Control & Guidance
Enginyeria Tècnica d'Aeronàutica
esp. en Aeronavegació
A
ió
Escola d'Enginyeria de Telecomunicació
i Aeroespacial de Castelldefels
Adeline de Villardi de Montlaur
Classical
Cl
i l control
t l
Control and guidance
Slide 1
2011
Cl
Classical
i l control
t l
1 P
1.
Parametric
t i estimation
ti ti
2 Steady state error
2.
3. Root locus
4. Controllers
5. Frequency response
6. Bode diagrams
Control and guidance
Slide 2
Study properties of the response of the system:
desired angle of attack αref
speed
d uref
LONGITUDINAL
CONTROL
SYSTEM
displacement
of elevator δe
actual angle
g of attack α
speed u
LONGITUDINAL
DYNAMICS
actual
t l angle
l off attack
tt k α
speed u
SENSORS:
INS,
Anemometer
1- Parametric estimation
Control and guidance
Slide 3
1- Parametric estimation
Temporal methods:
a. First
First--order systems
b. Second
Second--order systems
c. Higher
Higher--order systems
Control and guidance
Slide 4
1- Parametric estimation
a. FirstFi t-order
First
d systems
t
A first-order
first order system is defined by a first-order
first order
differential equation:

 y( t )  y( t )  Kr( t )
Y (s )
K

 G(s) 

R (s) 1  s
L
τ: system time constant
K: gain
Electrical/mechanical
examples
Control and guidance
Slide 5
1- Parametric estimation
a. FirstFi t-order
First
d systems
t
Impulse response
L[( t )]  R (s)  1
K
Y (s ) 
1  s
using the inverse Laplace transform, the impulse
response is:
K  t
y( t )  e

Control and guidance
Slide 6
t0
1- Parametric estimation
a First
a.
First--order systems
Impulse response
0.9
t0
Tangent
g
slope
p in 0:
dy( t )
K
 2
dt t 0

0.8
0.6
0.5
0.4
03
0.3
0.2
0.1
0
0
Control and guidance
Slide 7
y(t))
y(
0.7
y(t))
K  t
y( t )  e

1
r(t)
1
2
3
t
4
5
6
7
1- Parametric estimation
a. FirstFi t-order
First
d systems
t
Step
p response:
p
response
p
to a unit step
p function
1
L[ u( t )]  R (s) 
s
K
K
K
Y (s) 
 
(1  s)s s 1  s
using
i th
the iinverse L
Laplace
l
ttransform,
f
th
the step
t response
or indicial response is:

y( t )  K  1  e

Control and guidance
Slide 8
t



t0
1- Parametric estimation
a. FirstFi t-order
First
d systems
t
10
St response:
Step
y(t))
y(
9
8
y(t))
t 

y( t )  K  1  e  


Tangent slope in 0:
K
dy( t )

dt t 0 
Example
Control and guidance
Slide 9
7
6
5
4
3
r(t)
2
1
0
0
1
2
3
t
4
5
6
7
1- Parametric estimation
b.
b S
Second
Secondd-order
d systems
t
A second-order
second order system is defined by a second-order
second order
differential equation:
b2y( t )  b1y ( t )  b0 y( t )  a 0 r ( t )
Y (s )
a0
G (s ) 

2
R (s) b2s  b1s  b0
Electrical/Mechanical
ect ca / ec a ca e
examples
a p es
Control and guidance
Slide 10
1- Parametric estimation
b Second
b.
Second--order systems
It can be factorized to emphasize particular
parameters:
Y (s )
K
K

G (s ) 
 2
2
2
R (s )  s 
s  2ns  n
s
   2
1
n
 n 
2
n
with
K: system gain (corresponds to final value for a unit
step function)
ωn: undamped natural frequency
ζ: damping factor (ζ>0)
(ζ 0)
Control and guidance
Slide 11
1- Parametric estimation
b Second
b.
Second--order systems
Step
p response:
p
Y (s )
K2n
 2
R (s) s  2ns  2n
Response depends on the poles of the transfer function
s2  2ns  2n  0
 2n  4 22n  42n
s
2
let   4 22n  42n  42n  2  1
→ discriminant’s
di i i
t’ sign
i depends
d
d on ζ value
l
response's
s properties depend on ζ value
→ poles and response
Control and guidance
Slide 12
1- Parametric estimation
b Second
b.
Second--order systems
ζ>1 Over-damped movement (non-oscillatory modes)
Real and negative poles: s1, 2

 n     2  1

1
K2n
1
K2n
Y (s)   2
 
2
s s  2ns  n s s  s1   s  s2 
Development in simple fractions:
1




1
1
n

 
Y (s )  K  

 s 2  2  1  s1 s  s1  s2 s  s2   


Control and guidance
Slide 13
1- Parametric estimation
b Second
b.
Second--order systems
ζ>1 Over-damped movement (non-oscillatory modes)
St ep Respon se
1
Inverse Laplace transform:
Tangent slope in 0:
dy( t )
0
dt t 0
0.8
y(t)
0.6
Amp litud e
s1t
s2 t 




e
e
n




y( t )  K 1 
2
 2   1  s1
s2  

y(t)
0.4
=4
0.2
=1.5
0
0
5
10
15
Time (sec )
Control and guidance
Slide 14
20
25
30
1- Parametric estimation
b Second
b.
Second--order systems
ζ=1 Critically damped movement (non-oscillatory modes)
Double real negative poles:
s1, 2  n
1
K
Y (s )  
2
s s  n 
2
n
Development in simple fractions:
1
n
1 

