me677c6 frequencyanalysis t

Frequency Domain Analysis Lecture Notes by B.Yao

1

Frequency Domain Analysis of Feedback Systems System:

, ( , )

x Ax Buy Cx Du u t y

Assume (A, B) is controllable and (A, C) is observable. Sector Bounded Nonlinearities: A memoryless function : [0, ] p pR R is said to belong to the sector

[0 ∞] if ( , ) 0Ty t y

[ 1K ∞] if 1[ ( , ) ] 0Ty t y K y

2[0 ]K if 2[ ( , ) ] ( , ) 0Tt y K y t y where 2 2 0TK K

1 2[ ]K K with 2 1TK K K K if

1 2[ ( , ) ] [ ( , ) ] 0Tt y K y t y K y

SISO Illustration

-0r )(sG

( )

u y-

0r )(sG

( )

u y

y0

1K y

K

2K y


2

Absolute Stability: The above system is absolutely stable if the origin is globally uniformly asymptotically stable for any nonlinearity in the given sector. It is absolutely stable with a finite domain if the origin is uniformly asymptotically stable.

Circle Criterion: THEOREM 7.1:

The system is absolutely stable if 1

1 1[ , ) and ( ) [1 ( )]K G s K G s is SPR 1 2 2 1[ , ] with 0TK K K K K K , and 1

2 1[1 ( )][1 ( )]K G s K G s is SPR. If the sector condition is satisfied only on a set pY R then the above conditions ensure that the system is absolutely stable within a finite domain. #

Proof: For simplicity, only prove the theorem for [0, ] . For this case, 1 0K and G(s) is SPR by assumption. Then, by KYP lemma, 0TP P , L, W, and

0 such that T TPA A P L L P

T TPB C L W T TW W D D


3

Choose 1( ) .2

TV x x Px Then,

1 12 2

T T T T T T T TV x Px x Px x P Ax x P Bu x A Px u B Px

1 ( )2

T T Tx PA A P x x PBu

1 [ ]2 2

T T T T T Tx L Lx x Px x C L W u

( )

1 12 2 T

T T T T T T T

y Du

x L Lx x Px x C u x L Wu

1 12 2

T T T T T T Tx Px x L Lx x L Wu u Du y u

Noting 1 1( )2 2

T T T T Tu Du u D D u u W Wu , the above equation becomes

1 12 21 0, 0, as ( , ) 02

TT T

T T T

V x Px Lx Wu Lx Wu y u

x Px x y u y t y

which indicates that the origin is exponentially stable. #


4

SISO Systems:

Absolutely stable if 1 2 2 1[ ],k k k k and 2

1

1 ( )( )1 ( )

k G sZ sk G s

is SPR. #

Q: Can we check the SPR of Z(s) based on FR ( )G j only? A: Yes as follows As Z(s) is biproper, Z(s) is SPR if Z(s) is Hurwitz and Re ( ) 0,Z j Case 1: 2 1 0k k

(a) 2 2

1

1

1 ( )1 ( )Re 0 Re 011 ( ) ( )

G jk G j kk G j G j

k

the angle between 2

1 ( )G jk

and 1

1 ( )G jk

is less than 90 , i.e.,

2 1 90 the point ( )G j is outside the disk 1 2( , )D k k


5

Nyquist plot of ( )G j must be strictly outside the disk 1 2( , )D k k .

0

( )G j

1

1k

2

1k

1 2,D k k

2

1( )G jk

1

1( ) ( )G jk

(b) Z(s) is Hurwitz 1

1

( )1 ( )

k G sk G s

is Hurwitz.

Nyquist plot of ( )G j does not intersect 1

1k

and encircles it exactly

the same number of times pN in the counterclockwise direction as the number of unstable poles of G(s).


6

(a) + (b)

Absolutely stable if the Nyquist plot of ( )G j does not enter the disk 1 2( , )D k k and encircles it pN times in the counter clock direction.

Case 2: 2 1 0k k 2( ) 1 ( )Z s k G s

(a) ( )Z s is Hurwitz G(s) is Hurwitz.

