 27
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2- Steady state error2- Steady state error
)t(y)t(rlime )t(y)t(rlimetss
Moving to the Laplace space: Final value theorem:
)s(R)s(G1)s(G)s(Rslim)s(Y)s(Rslime
0s0sss )s(G1
)s(R Depends on the entry
)s(G1
)s(Rslime0sss
p y
+ on the system’s type
Control and guidance
Slide 28
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)...dscs1)...(bs1)(as1(K)(G2
2- Steady state error2- Steady state error
G(s)Y(s)+R(s)
)...ss1)...(s1)(s1(s)) ()((K)s(G 2N
G(s)-
1. Position error: error for a step function entry: r(t)=u(t)
1111
)s(Glim1
1)s(G1
1lims1
)s(G11slime
0s0s0sp
0type10s
Itype0
ypK1
Control and guidance
Slide 29
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1 Position error: error for a step function entry: r(t)=u(t)
2- Steady state error2- Steady state error
1
Step Response
1. Position error: error for a step function entry: r(t)=u(t)
0 8
1
Error
0.6
0.8
ude
0.4
Am
plitu
0.2
0 2 4 6 8 10 120
Ti ( )
Control and guidance
Slide 30
Time (sec )
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2- Steady state error2- Steady state error)...dscs1)...(bs1)(as1(K)(G2
G(s)Y(s)+R(s)
)...ss1)...(s1)(s1(s)) ()((K)s(G 2N
2 Speed error: error for a ramp function entry: r(t)= t
( )-
2. Speed error: error for a ramp function entry: r(t)= t
1lim1lim11slime
0type
)s(sGlim
)s(sGslim
s)s(G1slime
0s0s20sv
Itype10type
IItype0
ItypeK
Control and guidance
Slide 31
IItype0
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2- Steady state error2- Steady state error)...dscs1)...(bs1)(as1(K)(G2
G(s)Y(s)+R(s)
)...ss1)...(s1)(s1(s)) ()((K)s(G 2N
G(s)-
3. Acceleration error: error for a parabolic entry: r(t)= t2
1li1li11li
I0
)s(Gslim
)s(Gsslim
s)s(G1slime 20s220s30sv
IItype1Ior0type
IIItype0
IItypeK
Control and guidance
Slide 32
IIItype0
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2- Steady state error2- Steady state error
Error based on type + entryyp y
Input:Type:
step ramp parabolic
0 constant ∞ ∞
I 0 constant ∞
II 0 0 constant
Control and guidance
Slide 33
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2- Steady state error2- Steady state error
Example: G(s) θ(s)+θref(s) δe(s)K
C t th i t d t t f it t f ti
-
Compute the error in steady state for a unit step function entry and for a system with the following open loop transfer function: 10s2)s( function:
4s1.0s1.0s2
)s()s(
2e
• for K=1, K=10, K=100,
)(e
0 12sθ(s) • for K=1 and
40.1sss0.12s
(s)δθ(s)
2e
Control and guidance
Slide 34
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Design a simple proportional controller in order to satisfy some
desired angle of attack αrefd
actual angle of attack αdisplacement
constraints on the response of the system
LONGITUDINAL CONTROL LONGITUDINAL
DYNAMICS
speed urefg
speed udisplacement of elevator δe
CONTROL SYSTEM DYNAMICS
t l l f tt k
SENSORS:
actual angle of attack αspeed u
SENSORS: INS,
Anemometer
3- Root locus3- Root locusControl and guidance
Slide 35
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3- Root locus3- Root locus
• Root locus technique• Root locus technique
• Gain setting• Gain setting
• Effect of zeros and poles• Effect of zeros and poles
Control and guidance
Slide 36
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Root locus techniqueRoot locus technique
3- Root locus3- Root locus
Root locus techniqueRoot locus technique
• Introduced by W. R. Evans in 1949: developed a series of rules
that allow the control system engineer to quickly draw the
root locus diagram = locus of all possible roots of the
characteristic equation: 1+K G(s)= 0
= locus of all possible poles in closed loop= locus of all possible poles in closed loop
as K varies from 0 to infinity
• The resulting plot helps us in selecting the best value of K
G f f f• Gives information for the transitory part of the response
(stability, damping factor, natural frequency)
Control and guidance
Slide 37
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Root locus techniqueRoot locus technique
3- Root locus3- Root locus
Root locus techniqueRoot locus technique
Let
m21 zszszs)s(G
A d b tit t it i th h t i ti ti
n21 pspsps)(
And substitute it in the characteristic equation
zszszsk
Kkwhere0pspspszszszsk1n21
m21
Control and guidance
Slide 38
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Root locus techniqueRoot locus technique
3- Root locus3- Root locus
Root locus techniqueRoot locus technique
