svd (singular value decomposition) · 2016-08-25 · chosun university 170 nano system control lab....

Nano System Control Lab.Chosun University 170

SVD (Singular Value Decomposition)

1

2

0 0 0 00 0 0 00 0 0 00 0 0 0

0 0 0 0 0

n m K

1 2

; n m real matrixwhere 0K

H

H

S.V.D A UΣVU AV

Where

H -1

H -1matrix V

U : n n unitary matrix U =U

V : m m unitary =V

max SV

min SV


Where ; left singular vector of is right eigenvector of associated with

H 2: AA i i i

H1 2 m i jV v :v : v ; v v ij

H 2: A A i i i

HAA

i Hiλ (A A)

H1 2 i jU u , u u ; u un ij : Kronecker delta

Hλ (AA )i

Aiu

Where right singular vector of is right eigenvector of associated with

iv ; AHA A

i

Column vectors of U(V) are orthonormal


Geometrical Interpretation

(1) is a real matrix, exists n n -1AAn

2

n2

y Ax, x R x x x

y R y y y

T

T

Spectral norm of A

2

22 2X 0 X 1

2

AXA max max AX

X

Singular value relation

2

2 2

max 2 2X 1

min 2 1X 12

(A) max AX A

1(A) min AXA

: spectral norm


Visualization (2D)A

Unit Circle

0

1 1

2 2

Vector OA=v Vector OA =uVector OB=v Vector OB =u

maxH

min

1 2 1 2

0SVD : A U V ,

0

U u u , V v v

AB

B A

1y

2x

1x

2y

B

00


max 1

min 2

Length of vector OA σ σLength of vector OB σ σ

.

.

max SV, min SV, Range of gain


1) Singular values “size” of A

HA A 1, 2, Ki i i

2) SVD “Direction” of A,B

H

H

A U VU AV

max min, for

G(s)u(t) y(t)

MIMO Freq. Response

u(t) uy(t) yy G(jω)u

j t

j t

ee

Singular value

2

2

max max2 2u 1

min min2 2u 1

G(jω) max G jω u y jω

G(jω) min G jω u y jω


max maxy (jω)

min miny (jω) Singular value of y exists in this region.

S.V. Facts

(1) -1 -1max min

min max

1 1(A ) , (A )(A) (A)

(2) max max max(A) 1 (I+A) 1 (A)

(3) max max max(A+B) (A) (B)

(4) max max max(AB) (A) (B)


Feedback Performance Specs. In Frequency Domain

Use of SV to establish MIMO Performance specs.

Proof. Attributes

Review

r(s)

d(s)

u(s) y (s )

n(s)

e(s)K(s) G(s)

-Command Following-Disturbance Rejection-Insensitivity to Sensor noise-Robustness

Sinusoidal


1

1

t

Loop TFM : T(s) G(s)K(s)

Sens TFM : S(s) 1 T(s)

Closed-Loop TFM : C(s) 1 T(s) T(s)

e (s) S(s) r(s) d(s) C(s)n(s) e(s) r(s) y(s)

conflict S(s) C(s) I

1) Command Following : d(s) n(s) 0

tr(t) r e (t) ej t j te e : error vectore

Relation

2 2 2

max 2

S(jω)S(jω) r

S(jω) r

e re

True error


Define;

range of frequency that r has energyr

We have good command following by making max S(jω) 1, r

Interpretation

max2 2,maxr 1 S(jω)e

The worst error at is attained when points along the right singular vector associated with ,

rmax 1V

min2,minS(jω)e

The best error r VK

In General

min max2S(jω) S(jω)e

min


Good Command Following in terms of T(s)=G(s)K(s)

max S(jω) 1 ; r

1

maxmin

1(I+T) 1I+T

min min min1 T I+T 1, T(jω) 1

Overall Loop Gain

To visualize good C.F

max (S)

min (S)

dBLoop TFM

max (T)

min (T)ω

1

r

1

dB

11


2) Disturbance Rejection : r(s)=n(s) 0

Sinusoidal Disturbance

td(t) d e ej t j te e

Relation

max2 2

e S(jω) de S(jω) d

We have good disturbance rejection by

max

min

making S(jω) 1

also, T(jω) 1

; ω d

Define :d Range of frequency that d has energy

Interpretation ; 2d 1

Same on the C.F.


To visualize good C.F. & good D.R. P r d

dBdB

max (T)

min (T)ω

max (S)

min (S)

ω

P

1

:margin of design.