(b) 2Re ( ) Re 1 ( ) 0Z j k G j 2

1Re{ ( }G jk

Absolutely stable condition for Case 2:

G(s) is Hurwitz and the Nyquist plot lies to the right of the vertical line defined by

2

1Re[ ]sk


7

Case 3: 1 20k k

The same as Case 1 except with the following changes:

(a) 2

1

1 ( )Re { ( } 0 Re 01 ( )

G jkZ j

G jk

Thus, the absolute stability condition is changed to:

Absolute stable if G(s) is Hurwitz and the Nyquist plot of ( )G j lies in the interior of the disk 1 2( , ).D k k

0

( )G j

1

1k

2

1k

1 2,D k k


8

Popov Criterion System:

, (A, B) is controllable and (A, C) is observable

( ), 1 , time-invariant nonlinearitiesi i i

x Ax Buy Cxu y i p

Assume [0, ]i ik .

Theorem 7.3: The above system is absolutely stable if, for 1 , [0, ], 0i i ii p k k , and there exist constants 0i with (1 ) 0k i for every eigenvalue k of A such that ( ) ( )M I s G s is SPR where 1diag{ , , }p and

1

1 1diag , ,p

Mk k

. #

Proof:

Construct the following Lyapunov function and apply Lyapunov Theorems:

01

1 ( )2

ip yT

i ii

V x Px d

where 0P is the matrix in KYP Lemma for the SPR TF. #


9

SISO Case:

Absolutely stable for [0, ]k if 0 with (1 ) 0, ,j j and 1( ) (1 ) ( )Z s s G sk

is SPR.

Z(s) is SPR １ (a) G(s) is Hurwitz, and

(b) 1 Re 1 ( ) 0, [ , ]j G jk

(b) is equivalent to

1 Re Re{ } Im{ } Re{ } Im{ } 0G j G j G Gk

1 Re { ( )} Im{ ( )} 0G j G jk

which indicates that the system is absolutely stable if the Popov plot stays to the right of the line in the following figure.

１ For simplicity, assume k to avoid the special SPR condition for ( )Z s when its relative degree is not zero.


10

Popov Plot:

0

1k

Im ( )G j

1slope

For SISO systems, the system is absolutely stable if the Popov plot stays to the right of the line passing through the point 1

k with a slope of 1 .

Note:

When 0 , the Popov Stability Condition reduces to Circle Criterion.


11

Describing Function Method Goal:

Study the existence of periodic solutions of free response for the following system: where ( ) is a time-invariant, memoryless nonlinearity.

Method of Harmonic Balancing

Idea: represent a periodic solution by a Fourier series and seek a frequency ω and a set of Fourier coefficients that satisfy the system equation.

Assume:

2( ) ( ),y t y t T T

then

( ) j ktk

ky t a e

(D1)

-0r )(sG

( )

u y-

0r )(sG

( )

u y


12

where

0

1 ( )T jk t

k ka a y t e dtT

As ( ) is time-invariant, ( ( )) ( ( ))y t y t T is also periodic with the same frequency:

( ( )) jk tk

ky t c e

(D2)

where

0

1 ( ( ))

( ), 0, 1, 2,

T jk tk

k i

c y t e dtTc a i

(D3)

Let ( )( )( )

N sG sD s

. Then, for (D1) and (D2) to be a solution of the CL system, it

is required that they satisfy the differential equation as

( ) [ ( )] ( ) [ ]N p y D p y (D4) As

jk tk

k

dp y y a jk edt


13

[ ( )] [ ( )] jk tk

k

dp x y y c jk edt

(D4) is the same as:

( ) ( )jk t jk tk k

k k

D jk a e N jk c e

(D5)

Use the orthogonality of the basis function jk te for different value of k, i.e.,

1 2 1 2

01 2

011

T jk t jk t k ke e dt

k kT

(D5) is the same as:

( ) ( ) 0,( ) 0 ,

( ) 0 , 0

k k

k k

k k

D jk a N j k c kG jk c a k

G jk c a k

(D6)


14

Finite Dimension Approximation

Assume G(s) is strictly proper. Then, ( ) 0G j as . So assume that there is an integer q > 0 such that k q , ( ) 0G jk . Then (D6) becomes:

ˆ ˆ( ) 0 , 0,1, 2, ,k kG jk c a k q (D7) which is still a very difficult task to carry out.