The characteristic equation is complex and can be written in terms
of magnitude and angle as follows
zszszsk 21 1
pspspszszszsk
n21
m21
180)1q2(pszsn
1ii
m
1ii
)1mn(...,2,1,0qfor1i1i
Control and guidance
Slide 39
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Root locus technique: RulesRoot locus technique: Rules
3- Root locus3- Root locus
Root locus technique: RulesRoot locus technique: Rules
If we rearrange the magnitude criteria as
k1
pspspszszszs m21
Rule 1: The number of separate branches of the root locus plot is
kpspsps n21
equal to the number of poles of the transfer function (n)
Branches of the root locus originate at the poles of G(s) for k=0Branches of the root locus originate at the poles of G(s) for k=0
and terminate at either the open-loop zeroes or at infinity for k=+∞
n separate branches, n-m infinite branches, m finite branches
Control and guidance
Slide 40
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Root locus technique: RulesRoot locus technique: Rules
3- Root locus3- Root locus
Root locus technique: RulesRoot locus technique: Rules
Rule 2: Because the complex poles are always “conjugated”Rule 2: Because the complex poles are always conjugated ,
the root locus branches are symmetric with respect to the real axis
Control and guidance
Slide 41
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Root locus technique: RulesRoot locus technique: Rules
3- Root locus3- Root locus
Root locus technique: RulesRoot locus technique: Rules
Rule 3:Rule 3:
Segments of the real axis that are part of the root locus:
points on the real axis that have an odd number of poles and
zeroes to their right
Control and guidance
Slide 42
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Root locus technique: RulesRoot locus technique: Rules
3- Root locus3- Root locus
Root locus technique: RulesRoot locus technique: RulesRule 4: Asymptotes
The root locus branches that approach the open-loop zeroes atThe root locus branches that approach the open loop zeroes at
infinity do so along straight-line asymptotes that intersect the real
axis at the center of gravity of the finite poles and zeroesaxis at the center of gravity of the finite poles and zeroes
zpm
i
n
i
mn
p1i
i1i
i
The angle that the asymptotes make with the real axis is given by
)1(210f1q2º180
)1mn(...,2,1,0qformnq
a
Control and guidance
Slide 43
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Root locus technique: RulesRoot locus technique: Rules
3- Root locus3- Root locus
Root locus technique: RulesRoot locus technique: Rules
Rule 5: breakaway points
If a portion of the real axis is part of the root locus and a branch is
between two poles the branch must break away from the real axis p y
so that the locus ends on a zero as k approaches infinity. The
breakaway points on the real axis are determined by solvingbreakaway points on the real axis are determined by solving
kfor0zszszsk1 m21
and then finding the roots of the equation dk/ds=0
kfor0pspsps
1n21
and then finding the roots of the equation dk/ds=0
Only roots that lie on a branch of the locus are of interest
Control and guidance
Slide 44
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Root locus technique: RulesRoot locus technique: Rules
3- Root locus3- Root locus
Root locus technique: RulesRoot locus technique: Rules
Rule 6: Intersection with the imaginary axisRule 6: Intersection with the imaginary axis
Solve the characteristic equation for s=jω (equation of the
imaginary axis)
zjzjzjk 0
pjpjpjzjzjzjk1n21
m21
Control and guidance
Slide 45
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Root locus technique: RulesRoot locus technique: Rules
3- Root locus3- Root locus
Root locus technique: RulesRoot locus technique: Rules
Rule 7: for complex poles and zeroes only:Rule 7: for complex poles and zeroes only:
The angle of departure of the root locus from a pole of G(s) or
arrival angle at a zero of G(s) can be found by the following
expression
If you consider a test point t:nm
180ptztn
1ii
m
1ii
Control and guidance
Slide 46
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Root locus technique: examplesRoot locus technique: examples
3- Root locus3- Root locus
Root locus technique: examplesRoot locus technique: examples
Example 1:
Root Locus2
Example 1:
1
1.5
ary
Axi
s0
0.5
3s2s2s)s(G 2
Imag
in
1
-0.5
03s2s
-1.5
-1
Real Axis-4 -3 -2 -1 0
-2
Control and guidance
Slide 47
