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Jack H.Dynamic system modeling and control.2004.pdf

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state space analysis - 20.1

20. STATE SPACE ANALYSIS

Topics:

Objectives:

20.1INTRODUCTION

-state equations can be converted to transfer functions. The derivation follows.

state space analysis - 20.2

State equations as functions of time

·
x = Ax + Bu
y = Cx + Du
In the s-domain
sX – X0 = AX + BU
	X( sI – A) = BU + X0
	X = ( sI – A) –1BU + ( sI – A) –1X0
Y = CX + DU
	Y = C( ( sI – A) –1BU + ( sI – A) –1X0) + DU
	Y = ( C( sI – A) –1B + D) U + C( sI – A) –1X0
Assuming the system starts at rest,
Y = ( C( sI – A) –1B + D) U
Y	= ( C( sI – A)	–1	B + D)	(the transfer function)
---	= ( C( sI – A)		B + D)	(the transfer function)
U

- state equation coefficient matrices can be transformed to another equivalent for, if the state vector is rearranged.

state space analysis - 20.3

State equations as functions of time

x = Ax + Bu

y = Cx + Du

Map one state vector to another one

x = Tz

This can be used to calculate new coefficient matrices

–1

( Ax + Bu)

z =

–1

( ATz + Bu)

z = T

–1

ATz + T

–1

z = T

–1

ATz + T

A = T

z = T

z = Az + Bu

The output equation becomes

y = Cx + Du

y = CTz + Du

C = CT

y = Cz + Du

The equivalent set of state equations is,

–1

A =

z = Az + Bu

y = Cz + Du

C = CT

•

B = T–1B

D = D

B = T–1B

D = D

state space analysis - 20.4

The two following transfer functions are equivalents

G = C( sI – A) –1B + D

This can be shown by applying the transformation matrix

G = CTT–1( sI – A) –1TT–1B + TT–1D

G = ( CT) ( T–1( sI – A) –1T) ( T–1B) + D

G = ( CT) ( ( sI – A) –1) ( T–1B) + D

G = C( ( sI – A) –1) B + D = G

• The transfer function form can be put into a matrix form. In this case the denominator is the characteristic equation.

The transfer function can be said to be equivalant to the determinants of the matrix form.

( sI – A )

–B

(---------------------------------------------sI – A) D – ( –B) C

= ( sI – A) D + BC( sI – A)

sI – A

( sI – A)

–B

G = C ( sI – A) –1B + D = -----------------------------------

= poles-------------

sI – A

zeros

sI – A

characteristic equation = homogeneous

• The free (homogeneous) response of a system can be used to find the state transition matrix.

state space analysis - 20.5

The homogeneous equation can be written in the s-domain, and then converted to time.

Xh = sI – A X0

xh( t) = L–1[ sI – A X0]

( t)

= L

–1

+ …

+ --

+ -----

-----

xh( t) = eAtx0

eAt = transition matrix

aside: This expansion is a McLaurin (Taylor) series.

2 2

3 3

e = I + At +

----

A t + …

A t +

• The forced response (particular) response of the system can be found using convolution,

The homogeneous equation can be written in the s-domain, and then converted to time.

·
x = Ax + Bu
x( t) = eAtx( 0) + ∫t		eA( t – τ ) Bu( τ ) dτ
0
y = Cx + Du
y( t) = CeAtx( 0) + ∫t			CeA( t – τ ) Bu( τ ) dτ + Du( t)
		0

	initial		impulse
	response		response

• As an example the homogeneous/free response of the system is shown below.

state space analysis - 20.6

Given a second order differential equation, the state matrix can be found,
··
F = Mx
·	= v
·
x				d	x	=		0 1				x	+		0	F								0 1
·	F		----		x	=		0 1				x	+		1	F			A =					0 1
·	F		dt		v			0	0			v			1				A =					0 0
v	= ----				v			0	0			v			----									0 0
	M														M


This can be used to find the inverse matrix,
																	s 1			s		1			1		1
						–1								–1			0 s		----			----			--		----
						–1								–1					----			----			--		----
		1 0		0	1						s –1								s		2	s	2			s		2
( sI – A) –1 = s			–				=								=		-------------	=						=			s
( sI – A) –1 = s			–				=								=		-------------	=						=			s
		0 1		0 0						0 s						s2 – 0				0 s							1
																			----			----			0		--
																			----			----			0		--
																			s2			s2					s