Y(s)  K 

2
 s s  n  s  n 
Control and guidance
Slide 15
1- Parametric estimation
b Second
b.
Second--order systems
ζ=1 Critically damped movement (non-oscillatory modes)
St ep Respon se
1
Inverse Laplace transform:
y( t )  K 1  1  n t e
 n t

y(t))
y(
0.8
Tangent slope in 0:
dy( t )
0
dt t 0
Amp litud e
06
0.6
0.4
0.2
0
0
2
4
Time (sec )
Control and guidance
Slide 16
6
8
1- Parametric estimation
b Second
b.
Second--order systems
ζ<1 Under-damped movement (oscillatory modes)
Conjugated complex poles:

s1, 2  n    j 1   2
1
K
Y (s )  
2
s s s  n   2n 1   2 

Development
p
in simple
p fractions…
Control and guidance
Slide 17
2
n


1- Parametric estimation
b Second
b.
Second--order systems
ξ<1 Under-damped movement: Inverse Laplace transform:






t


2

n 
y( t )  K 1  e
cos n 1   t 
sin n 1   2 t
2


1 


1.8
S ep Respon
St
R
se





=0.7
=0.1
1.6
14
1.4
Tangent slope in 0:
dy
y( t )
0
dt t 0
Amp litud e
1.2
1
0.8
0.6
04
0.4
0.2
0
0
10
20
Time (sec )
Control and guidance
Slide 18
30
40
1- Parametric estimation
b Second
b.
Second--order systems
Characteristic parameters
K: gain output final value
input final value
M: maximum overshoot : represents the value of the
highest peak of the system response measured with
respectt to
t the
th reference
f
value
l (final
(fi l value)
l )
tp: p
peak time: time needed for the response
p
to arrive at its
first peak
T: period
ts: settling time
Control and guidance
Slide 19
1- Parametric estimation
b Second
b.
Second--order systems
Characteristic parameters: for second-order systems
K
Me
tp 

tan 
e
 
1  2
(
(damped
p natural
frequency)

n 1  
2
2
2
T

d n 1   2
Control and guidance
Slide 20
ωd  ω n 1  ζ 2
1- Parametric estimation
b Secondb.
Second-order systems
Obtain:
K
M
tp
ts5%
5%
T
St ep Response
3,5
3
r(t)
Amplitude
2,5
2
1,5
Deduce:
ζ
ωn
1
0,5
0
y(t))
y(
0
1
2
3
4
5
6
Time (sec)
Control and guidance
Slide 21
7
8
Calculate:
9 10 11 12 G(s)=Y(s)/R(s)
( ) ( ) ( )
1- Parametric estimation
c. HigherHi h -order
Higher
d systems
t
→ 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:
• real poles: non- oscillatory modes, exponential term in
the response
• complex poles: oscillatory modes, exponential term
multiplied
p
by
y sine or cosine
Control and guidance
Slide 22
1- Parametric estimation
c. HigherHi h -order
Higher
d systems
t
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,
pole this pole contribution
will be weak
Control and guidance
Slide 23
Study error of the response of the system:
desired angle of attack αref
speed uref
LONGITUDINAL
CONTROL
SYSTEM
displacement
of elevator δe
actual angle
g of attack α
speed u
LONGITUDINAL
DYNAMICS
actual
t l angle
l off attack
tt k α
speed u
SENSORS:
INS,
Anemometer
2- Steady state error
Control and guidance
Slide 24
2- Steady state error
R(s) +
Ess
-
G(s)
Y(s)
Steady State error:
ess= difference between the entry signal and the exit signal
ess = “what
what we want minus what we get”
get
Control and guidance
Slide 25
2- Steady state error
System’s
y
type:
yp
Given the transfer function:
(1  as)(1  bs)...(1  cs  ds2 )...
G (s )  K N
2
s (1  s)(1  s)...(
) (1  s  s )...
)
with
K: system gain,
and
N: number of poles in the origin
→N
Control and guidance
Slide 26
= system’s type
2- Steady state error
Definition of Steady State error:
ess  lim r ( t )  y( t )
t

ess > 0 : exit signal has not reached the entry reference
ess < 0 : exit signal
g
is higher
g
than the entry
y
Control and guidance
Slide 27
2- Steady state error
ess  lim r ( t )  y( t )
t  
Moving to the Laplace space: Final value theorem:
 

G (s )
R (s) 
ess  limsR (s)  Y (s)  lim s  R (s) 
s 0
s0
1  G (s )

 

R (s ) 
ess  lim s 

s0
 1  G (s ) 
Control and guidance
Slide 28
Depends
p
on the entry
y
+ on the system’s type
2- Steady state error
(1  as)(1  bs))...((1  cs  ds2 ))...
G (s )  K N
s (1  s)(1  s)...(1  s  s2 )...
R(s) +
-
G(s)
Y(s)
1. Position error: error for a step function entry: r(t)=u(t)

