Control Crash Course

8/2/2019 Control Crash Course

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Astro 250

Crash course on Control Systems

Part I, March 3, 2003

Andy Packard, Zachary Jarvis-Wloszek, Weehong Tan, Eric Wemhoff

[email protected]

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Feedback Systems Motivation

Process to be controlled

-l

-

P-

l -

? ?

d1

u

d2

y

P

Goal: regulate y as desired, by freely manipulating u

Problem: Effect ofu on y is partially unknown:

External disturbances, d1 and d2 act

Process behavior, P is somewhat unknown, and may drift/changewith time

Note: arrows indicate cause/effect relationships, not necessarily power,

force, flow, etc.

Open-loop regulation: Make H the inverse of P

- H - l - P - l -? ?

ydes u

d1 d2

yP

If the unknown effects (d1, d2, ) are small, this calibration strategy

may work. Well not focus on this.

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Feedback Systems Motivation

Feedback regulation:

-

C-

l-

P-

l-

?

F?l

-

? ?ydes

u

d1 d2

y

P

Benefits of feedback

1. Strategy C turns ydes and ymeas into u, so u depends on d1, d2,

(and ). Automatic compensation for the unknowns occurs, but

it is corrupted by .

2. If C is properly designed, the feedback mechanism yields several

benefits:

(a) Effects ofd1 and d2 on y are reduced, and modestly insensitive

to Ps behavior

(b) The output y closely follows the desired trajectory, ydes, per-

haps responding faster than the process naturally does on its

own.

(c) If the process P is inherently unstable, the feedback provides

constant, corrective inputs u to stabilize the process

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Feedback Systems Motivation/Nomenclature

-C - l - P - l -

?

F?l

-? ?ydes

u

d1 d2

y

P

Drawbacks of using feedback

1. A feedback loop requires a sensing element, F, which may be F

2. Measurements potentially introduce additional noise, , into

process

3. System performance or even stability can be degraded if strategy

C is not appropriate for P

Nomenclature

If keeping the mapping from d1 and d2 to y small is the focus,

then the problem is a disturbance rejection problem.

If keeping the mapping from r to y approximately unity is thefocus, then the problem is a reference tracking problem.

In either case, the ability for C to augment the systems performance is

dependent on the dynamics ofF, the noise level , and the uncertainty

in the process behavior, P.

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Feedback Loops Arithmetic

Many principles of feedback control derive from the arithmetic rela-

tions (and their sensitivities) implied by the figure below.

g C G g

H

F

Sg

- - - -

-

? -

?

?

6

Process to be controlled

Sensor

Controller

Filter

r

d

e u y

nymyf

Analysis is oversimplified, not realistic, but relevant.

Lines represent variables, arrows give cause/effect direction, rect-

angular blocks are multiplication operators. Finally, circles are

summing junctions (with subtraction explicitly denoted).

(r,d,n) are independent variables; (e,u,y,ym, yf) are dependent,

being generated (caused) by specific values of (r,d,n).

Writing each operation in the loop gives

e = r yf generate the regulation error

u = Ce control strategyy = Gu + Hd process behavior

ym = Sy + n sensor behavior

yf = F ym filtering the measurement

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Goals Implications

The first two goals are:

1. Make the magnitude of (d y)CL significantly smaller than the

uncontrolled effect that d has on y, which is H.2. Make the magnitude of (n y)CL small, relative to

1S.

Implications

Goal 1 implies

H1 + GCFS 1.

Goal 2 implies that any noise injected at the sensor output shouldbe significantly attenuated at the process output y (with proper

accounting for unit changes by S). This requires GCF1 + GCFS


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Conflict Impact on achieving Goal 3

3. Make (r y)CL response approximately equal to 1

Depending on which of Goal 1 or Goal 2 is followed, Goal 3 is accom-plished in different manners. By itself, goal 3 requires

GC

1 + GCFS 1.

If Goal 1 is satisfied, then |GCFS| is large (relative to 1), so

GC

1 + GCFS

GC

GCFS =

1

F S.

Therefore, the requirement of Goal 3 is that

1

F S 1, |GC| >> 1.

On the other hand, if Goal 2 is satisfied, then |GCFS| is small

(relative to 1), so

GC1 + GCFS

GC

Therefore, the requirement of Goal 3 is that

F S


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Tradeoffs Arithmetic

Let T(G,C,S,F) denote the factor

that relates r to y

T(G,C,S,F) = GC1 + GCFS

.

d C G dH

F

S

d

- - - -

-

? -?