Quantity Relation

Let : 0 1

If max S(jω) , ω P

Then, min11 T(jω) ; ω P

and min max1 C(jω) C(jω) 1

11

C(jω) I “Plant Inversion”


Visualization

dB

P max S( jω)

Sensitivity TFM

dB

P

min T(jω)1

Loop TFM

dB max C(jω)

min C(jω)

1

1 Closed-Loop TFM


Example of the FP (FP : Feedback Performance)

(Temp. regulation in a leaky pipe line)

(1) Problem ; monitor & control temperature.

(2) Model of the system

• Dynamics of medium Flow in the pipeConstitutive relation of the medium

• Dynamics of leak• Dynamic of a heater (time-lag)

ideal gas

Assumptions what? Why? Validity, MIMO, TITO

(3) Solution-Further assumption, linearization

Application of tools & concepts

(4) Evaluation of your solution & problem

heater leak (pressure drop)

FlowPV=nRT


Quantitative Relations

max

min

min max

: 0 1; S(jω) 1 , ω

;1(1) 1 T(jω) ;

(2) 1 C(jω) C(jω) 1

p

p

Let

If

Then

Proof of (1)

1max max

min min

max

min

min min

1 1S(jω) I+T(jω)(I+T(jω)) 1 T(jω)

Given , S(jω)1

1 T(jω)1 1 11 T(jω) , or T(jω) 1 1


Proof of (2)

max

1

-1 -1 -1

-1 -1 -1max max max

min

min

(1 ) C(jω)

C(jω) I+T(jω) T(jω)

C (jω) T (jω) I+T(jω) I+T (jω)1 C (jω) I+T (jω) 1 T (jω)C(jω)

11T(jω)

①

Since min1T(jω) from

(1)

min

min min

min

1T(jω) 1

1 1 11 1C(jω) T(jω) 1 1

C(jω) 1


②

max

max max max

max max

max

C(jω) 1C(jω) I S(jω)

C(jω I S(jω) 1 S(jω)

1 S(jω) , since S(jω)

C(jω) 1

Summary for good C.F & D.R

pω

• Large Loop Gain

• Small Sensitivity

• Flat CL response

min T(jω) 1

max S(jω) 1

min maxC(jω) C(jω) I

for

Constraint


3) Insensitivity to S.N. (Sinusoidal Noise) ; r(s) d(s) 0

n( ) n ( ) ej t j ttt e e t e

Relation

max2 2

e C(jω)nC(jω)e n

Define range of frequency of noise with singnificant energyn

We can desensitize the system from S.N. by making for all n

Large separatebetween the two

( , )p n max C(jω) 1 for all n

suppose max C(jω) ; 0 1

① Then, min T(jω) 1 ;(low loopgain)1

② max1 1 S(jω) ; high sensitivity


Proof: ① -1 -1min min

max max

1 1C I+T 1(C) (T)

maxmax

maxmax

1 1Since (C) ,(C)

1 11 (T)(T) 1

Proof ②1

max max minmin min max

maxmax

1 1 1(S) (I+T) 1 (T)(I+T) 1 (T) (S)

11 1 (S)1 (S)


Design Implication

We need wide frequency separation between sets.

andp r d n

P

n

ω

dB

min S(jω)

1

max S(jω)


dB

1

1 P

P max C(jω)

min C(jω)

n

ω

dB

1

P

n

1 min T(jω)

max T(jω)

ω


Direction information in SV plot

SVDA

m m

m m (A)i

-1AComplex matrix , exists

Real diagonal matrix of

1 maxH

min

0 00 0 (A) (A A)0 0

i i

m

H HA U V : Σ=U AV

• right singular vector , orthonormal• left singular vector , orthonormal

iv ;iu ;

miv C

miu C


For

AX Y Y=AX

SVD HY U V X

Suppose ;H

i iX=v y=U V v

Since

H Hi

0 00

0V v , V v1

0 0

0 0

i i

rowthi


1 i

0

y u0

0

ii m iu u u

Directional Information

Max : 1 max , 1 max max 2

1 max max 2

v V , V 1

u U , U 1

If maxX V

max max max 2

max2

Y= U ; U 1

Y (A) max. .Amp

Min : min ,m m min min 2

m min min 2

v V , V 1

u U , U 1


If minX V

min min min2Y σ U Y σ (A) min. .Amp

Input

minV

Unit sphere

min (A)

maxU

minU

maxy

miny

max (A)

“ can be less than 1”max miny , y

.

maxV

output


Analytical Expression

Real Amp.Complex sinusoid

x(t) A j te

Re x(t) A cosωt

Im x(t) Asinωt

Complex Amp.complex sinusoid

x(t) B j te B=B j

je

Re x(t) B cos (ωt )