 1 
1
1
1
e p  lim s 
   lim
 
s 0
G (s)
 1  G (s) s  s0  1  G (s)  1  lim
s0
 1
type
yp 0

 1  K
0
type  I
Control and guidance
Slide 29
2- Steady state error
1 Position error: error for a step function entry: r(t)=u(t)
1.
St ep Response
1
Error
Amplitu
ud e
08
0.8
0.6
0.4
0.2
0
0
2
4
6
Ti
Time
((sec )
Control and guidance
Slide 30
8
10
12
2- Steady state error
R(s) +
(1  as)(1  bs))...((1  cs  ds2 ))...
G (s )  K N
s (1  s)(1  s)...(1  s  s2 )...
-
G(s)
( )
Y(s)
2 Speed error: error for a ramp function entry: r(t)= t
2.



 1 
1
1
1
e v  lim s 
 2   lim
  lim

s0
 1  G (s) s  s0  s  sG (s)  s0  sG (s) 
type 0

 1
  type I
K
type  II
0
Control and guidance
Slide 31
2- Steady state error
(1  as)(1  bs))...((1  cs  ds2 ))...
G (s )  K N
s (1  s)(1  s)...(1  s  s2 )...
R(s) +
-
G(s)
Y(s)
3. Acceleration error: error for a parabolic entry: r(t)= t2



 1 
1
1
1
e v  lim
li  s 
 3   lim
li  2 2
li  2
  lim

s0
 1  G (s ) s  s0  s  s G (s )  s0  s G (s ) 
type 0 or I

 1
  type II
K
type  III
0
Control and guidance
Slide 32
2- Steady state error
Error based on type
yp + entry
y
Input:
Type:
step
ramp
parabolic
0
constant
∞
∞
I
0
constant
∞
II
0
0
constant
Control and guidance
Slide 33
2- Steady state error
Example:
θref(s) +
K
-
δe(s)
G(s)
C
Compute
t the
th error iin steady
t d state
t t ffor a unit
it step
t function
f
ti
entry and for a system with the following open loop transfer
function:
(s)
2s  0.1
 2
e (s) s  0.1s  4
• for K=1, K=10, K=100,
• for K=1 and
Control and guidance
Slide 34
θ(s)
2s  0.1
01
 2
δ e (s) s s  0.1s  4


θ(s)
Design a simple proportional controller in order to satisfy some
constraints on the response of the system
desired angle of attack αref
speed
d uref
LONGITUDINAL
CONTROL
SYSTEM
actual angle
g of attack α
speed u
displacement
of elevator δe
LONGITUDINAL
DYNAMICS
actual
t l angle
l off attack
tt k α
speed u
SENSORS:
INS,
Anemometer
3- Root locus
Control and guidance
Slide 35
3- Root locus
• Root locus technique
• Gain setting
• Effect of zeros and poles
Control and guidance
Slide 36
3- Root locus
Root 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
as K varies from 0 to infinity
• The resulting plot helps us in selecting the best value of K
• Gives
G
information
f
for
f the transitory part off the response
(stability, damping factor, natural frequency)
Control and guidance
Slide 37
3- Root locus
Root locus technique
Let
s  z1 s  z 2 s  z m 
G (s) 
s  p1 s  p2 s  pn 
A d substitute
And
b tit t it in
i th
the characteristic
h
t i ti equation
ti
k s  z1 s  z 2 s  z m 
1
0
s  p1 s  p2 s  pn 
Control and guidance
Slide 38
where k  K
3- Root locus
Root locus technique
The characteristic equation is complex and can be written in terms
of magnitude and angle as follows
k s  z1 s  z 2 s  z m
1
s  p1 s  p2 s  p n
n
m
 s  z    s  p   (2q  1)  180
i 1
i
i 1
i
for q  0, 1, 2 ..., ( n  m  1)
Control and guidance
Slide 39
3- Root locus
Root locus technique: Rules
If we rearrange the magnitude criteria as
s  z1 s  z 2 s  z m
1

s  p1 s  p2 s  p n
k
Rule 1: The number of separate branches of the root locus plot is
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=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
3- Root locus
Root locus technique: Rules
Rule 2: Because the complex poles are always “conjugated”
conjugated ,
the root locus branches are symmetric with respect to the real axis
Control and guidance
Slide 41
3- Root locus
Root locus technique: Rules
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
3- Root locus
Root locus technique: Rules
Rule 4: Asymptotes
The root locus branches that approach the open-loop
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 zeroes
m
n