?

6r

d

y

n

Use T to denote T(G,C,F,S) for short, and consider two sensitivities:

sensitivity of T to G, and sensitivity of T to the product F S.

Obtain

STG =

1

1 + GCFS, STFS =

GCFS

1 + GCFSNote that (always)

STG = 1 + STFS

Hence, if one of the sensitivity measures is very small, then the other

sensitivity measure will be approximately 1. So, if T is insensitive to

G, it will be sensitive to F S and visa-versa.

Defn: For a function F of a many variables (say two) the sensitivity

ofF to x is defined as the percentage change in F due to a percentage

change in x. denoted as SFx . For infinitesimal changes in x, the

sensitivity isSFx =

x

F(x, y)

F

x

Other more interesting conservation laws hold.

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Systems time, signals

Time: dominant (only) independent variable

usual notation, t (and , , ,...)

real number, so t R, often time starts from 0, so t R+

Signals: real-valued, functions on time variable

usual notation, u,y,x,w,v

explicit example: u(t) = e3t

sin4t for all t R+.

Systems: mapping from signal to signal (often called op-

erator)

usual notation, L for mapping, and Lu for L acting on u.

explicit example: (Lu)(t) := t0 u2()d t exp4 u()dA system L is linear if for all signals u and v, and all scalars ,

L(u + v) = Lu + Lv

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Linear systems Examples/Non-Examples

Examples

(Lw)(t) =t

0

e2(t)w() 12 + 1

w( 4)d

(Lw)(t) = 5tw(t)

(Lw)(t) =

t0

w()d

(Lw)(t) = 3w(t 4)

Non-Examples

(Lw)(t) =

w2(t) + 1

(Lw)(t) =

t0

sin(w())d

(Lw)(t) = 3e|w(t4)|

Alot is learned by considering feedback configurations of linear sys-

tems, and then studying how nonideal aspects affect the conclusions

Direct consideration of nonlinear systems is also possible, but is not

how we structure this crash-course.

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Causality Time Invariance

A system L is causal if for any two inputs u and v,

u(t) = v(t) for all t T

implies that

(Lu)(t) = (Lv)(t) for all t T

ie. output at t only depends on past values of input.

Anything operating in real-time produces outputs that are causally

related to its inputs.

Off-line filtering (ie., what is the best estimate of what was happening

at t = 2.3 given the data on the window [0 10]) is not necessarily

causal.

A system L is time-invariant if the systems input/output behavior

is not explicitly changing/varying with time.

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3 Representations Linear, Time-Invariant, Causal Systems

Convolution: given a function g (on time), define a system (rela-

tionship between input u and output y as

y(t) = t0

g(t )u()d

g can also be a matrix-valued function of time, and the u and y are

vector-valued signals. g is called the convolution kernel.

Linear Ordinary Differential Equations: Given constants aiand bi, define a system (relationship between input u and output y as

y[n](t) + a1y[n1](t) + + an1y

[1](t) + any(t)

= b0u[n](t) + b1u

[n1](t) + + bn1u[1](t) + bnu(t)

with given initial conditions on y and its derivatives.

State-Space (coupled, first-order LODEs): Given matrices

A, B, C , D of appropriate dimensions, define a system (relationshipbetween input u, output y, and internal state x as

x(t)

y(t)

=

A B

C D

x(t)

u(t)

with given initial conditions on x. Well focus on these types of de-

scriptions on Wednesday.

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Linear, Time-Invariance and Convolution Equivalence

Fact: Give a linear, time-invariant system. If for all T > 0, there is

a number MT such that

max0tT |u(t)| 1 max0tT |y(t)| MT

then the system can be represented as a convolution system, withT0

|g()| d <

for all T.

If the convolution kernel, g, is the finite sum of exponentially weightedsines, cosines, and polynominals in t, then it can also come from a

linear ODE, or a system of coupled, 1st order linear ODEs.

Translation between the representations, when possible, is easy...

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Stability Things to know

A system L is BIBO (Bounded-Input, Bounded-Output) stable if

there is a number M < such that

maxt |y(t)| Mmaxt |u(t)|

for all possible input signals u, starting from 0 initial conditions.