Im x(t) B sin (ωt )

2 2

1 ( )

B

tan x( ) B j tt e

Complex vectors 1

x( ) x , xj ti

n

x

t e x

x


Summary of SVD

System : Y(s)=G(s)X(s)

Pick ω , calculate G(jω) SVD

• Max - direction

Find max max max, V , U

Set max max

( )max

x ( ) V a , U b

y b

j ji i ii i

j ti i

t e e

e

• Min-minimum direction

Find min min min, V , U0

0

& : 90 difference*

& : 90 difference

min min

( )min

x ( ) V C , U d

y ( ) d

j ii i ii i

j ti i

t e e

t e


F-404 turbofan engine control (GE)

1 11

2 22

3 3

x xu

x * x *u

x x

1

2

3

: LP roto speed: HP roto speed

: Turbine temper

xx

x

1

2

u : Fuel Flowu : Nozzle Area

11

32

1

xyy

xy

G(s) C(sI A) B at ω 0.1 rad/sec0.863 0.11 4.609 0.47

G(j0.1)0.9347 0.25 0.88 0.32

j jj j

G(0) :steadystate

max

min

4.846 (13.7 )0.757 ( 2.4 )

DBDB

G(s)K(s)


Open loop TR function

1 rad/sec

2.4

13.70

max

min

( )G j

From Matrix X

0

0

max 0.22

180

min 0.23

0.22 0 0.22V

0.975 0.009 0.98

0.98V0.22

j

j

j

j

j ej e

ee

*

6.15

max 19.38

0

min 166.6

0.94U

0.22

0.23U

0.97

j

j

j

j

ee

ee

( )i max iy b j te

0.1

,

,

,( )

i min iy d ; y = Gxj te


For max ;

01

max 02

u 0.22sin (0.1 0) 4.55sin(0.1 6.15 )x y

u 0.98sin (0.1 0.22) 1.115sin(0.1 19.38 )t t

t t

Input to study Max. amplitude

Phases are differentFor min

min min min 0

0.174sin(0.1 0)y U

0.7341sin(0.1 166.6 )t

t

Min amplitude

0

min 0

0.98sin(0.1 180 )x u V

0.22sin(0.1 0.23 )tt

max max max maxx V , y U


SVD problem

x

F

v

0x

Car suspension

0M F K( ) b( v)x x x x

0If 0x

0 1 0 0 FK b 1 b v- -M M M M

x xx x

1

2 2

1 0 0 0 Fy

v 0 1 0 1 v

1 b- -1 M MG(s) C(sI A) B D b K s Ks s -s -M M M M

x xx x

M

b k


For n0.5 , ω 10, M 1

At ω 10.01 0.001 0.1 0.01

G(j1)0.001 0.01 0.99 0.1

kg

j jj j

max min

max min max min

U. .V SVD G(j1)1.002, 0.01

U= U U V= V V

If input u (not U) is chosen as

(1) max max max maxu V , then y Uj t j te e

(2)min min min minu V , then y Uj t j te e

1 1 1 1max min

2 2 2 2

a sin(ωt ) b sin (ωt )U U

a sin(ωt ) b sin (ωt )

,

,


u1u

2u

y


Stability of MIMO Feedback system Nominal stability

• Time domain - locations of poles• Frequency domain- Nyquist criterion

For

1T(s) C(sI A) B T(s)r(s) e(s) y(s)

1CL: C(s) I+T(s) T(s)

For u(t) r(t) y(t) r(t) Cx(t)

x (A BC)x Bry=Cx

“Close” Loop TF1C(s) C(sI A BC) B

Stability Re (A BC) 0 for alli i


Key relation

0 (s) det(sI A) ; T(s)L

(s) det(sI A+BC) ; C(s)CL

0* (s) ( ) det( I T(s))CL L m m m ms

Proof ;

1m

1m

1m

m

(s) det (sI A BC) det(sI A) det(I C(sI A) B)

det (sI A) (I C(sI A) B)

det (sI A) I (sI A) BC

(s) I +T(s)

CL

OL


Complex variable theory facts

• = Complex scalar variable ; “Lives” on s-plane• = Scalar-valued analytic function : maps

s-plane to another complex plane

(s)fs

(s)f Analytic in R

(1) Derivative exist at each point of R

(2) Unique in R


Ex)If

s+1(s)= , s 1 j2s

f

then2+j2 6 2(s)= j1+j2 5 5

f

jω

..(s)f

Im

Re

The principle of the Argument

A theorem in complex variable theory

Suppose

C = closed contour in the s-plane.