 pi   z i 
i 1
i 1



nm
The angle that the asymptotes make with the real axis is given by
180º 2q  1
a 
ffor q  0, 1, 2 ..., ( n  m  1)
nm
Control and guidance
Slide 43
3- Root locus
Root 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 p
poles the branch must break away
y from the real axis
so that the locus ends on a zero as k approaches infinity. The
breakaway points on the real axis are determined by solving
k s  z1 s  z 2 s  z m 
1
0
s  p1 s  p2 s  pn 
for k
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
3- Root locus
Root locus technique: Rules
Rule 6: Intersection with the imaginary axis
Solve the characteristic equation for s=jω (equation of the
imaginary axis)
k  j  z1  j  z 2  j  z m 
1
0
 j  p1  j  p2  j  pn 
Control and guidance
Slide 45
3- Root locus
Root locus technique: Rules
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:
m
n
 t  z    t  p   180
i 1
Control and guidance
Slide 46
i
i 1
i
3- Root locus
Root locus technique: examples
Root Locus
2
Example 1:
1.5
s2
G (s)  2
s  2s  3
Imagin ary Axis
1
0.5
0
-0.5
-1
1
-1.5
-2
-4
-3
-2
Real Axis
Control and guidance
Slide 47
-1
0
3- Root locus
Root locus technique: examples
Root Locus
4
Example 2:
3
1
G (s) 
ss  1s  2 
I magina
ary Axis
2
1
0
-1
-2
2
-3
-4
-6
-4
-2
Real Axis
Control and guidance
Slide 48
0
2
3- Root locus
Setting of the gain and natural frequency
Basic operation: to adjust the gain K to obtain a damping factor
given by the poles in closed
closed-loop
loop and fixed by the damping factor ζ
Cf second-order systems:
Re((si )
R

si
straight
g line doing
g an angle
g φ with the real axis (cos
(
φ= ζ) sets an
φ
intersection point with the poles position, and k (and then K) is
obtained solving the characteristic equation
Natural frequency for a second-order system: ωn=|si|
Control and guidance
Slide 49
3- Root locus
Gain setting
1  KG(s)  0
s  z1 s  z 2 s  z m 
0
1 K
s  p1 s  p2 s  pn 
The system total gain is computed thanks to the module condition
p1s  p 2s  p ms
k  K 
z1s  z 2s  z ns
“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
3- Root locus
Gain setting
1  KG(s)  0