A system L is internally stable all homogeneous solutions (ie., u 0,

nonzero initial condiitons) decay to zero as t .

Ignoring mathematically relevant, but physically artificial situations,

these are the same, and are equivalent to:

for a convolution description:0

|g()| d <

for a LODE description: All roots of

n + a1n1 + + an1 + an = 0

(which may be complex) have negative real-part

for a state-space description: All eigenvalues of A have negative

real-part

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Simple tools

Frequency Response for stable systems, derived from model and/or

obtained from experiment

Behavior (model) of interconnection of a collection of linear sys-

tems, from the individual behaviors

Quantitative, qualitative reasoning about 1st, 2nd, and 3rd order

linear differential equations

Decrease in sensitivity and linearizing effect of feedback

Destabilizing potential of time-delays in feedback path

A few relevant architectures for control of simple dynamic pro-

cesses

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Frequency Response Convolution

Assume convolution system is BIBO,0

|g()| d <

The tail of the integral satisfies

limt

t

|g()| d = 0

and for all R,

G() :=

0

g(t)ejtdt

is well defined. Let R, and u C. Apply the complex sinusoidalinput u(t) = uejt. The output is

y(t) =

t0

g(t )u()d

=

t0

g(t )uejd

= t0

g()ej(t)d u using := t

= ejtt

0

g()ejd u

= ejt

0

g()ejd

t

g()ejd

u

= G()uejt + ejt

t

g()ejd u

yd(t)Clearly, limt y

d(t) = 0, and the response tends to a complex sin at

same frequency of input. For stable, linear time-invariant systems,

u(t) = ejt yss(t) = H()ejt

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Frequency Response Other representations

If system is given in convolution form,

y(t) = t

0

g(t )u()d

then H() = G()

If system is given in linear ODE form,

y[n](t) + a1y[n1](t) + + an1y

[1](t) + any(t)

= b0u[n](t) + b1u

[n1](t) + + bn1u[1](t) + bnu(t)

then

H() :=b0(j)

n + b1(j)n1 + + bn1(j) + bn

(j)n + a1(j)n1 + + an1(j) + an

Finally, if system is given in 1st-order form,

x(t) = Ax(t) + Bu(t)y(t) = Cx(t) + Du(t)

then

H() = D + C(jI A)1 B

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Complex Arithmetic Review

Suppose G C is not equal to zero. The magnitude of G is denoted

|G| and defined as

|G| := [ReG]2 + [ImG]21/2The quantity G is a real number, unique to within additive 2, which

has the properties

cos G =ReG

|G|, sin G =

ImG

|G|.

Then, for any real ,

Re

Gej

= Re[(GR +jGI)(cos + j sin )]

= GR cos GIsin

= |G|GR|G|

cos GI|G|

sin

= |G| [cos G cos sin G sin ]

= |G| cos( + G)

Im Gej = = |G| sin( + G)

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Complex Arithmetic Real-valued Interpretation

g is real, u is complex, so

y(t) := t

0

g(t )u()d

is complex. The linearity, obvious from the integral form, implies that

the real part of u leads to/causes the real part ofy, and

the imaginary part of u leads to the imaginary part of y

namely

y(t) =

t0

g(t )u()d yR(t) =

t0 g(t )uR()d

yI(t) =t

0 g(t )uI()d

In steady-state (after transients decay), we saw

u(t) = ejt y(t) = G()ejt

The real and imaginary parts mean

u(t) = cos t y(t) =G() cost + G()

u(t) = sin t y(t) =G() sint + G()

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A most important feedback loop... Feedback around integrator

Diagram and equations:

l l- - - - -?l

6

?x x

d

r y

n

x(t) = r(t) y(t) n(t)

y(t) = d(t) + x(t)

Eliminate x to yield

y(t) + y(t) = r(t) + d(t) n(t)

Frequency Response Functions (r y, d y and n y):

GRY() = GNY =

j + GDY() =

j

j +

Properties:

Ifr(t) r, d(t) d, then

y(t) = r + et(x0 + d y0

r)

d, not d itself, affects y. Slowly varying d has little affect ofy

The bandwidth of the closed-loop system is , the time constant

is 1.

r and n, though interpreted differently, enter in essentially the

same manner; feedback loop and integrator combine to a low-pass

filter to y.

Apparent from time simulations and frequency response function plots.