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Root locus technique: examplesRoot locus technique: examples
3- Root locus3- Root locus
Root locus technique: examplesRoot locus technique: examples
Example 2: 4Root Locus
Example 2:
2
3
2s1ss1)s(G
0
1
ary
Axi
s
2s1ss
2
-1
0Im
agin
a
-3
-2
-6 -4 -2 0 2-4
Real Axis
Control and guidance
Slide 48
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Setting of the gain and natural frequencySetting of the gain and natural frequency
3- Root locus3- Root locus
Setting of the gain and natural frequencySetting of the gain and natural frequencyBasic operation: to adjust the gain K to obtain a damping factor
given by the poles in closed loop and fixed by the damping factor ζgiven by the poles in closed-loop and fixed by the damping factor ζ
)R (Cf second-order systems: i
s)sRe(
straight line doing an angle φ with the real axis (cos φ= ζ) sets an
isg g g φ ( φ ζ)
intersection point with the poles position, and k (and then K) is
obtained solving the characteristic equationobtained solving the characteristic equation
Natural frequency for a second-order system: ωn=|si|
Control and guidance
Slide 49
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain setting
0)s(KG1 0
pspspszszszsK1 m21
The system total gain is computed thanks to the module condition
pspsps n21
szszszspspspKk m21
szszsz n21
“total gain” = product of the distances from the poles of G(s) to the
intersection point (= target pole) divided by the product of the
distances from zeros of G(s)
Control and guidance
Slide 50
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain setting
0)s(KG1
0pspsps
K1
If there are no zeros:
pspsps n21
spspspK m21 “total gain” =product of the distances between the poles of G(s)
( )and the intersection point (= target pole)
Examples
Control and guidance
Slide 51
p
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain setting
40.5
Root Locus
1
2
3
0.5
2s1ss1)s(G
0
1
ary
Axi
sDesigner requirement:we want ζ=0.5
2
-1
0
Imag
ina
ζ=0.5=cos φ → φ=60º
-3
-2
0.5
ζ φ φ
Examples
-6 -5 -4 -3 -2 -1 0 1 2-4
Real Axis
Control and guidance
Slide 52
p
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain setting
1
Root Locus3 0.5
S t
2s1ss1)s(G
1
2System: sys
Gain: 1.04Pole: -0.332 + 0.577i
Damping: 0.499Overshoot (%): 16.4
Designer requirement:we want ζ=0.5
gin
ary
Axi
s
0
1 Frequency (rad/sec): 0.666
ζ=0.5=cos φ → φ=60º
Imag
2
-1
ζ φ φ
3 5 3 2 5 2 1 5 1 0 5 0 0 5 1 1 5 2
-3
-2
0.5
Examples
Real Axis-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Control and guidance
Slide 53
p
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain setting
1
Root Locus3 0.5
2s1ss1)s(G
1
2System: sysGain: 1.04Pole: -0.332 + 0.577iDamping: 0.499Overshoot (%): 16.4
Sys tem: sysGain: 1.04Pole: -2.33
Designer requirement:we want ζ=0.5 in
ary
Axi
s
0
1( )
Frequency (rad/sec): 0.666Damping: 1Overshoot (%): 0
Frequency (rad/sec): 2.33
This corresponds to Im
ag-1 System: sys
Gain: 1.04Pole: -0.332 - 0.578iDamping: 0.498p
k=1.04
-3
-2
0.5
Damping: 0.498Overshoot (%): 16.5Frequency (rad/sec): 0.667
Examples
Real Axis-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Control and guidance
Slide 54
p
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain setting
2
Root Locus
1 5
20.7
3s2s2s)s(G 2
0 5
1
1.5
Designer requirement:we want ζ=0.7
agin
ary
Axi
s
0
0.5
Ima
-1
-0.5
-4 -3 -2 -1 0-2
-1.50.7
Examples
Real Axis-4 -3 -2 -1 0
Control and guidance
Slide 55
p
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain settingRoot Locus
20.7
2
1
1.5 System: sysGain: 1.32Pole: -1.66 + 1.69iDamping: 0.7O h t (%) 4 6
3s2s2s)s(G 2
Designer requirement:we want ζ=0.7
nary
Axi
s
0
0.5Overshoot (%): 4.6Frequency (rad/sec): 2.37
This corresponds to
Imag
in
-1
-0.5 System: sysGain: 1.32Pole: -1.66 - 1.69ip
k=1.32
2
-1.5
1
0.7
Damping: 0.7Overshoot (%): 4.6Frequency (rad/sec): 2.37
ExamplesReal Axis
-4 -3 -2 -1 0-2
Control and guidance
Slide 56
p
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R l ti t bilit i iR l ti t bilit i i
3- Root locus3- Root locus
Relative stability: gain marginRelative stability: gain margin
“Re(s)<0” criterion informs about the absolute stability of a system ( ) y y
but it says nothing about its relative stability
h f i i f h i bili h= how far it is from the instability → system strength
Gain margin: maximum proportional factor that can be introduced
into the control loop until the system becomes critically stable.
