lecture 19. systems and feedback reading assignment: tmb2 section 3.1 and 3.2

1Michael Arbib CS564 - Brain Theory and Artificial Intelligence, USC, Fall 2001. Lecture 19. Systems Concepts

Michael Arbib: CS564 - Brain Theory and Artificial Intelligence

University of Southern California, Fall 2001

Lecture 19. Systems and Feedback

Reading Assignment:

TMB2 Section 3.1 and 3.2

Note: Self-study on eigenvalues: TMB2, pp.107,108,111-115.


A System is Defined by Five Elements

The set of inputs The set of outputs (These involve a choice by the modeler)The set of states: those internal variables of the system — which may or may not also be output variables — which determine the relationship between input and output. state = the system's "internal residue of the past"

The state-transition function : how the state will change when the system is provided with inputs.

The output function : what output the system will yield with a given input when in a given state.


From Newton to Dynamic Systems

Newton's mechanics describes the behavior of a system on a continuous time scale.

Rather than use the present state and input to predict the next state,

the present state and input determine the rate at which the state changes.

Newton's third law says that the force F applied to the system equals the mass m times the acceleration a.

F = maPosition x(t)Velocity v(t) = Acceleration a(t) = x

)(tx)(tx


Newtonian Systems

According to Newton's laws, the state of the system is given by the position and velocity of the particles of the system.

We now use u(t) for the input = force; and y(t) (equals x(t)) for the output = position.Note: In general, input, output, and state are more general than in the following, simple example.


State Dynamics

With only one particle, the state is the 2-dimensional vector

q(t) =

Then

yielding the single vector equation

The output is given by

y(t) = x(t) =

The point of the exercise: Think of the state vector as a single point in a multi-dimensional space.

)(

)(

tx

tx

xmtucesintutxandtxtxmdt

d

dt

d )()()()()(1

)(0

)(

)(

00

10

)(

)()(

1tu

tx

tx

tx

txtq

m

)(

)(

0

1

tx

tx


Linear Systems

This is an example of a Linear System: = A q + B u y = C qwhere the state q, input u, and output y are vectors (not necessarily 2-dimensional) and A, B, and C are linear operators (i.e., can be represented as matrices).

Generally a physical system can be expressed byState Change: q(t) = f(q(t), u(t))Output: y(t) = g(q(t))where f and g are general (i.e., possibly nonlinear) functionsFor a network of leaky integrator neurons:the state mi(t) = arrays of membrane potentials of neurons,

the output M(t) = mi(t)) = the firing rates of output neurons,

obtained by selecting the corresponding membrane potentials and passing them through the appropriate sigmoid functions.

q


Attractors

1. The state vector comes to rest, i.e. the unit activations stop changing. This is called a fixed point. For given input data, the region of initial states which settles into a fixed point is called its basin of attraction.

2. The state vector settles into a periodic motion, called a limit cycle.

For all recurrent networks of interest (i.e., neural networks comprised of leaky integrator neurons, and containing loops), given initial state and fixed input, there are just three possibilities for the asymptotic state:


Strange attractors

3. Strange attractors describe such complex paths through the state space that, although the system is deterministic, a path which approaches the strange attractor gives every appearance of being random.

Two copies of the system which initially have nearly identical states will grow more and more dissimilar as time passes.

Such a trajectory has become the accepted mathematical model of chaos,and is used to describe a number of physical phenomena such as the onset of turbulence in weather.


Stability

The study of stability of an equilibrium is concerned with the issue of whether or not a system will return to the equilibrium in the face of slight disturbances:

A is an unstable equilibrium B is a neutral equilibrium C is a stable equilibrium, since small displacements will tend to disappear over time.

Note: in a nonlinear system, a large displacement can move the ball from the basin of attraction of one equilibrium to another.


Negative Feedback Controller (Servomechanism)


Using Spindles to Tell -Neurons if a Muscle Needs to Contract

What’s missing in this Scheme?


Using -Neurons to Set the Resting Length of the Muscle


Self-study: theelegant extensionof this scheme to anagonist-antagonistpair of muscles


Discrete-Activation Feedforward

“Cortex”

“Spinal Cord”


“Ballistic” Correction then Feedback

This long latency reflex was noted by Navas and Stark.


The Mass-Spring Model of Muscle


Hooke’s Law – The Linear Range of Elasticity


The Mass-Spring Model of Muscle

The dynamics of muscle length: mx = mg - u(t) - k(x - xo) - bx

mg gravitational force on the mass m

u(t) controllable force of muscle contraction

-k(x-xo) for a constant k > 0 represents therestoring force of muscle for moderate displacementsof length x from resting value xo [Hooke'slaw]-bx viscosity

Use as state the two-dimensional vector

q(t) =

x(t)

x.(t)

so that

q.(t) =

0 1

- km -

bm

x(t)

x.(t) +

0

g - u(t)m +

kxom


An equilibrium state is one which (for fixed values

of the input) does not change, i.e., for which q. (t) = 0.

Fix the contractile force u(t) to constant u.

x.. = x

. = 0

so

mx.. = mg - u(t) - k(x - xo) - bx

.

yields0 = mg - u - k(x - xo)

Thus, the equilibrium length xe(u) for given force usatisfies

xe(u) = xo + mg - u

k (3,4)

... in the domain of linearity of the spring


Linear Systems

q. (t) = Aq(t) + (t)

where A is a matrix, and (t) is a general input vector.