1 K
0
s  p1 s  p2 s  pn 
If there are no zeros:
K    p1s  p2s  p ms
“total gain” =product of the distances between the poles of G(s)
and the intersection point (=
( target pole))
Examples
p
Control and guidance
Slide 51
3- Root locus
Gain setting
Root Locus
4
1
G (s) 
ss  1s  2 
0.5
3
Designer requirement:
we want ζ=0.5
ζζ=0.5=cos φ → φ
φ=60º
Imag ina
ary Axis
2
1
0
-1
-2
2
-3
-4
-6
0.5
-5
-4
-3
-2
-1
0
1
Real Axis
Examples
p
Control and guidance
Slide 52
2
3- Root locus
Gain setting
Root Locus
3
1
G (s) 
ss  1s  2 
0.5
System
S
t
: sys
Gain: 1.04
P ole: -0.332 + 0.577i
Damping: 0.499
Overshoot (%): 16.4
Frequency (rad/s ec): 0.666
2
Designer requirement:
we want ζ=0.5
ζζ=0.5=cos φ → φ
φ=60º
Imag
g inary Axis
1
0
-1
-2
2
0.5
-3
35
-3.
-3
3
-2.5
25
-2
2
-1.5
15
-1
1
-0.
05
0
05
0.5
1
15
1.5
Real Axis
Examples
p
Control and guidance
Slide 53
2
3- Root locus
Gain setting
Root Locus
3
Designer requirement:
we want ζ=0.5
This corresponds
p
to
k=1.04
S ystem : sys
Gain: 1.04
P ole: -0.332 + 0.577i
Damping: 0. 499
Overshoot ((%):
) 16.4
Frequency (rad/s ec): 0.666
2
Sys tem: sys
Gain: 1.04
Pole: -2.33
1
Damping: 1
Overshoot (%): 0
Frequency (rad/sec): 2.33
0
Imag inary Axis
1
G (s) 
ss  1s  2 
0.5
S ystem : sy s
Gain: 1.04
P ole: -0.332 - 0.578i
Damping: 0. 498
Overshoot (%): 16.5
Frequency (rad/s ec): 0.667
-1
-2
0.5
-3
-3. 5
-3
-2.5
-2
-1.5
-1
-0. 5
0
0.5
1
1.5
Real Axis
Examples
p
Control and guidance
Slide 54
2
3- Root locus
Gain setting
Root Locus
2
15
1.5
s2
G (s)  2
s  2s  3
1
Ima
aginary Axis
Designer requirement:
we want ζ=0.7
0.7
05
0.5
0
-0.5
-1
-1.5
-2
0.7
-4
-3
-2
-1
Real Axis
Examples
p
Control and guidance
Slide 55
0
3- Root locus
Gain setting
Root Locus
2
s2
s2  2s  3
Designer requirement:
we want ζ=0.7
This corresponds
p
to
k=1.32
1.5
Sys tem: sys
Gain: 1.32
Pole: -1.66 + 1.69i
Dam ping: 0.7
O
Overshoot
h t (%):
(%) 4
4.6
6
Frequenc y (rad/sec): 2.37
1
Imagin
nary Axis
G (s) 
0.7
0.5
0
-0.5
Sys tem: sys
Gain: 1.32
Pole: -1.66 - 1.69i
Dam ping: 0.7
Overshoot (%): 4.6
Frequenc y (rad/sec): 2.37
-1
1
-1.5
-2
2
0.7
-4
-3
-2
-1
Real Axis
Examples
p
Control and guidance
Slide 56
0
3- Root locus
R l ti stability:
Relative
t bilit gain
i margin
i
“Re(s)<0”
( )
criterion informs about the absolute stability
y of a system
y
but it says nothing about its relative stability
= how
h
ffar it
i iis ffrom the
h iinstability
bili → system strength
h
Gain margin: maximum proportional factor that can be introduced
into the control loop until the system becomes critically stable.
k Cr
MG 
k actual
Control and guidance
Slide 57
Examples
p
3- Root locus
Gain setting
Root Locus
3
1
G (s) 
ss  1s  2 
0.5
2
System : sys
Gain: 5.97
Pole: 0.000761 + 1.41i
Damping: -0. 00054
Overshoot (%): 100
Frequency (rad/sec): 1
1.41
41
Designer requirement:
we want ζ=0.5
This corresponds to
k=1
kcr=6
MG=6
Imag inary Axis
1
0
-1
-2
0.5
-3
-3. 5
-3
-2.5
-2
-1.5
-1
-0. 5
0
0.5
1
1.5
Real Axis
Examples
p
Control and guidance
Slide 58
2
3- Root locus
Roots locus exercise
2
1
G (s) 
ss  2 s2  2s  2 
1 .5
1
0 .5
0
-0 .5
-1
-1
1 .5
5
-2
-3
-2 .5
-2
-1.5
-1
Real Axis
Control and guidance
Slide 59
-0.5
0
0 .5
1
3- Root locus
Roots locus exercise
2
0.5
1 .5
5
1
Imagin ary A xis
0 .5
2
0
-0 .5
-1
-1
1 .5
5
0.5
-2
-3
-2 .5
-2
-1.5
-1
Real
ea Axis
s
Control and guidance
Slide 60
-0.5
0
0 .5
1
3- Root locus
Roots locus exercise
2
0.5
1 .5
5
1
Imagin ary A xis
0 .5
2
0
-0 .5
System: sys
S
Gain: 1.62
Pole: -0.373 - 0.627i
Damping: 0.511
) 15.4
Overshoot ((%):
Frequency (rad/sec): 0.73
-1
-1
1 .5
5
0.5
-2
-3
-2 .5
-2
-1.5
-1
Real
ea Axis
s
Control and guidance
Slide 61
-0.5
0
0 .5
1
3- Root locus
Roots locus exercise
Root Locu s
2
0.5
15
1.5
1
Imagin arry A xis
0.5
2
0
-0.5
-1
-1.5
0.5
-2
-3
-2. 5
-2
-1.5
-1
Real Axis
Control and guidance
Slide 62
-0.5
0
0 .5
1
3- Root locus
Gain 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 magnitude
However,
order, the influence of poles and zeros on the response is seen in
the following study
Control and guidance
Slide 63
3- Root locus
Additional pole
1. A second-order system
y
is considered
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 is
the 2nd order system entry
→ exit has less overshoot and it takes more time to reach its
final value
Control and guidance
Slide 64
3- Root locus
Additional pole
f1 
2.4
(0.5s  1)( s 2  3.2s  5.4)
2.4
f2  2
s  3.2 s  5.4
Control and guidance
Slide 65
3- Root locus
Additional zero
1.2( s  2)
f1  2
s  3.2 s  5.4
2.4
f2  2
s  3.2 s  5.4
Zeros ((negative)
g
)
• increase the initial slope,
• make the system faster so it
reaches its final value earlier,
• can produce overshoot
Control and guidance
Slide 66
3- Root locus
Effect of an additional p
pole in the roots locus
Transfer function of a
8
vehicle cruise-control
6
system:
4
Root Locus
2
0
G (s ) 
2.48
(s  0.06)(s  1)(s  3.33) -2
-4
4
-6
-8
8
-10
-8
-6
-4
-2
Real Axis
Control and guidance
Slide 67
0
2
4
3- Root locus
Effect of an additional pole in the roots locus
Root Locus
6
A pole is added on 0
4
2.48
G (s) 
s(s  0.06)(s  1)(s  3.33)
2
Roots locus moved
to the right
 + unstable
I maginary Axis
(i t
(integrator)
t )
0
Root Locus
0 15
0.15
-2
0.1
-4
-6
-8
0 05
0.05
0
-6
-4
-2
-0.05
0
2
4
6
R l A
Real
Axis
i
-0.1
-0.15
-0.2
02
Control and guidance
Slide 68
-0.15
0 15
-0.1
01
-0.05
0 05
0
0 05
0.05
01
0.1
0 15
0.15
3- Root locus
Effect of an additional zero in the roots locus
Root Locus
3
A zero is added
i -0.12
in
0 12
2
G (s) 
I mag
ginary Axis
1
2.48(s  0.12)
s(s  0.06)(s  1)(s  3.33)
0
-1
-2
-3
-5
5
-4
4
-3
3
-2
2
-1
1
0
1
Real Axis
Root locus moved to the left  + stable, and faster
Control and guidance
Slide 69
Design a controller/compensator in order to satisfy some
constraints on the response of the system
desired angle of attack αref
speed
d uref
LONGITUDINAL
CONTROL
SYSTEM
actual angle
g of attack α
speed u
displacement
of elevator δe
LONGITUDINAL
DYNAMICS
actual
t l angle
l off attack
tt k α
speed u
SENSORS:
INS,
Anemometer
4- Controllers
Control and guidance
Slide 70
4- Controllers
• Proportional controller:
P:
KP
• Integral
I t
l controller
t ll :
I
I:
• Derivative controller:
D:
KI
s
tp
M
K Ds
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
accurate, because KP, KI, and KD
are dependent of each other
• changing 1 of these variables can change the effect of the other 2
Control and guidance
Slide 71
4- Controllers
• 2 kinds of controllers improve the transitory response:
Lead Compensator:
s  z0
G C (s ) 
with p0  z 0
s  p0
adds 1 zero and 1 pole, but zero is more important: it moves the
roott locus
l
tto the
th left:
l ft iimproves stability
t bilit (system
(
t
is
i faster
f t and
d has
h
less overshoot)
Proportional Derivative Compensator:
adds
dd 1 zero: improves
i
stability
t bilit
Control and guidance
Slide 72
G C (s)  K P  K Ds
4- Controllers
• 2 kinds of controllers improve the steady state response:
Lag Compensator:
s  z0
G C (s ) 
with z 0  p0
s  p0
add 1 zero and 1 pole, but pole is more important: it moves the
roott locus
l
tto the
th right:
i ht decreases
d
stability
t bilit ((system
t
iis slower
l
and
d
has more overshoot), but decreases the steady state error
Proportional Integral
g Compensator:
Control and guidance
Slide 73
KI
G C ((s))  K P 
s
4- Controllers
• 2 kinds of controllers improve both transitory and steady state
response:
Lead - Lag Compensator:
s  z 0 s  z1
G C (s) 