CrG
kM actual
G kM
Examples
Control and guidance
Slide 57
p
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain setting
1
Root Locus3 0.5
2s1ss1)s(G
1
2
System: sys
Designer requirement:we want ζ=0.5 in
ary
Axi
s
0
1 System: sysGain: 5.97Pole: 0.000761 + 1.41iDamping: -0.00054Overshoot (%): 100Frequency (rad/sec): 1 41
This corresponds to k=1
Imag
-1
Frequency (rad/sec): 1.41
kcr=6
MG=6 -3
-2
0.5
Examples
Real Axis-3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Control and guidance
Slide 58
p
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Roots locus exerciseRoots locus exercise
3- Root locus3- Root locus
Roots locus exerciseRoots locus exercise
2
1
1
1.5
2s2s2ss1)s(G 2
0.5
-0.5
0
1 5
-1
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2
-1.5
Real Axis
Control and guidance
Slide 59
Real Axis
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Roots locus exerciseRoots locus exercise
3- Root locus3- Root locus
Roots locus exerciseRoots locus exercise
1 5
2
0.5
1
1.5
0
0.5
2ary
Axi
s
-0 .5
0
Imag
in
-1 5
-1
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2
1.5
0.5
Real Axis
Control and guidance
Slide 60
ea s
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Roots locus exerciseRoots locus exercise
3- Root locus3- Root locus
Roots locus exerciseRoots locus exercise
1 5
2
0.5
1
1.5
ary
Axi
s
0
0 .5
2
Imag
in
-0 .5
0
S
-1 5
-1
System: sysGain: 1.62Pole: -0.373 - 0.627iDamping: 0.511Overshoot (%): 15.4
Real Axis
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2
1.5
0.5
( )Frequency (rad/sec): 0.73
Control and guidance
Slide 61
ea s
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Roots locus exerciseRoots locus exercise
3- Root locus3- Root locus
Roots locus exerciseRoots locus exercise
1 5
2
0.5
Root Locus
1
1.5
0
0.5
2ry A
xis
-0.5
0
Imag
inar
-1
-3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1-2
-1.5
0.5
Control and guidance
Slide 62
Real Axis
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Gain settingGain setting
3- Root locus3- Root locus
Gain settingGain setting
Note that even though the closed-loop poles have this value of
damping factor, the transitory response is not exactly sub-damped
with that characteristic, because the ζ formula has been used as if
it was a 2nd order system.
However the approximation is valid to obtain a good ζ magnitudeHowever, the approximation is valid to obtain a good ζ magnitude
order, the influence of poles and zeros on the response is seen in
the following studythe following study
Control and guidance
Slide 63
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Additional poleAdditional pole
3- Root locus3- Root locus
Additional poleAdditional pole
1. A second-order system is consideredy
2. A pole is added in s=-p
• system reference signal first affected by a first-order system and then by a 2nd order one
• for a step function, signal attenuated by an exponential, which isfor a step function, signal attenuated by an exponential, which is the 2nd order system entry
→ exit has less overshoot and it takes more time to reach its→ exit has less overshoot and it takes more time to reach its final value
Control and guidance
Slide 64
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Additional poleAdditional pole
3- Root locus3- Root locus
Additional poleAdditional pole
)4.52.3)(15.0(4.2
21
sssf
45234.2
22
ssf
4.52.3 ss
Control and guidance
Slide 65
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Additional zeroAdditional zero
3- Root locus3- Root locus
Additional zeroAdditional zero
4523)2(2.1
21
ss
sf4.52.3 ss
4.2f
Zeros (negative)
4.52.322
ssf
( g )
• increase the initial slope,
• make the system faster so it reaches its final value earlier,
• can produce overshoot
Control and guidance
Slide 66
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Effect of an additional pole in the roots locusEffect of an additional pole in the roots locus
3- Root locus3- Root locuspp
Transfer function of a 8Root Locus
vehicle cruise-control system: 4
6
4820
2
)33.3s)(1s)(06.0s(48.2)s(G
-4
-2
8
-6
4
-10 -8 -6 -4 -2 0 2 4-8
Real Axis
Control and guidance
Slide 67
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Effect of an additional pole in the roots locusEffect of an additional pole in the roots locus3- Root locus3- Root locus
A pole is added on 0 (i t t )
Root Locus
4
6
(integrator)
48.2)s(G is
2
4
)33.3s)(1s)(06.0s(s)(
Roots locus moved Imag
inar
y A
x
0
0 15Root Locus
to the right
+ unstableI
-4
-2
0 05
0.1
0.15
+ unstable
R l A i-8 -6 -4 -2 0 2 4 6
-6-0.05
0
0.05
Real Axis
0 2 0 15 0 1 0 05 0 0 05 0 1 0 15-0.15
-0.1
Control and guidance
Slide 68
-0.2 -0.15 -0.1 -0.05 0 0.05 0.1 0.15
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Effect of an additional zero in the roots locusEffect of an additional zero in the roots locus
3- Root locus3- Root locus
A zero is added i 0 12
Root Locus3
)333)(1)(060()12.0s(48.2)s(G
in -0.12
1
2
)33.3s)(1s)(06.0s(s
gina
ry A
xis
0
1
Imag
-1
5 4 3 2 1 0 1-3
-2
Root locus moved to the left + stable, and fasterReal Axis
-5 -4 -3 -2 -1 0 1
Control and guidance
Slide 69
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Design a controller/compensator in order to satisfy some
desired angle of attack αrefd
actual angle of attack αdisplacement
constraints on the response of the system
LONGITUDINAL CONTROL LONGITUDINAL
DYNAMICS
speed urefg
speed udisplacement of elevator δe
CONTROL SYSTEM DYNAMICS