In a feedback system, (t) will itself be a function ofthe state q — so will in general depend on q() fortimes < t.

Assume that A is non-singular and thus has inverse

A-1.Then an equilibrium state qe for fixed (t) = mustsatisfy

0 = Aqe + so that A-10 = A-1Aqe + A-1i.e.

qe + A-1; and so qe = -A-1,a unique

equilibrium pointLet p = q - qe be the displacement of the system statefrom this equilibrium.

p . =

ddt (q - qe) = q

. = Aq +

= A(p + qe) + = Ap + Aqe + = Ap


Eigenvectors and Eigenvalues

See TMB2, pp.107,108,111-115 for an expositionof their use in characterizing the stability of alinear system.The key result is:

Moreover, the system will oscillate if the eigenvalues are complex.We now consider how that theory is appliedto the mass-spring model.

Theorem. The system q. = Aq is stable

i.e., q(t) 0 as t for each initial state q(0)

if both eigenvalues of A have negative real parts.


Stability of the Mass-Spring Model

Our mass-spring model

q . (t) =

0 1

-km -

bm

q(t) +

0

g - u(t)m +

kxom

has equilibrium state

qe(u) =

xe(u)

0 where xe(u) = xo +

mg - uk

The eigenvalues are

1,2 = 12

-bm +

b

m2 -

4km =

12m

-b + b2 - 4km

which are real if b2 4km.

But since b, k and m are all positive, in this case

0 b2 - 4km < b,and so both 1 and 2 must be negative.

Again, if 1 and 2 are complex, their real part is -b/ 2m which is always negative.


Theorem: The mass-spring system with each b, mand k positive is always stable and undergoesdamped oscillations just in case b2 < 4km.

Large b will prevent the system oscillating, unless it isdominated by a large spring term (large k), for fixedmass m.

For b2 < 4km, the oscillations are quickly dampedeither if b is large (high friction or viscosity) or m issmall.


For a given m, and a desired resting length xe, theappropriate force u(xe) must satisfy

xe = xo + mg - u(xe)

k

kxe = mg - u(xe) + kxo

so thatu(xe) = mg + k(xo-xe)

If the mass is m + m then the length achieved will be

xo + (m+m)g - u(xe)

k = xe + m.g/ k


A simple linear model: the mass-spring model with

-style feedback


Substituting u(xe) = mg + k(xo-xe), we obtain

q.(t) =

0 1

-(k+h)

m -bm

q(t) +

0

(k+h)xe

m

Thus at equilibrium,

(k+h)x

m = (k+h)xe

m

and so our new model has the expected property thatthe equilibrium length is still xe.


If the mass changes unexpectedly from m to m+mwhile xe and u(xe) remain fixed, then equation (10)becomes

q . (t) =

0 1

- k

m+m - b

m+m

x(t)

x.(t)

+

0

g - u(xe) + h(x-xe)

m+m + kxo

m+m

However, when we now substitute for u(xe) wemust note that this control value was based on theoriginal unperturbed value of m, and so we get

q . (t) =

0 1

- (k+h)m+m -

bm+m

q(t) +

0

m.g+(k+h)xe

m+m

Thus at equilibrium

(k+h)xm+m =

m.g+(k+h)xem+m

and so the new equilibrium length is

x = xe + m.gk+h


The feedback with gain h has served to reduce the

perturbation from m.g

k to m.gk+h , so that the larger

is h

the smaller is the effect of the change of load.

Feedback reduces the effect of small disturbancesto the point where they may be negligible.

Unless h is inordinately large, a large disturbancewill yield a significant perturbation.

In this case, the brain must, essentially, recomputeu(xe) using an estimate of the actual load m+mrather than the "standard load" m.

This is our explanation of the long latency reflex

noted by Navas and Stark.


What is the price we pay for using a high value of h?

We must examine the stability of the new system.

We replace k in the system matrix by k+h, which isstill positive, and possibly perturb the value of m.

Thus the system remains stable.

However, for fixed b, k, and m, we can make

b2 < 4(k + h)m

by making the "feedback gain" h large enough.

This happens as soon as

h > (b2 / 4m) - k.


Theorem: For the mass-spring model with " "coactivation with u(t) = u(xe) and feedback term h(x -xe) with gain h for desired equilibrium length xe:

(i) the perturbation of the equilibrium lengthoccasioned by a change in load m will be reduced by

the feedback from m.g

k to m.gk+h ;

(ii) the system will be stable; but

(iii) even if the system without feedback exhibits nooscillations (b2 > 4km), the feedback system willexhibit oscillations for sufficiently high gain, h > (b2

/ 4m) - k.


This model can only be part of the story:

• Our model is only valid for a range of x in which the linearmodel of the spring is valid; while the spindle can onlymonitor x-xe when this value is positive.

• A complete model would take into account those musclespindles which monitor velocity, and the way in which theGolgi tendon organ monitors force.

• The efficacy of feedback depends crucially on whether or notthe correction signal arrives while it is still valid, and thusdelays in feedback paths can have a great effect on the validityand stability of a control system.

• Nonlinearities may prove crucial in understanding muscletremor in terms of limit cycles rather than the harmonicoscillations of a linear system.

lecture 19. systems and feedback reading assignment: tmb2 section 3.1 and 3.2

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