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Lecture 15: Stability and Control III — Control

Philosophy of control: closed loop with feedback

Ad hoc control thoughts

Controllability

Three link robot control

Holonomic systems with N degrees of freedom and K distinct inputs

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PLANT

CONTROL

GOALINVERSEPLANT

+

-

+

-

Ideal output

input

feedback

Actual output

error

Mechanical/electromechanicalsystem

disturbance

+

K dimensional vector

3

SIMULATION

LINEAR CONTROL

GOALEQUILIBRIUM

+

-

+

-

Ideal output

input

feedback

Actual output

error

disturbance

+

Q0

4

LINEAR CONTROL

+

-

+

-

Ideal output

input

feedback

Actual output

error

disturbance

+

Q0 €

qi = q0i , ˙ q i = 0

€

˙ p i = ∂L∂qi + Qi

€

˙ q j = M ji pi

€

0 = ∂L∂qi + Q0i

5

Let’s think about control in the context of the simple inverted pendulum

q

add a small, variable torque at the pivot

6

€

˙ q = 1ml2 p, ˙ p = ∂L

∂θ= mglsinq + Q

There’s a change of sign from the simple pendulum from last timebecause 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

Hamilton’s equations

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€

x =qp ⎧ ⎨ ⎩

⎫ ⎬ ⎭⇒ ˙ x =

1ml2 p

mglsinq

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪+

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭

εQ

(I’ve put in the e because Q is zero at equilibrium)

€

˙ x ⇒ ε˙ ′ q ˙ ′ p ⎧ ⎨ ⎩

⎫ ⎬ ⎭=

1ml2 ε ′ p

mglsin q0 + ε ′ q ( )

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪+

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭

εQ

€

˙ ′ q ˙ ′ p ⎧ ⎨ ⎩

⎫ ⎬ ⎭=

1ml2 ′ p

mglcosq0 ′ q

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪+

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭Q = 0 1

ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪′ q ′ p

⎧ ⎨ ⎩

⎫ ⎬ ⎭+

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭

Q

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€

˙ x = 0 1ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x +

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭Q

If q starts to increase, we feel intuitively that we ought to add a torque to cancel it

€

Q = −g1q

€

˙ x = 0 1ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x −

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭g1q

€

˙ x = 0 1ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x +

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭−g1 0{ }

qp ⎧ ⎨ ⎩

⎫ ⎬ ⎭

We can expand the feedback term

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€

˙ x = 0 1ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x +

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭−g1 0{ }x

€

˙ x = 0 1ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x +

0 0−g1 0 ⎧ ⎨ ⎩

⎫ ⎬ ⎭x

€

˙ x = 0 1ml2

−g1 + mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x

multiply the column vector and the row vector

combine the forced system into a single homogeneous system

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The characteristic polynomial for this new problem can be solved for

€

s2 = 1ml2 −g1 + mglcosq0( )

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 . . .

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Suppose we feed back the speed of the pendulum as well as its position?

€

Q = −g1q − g2 p

€

˙ x = 0 1ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x −

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭g1q + g2 p( )

€

˙ x = 0 1ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x +

01 ⎧ ⎨ ⎩

⎫ ⎬ ⎭−g1 −g2{ }

qp ⎧ ⎨ ⎩

⎫ ⎬ ⎭

€

˙ x = 0 1ml2

mglcosq0 0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x +

0 0−g1 −g2

⎧ ⎨ ⎩

⎫ ⎬ ⎭x

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€

˙ x = 0 1ml2

−g1 + mglcosq0 −g2

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪x

And now the characteristic polynomial comes from

€

det s − 1ml2

g1 − mglcosq0 s + g2

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪= 0

Combining everything again we get

€

s s + g2( ) + 1ml2 g1 − mglcosq0( ) = 0

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€

s s + g2( ) + 1ml2 g1 − mglcosq0( ) = 0

€

s = − 12

g2 ± 12

g22 − 4 1

ml2 g1 − mglcosq0( )

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 ratiothen you might like to set the control problem up in that language

The linear term is the key — the feedback from the derivative

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€

s s + g2( ) + 1ml2 g1 − mglcosq0( ) = 0

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

€

s2 + 2ζωns + ωn2 = 0

can be made the same as the one degree of freedom mass-spring equation

€

g2 = 2ζωn , g1 = mglcosq0 + ml2ωn2

by setting

€

s = −ζωn ± ωn ζ 2 −1giving

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??

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This suggests a bunch of questions

Is this generalizable to more complicated systems?

Is there a nice ritual one can always employ?

Is this always possible?

Will the linear control control the nonlinear system?

How much of this does it make sense to include in this course?

YES

SOMETIMES

NO

SOMETIMES

??

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The question of possibility is really importantso I’m going to address that as soon as I can develop some more notation

The general perturbation problem for control will be

€

˙ x = A{ }x + B{ }fE ⇔ ˙ x i = A ji x j + B j

i fEj

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

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For a single input system like the one we just sawB will be a column vector and fE a scalar and the equation is

€

˙ x i = A ji x j + B i fE

We’ll eventually see how we get from Euler-Lagrange to state space including B

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We want fE (or fE for one input) to be proportional to x

€

fEj = −Gk

j x k

€

˙ x i = A ji x j − B j

i 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

the minus sign is conventional

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Rename dummy indices (j —> m) to make it possible to combine terms

€

˙ x i = A ji − Bm

i G jm( )x j

€

˙ x i = A ji − B iG j( )x jWe have for the single input case

Our control characteristic polynomial will come from

€

det sI ji − A j

i + Bmi G j

m( ) = 0

and the question is:is it always possible to find G such that the roots are where we want them?