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MIFL Step/Frequency Responses

l l- - - - -?l

6

?x x

d

r y

n

y(t) + y(t) = r(t) + d(t) n(t)

GRY() = GNY =

j+

GDY() =j

j+

Time Responses

0 2/Beta 4/Beta 6/Beta 8/Beta 10/Beta 12/Beta0.2

0

0.2

0.4

0.6

0.8

1

1.2

TIME

REFERENCE R, and OUTPUT Y

Reference, rOutput, y

0 2/Beta 4/Beta 6/Beta 8/Beta 10/Beta 12/Beta

0.5

0

TIME

DISTURBANCE D

0 2/Beta 4/Beta 6/Beta 8/Beta 10/Beta 12/Beta0.2

0

0.2

0.4

0.6

0.8

1

1.2

1.4

1.6

TIME

STATE X

Frequency Responses

Beta/100 Beta/10 Beta 10Beta 100Beta0.01

0.1

1

10

FREQUENCY

FREQUENCY MAGNITUDE RESPONSE from R > Y

Beta/100 Beta/10 Beta 10Beta 100Betapi/2

pi/4

0

FREQUENCY

FREQUENCY PHASE RESPONSE from R > Y

Beta/100 Beta/10 Beta 10Beta 100Beta0.01

0.1

1

10

FREQUENCY

FREQUENCY MAGNITUDE RESPONSE from D > Y

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MIFL Application Integral Control

Process model:

y(t) = Hu(t) + d(t)

with H uncertain gain of process, and d an exogenous disturbance.Goal: Regulate y to a given value r, even in the presence of slowly-

varying d and measurement noise n.

A Solution: Integral control action:

u(t) = KIt

0

e()d

(equivalently: u(0) = 0, u(t) = KIe(t)).

KI Hl l- - - - - -

?6

l

?e x u

d

r y

n

After KI is chosen, certain properties of the closed-loop system are

insensitive to H, others are still 1-1 sensitive to H...

Description Value Sensitivity

Time constant 1HKI

1

Time-Delay for instability 2HKI1

(r y)ss for r(t) r, d(t) d 0 0

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MIFL Application Some plots

2 1.5 1 0.5 0 0.5 1 1.5 22

1.5

1

0.5

0

0.5

1

1.5

2

Experimental Process Data Runs

U

Y

Shown are several plots of the

(u, y) relationship

y = Hu + d

for different values of H and d.

Fix these, and try the integral

control solution to regulate y

0 15/KI 30/KI 45/KI 60/KI

0.5

0

0.5

1

1.5

2

2.5

3

TIME



Closed-loop time-responses ofy(t) for staircase r: Note that

time-constant is affected by the

variability in H, but the steady-

state tracking (y = r) is not.

0 15/KI 30/KI 45/KI 60/KI

0

2

4

6

8

10

12

TIME

U/KI

CONTROL ACTION, U

Corresponding value of control

input u: Even though regula-

tory strategy is fixed, namely

u(t) = KIt

0

e()d

the value clearly depends on the

specific d and H.

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What limits bandwidth? Discussion

Without loss in generality,

Take Hnominal := 1;

Drop r from the discussion (r = 0, or recenter variables around

the value of r).

Give sensor a model, S.

Then, closed-loop (nominal) appears as

KI Hf f- - -

?

d

- -

?

S?

6

f

e x u y

n

H = 1

Here, KI sets the bandwidth of the system. What limits our choice?

Time-delay in feedback path Tradeoff between effect of d and n on y

H may actually not have constant gain at all frequencies.

We know this, and use a more complex corrective strategy

(Wednesday)

We dont know this, or choose not to figure out (for instance,too difficult and/or unreliable to predict) Hs behavior at

high frequencies

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What limits bandwidth? Case 1: Lag/Delay in Feedback Path

If the feedback signal is subject to a time delay of magnitude T, some

of the properties are adversely effected.

Diagram and equations:l l- - - - -

delay, T

6

?x x

d

r y

x(t) = r(t) y(t T)

y(t) = d(t) + x(t)

Eliminate x to yield

y(t) = d(t) + [r(t) y(t T)]

or

y(t) + y(t T) = r(t) + d(t)

Properties: If 0 T < 2, then the system is stable. Time re-

sponses for T = {0, .1, .3, .5, .7, .9} 2

are shown on left. Time

delay in feedback loop degrades the systems ability to reject a rapidly

changing disturbance.