t l l f tt k
SENSORS:
actual angle of attack αspeed u
SENSORS: INS,
Anemometer
4- Controllers4- ControllersControl and guidance
Slide 70
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4- Controllers4- Controllers
• Proportional controller: P: KP
I t l t ll I K• Integral controller : I:
• Derivative controller: D: sKI
sKD
tp M ts steady-state error
P decreases increases small changes decreases
I decreases increases increases eliminates (=0)
D small changes decreases decreases small changes
• these correlations may not be exactly accurate because K K and K• these correlations may not be exactly accurate, because KP, KI, and KDare dependent of each other
• changing 1 of these variables can change the effect of the other 2Control and guidance
Slide 71
changing 1 of these variables can change the effect of the other 2
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4- Controllers4- Controllers
• 2 kinds of controllers improve the transitory response:
Lead Compensator:00
0C zpwith
pszs)s(G
adds 1 zero and 1 pole, but zero is more important: it moves the t l t th l ft i t bilit ( t i f t d h
0ps
root locus to the left: improves stability (system is faster and has less overshoot)
Proportional Derivative Compensator:
dd 1 i t bilit
sKK(s)G DPC
adds 1 zero: improves stability
Control and guidance
Slide 72
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4- Controllers4- Controllers
• 2 kinds of controllers improve the steady state response:
Lag Compensator:00
0C pzwith
pszs)s(G
add 1 zero and 1 pole, but pole is more important: it moves the t l t th i ht d t bilit ( t i l d
0ps
root locus to the right: decreases stability (system is slower and has more overshoot), but decreases the steady state error
Proportional Integral Compensator: KK(s)G I
PC gs
( ) PC
Control and guidance
Slide 73
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4- Controllers4- Controllers
• 2 kinds of controllers improve both transitory and steady state response:
Lead - Lag Compensator:
110010
C zpandpzwithzszs)s(G
110010
C pppsps
)(
Proportional Integral Derivative (PID) Compensator:
K sDIPC K
sKK(s)G
Control and guidance
Slide 74
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5- Frequency response5- Frequency response
11 FourierFourier transformstransforms andand propertiesproperties
22 FrequencyFrequency responseresponse
33 ExamplesExamplespp
Control and guidance
Slide 75
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1 Fourier transforms and properties1 Fourier transforms and properties
5- Frequency response5- Frequency response
1 Fourier transforms and properties1 Fourier transforms and properties
The Fourier transform of a function x(t) is a function of theThe Fourier transform of a function x(t) is a function of the pulsation ω:
dt)t(xe)(X)]t(x[F tj
dt)t(e)()]t([
→ It transforms a signal from the time domain to the frequency domain
Control and guidance
Slide 76
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1 Transforms and properties1 Transforms and properties
5- Frequency response5- Frequency response
The inverse Fourier transform recovers the original function
1 Transforms and properties1 Transforms and properties
The inverse Fourier transform recovers the original function x(t):
d)(Xe
21)](X[F)t(x tj1 2
This is true for an absolutely integrable signal:
dt)t(x 2 dt)t(x
Control and guidance
Slide 77
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1 Transforms and properties1 Transforms and properties
5- Frequency response5- Frequency response
Linearity:
1 Transforms and properties1 Transforms and properties
Linearity:
)(Y)(X)]t(y)t(x[F )()()]t(y)t([
)t(dx Derivation:
)(Xjdt)t(dxF
)(Xj)t(xdF nn
)(Xjdt)(F n
Control and guidance
Slide 78
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1 Transforms and properties1 Transforms and properties
5- Frequency response5- Frequency response
Additional properties
1 Transforms and properties1 Transforms and properties
Additional properties
X1)at(xF
a
Xa
)at(xF
Duality Xtx F
x2tXXtx
F x2tX
Control and guidance
Slide 79
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1 Transforms and properties1 Transforms and properties
5- Frequency response5- Frequency response
C l i h
1 Transforms and properties1 Transforms and properties
Convolution theorems
FFtff F
FFtff
FFtffF
21F
21
FFtff 2121
dsstfsftffwhere 2121
Control and guidance
Slide 80
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1 Transforms and properties1 Transforms and properties
5- Frequency response5- Frequency response
1 Transforms and properties1 Transforms and properties
Time delay
XeTtxF Tj
Control and guidance
Slide 81
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1 Transforms and properties1 Transforms and properties
5- Frequency response5- Frequency response
Important pairs of transforms
1 Transforms and properties1 Transforms and properties
δ(t-t0), unit impulse )t(f )(F
0tje
2π δ(ω-ω0)tj 0e e
u(t), unit step1
δ(ω)
)t(ue at
a)t(ue t2cos 0 002
1 0
t2sin 02
00j21
Control and guidance
Slide 82
j2
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2 Frequency response2 Frequency response
5- Frequency response5- Frequency response
In previous examples we examined the free response of an airplane with step changes in control input
2 Frequency response2 Frequency response
step changes in control input
Other useful input function is the sinusoidal signal. Why?
