Lecture 15: Stability and Control III — Control
Holonomic systems with N degrees of freedom and K distinct inputs
Philosophy of control: closed loop with feedback
Ad hoc control thoughts
Controllability
Three link robot control
1
K dimensional vector
Ideal output
INVERSE
PLANT
disturbance
Mechanical/electromechanical
system
+
+
GOAL
input
PLANT
-
Actual output
+
error
feedback
CONTROL
2
Ideal output
EQUILIBRIUM
disturbance
+
+
GOAL
Actual output
Q0
input
+
SIMULATION
-
error
feedback
LINEAR CONTROL
3
Ideal output
0

disturbance
+
+
L
 Q0i
i
q
q i  q0i , qÝi  0

Q0
Actual output
qÝj  M ji pi
input
-
pÝi 
L
 Qi
q i

feedback
+
error

LINEAR CONTROL
4
Let’s think about control in the context of the simple inverted pendulum
add a small, variable torque at the pivot
q
5
Hamilton’s equations
qÝ
1
L
Ý
p,
p

 mglsin q  Q
2
ml
q
There’s a change of sign from the simple pendulum from last time

because
I have chosen a different definition of q
We have equilibrium at q = 0, and Q = 0 there as well.
We know that this will be unstable if it is perturbed with Q remaining zero
Let’s see how this goes in a state space representation
6

q 
 1 p 
 0
2
x    xÝ  ml
  Q
p
1
 

mglsin q
  

(I’ve put in the  because Q is zero at equilibrium)
  
1
qÝ 

p  0
2
xÝ   
  Q
ml
Ý

p
1
  
mglsin q0  q
  
q
Ý 
 1 p 
 0 
 0
2
   ml
  Q  
Ý
1
p 
mglcosq0q
   
mglcosq0

1 
q 0
2    Q
ml
p
1
0 
   
7

 0
Ý
x  

mglcosq0
1 
 0
2 x   Q
ml
1
0 
  
If q starts to increase, we feel intuitively that we ought to add a torque to cancel it

Q  g1q

 0
xÝ 


mglcosq0
1 
 0
2 x   g q
ml
1
1



0 
We can expand the feedback term


 0
Ý
x  

mglcosq0
1 
q 
 0
2 x   
g1 0 
ml
1
p
0 
  
8

 0
xÝ 

mglcosq0
1 
 0
2 x   
g1 0x
ml
1
0 
  
multiply the column vector and the row vector


 0
xÝ 

mglcosq0
1 
  0 0
2 x  
x
ml
g 0
0 
  1 
combine the forced system into a single homogeneous system



0
xÝ 

g1  mglcosq0

1 

2 x
ml
0 

9
The characteristic polynomial for this new problem can be solved for
s2 
1
g1  mglcosq0 
2 
ml
and so I can make s2 negative by applying some gain g1.

So this very simple feedback can make an unstable system marginally stable
We can do better . . .
10
Suppose we feed back the speed of the pendulum as well as its position?
Q  g1q  g2 p

 0
xÝ

mglcosq0



 0
Ý
x  

mglcosq0
1 
 0
2 x    g q  g p
1 2 
ml
1
0 
  
1 
q 
 0
2 x   
g1 g2 
ml
1
p
0 
  

 0
xÝ 

mglcosq0
1 
  0
0 
2 x  
x
ml
g g2 
0 
  1
11
Combining everything again we get


0
xÝ 

g1  mglcosq0
1 

2 x
ml
g2 

And now the characteristic polynomial comes from


1 

s
 2 
det 
ml  0

g1  mglcosq0 s  g2 


ss  g2  
1
g  mglcosq0   0
2  1
ml
12
ss  g2  
1
g1  mglcosq0   0
ml2
The linear term is the key — the feedback from the derivative

1
1 2
1
s   g2 
g2  4 2 g1  mglcosq0 
2
2
ml
We can adjust this to get any real and imaginary parts we want

If you are familiar with the idea of a natural frequency and a damping ratio
then you might like to set the control problem up in that language
13
ss  g2  
1
g1  mglcosq0   0
ml2
can be made the same as the one degree of freedom mass-spring equation

s2  2z n s   n2  0
by setting
 g2  2z n , g1  mglcosq0  ml2 n2
s  z n   n z 2 1
giving

The real part is always negative. If z is less than unity, there is an imaginary part.

If z equals unity the system is said to be critically damped
14
??
15
This suggests a bunch of questions
Is this generalizable to more complicated systems?
Is there a nice ritual one can always employ?
YES
SOMETIMES
Is this always possible?
Will the linear control control the nonlinear system?
NO
SOMETIMES
How much of this does it make sense to include in this course?
??
16
The question of possibility is really important
so I’m going to address that as soon as I can develop some more notation
The general perturbation problem for control will be
xÝ Ax  BfE  xÝi  A ij x j  B ij f Ej
In the ad hoc
 example we worked, we used Q, the generalized force, as input
We cannot do that in general, so I have put in fE as the actual external force vector
17
For a single input system like the one we just saw
B will be a column vector and fE a scalar and the equation is
xÝi  A ij x j  B i f E