Time responses for T = {0, .1, .3, .5, .7, .9} 2

are shown:


0

0.5

1

1.5

2

TIME




0

0.5

1

1.5

2

TIME



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What limits bandwidth? Case 2: Sensor Noise

KI Hf f- - -

?

d

- -

?

S?

6

f

e x u y

n

H = 1

Consider given power spectral densities for independent d and n

d() =2

2, n() =

2

With integral feedback, the PSD of y and variance of a weighted (by

a scalar q) multiple ofy are

y() =2 + K2I

2

2 + K2IS2

, E(q2y(t)2) = q22 + K2I

2

2KIS

The integral gain which minimizes the variance is KI =

, leading to

a closed-loop bandwidth of BW = S

, and variance

E(qy)2

(t) = q2

S .

A specification imposes a lower bound on bandwidth. If E(qy)2(t)

M is a requirement, then we must have

M S

q2,

and relating this to bandwidth gives

BW q22

M.

This has implications on how much the actual process H can deviate

from its idealized model within the frequency range [0, BW].

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What limits bandwidth? Case 3: Process Uncertainty

KI Hf f- - - - -

?

S?

6

f

?e x u

d

y

n

6perception

KI Hf f- - - - -

?

S?

6

f

?e x u

d

y

n

6reality

How much can process H change if

G() := GDY() =1

1 + HKIj S=

j

j + HKIS

is not to degrade significantly? Let BW denote bandwidth here

BW = HKIS. Pick R > 0. Easy-to-show that for all stable Hsatisfying H(j) HH

R1 + Rj + BWBW

it follows that

G()

(1 + R) |G()|.

0.01 BW 0.1 BW BW 10 BW 100 BW10

2

101

100

101

102

R=10

R=0.1

R=1

Take R = 0.1 (for example).

For a guarantee of no surprises

10% degradation across frequency

then one should be able to say

thatH(j) H < |H| for

[0, 10 BW].

Statement above is general, and tight, in that no stronger statement

can be made.

There probably is a better way to get the take-home-message across...

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Effect of Process/Model Mismatch Toy Example

Take = 1, S = 1, q = 1 and from 0.5 3, giving

BW =S

= 2

1

3, E(qy)2(t) = q2

S= 0.5 3

Suppose that the (u, d) y relationship is not simply y(t) = u(t) +

d(t), but rather y(t) = f(t) + d(t), where f behaves

f(t) + 2nf(t) + 2nf(t) =

2nu(t)

with n = 10 and = 0.1.

100

101

102

102

101

100

101

MAGNITUDE

FREQUENCY10

0

101

102

180

160

140

120

100

80

60

40

20

0

PHASE

ANGLE

(degrees)

FREQUENCY

Frequency re-sponses of H(= 1)

and H. Similar

over [0 1], differ by

about 10% at 3,

and 100% at 8.

As decreases (and is exploited by increasing the bandwidth) the

output variance decreases. But, for large bandwidths (about 1.4 and

higher), the performance actually degrades as ones attempts to exploit

the sensor quality.

0.5 1 1.5 2 2.5 30.5

0

0.5

1

1.5

2

2.5

3

3.5

VARIANCE

Noise

Actual

Expected

100

101

102

103

102

101

100

101

Magnitude

FREQUENCY

Percentage Mismatch

R=5 Bounds

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Multi-Input, Multi-Output MIFL

Cf f- - - Q -?

S?

6

f

?e u

d

y z

n

Many control inputs, many disturbances, many sensors (not the regu-

lated variables)

Process, Measurement, Error criterion:

y(t) = u(t) + d(t)

ym(t) = Sy(t) + n(t)z(t) = Qy(t)

For now, assume all are the same dimension, so S is a square matrix.

Statistical descriptions of d and n: say, for instance

d() =1

2, n() = N N

Note that everything is a matrix: , N , S , Q. Each component of theproblem has its own prefered directions, and these will interact...

Goal: Find best feedback strategy, minCEzT(t)z(t).

Solution: Easy to use singular value decomposition to reduce to

many scalar problems (exercise) ... Facts:

Optimal control is integral control, u(t) = KIx(t), x(t) = e(t).

KI depends in a complicated way on the directionality/magnitudes

of the matrices , S , N (though not on Q).

Feedback loop has many bandwidths (eigenvalues of matrix SKI).