with z 0  p0 and p1  z1
s  p0 s  p1
Proportional Integral Derivative (PID) Compensator:
KI
G C (s)  K P 
 KDs
s
Control and guidance
Slide 74
5- Frequency response
1 Fourier transforms and properties
2 Frequency response
3 Examples
p
Control and guidance
Slide 75
5- Frequency response
1 Fourier transforms and properties
The Fourier transform of a function x(t) is a function of the
pulsation ω:

F[ x ( t )]  X ( )   e

 jt
x ( t )dt
→ It transforms a signal from the time domain to the
frequency domain
Control and guidance
Slide 76
5- Frequency response
1 Transforms and properties
The inverse Fourier transform recovers the original function
x(t):
1  jt
e
X
(

)
d

x ( t )  F [X()] 
2 
1
This is true for an absolutely integrable signal:



Control and guidance
Slide 77
2
x ( t ) dt  
5- Frequency response
1 Transforms and properties
Linearity:
F[x ( t )  y( t )]  X()  Y()
Derivation:
 dx ( t ) 
F

j

X
(

)

dt


 d x( t) 
n
F
  j X()
n 
 dt 
n
Control and guidance
Slide 78
5- Frequency response
1 Transforms and properties
Additional properties
1  
Fx (at )  X  
a a
Duality
x t  

 X 
F
X t  

 2  x  
F
Control and guidance
Slide 79
5- Frequency response
1 Transforms and properties
C
Convolution
l i
theorems
h
f1  f 2 t  
 F1   F2 
F
f1  f 2 t  
F1  F2 
F

where f1  f 2 t    f1 s f 2 t  s ds

Control and guidance
Slide 80
5- Frequency response
1 Transforms and properties
Time delay
Fx t  T   e
Control and guidance
Slide 81
 jT
X 
5- Frequency response
1 Transforms and properties
Important pairs of transforms
f (t)
δ(t-t0), unit impulse
e
j0 t
u(t), unit step
 at
e u( t )
cos20 t 
sin 20 t 
Control and guidance
Slide 82
F ( )
e jt0
2π δ(ω-ω0)
δ(ω)
1
 a
1
  0     0 
2
1
  0     0 
2j
5- Frequency response
2 Frequency response
In previous examples we examined the free response of an airplane with
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
5- Frequency response
2 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
5- Frequency response
1.5
2 Frequency response
1
0.5
Arbitrary wind gust profiles:
0
-0.5
1
0.9
-1
0.8
-1.5
0
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.1
0.2
0.3
Control and guidance
Slide 85
0.4
0.5
0.6
0.7
0.8
0.9
1
1
2
3
4
5
6
7
8
9
10
5- Frequency response
2 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
5- Frequency response
2 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
5- Frequency response
2 Frequency response
Given any system:
R(s)
G(s)
C(s)
Hypothesis: stable system
Sinusoidal input

r t   sin t   R s   Lsin t   2
s  2
Control and guidance
Slide 88
5- Frequency response
2 Frequency response
It can be demonstrated that the steady state response is:
c( t )  G  j sin t   
with G  j 
and
Control and guidance
Slide 89
ReG  j  ImG  j
 ImG  j
  arg G  j  arctan 

 ReG  j
2
2
5- Frequency response
Parametric estimation
a. First
First--order
Frequency response:
Gain:
Delay:
y
Control and guidance
Slide 90
G  j 
K
K 1  j
G  j 