(s)f = complex-valued scalar function

(1) Analytic in C(2) z zeros inside C(3) p poles inside C


Then, The image of C under mapping (s)f

(A) Generates a closed contour in the C.P.(B) encircle the origin 0, z-p times in a clockwise direction.

Im

Re

o

o

j

Notation

N( , (s), ) #a f C of clockwise encirclement of point by the image of under mapping (s)f

a C


The principle of Argument

N(0, (s),C) z pf

z p 3

Suppose , 1 2(s) (s) (s)f f f

1 1,z p 2 2,z p

Then, 1 2N(0, (s), C) N(0, (s), C) N(0, (s),C)f f f


S-plane

Scalar mapping plane to plane det I T(s)

Im

Re

MIMO Nyquist contour,

R

jω

Semi circle with radius

Define RD ;

R0

RD


The MIMO CL system is stable iff

R u R uN 0, det I T(s) , D P , or N 1, 1 det I T(s) , D P

uP # of unstable poles of T(s) (open-loop)

Proof)

Recall (s) det I+T(s)CL OL

and R R RN 0, (s), D N 0, (s), D N 0, det I T(s) ,DCL OL

MIMO Nyquist Criterion

T(s)=G(s)K(s)e(s)r(s) y(s)

T(s)

0 0 0(z p ) ?


Let g(s) det I+T(s)

2 2

kT(s)s(s 2 s )n n

A B

C

1

j

s-plane


( )g j g(j ) je

( )

. A ; s j 0g(jω) g(jω) j

sege

. B ; s Re , RT(s) 0

g(s) det I+T(s) 1

jseg

Re

Im

1

B .

A

0

C

.

. C ; s j 0seg


* ( ) G(s) K(s)t s

( j )K(jω)

G(jω)K(jω) ( j ) G(jω)

t

t

For SISO case

u R

R R

T(s) ( ) det I+T(s) 1 ( )

-P N(0, det I T(s) , D

N 0,1 (s), D N 1, (s), D (open- loop circlement)

t s t s

t t

G(jω) = Unstable plant

T(s): desired

1mPositive phase

margin

1

1 1G( jω)

Positive gain margin


MIMO Case

0

det I+G(jω)

det I+T(jω) : desired T jω G(jω)K(jω)

1


MIMO Nyquist Criterion

• SISO 1C(s) 1 ( ) ( )t s t s

1R R RN 0, ( ), D N 0, 1 ( ) , D N 0, ( ), DC s t s t s

C C C R 0 0

C R 0 0

Z P N 0, 1 ( ), D Z P

N 1, ( ), D Z P

t s

t s

We want CP 0

C 0 0 C R

0 C R

Z Z +P =N 1, ( ), D

P =N 1, ( ), D

t s

t s

: Nyquist criterion for SISO (ccw)

cZ Pc 0 0Z Pccw


• MIMO

1

1 21 2 1 0

11 1 0

C(s) I+T(s) T(s)

(s)det I+T(s)

( ) det sI A BC s s s s

only eros poles of CL(s) det(sI A) s s s

eros poles of OL

CL OL

n n nCL n n

n nOL n

s

z

z

R R R

SC UC

SO UO

N 0, (s), D N 0, (s), D N 0, det I+T(s) , D

#of Z( ) Z Z#of Z( ) Z Z

CL OL

CL

OL

nn

Stable Unstable open loop


UC UO R

UC UO R

Z =Z N 0, det I T(s) , D

P =P N 0, det I T(s) , D

Counter clockwise

0

det I T(s)

Stable

: roots of characteristic eq.

: poles of closed loopC(s) T(s)

UC

R UO

C R UO

We want "P 0"

N 0, det I+T(s) , D P

N 0, det I+T(s) , D P


Stability Robustness : SISO

Central theme- to provide guarantee that compensator designed on the basis of nominal plant model, will not yield unstable feedback system

Assume

Question: then,

Ng (s) ,NK (s)

Ng (s)Ne (s) Ny (s)NK (s) Nu (s)

stable

Is this stable?

Ag (s)e(s) y(s)NK (s)

u(s)


A Ng (s) g(s) g (s)

Source of g(s)

• Low order approximation of high-order system.

Directional Uncertainty

At (jω) >1g( jω) ?

1

Nt (jω)At (jω)

g( jω) has 180 uncertaintyo

• actuator of sensor dynamics.

• “Fast” mechanical dynamics : bending, torsional.

• “Small” time delay.