1. Input to many physical systems takes the form or either a step change or sinusoidal signal
2. An arbitrary function can be represented by a series of step changes or a periodic function can be decomposed by means of Fourier analysis into a series of sinusoidal waves
→ if we know the response of a linear system to either a step or sinusoidal input then we can construct the system's response to an arbitrary input by the principle of superposition
Control and guidance
Slide 83
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2 Frequency response2 Frequency response
5- Frequency response5- Frequency response
2 Frequency response2 Frequency response
Example:
Examine the response of an airplane subjected to an external disturbance such as a wind gust
Wind gust can be a sharp edged profile or a sinusoidal profile (these 2 types of gust inputs occur quite often in nature) + arbitrary gust profile can be constructed by step and sinusoidal functions
Control and guidance
Slide 84
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1.5
2 Frequency response2 Frequency response
5- Frequency response5- Frequency response
0.5
1
2 Frequency response2 Frequency response
Arbitrary wind gust profiles:0
0.9
1
-1
-0.5
0.6
0.7
0.8
0 1 2 3 4 5 6 7 8 9 10-1.5
0.3
0.4
0.5
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
0.1
0.2
Control and guidance
Slide 85
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2 Frequency response2 Frequency response
5- Frequency response5- Frequency response
2 Frequency response2 Frequency response
Definition of “ frequency response”:
Response in steady state to a sinusoidal input
We will demonstrate that the steady state response is another
sinusoidal with the same frequency
Control and guidance
Slide 86
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2 Frequency response2 Frequency response
5- Frequency response5- Frequency response
2 Frequency response2 Frequency response
Similarity with Laplace functions with regard to the
operational properties (ex: differentiation)
→ the transfer function models can be transformed from one
method to the other replacing jω with s (or s with jω). (for
causal signals: signals defined for positive time)
Control and guidance
Slide 87
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2 Frequency response2 Frequency response
5- Frequency response5- Frequency response
2 Frequency response2 Frequency response
Given any system:
R(s) G(s) C(s)
Hypothesis: stable system
Sinusoidal input
22stsinLsRtsintr
Control and guidance
Slide 88
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2 Frequency response2 Frequency response
5- Frequency response5- Frequency response
2 Frequency response2 Frequency response
It can be demonstrated that the steady state response is:
tsinjG)t(c
jGImjGRejGwith 22
jGRejGImarctanjGargand jGRe
Control and guidance
Slide 89
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Parametric estimationParametric estimation
5- Frequency response5- Frequency response
Parametric estimationParametric estimation
a. Firsta. First--orderorder
a. Firsta. First orderorder
Frequency response: 21j1K
j1KjG
1j1
Gain: KjGGain: 21
jG
Delay: arctany arctan
Control and guidance
Slide 90
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5- Frequency response5- Frequency response
b. Secondb. Second--orderorder
KjG
2n j2
jGn
22n
Gain:
222
21
KjG
nn
21
D l
nωω2ξ
arctanDelay:
2
nωω1
arctan
Control and guidance
Slide 91
n
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5- Frequency response5- Frequency response
For a system composed by series of blocks:c. c. HigherHigher--orderorder
=
C(s)G1(s)
R(s)G2(s) G3(s)
=
G(s)C(s)R(s)
GGGGi h
jGjGjGjG
sGsGsGsGwith 321
jGjGjGjG 321
321 jGjGjGjG 321
321 jGjGjGjG
Control and guidance
Slide 92
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6- Bode diagram6- Bode diagram
1 Introduction1 Introduction1. Introduction1. Introduction
2. Construction rules2. Construction rules
3. Stability3. Stability
Control and guidance
Slide 93
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G l f th B d di
6- Bode diagrams6- Bode diagrams
Goals of the Bode diagrams:
To show the frequency response characteristics in a graphical form
2 graphics for the frequency using a logarithmic scale:
• one for the logarithm of a function magnitude (in decibels):
• one for the phase angle (in degrees):
dB)j(G
)j(Garg one for the phase angle (in degrees):
The decibel is a unit measure used to compare a certain value with a
)j(Garg
reference one. It is basically used to measure a signal power, and it is
defined as:
)j(GP2 )j(Glog20
1)j(G
log10PPlog10)j(G
ref
medidadB
Control and guidance
Slide 94
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6- Bode diagrams6- Bode diagrams
Semi-logarithmic axes: with lineal scale for the magnitude or the phase, and logarithmic for the frequencyp , g q y
Represents the complex transfer function adding each pole or zero effect, which compose this function (adding property of the log)
Control and guidance
Slide 95
effect, which compose this function (adding property of the log)
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GainGain