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Linear Algebra Singular Value Decomposition (SVD)

Theorem: Given M Fnm. Then there exists

U Fnn, with UU = In,

V Fmm, with VV = Im,

integer 0 k min(n, m), and

real numbers 1 2 k > 0

such that

M = U 00 0

Vwhere Rkk is

=

1 0 0

0 2 0... ... . . . ...

0 0 k

We need to apply it to real, square, invertible matrices...

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Multi-Input, Multi-Output MIFL Solution

Cf f- - - Q -?

S?

6

f

?e u

d

y z

n

Process, Measurement, Error criterion:

y(t) = u(t) + d(t)

ym(t) = Sy(t) + n(t)

z(t) = Qy(t)

Statistical descriptions of d and n: say, for instance

d() =1

2, n() = N N

Solution:

1. Calculate SVD of =: UVT

2. Calculate SVD of N =: UNNVTN

3. Calculate SVD of 1N UTNSU =: UV

T

4. Define KI := UV UT1N U

TN

This is a special case of the LQG problem.

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Linear-Quadratic Gaussian Problem Statement

General dynamical system setup for process:

x(t)

e(t)y(t)

= A B1 B2

C1 0 D12C2 D21 D22

x(t)

d(t)u(t)

Assumptions:

All matrices known

d zero mean, d() = I (absorb actual PSD into process model)

Measure y, manipulate u

Goal: Find the best dynamic, linear control strategy (matrices F,G,H,L)(t)

u(t)

=

F G

H L

(t)

y(t)

to minimize EeT(t)e(t).

Solution: Well-known since 1960s (Kalman, Bucy, Kushner, Won-

ham, Fleming, and others).

Computation: Easy to compute controller matrices, solving 2 quadratic

matrix equations. Ordered Schur decomposition is the main tool.

Issues (1978): There are no guarantees as to how sensitive theachieved closed-loop performance is to variations in the process be-

havior. [Doyle, IEEE TAC].

Robust Control (1978-199X): Tempering the optimization based

on description of what is possibly unreliable in process model.

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2nd MIFL Controlling the position of an inertia

Diagram and equations:

KI

1m

KD

KP

-f - - -f -f - - - -

?f

?f

6

-

6

-

?

?r u x x

n2

n1

d

y2 = x + n2

y1 = x + n1

mx(t) = u(t) + d(t)mx(t) = u(t) + d(t)

Controller equations:e(t) = r(t) y1(t)

z(t) = e(t)

u(t) = Kpe(t) + KIz(t) KDy2(t)

Eliminating z and u:

m...x(t) + K

Dx(t) + K

Px(t) + K

Ix(t)

= KIr(t) + KPr(t) + d(t) KPn1(t) KIn1(t) KDn2(t)

Facts:

Knowledge of m implies characteristic polynomial can be set

with ni 0, r(t) r, d(t) d, x(t) r.

KP gives initial control reaction to error

KI keeps fighting low frequency biases

KD adds damping

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2nd MIFL Design Equations

Characteristic polynomial is

p() = 3 +KD

m2 +

KP

m +

KI

m

Parametrize roots with positive real numbers , n,

n jn

1 2, n

which implies

p() = 2 + 2n +

2n ( + n)= 3 + n(2 + )2 + 2n(2 + 1) + 3n

Matching coefficients yields the design equations

KI = m3n

KP = m2n(2 + 1)

KD = mn(2 + )

Look at results for = 0.707, and = 0, 0.4, 2.5. Start with a

robust stability calculation - how much variation can be tolerated in

the process behavior, which nominally is mx(t) = u(t).

0.01 wn 0.1 wn wn 10 wn 100 wn

100

101

102

PERCENTAGEVARIATION

FREQUENCY

The maximum allowable percent-

age variation in U X (described

in terms of Frequency Response)for which closed-loop stability is

guaranteed to be maintained.

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Results Frequency Response Functions

0.01 wn 0.1 wn wn 10 wn 100 wn10

4

103

10

2

101

100

101

MAGNITUDE

FREQUENCY

Magnitude of frequency response

from R X,

KP(j) + KI

m(j)3 + KD(j)2 + KP(j) + KI

0.01 wn 0.1 wn wn 10 wn 100 wn180

160

140

120

100

80

60

40

20

0

PHASE

FREQUENCY

Phase of frequency response from

R X

0.01 wn 0.1 wn wn 10 wn 100 wn10

4

103

102

101

100

MAGNITUDE

FREQUENCY


from D X (normalized).