We’ll eventually see how we get from Euler-Lagrange to state space including B
18
We want fE (or fE for one input) to be proportional to x
the minus sign is conventional
f Ej  Gkj x k

xÝi  A ij x j  B ij Gkj x k
We see that G has as many rows as there are inputs
and as manycolumns as there are state variables
G is a row vector for single input systems
19
Rename dummy indices (j —> m) to make it possible to combine terms
xÝi  Aij  BmiGmj x j


We have xÝi  Aij  BiG j x j for the single input case

Our control characteristic polynomial will come from

det sI ij  Aij  BmiGmj  0
and the question is:
is it always
 possible to find G such that the roots are where we want them?
20
There are always at least as many gains as there are roots, so you’d think so
But it isn’t.
The controllability criterion, which I will state without proof, is that the rank of
W ji  Bij , Aki Bkj , Aki Amk Bmj ,

must be equal to the number of variables in the state

There are as many terms in W as there are variables in the state
21
W has as many rows as there are variables.
The number of columns in W is equal to the number of variables times the number of inputs
In the single input case W is a square matrix
AND there is a nice simple way to figure out what the gains must be for stability
We are not going to explore this — we haven’t the time —
and it is covered in most decent books on control theory
We can get by with guided intuition.
22
??
23
Single input systems are much simpler than multi-input systems
but we have need of multi-input systems frequently
I will outline the intuitive approach to multi-input systems
which works best (at least for me) through the Euler-Lagrange equations
We are working on the yellow box in the block diagram
(This may be a bit hard to follow; we’ll do an example shortly.)
24
Euler-Lagrange equations
d L  L
 Qi
 
dt qÝi  q i
which we can rewrite

Ýj 
M ijqÝ
1 d
L
j
Ý
M
q

 Qi

ij 
i
2 dt
q
1 M ij j k L
Ý
M ijqÝ
qÝqÝ  i  Qi
k
2 q
q
j


25
For a steady equilibrium, which is what we are learning how to do
L
L

Q

0

Q


i
0i
q i
q i
qÝk  0
perturbation


j
Ýj  qÝ
Ý
q j  q0j  q1j  qÝj  qÝ1j , qÝ
1
Qi  Q0i  Qi
1 M ij 2 j k L
Ý 
M ijqÝ
 qÝ qÝ  i  Q0i  Qi
k
2 q
q
j
We can drop this term because of the 2.

26
Ýj 
M ijqÝ
L
 Q0i  Qi
i
q
and we need to perturb the gradient of the Lagrangian to finish the linearization

L L
 i
i
q q
2 L
 i k
q q
q k  0
2 L
Ý  i k
M ijqÝ
q q
qk  O 2 
q k  0
qk  Qi
j

q k  0
or

2 L
Ý  M
qÝ
qiqk
m
qk  M miQi
mi
q k  0
27
We can use our old method of converting to first order odes on this
and decide controllability (before we knock ourselves out trying to control it)
qÝm  u m
2 L
uÝ  M
i
k
q
q

m
vector is
The state
qk  M miQi
mi
q k  0
qm 
x   m 
u 
and the A matrixis straightforward
28
The meaning of B is not immediately clear
B is a matrix connecting the actual inputs to the system
It is not the connection between the generalized forces and the system
We get that by going back and remembering how the generalized forces are defined
WÝ
Qi  i
qÝ

29
Recall
WÝ  F1  v1 
 1  1 
The vector forces and torques are made of components
and wearrange the components into the vector fE
The velocities and rotation rates are linear functions of the qÝi
The terms in WÝ are linear in the components of fE

We can combine all this to write

WÝ  aij qk qÝi f Ej

30
The coefficients will depend on the individual problem
but in any case we will have
WÝ
Qi  i  aij qk f Ej
qÝ
We can substitute this into the momentum equation

2 L
uÝ  M
qiqk
m

2 L
uÝ  M
qiqk
m
qk  M miQi
mi
q k  0
qk  M miaij qk f Ej
mi
q k  0
31
Now we can go back to the state picture with a better idea of what A and B are
qm 
x   m 
u 
state vector
state equations
Explicit appearance of the actual forces
in the differential equations


0

2

L
mi
xÝ M
i
k

q
q

q k  0
I m  

0
 q
0um  M mi a qk fE
ij
  




32
The A matrix is

0

2

L
mi
A  M
i
k

q
q

q k  0
I

0


and the B matrix
is



0
B   mi
k 
M
a
q
ij  


33
Can we make sense out of the whole thing — how do we do a problem?!
Suppose we are given a mechanism and a task for it to do
Consider only the simple task of going to some configuration and stopping there
We can treat the final configuration as our equilibrium
and design a control that will bring us to that equilibrium
j
Ýj  qÝ
Ý
q j  q0j  q1j  qÝj  qÝ1j , qÝ
1
Qi  Q0i  Qi
We want to make q1j  0, Qi 0

34
Start from square one using what we know so far
all the holonomic constraints
generalized coordinates
Lagrangian
generalized forces
goal
Euler-Lagrange track
Hamilton track
linearization
equilibrium forces
control design
simulation
(nonlinear)
35
Ideal output
Hamilton
0

disturbance
+
+
L
 Q0i
i
q
q i  q0i , qÝi  0

Q0
qÝj  M ji pi
input
Actual output
Hamilton
-
pÝi 
L
 Qi
q i

feedback
+
error

LINEAR CONTROL
Euler-Lagrange
36
??
37