6- Bode diagrams6- Bode diagramsGainGain
The gain is a factor that only modifies the magnitude and its angular value is 0º; that is the gain value remains constant for any frequencyvalue is 0 ; that is, the gain value remains constant for any frequency value, because it does not depend on it.35
33 5
34
34.5
agn
itud
e (d
B)
Bode diagram for a K 50 i
32.5
33
33.5
Ma
1 K=50 gain
0
0.5
1
e (d
eg)
-1
-0.5
0
Pha
se
Control and guidance
Slide 96
100 10 11
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Integral and derivative factorsIntegral and derivative factors6- Bode diagrams6- Bode diagrams
gg
An integral factor or a pole centered in zero, has a transfer function of:
j11
Its magnitude is therefore:
j
j1)j(G
s1)s(G
Its magnitude is, therefore:
)log(201log20)j(GdB
For a logarithmic frequency axis: it corresponds to a straight negative
line of -20 dB per decade
line of -20 dB per decade
The phase is:1j1
900
1arctanjarg
j1arg)j(Garg
Control and guidance
Slide 97
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I t l d d i ti f tI t l d d i ti f t
6- Bode diagrams6- Bode diagrams
Integral and derivative factorsIntegral and derivative factors
For a derivative factor or a zero centered in zero, the results are
deduced using a similar development:
)log(20)j(GdB
90)j(Garg
Control and guidance
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Integral and derivative factorsIntegral and derivative factors
6- Bode diagrams6- Bode diagrams
Integral and derivative factors Integral and derivative factors 20
B ode Diagram
Bode diagrams of the0
10
tude
(dB
)
Bode diagrams of the derivative and the
integrator-10
Mag
ni
g-20
90
0
Pha
se (
deg)
100 101
-90
Control and guidance
Slide 99
Frequency (rad/sec)
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FirstFirst--order factors: firstorder factors: first--order poleorder pole
6- Bode diagrams6- Bode diagrams
FirstFirst order factors: firstorder factors: first order poleorder pole
j11)j(G
2211log20Its magnitude is:
j1 1
01log201log20)j(G11for 22 It seems more complicated, but approximations are made:
01log201log20)j(G,1fordB
log201log20)j(G,11for 22dB
gg)j(,dB
• substitute the curve by its two asymptotes
• magnitude is 0 dB until it reaches the point where both asymptotesg p y pmeet: ωτ=1, this point is called cut frequency
• from there: other asymptote with a 20 dB per decade slope• from there: other asymptote, with a -20 dB per decade slope.
• point where approximation error is maximum corresponds to the cutfrequency and the error is 3 dB
Control and guidance
Slide 100
frequency and the error is 3 dB.
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FirstFirst--order factors: firstorder factors: first--order poleorder pole
6- Bode diagrams6- Bode diagrams
similar for the phase, the phase real value is:
FirstFirst order factors: firstorder factors: first order poleorder pole
1arctan
1j1arg
j11arg)j(Garg 22
However the approximation in this case is:
0)j(Garg11for
•
90)j(Garg11for
)j(g
90)j(Garg1for
Control and guidance
Slide 101
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FirstFirst--order factors: firstorder factors: first--order poleorder pole
6- Bode diagrams6- Bode diagrams
1
FirstFirst order factors: firstorder factors: first order poleorder pole0
Bode Diagram
s11)s(G
-20
-10
nitu
de (
dB)
Bode diagram of a -40
-30Mag
n
gfirst-order system
-400
deg)
-45
Pha
se (
d
10-2
10-1
100
101
102
-90
Frequency (rad/sec)
Control and guidance
Slide 102
Frequency (rad/sec)
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FirstFirst--order factors: firstorder factors: first--order poleorder pole
6- Bode diagrams6- Bode diagrams
For the study of the first-order zeros, similar development, there is only a sign change: j1)j(G
FirstFirst order factors: firstorder factors: first order poleorder pole
a sign change: j1)j(G
2222 1l101l20M it d 2222 1log101log20 Magnitude:
Angular contribution:
1
arctanj1arg)j(Garg
Control and guidance
Slide 103
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FirstFirst--order factors: firstorder factors: first--order poleorder pole
6- Bode diagrams6- Bode diagrams
Example:
FirstFirst order factors: firstorder factors: first order poleorder pole
20Bode Diagram
Example:
10)s(G 10
nitu
de (
dB)
1s2)s(G
-10
0
Mag
n010
45
0(d
eg)
90
-45
Pha
se
10-2
10-1
100
101
-90
Frequency (rad/sec)
Control and guidance
Slide 104
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SecondSecond order factorsorder factors
6- Bode diagrams6- Bode diagrams
SecondSecond--order factors order factors
22 1
22nn
2
2n
2nn
2
2n
2j1
12j
)j(Gs2s
)s(G
nn
The magnitude is:
222
21log201log20)j(G
nn
2
n
22
n
21log20
21
log20)j(G
nn
Control and guidance
Slide 105
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SecondSecond--order factorsorder factors 222
6- Bode diagrams6- Bode diagrams
SecondSecond order factors order factors
it can be approximated this way:
nn
21log20)j(G
2n
dB0)1log(20)j(G,1for
2
nn
log20)j(G,1for
F l f i t i ht li t 0dBFor low frequencies: straight line at 0dB
For high frequencies: straight line with a –40 dB per decade slope.
Both asymptotes cross on ω=ωn.