0.01 wn 0.1 wn wn 10 wn 100 wn10

4

103

102

101

100

101

MAGNITUDE

FREQUENCY


from N1, N2 X.

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Results Time Responses

0 10 20 30 400.5

0

0.5

1

1.5

2

2.5

R

eferenceR

Normalized Time, t/wn

Applied reference signal r.

0 10 20 30 401

0.5

0

0.5

NormalizedD

isturbanceD


Applied disturbance signal d.

0 10 20 30 401

0.5

0

0.5

1

1.5

2

2.5

OutputResp

onse


Output (x) response.

0 10 20 30 4010

5

0

5

ControlAction


Control action u.

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Reduction in sensitivity from feedback

Ll l- - - -6

?e

d

r y

Constraints are

y = d + Le, e = r y

Eliminating e (for instance) gives

y =L

1 + L

T(L) orTr +

1

1 + L

S(L) orSd

Obviously, L > 0 (which is negative feedback) means 1

1+L

< 1.Suppose L changes to L + . Obviously T changes as well. Compare

percentage change in T to percentage change in L,

% change in T

% change in L =

T(L+)T(L)T(L)

L+LL

=

T(L + ) T(L)

L

T(L)

Compute for differential changes in L, so take lim0, giving

% change in T

% change in L=

dT

dL

L

T(L)=

1

(1 + L)2L

T(L)= S

This is why Bode called S the sensitivity function.

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Linearizing effect of Feedback

K ()l l- - - - -6

?e

d

r y

Constraints are

y = d + (e), e = K(r y) y = r 1K

e

whose solution (for certain ) implicitly defines a function y(r, d). The

chain rule implies

y

r =K

(e

) 1 yr , yd = 1 K(e)ydwhere e = K(r y(r)). Rearranging, gives

y

r=

K(e)

K(e) + 1,

y

d=

1

K(e) + 1

Note, ifK >> 1 everywhere, then the function y is more linear in

r than , and nearly unaffected by d.

10 8 6 4 2 0 2 4 6 8 1015

10

5

0

5

10

15

20

25

Solution of y

e

y

y = (e)

y=r e/Kr

y(r)

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Linearizing effect of Feedback Dynamic Example

()l l- - - - -

6

?e

d

r y

Replace K by an integrator and inject sine wave, r = 10sin0.01t. At

= 0.01, the gain from the integrator is 100.

10 8 6 4 2 0 2 4 6 8 1020

15

10

5

0

5

10

15

20

Nonlinear function 1

x

e

y = x + 0.01x3

10 8 6 4 2 0 2 4 6 8 1020

15

10

5

0

5

10

15

20

Nonlinear function 2

e

y

y = 2e for e < 0

y = 4e for 0


41/41

Linearizing effect of Feedback Dynamic Example (contd)

()l l- - - - -

6

?e

d

r y

However, if r = 10sin0.1t, the gain from the integrator is 10, and the

time responses for this system with the same nonlinear functions are

shown.

0 10 20 30 40 50 60 70 80 90 10020

15

10

5

0

5

10

15

20

Function 1: Ref(o), Output(), Openloop(.)

Time (sec)

Output

y

Ref &Closedloop

0 10 20 30 40 50 60 70 80 90 10020

15

10

5

0

5

10

15

20


Time (sec)

Output

y

Ref &Closedloop

0 10 20 30 40 50 60 70 80 90 10020

15

10

5

0

5

10

15

20


Time (sec)

Output

y

Ref &Closedloop

20 15 10 5 0 5 10 15 2010

8

6

4

2

0

2

4

6

8

10

Reference vs Input, compared to inverse of function 1

Reference r

I

npute

20 15 10 5 0 5 10 15 2010

8

6

4

2

0

2

4

6

8

10


Reference r

I

npute

20 15 10 5 0 5 10 15 2010

8

6

4

2

0

2

4

6

8

10


Reference r

I

npute

Notice that the output, y, does not track the reference, r, as well as

when r = 10sin0.01t. Also, the scatter plots of r vs e have more

dispersion and indicate that e does not invert (.) as well as in the

previous case.

This example shows that feedback can have a linearizing effect when

the gain is large enough

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