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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 = B j

i , Aki B j

k,Aki Am

k B jm ,L{ }

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

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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 matrixAND 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.

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??

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Single input systems are much simpler than multi-input systemsbut we have need of multi-input systems frequently

I will outline the intuitive approach to multi-input systemswhich works best (at least for me) through the Euler-Lagrange equations

(This may be a bit hard to follow; we’ll do an example shortly.)

We are working on the yellow box in the block diagram

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Euler-Lagrange equations

€

ddt

∂L∂˙ q i ⎛ ⎝ ⎜

⎞ ⎠ ⎟= ∂L

∂qi + Qi

which we can rewrite

€

M ij˙ ̇ q j + 12

ddt

M ij( ) ˙ q j = ∂L∂qi + Qi

€

M ij˙ ̇ q j + 12

∂M ij

∂qk ˙ q j ˙ q k = ∂L∂qi + Qi

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For a steady equilibrium, which is what we are learning how to do

€

M ijε˙ ̇ ′ q j + 12

∂M ij

∂qk ε 2 ˙ ′ q j ˙ ′ q k = ∂L∂qi + Q0i + ε ′ Q i

€

∂L∂qi + Qi = 0⇒ Q0i = − ∂L

∂qi˙ q k = 0

€

q j = q0j + ε ′ q 1

j ⇒ ˙ q j = ε ′ ˙ q 1j , ˙ ̇ q j = ε ′ ˙ ̇ q 1

j

Qi = Q0i + ε ′ Q i

perturbation

We can drop this term because of the e2.

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€

M ijε˙ ̇ ′ q j = ∂L∂qi + Q0i + ε ′ Q i

and we need to perturb the gradient of the Lagrangian to finish the linearization

€

∂L∂qi = ∂L

∂qi′ q k = 0

+ ∂2L∂qi∂qk

′ q k = 0

ε ′ q k + O ε 2( )

€

M ijε˙ ̇ ′ q j = ∂2L∂qi∂qk

′ q k = 0

ε ′ q k + ε ′ Q i

or

€

˙ ̇ ′ q m = M mi ∂2L∂qi∂qk

′ q k = 0

′ q k + M mi ′ Q i

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We can use our old method of converting to first order odes on thisand decide controllability (before we knock ourselves out trying to control it)

€

˙ ′ q m = um

€

˙ u m = M mi ∂2L∂qi∂qk

′ q k = 0

′ q k + M mi ′ Q i

The state vector is

€

x =′ q m

um

⎧ ⎨ ⎩

⎫ ⎬ ⎭

and the A matrix is straightforward

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The meaning of B is not immediately clear

B is a matrix connecting the actual inputs to the systemIt 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

€

Qi = ∂ ˙ W ∂˙ q i

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Recall

€

˙ W = F1 ⋅v1 +L + τ1 ⋅ω1 +L

The vector forces and torques are made of componentsand we arrange the components into the vector fE

The velocities and rotation rates are linear functions of the

€

˙ q i

The terms in are linear in the components of fE

€

˙ W

We can combine all this to write

€

˙ W = aij qk( ) ˙ q i fEj

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The coefficients will depend on the individual problembut in any case we will have

€

Qi = ∂ ˙ W ∂˙ q i

= aij qk( ) fEj

€

˙ u m = M mi ∂2L∂qi∂qk

′ q k = 0

′ q k + M mi ′ Q i

We can substitute this into the momentum equation

€

˙ u m = M mi ∂2L∂qi∂qk

′ q k = 0

′ q k + M miaij qk( ) fEj

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Now we can go back to the state picture with a better idea of what A and B are

€

x =′ q m

um

⎧ ⎨ ⎩

⎫ ⎬ ⎭

state vector

state equations

€

˙ x =0 I

M mi ∂2L∂qi∂qk

′ q k = 0

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪′ q m

um

⎧ ⎨ ⎩

⎫ ⎬ ⎭+

0M miaij qk( )

⎧ ⎨ ⎩

⎫ ⎬ ⎭fE

Explicit appearance of the actual forcesin the differential equations

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and the B matrix is

The A matrix is

€

A =0 I

M mi ∂2L∂qi∂qk

′ q k = 0

0

⎧ ⎨ ⎪

⎩ ⎪

⎫ ⎬ ⎪

⎭ ⎪

€

B =0

M miaij qk( )

⎧ ⎨ ⎩

⎫ ⎬ ⎭

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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 equilibriumand design a control that will bring us to that equilibrium

€

q j = q0j + ε ′ q 1

j ⇒ ˙ q j = ε ′ ˙ q 1j , ˙ ̇ q j = ε ′ ˙ ̇ q 1

j

Qi = Q0i + ε ′ Q i

We want to make

€

′ q 1j → 0, ′ Q i → 0

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all the holonomic constraints

generalized coordinates

Lagrangian

generalized forces

Euler-Lagrange track Hamilton track

goal

equilibrium forceslinearization

control design simulation(nonlinear)

Start from square one using what we know so far

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LINEAR CONTROL

+

-

+

-

Ideal output

input

feedback

Actual output

error

disturbance

+

Q0 €

qi = q0i , ˙ q i = 0

€

˙ p i = ∂L∂qi + Qi

€

˙ q j = M ji pi

€

0 = ∂L∂qi + Q0i

Euler-Lagrange

Hamilton

Hamilton

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??