1  j 1  2
K
1  
2
   arctan 
5- Frequency response
b. SecondSecond-order
K
G  j  2
2
n     2n j
Gain:
D l
Delay:
Control and guidance
Slide 91
G  j 
2
n
K
  
1   
  n 

2 2
2
 
   2  
  n 



ω
 2ξ


ωn 
  arctan
2 
  ω 
 1   ω  
  n 
5- Frequency response
c. Higher
Higher--order
For a system composed by series of blocks:
R(s)
G1(s)
G3(s)
G2(s)
C(s)
=
R(s)
G(s)
C(s)
with
i h G s   G1 s   G 2 s   G 3 s 
G  j  G1  j  G 2  j  G 3  j
G  j  G1  j  G 2  j  G 3  j
  1  2  3
Control and guidance
Slide 92
6- Bode diagram
1.
1 Introduction
2. Construction rules
3. Stability
Control and guidance
Slide 93
6- Bode diagrams
G l off the
Goals
th Bode
B d diagrams:
di
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):
argG ( j)
G ( j) dB
The decibel is a unit measure used to compare a certain value with a
reference one. It is basically used to measure a signal power, and it is
defined as:
G ( j) dB
2

 Pmedida 
 G ( j) 
 10 log 
  10 log 
  20 logG ( j) 
 1 
 Pref 
Control and guidance
Slide 94
6- Bode diagrams
Semi-logarithmic axes: with lineal scale for the magnitude or the
phase,, and logarithmic
p
g
for the frequency
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
6- Bode diagrams
Gain
The gain is a factor that only modifies the magnitude and its angular
value is 0º;
0 ; that is,
is the gain value remains constant for any frequency
value, because it does not depend on it.
35
Ma
agn itud e (dB )
3 4. 5
34
3 3.
3 5
33
Bode diagram for a
K 50 gain
K=50
i
3 2. 5
1
Pha se
e (d eg)
0.5
0
-0. 5
-1
0
10
Control and guidance
Slide 96
10
1
6- Bode diagrams
Integral
g and derivative factors
An integral factor or a pole centered in zero, has a transfer function of:
1
1
j

G (s)   G ( j) 
j

s
Its magnitude is,
is therefore:
G ( j) dB
1
 20 log   20 log()
 
For a logarithmic frequency axis: it corresponds to a straight negative
line of -20 dB per decade
The phase is:
1 

1

j
 
argG ( j)  arg    arg    arctan     90
 0 

 j 


Control and guidance
Slide 97
6- Bode diagrams
I t
Integral
l and
d derivative
d i ti ffactors
t
For a derivative factor or a zero centered in zero, the results are
deduced using a similar development:
G ( j) dB  20 log()
argG ( j)  90

Control and guidance
Slide 98
6- Bode diagrams
Integral and derivative factors
B ode D iagra m
20
Magnitude (dB)
10
0
Bode diagrams of the
derivative and the
-10
integrator
g
-20
Phase (deg)
90
0
-90
10
0
10
Frequ ency (rad /sec)
Control and guidance
Slide 99
1
6- Bode diagrams
First--order factors: first
First
first--order pole


1
1
Its magnitude is: 20 log 
G ( j) 
2 2 
1  j
 1   
It seems more complicated, but approximations are made:




1
for      1 , G ( j) dB  20 log 1  2  2  20 log1  0

1
for      1 , G ( j) dB  20 log
g 1  2 2  20 logg

• substitute the curve by its two asymptotes
• magnitude
g
is 0 dB until it reaches the p
point where both asymptotes
y p
meet: ωτ=1, this point is called cut frequency
• from there: other asymptote,
asymptote with a -20
20 dB per decade slope.
slope
• point where approximation error is maximum corresponds to the cut
frequency and the error is 3 dB.
dB
Control and guidance
Slide 100
6- Bode diagrams
First--order factors: first
First
first--order pole
similar for the phase, the phase real value is:
 1 
 1  j 
   


arg
arctan
argG ( j)  arg 




2 2



1

j

1
1






However the approximation in this case is:
•
1


for       1 arggG ( j)  0

for   1    1 argG ( j)  90


Control and guidance
Slide 101
6- Bode diagrams
First--order factors: first
First
first--order pole
B ode Diagram
1
G (s ) 
1 s
Magn
nitude ( dB)
0
-10
-20
-30
Phase (d
deg)
Bode diagram
g
of a -40
0
first-order system
-45
-90
-2
10
-1
10
10
0
Frequency (rad/sec)
Control and guidance
Slide 102
10
1
10
2
6- Bode diagrams
First--order factors: first
First
first--order pole
For the study of the first-order zeros, similar development, there is only
a sign change:
M
Magnitude:
it d
G ( j)  1  j


20 log
l
1  2 2  10 log
l 1  2 2 
Angular contribution:
  
argG ( j)  arg1  j  arctan  
1 
Control and guidance
Slide 103
6- Bode diagrams
First--order factors: first
First
first--order pole
Magn
nitude ( dB)
Example:
10
G (s ) 
2s  1
B ode Diagram
20
10
0
-10
10
Phase (deg )
0
-45
45
-90
90
-2
10
-1
0
10
10
Frequency (rad/sec)
Control and guidance
Slide 104
1
10
6- Bode diagrams
Second--order factors
Second
2n
2n
1