Describing Function Approximation

Choose q=1 and neglect all higher harmonies, i.e., assume 0, 0, 2k kc a k . Then (D7) becomes:

0 0 1 0

1 0 1 1

ˆ ˆ ˆ ˆ(0) , 0

ˆ ˆ ˆ ˆ( ) , 0

G c a a a

G j c a a a

(D8)

which has four unknowns: 1ˆ ˆ, , complexoa a , but only three equations. This is reasonable as the time origin of a periodic solution is not defined. So assume that first harmonic of y(t) to be of sinA t :

11ˆ ˆ( ) sin2 2 2 2

j t j t j t j tA A Ay t A t A e e e e aj j j j


15

Conventionally, we define the describing function of a nonlinearity ( ) as:２

01

1

0

0 0

1 sinˆ( )

ˆ2

2 sin cos( ) sin

2 2 sin sin sin cos

T j t

T

T T

A t e dtc TN A Aaj

j A t t j t dtTA

jA t t dt A t t dtTA TA

Let t :

2 2

0 0

1 1( ) sin sin sin cosN A A d j A dA A

(D9)

With the above definition, (D8) becomes:

( ) ( ) 1 0G j N A (D10)

２ This equivalent gain concept (or equivalent linearization) can be generalized to general time-varying nonlinearities with memory, like hysteresis, in which the describing function might be complex and dependent on both Aand , i.e., ( , )N A


16

Odd Nonlinearities:

0ˆ( ) ( ) 0y y c

From (D8), 0ˆ 0a . Furthermore, ( )N A is real given by:

0

2( ) sin sinN A A dA

(D10) becomes: 1( ) ( ) 1 0 ( )( )

G j N A G jN A

(D11)

which can be solved graphically:

0

( )G j 1

N A

A 0A

s

sA A


17

0

1k

2k

0

0 t

sin( )y A t

Ex 7.7: Describing Function of Piece-wise Linear Function.

sinA 1sinA

0

2( ) ( sin )sinN A A dA

10

22 1

2 sin sin2

2 sin sin

k A d

A k A k d

1 22

211 2

2

2( ) cos

2( ) sin 1

k k kA

k k kA A A


18

For saturation nonlinearity:

When 1 21 , 0k k :

2

1

1

2 sin 1 , for ( )

1, for

AN A A A A

k A

(D14)

For a dead-zone nonlinearity:

212

2

1

( )

2 sin 1 , for

0, for

N A

k k AA A A

k A

(D15)

0 1 0k

2k

0

1k

2 0k


19

Ex 7.10

2( )0.8 8

sG ss s

2 2

2 2

0.8 (8 )( )0.64 (8 )

jG j

For odd nonlinearities, N(A) is real. Then, (D11) can be satisfied only for

2 2s

and 1 1( ) 0.8

( ) 1.25s

N AG j

If is the saturation nonlinearity with 1 , then, from (D14): 1.455A

If is a dead-zone non linearity with 2 0.5, 1k , then from (D15), no solution is found. This indicates that “probably” no periodic solution. This statement can be proved using the Circle Criterion as follows:

-1.4 -1.2 -1 -0.8 -0.6 -0.4 -0.2 0-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1Nyquist Diagram

Real Axis

Imag

inar

y Ax

is


20

Re ( ) 1.25,G j

The system is absolutely stable for any nonlinearity within the

sector 10, 0.8

1.25

.

As the dead-zone nonlinearity is within this sector, the origin is globally uniformly asymptotically stable. Therefore, no periodic solution exists.


21

Ex 7.11 Consider Raleigh’s equation:

21 0v v v v

2( )1

sG ss s

, 2

2 2 2 2

(1 )( )

(1 )j

G j

3 3

0

22 2

0

2, ( ) ( sin ) sin

3 1 cos(2 )2 4

y y N A A dA

A d A

For odd nonlinearities, N(A) is real. Then, (D11) can be satisfied only for

1s and

1 1( ) 1( ) 1s

N AG j

2 / 3A

Thus, we expect that Raleigh’s equation has a periodic solution of frequency near 1 rad/sec and amplitude of oscillation in v around 2 / 3 .

me677c6 frequencyanalysis t

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