SISO Model Error (cascade Representation)

; Error reflected at the plant output

A 0 N N 1, g (s) (s) g (s) g (s) (s)l l

0 (s)lu(s) y(s)

Ng (s)

1(s)lu(s) y(s)

Ng (s)

For SISO

11ω

2ωAt (jω)

Nt (jω)

; modelling error0l

0


Test 1 (For Robustness)

N N

N N

N N

Vector sum-1 d (ω) t (jω)

d (ω) 1 t (jω)

d (ω) 1 t jω

" make this resonable big"

Test 2 NN

N

t (s)C (s)1+t (s)

Assume NC (s), (s)l are stable.

Then, NN

1 1If ( jω) 1 or C (jω)C ( jω) ( jω) 1

ll

The actual Loop is still stable Sufficient condition

Nt (jω)

1

Nd (ω)0

A N; t (s) (s)t (s)l


Proof)

A N

A N N

A N

N

t (s) (s)t (s)t (s) t (s) (s) 1 t (s)

t (s) t (s)(s) 1t (s)

ll

l

A N Nt ( jω) t ( jω) 1 t ( jω)

“The inverse is true”

1

N1 t (jω)

Nt (jω)

A Nt (jω) t (jω)

At (jω)

0

N

N N

1 t ( jω) 1( jω) 1t (jω) C ( jω)

l

divided by at the both sidesNt ( jω)


Alternate model

Percent deviation

m

m

m

(s) 1 e (s) (s)(s) e (s) (s)

(s) 1 e (s)

A N

N

g gg g

l

u(s) y(s)Ng (s)

me (s)

12

unstable region N

1C (jω)

me (jω)


N

1C (jω)

me (jω)stable

NC (jω)

Robustnessboundary

m

1e (jω)


Design Implication

Need ; (1) Ng (s)

(2) Some upper bound of m maxe (jω) e (ω)

Conservatism.“sufficient condition”

maxe

max

1e

NC (jω)minimize the bandwidth NC (jω)


Robustness Condition for SISO case

mN

Nm

1e (jω)C (jω)

1C (jω)e (jω)

; “MIMO Nyquist criterionNec. Sufficient condition”

NC (jω)

1me (jω)

Bode plot

m maxe (jω) e (ω)


MIMO Stability Robustness

Assume : Nominal Feedback Loop is stable

Need : Actual Feedback Loop

Nominal plant ; NG (s)Actual plant ; AG (s)

• Plant Output Error

• Plant input Error

oL (s)u(s) y(s)

NG (s)

A 0 NG (s)=L (s)G (s)

NG (s)y(s)

IL (s)

A N IG (s)=G (s)L (s)

u(s)

0 IL (s) L (s) for MIMO


Robustness Test (MIMO Case)

For a stable nominal loop,

L(s)NK (s) NG (s)

The actual loop is stable if

(1) L(s) is stable

and

-1max min N

N N N

-1max min N

1

L(jω) I I T (jω) , ω

T (jω) G (jω)K (jω)or

L(jω) I C (jω)

C I+T T


NT (s)

mE (s)

mL(s)= I+E (s)

C

Alternate Test

1max N max m

max m

1C (jω) E (jω)E (jω)

max NC (jω)

1max mE (jω)


Robustness Test proof;

Step ;

(1) Assume is stable

(2) Find “Smallest” model error in “worst” direction that will cause actual Feedback System to be on the verge of instability

(3) Express (2) on Nyquist plot.

(4) Use a SV fact to obtain a sufficient condition for stability

NC (s)


SV factmax minIf ( B ) ( A )

Then , det(A+B) 0n n n n

0. 1

N

N u

det I+T (jω)

N 0, det I T (jω) ,D PR

Adet I+T ( jω)

i

A

at ω ω*det I T ( jω) 0

For stability ;

Adet I T (jω) 0, ω

Set

-1m N

-1 -1 -1 -1m N m N N N N N

-1m N N N

B E ( ) , A=I+T ( )

0 det E +I+T det E T T +T T +T

det E T +T +I det T

s s

-1Nsince det(T ) 0,


A m NT I+E T

m N N m Ndet E T +T +I det I+(I+E )T 0

AT (s)

NT

mE

* Sufficient condition for stability Robustness

-1max m min N

-1min N

max N

E (jω) I+T (jω)

C (jω)

1C (jω)


max Nmax m

1C (jω)E (jω)

max NC (jω)

1max mE (jω)

max NT (jω)

svd (singular value decomposition) · 2016-08-25 · chosun university 170 nano system control lab....

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