G (s)  2
G
(
j
)
 2  j2n   2n    2 
s  2ns  2n
1     
  n  
 
j2 
 n 
The magnitude is:
  
1
G ( j)  20 log
 20 log 1   
2
2 2
     
  n 
1      2 
  n    n 
Control and guidance
Slide 105
2 2
   2
  2 
  n 
6- Bode diagrams
Second--order factors
Second
  
G ( j)  20 log 1   
  n 
it can be approximated this way:

for
 1 , G ( j)  20 log(1)  0 dB
n
 

for
 1 , G ( j)  20 log 
n
 n 
2 2
   2
  2 
  n 
2
F low
For
l
ffrequencies:
i
straight
t i ht line
li att 0dB
For high frequencies: straight line with a –40 dB per decade slope.
Both asymptotes cross on ω=ωn.
However, in the second-order
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 value.
value
Control and guidance
Slide 106
6- Bode diagrams
Second--order factors
Second

2 
n 

F the
For
h phase:
h
tan   
2
 
1   
 n 


For
 1 , generally
 0.2 ,    Arc tan(0)  0
n
n
For


 1 , generally
 5,
n
n
   Arc tan(0 )  180

For
 1 ,    Arc tan( )  90
n
The phase graphic form depends also on ζ
Control and guidance
Slide 107
6- Bode diagrams
Second--order factors
Second
Bode diagram of
a second-order
pole for different
ζ values
Control and guidance
Slide 108
6- Bode diagrams
Second--order factors
Second
B ode Diagram
Example:
Mag
gnitude ( dB)
2.5
G (s )  2
s  3s  25
-20
-40
-60
Phas e (d eg)
-80
0
-45
-90
90
-135
-180
-1
10
10
0
10
Frequency (rad/sec)
Control and guidance
Slide 109
1
10
2
6- Bode diagrams
Stability condition
Notation:
G  0 for   ( G 0 ) and   180º for   ( G 180 º )
The Bode diagram in open loop is studied
Stability condition:
If for   ( G 0 ) ,   180º
If for   ( G 0 ) ,   180º
And for   ( 180 º ) , G  0
Or for   ( 180 º ) , G  0
THEN the system is STABLE
THEN the system is UNSTABLE
in closed loop
in closed loop
Control and guidance
Slide 110
6- Bode diagrams
Stability condition
STABLE
Control and guidance
Slide 111
6- Bode diagrams
Stabilityy condition
UNSTABLE
Control and guidance
Slide 112
6- Bode diagrams
Stability margins
Common values:
Minimum gain
margin: 10 a 12 dB,
Minimum phase
margin: 45 a 50º
Control and guidance
Slide 113
6- Bode diagrams
Stability margins
B ode Diagram
Mag nitud e (dB)
50
0
-50
-100
1
( )
G(s)
s(s
(  1)(s
1)(  2)
Phas e (d eg)
-150
90
-90
-135
-180
180
-225
-270
270
-2
10
10
-1
10
0
Frequency (rad/sec)
Control and guidance
Slide 114
10
1
10
2
6- Bode diagrams
Stability margins
G(s) 
1
s(s  1)(s  2)
MG=14.5dB
with roots locus:
MG=6,,
and note that:
20log(6)=15.5dB
Control and guidance
Slide 115
6- Bode diagrams
Stability margins
1
G(s) 
s(s  1)(s  2)
MΦ=51º
Control and guidance
Slide 116
6- Bode diagrams
Stability margins
Transfer function of a vehicle cruise-control system G (s) 
B ode Diagram
Phas e (d eg
g)
Mag nitud e (dB )
50
0
-50
-100
-150
0
-45
-90
90
-135
-180
-225
225
-270
10
-2
10
0
Frequency (rad/sec)
Control and guidance
Slide 117
2.48
(s  0.06)(s  1)(s  3.33)
10
2
6- Bode diagrams
Stability margins
A pole is added on 0 (integrator): Bode diagram shifted downward : + unstable
Control and guidance
Slide 118
6- Bode diagrams
Stability margins
A zero is added in -0.12: Bode diagram shifted upward : + stable
B ode Diagram
Mag nitud e (dB )
100
50
0
-50
-100
100
Phas e (d eg))
-150
-90
135
-135
-180
-225
-270
10
-2
10
0
F
Frequency
(rad/sec)
( d/
)
Control and guidance
Slide 119
10
2
REFERENCES
G.F Franklin, J.D. Powell, A. Emani-Naeini, Feedback Control of Dynamic
Systems, 4a Edición, Prentice-Hall, 2002
P. Lewis, Sistemas de Control en Ingeniería, Prentice Hall, 1999
W Bolton,
W.
B lt
C t l Engineering,
Control
E i
i
2ª Edi
Edición,
ió Longman,
L
1998
D. Arzelier, D. Peaucelle, Représentation et analyse des systèmes linéaires,
Tomes 1 et 2, Version 1, ENSICA, 1999
Control and guidance
Slide 120