However, in the second-order poles a resonance effect can appear.However, in the second order poles a resonance effect can appear.
In the frequency domain the resonance is shown as a peak close to the cut frequency; the resonance peak value is conditioned to the ζ valueControl and guidance
Slide 106
cut frequency; the resonance peak value is conditioned to the ζ value.
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SecondSecond--order factorsorder factors
6- Bode diagrams6- Bode diagrams
SecondSecond order factors order factors
F h h n
2
For the phase: 2n
1
tan
n
0)0tan(Arc,2.0generally,1For
180)0tan(Arc,5generally,1For
nn
90)tan(Arc,1For
nn
n
The phase graphic form depends also on ζ
Control and guidance
Slide 107
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SecondSecond--order factors order factors 6- Bode diagrams6- Bode diagrams
Bode diagram of a second-order pole for different ζ values
Control and guidance
Slide 108
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SecondSecond--order factors order factors 6- Bode diagrams6- Bode diagrams
Example: -20
Bode Diagram
25s3s5.2)s(G 2
-40
gnitu
de (
dB)
25s3s
-80
-60Mag
0
90
-45
0
(deg
)
-135
-90
Pha
se
10-1
100
101
102
-180
Frequency (rad/sec)
Control and guidance
Slide 109
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Stability conditionStability condition
6- Bode diagrams6- Bode diagrams
Stability condition Stability condition
)º180G()0G( forº180andfor0G Notation: )180G()0G(
The Bode diagram in open loop is studied
Stability condition:
0GforAndº180,forIf )0G(
0GforOrº180,forIf )0G(
0G,forAnd )º180( 0G,forOr )º180(
THEN the system is STABLE THEN the system is UNSTABLETHEN the system is STABLE THEN the system is UNSTABLE
in closed loop in closed loop
Control and guidance
Slide 110
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Stability conditionStability condition
6- Bode diagrams6- Bode diagrams
Stability condition Stability condition
STABLE
Control and guidance
Slide 111
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Stability condition Stability condition 6- Bode diagrams6- Bode diagrams
yy
UNSTABLE
Control and guidance
Slide 112
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Stability marginsStability margins
6- Bode diagrams6- Bode diagrams
Stability margins Stability margins
Common values:Common values:
Minimum gain
margin: 10 a 12 dB,
Minimum phaseMinimum phase
margin: 45 a 50º
Control and guidance
Slide 113
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Stability marginsStability margins
6- Bode diagrams6- Bode diagrams
Stability margins Stability margins
50Bode Diagram
-50
0
nitu
de
(dB
)
2)1)((1G(s) -150
-100Mag
90 2)1)(ss(s( )
-180
-135
-90
e (d
eg)
-270
-225
180
Pha
se
10-2
10-1
100
101
102
270
Frequency (rad/sec)
Control and guidance
Slide 114
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Stability margins Stability margins 6- Bode diagrams6- Bode diagrams
2)1)(ss(s1G(s)
MG=14.5dB
with roots locus:
MG=6, G ,
and note that:
20log(6)=15.5dB
Control and guidance
Slide 115
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Stability margins Stability margins 6- Bode diagrams6- Bode diagrams
2)1)(ss(s1G(s)
MΦ=51º
Control and guidance
Slide 116
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Stability margins Stability margins 6- Bode diagrams6- Bode diagrams
50Bode Diagram
)33.3s)(1s)(06.0s(48.2)s(G
Transfer function of a vehicle cruise-control system
0
50
e (d
B)
-100
-50
Mag
nitu
de
-150
90-45
0
g)
225-180-135
-90
Pha
se (d
eg
10-2
100
102
-270-225
Frequency (rad/sec)
Control and guidance
Slide 117
Frequency (rad/sec)
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Stability margins Stability margins 6- Bode diagrams6- Bode diagrams
A pole is added on 0 (integrator): Bode diagram shifted downward : + unstable
Control and guidance
Slide 118
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Stability margins Stability margins 6- Bode diagrams6- Bode diagrams
A zero is added in -0.12: Bode diagram shifted upward : + stable
Bode Diagram
0
50
100(d
B)
Bode Diagram
-100
-50
0
Mag
nitu
de
-150
100
135
-90
)
-225
-180
-135
Pha
se (d
eg)
10-2
100
102
-270
-225
F ( d/ )
Control and guidance
Slide 119
Frequency (rad/sec)
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REFERENCESREFERENCESREFERENCESREFERENCES
G.F Franklin, J.D. Powell, A. Emani-Naeini, Feedback Control of DynamicSystems, 4a Edición, Prentice-Hall, 2002
P. Lewis, Sistemas de Control en Ingeniería, Prentice Hall, 1999
W B lt C t l E i i 2ª Edi ió L 1998W. Bolton, Control Engineering, 2ª Edición, Longman, 1998
D. Arzelier, D. Peaucelle, Représentation et analyse des systèmes linéaires, Tomes 1 et 2, Version 1, ENSICA, 1999
Control and guidance
Slide 120