The function of time can be found assuming an initial position of 10 and velocity of 5.


eAt = L–1[ ( sI – A) –1] =						1 t
						0 1

xh( t) =	x	= e	At	10	=		1	t	10	=	10 + 5t
xh( t) =	v	= e		5	=		0	1	5	=	5
	v			5			0	1	5		5

• The forced/particular solution is shown below,

Given a second order differential equation, the state matrix can be found,

··

F = Mx

• If a matrix is diagonizable, the diagonal matrix can be found with the following technique. This can be used for more advanced analysis techniques to create diagonal (and

state space analysis - 20.7

separable) system matrices.

Given a matrix A, a purely diagonal matrix may be found,

T–1AT = Λ

A = TΛ T–1

Λ =

λ1

…

… …

λn

This may also be written as,

eAt = TeΛ tT–1

eΛ t =

eλ1t …

…

This can also be applied to the

eλnt

e = I + At +

----

2!A t + …

–1 2 2

–1

e = I + TΛ T t +

----

( T

Λ T

) t + …

Eigenvalues can be found using the following relationship. If any of the Eigenvalues are repeated the Jordan normal form is required.

λI – A	= 0	Note: the Eigenvalues are the values


		found in the characteristic/homoge-
		neous equation solution.

The Eigenvalues are then used to find the Eigenvectors,

( λI – A) v = 0

Av = λv

The Eigenvectors are then combined into a single matrix, this is the transition matrix.

• example,

state space analysis - 20.8

Given the state matrix

A = –1 2 –3 4

The eigenvalues can be calculated,

λI – A

1 0

–

–1 2

λ + 1 –2

0 1

–3 4

3 λ – 4

= λ( + 1) (λ – 4) – ( –6) = λ2 – 3λ + 2 = ( λ

Therefore the Eigenvalues are,

λ1 = 2

An eigenvector can be calculated using the Eigenvalue 2.

–1 2

u11

= λ

u11

= 2

u11

–3 4

u21

2u21 = 3u11

– u11 + 2u21 = 2u11

Another eigenvector can be calculated using the Eigenvalue 1.

–1 2

u12

= λ

u12

= 1

u12

–3 4

u22

u22 = u12

– u12 + 2u22 = u12

The Eigenvectors can be combined to a single matrix.

T =

2 1

v1 v2

3 1

–2) (λ – 1)

λ2 = 1


		u11
v1	=	u11	=	2
v1	=	u21	=	3
		u21


		u12
v2	=	u12	=	1
v2	=	u22	=	1
				1

To check,				Avi = λivi
				Avi = λivi

A		u11		, A		u12			= 2			2		,	1	1
A		u21		, A		u22			= 2			3		,	1	1
												3				1


	–1 2			2	,		–1 2				1		=			4	,	1
	–3 4			3			–3 4				1					6		1

	4	,	1	=			4	,		1									verified
	6		1				6			1

• If there are repeated Eigenvalues in the system the Jordan Form can be used.

state space analysis - 20.9


	λ1		1	0	…	0	0
		0	λ2	1	…	0	0
J =		0	0	λ3	…	0	0	Jordan block
	…	…	… …	…	…
		0	0	0	… λ	n – 1	1
		0	0	0	…	0	λn

Consider the example below with repeated eigenvalues.

–2

Find a matrix, that has repeated eigenvalues, and is not singular

A =

–2

λI – A

λ – 2

–1

= ( λ + 3) 2( λ – 0)

–1 λ – 2 –1

–1

λ – 2

state space analysis - 20.10

Solve for the Jordan block one column at a time,

T11

–2T11 + T21 + T31

–2

T21

–2

T21

T11 – 2T21 + T31

FIX

T31

–2

T31

T11 + T21 – 2T31

T11

= 2T11 + T21 + T31

T21

= T11 + 2T21 + T31

T21 =

T31

= T11 + T21 + 2T31

T31

T12

–2T12 + T22 + T32

–2

T22

–2 1

T22

T12 – 2T22 + T32

T32

–2

T32

T12 + T22 – 2T32

T12

= 2T12 + T22 + T32

T22

= T12 + 2T22 + T32

T32

= T12 + T22 + 2T32

T13

–2T13 + T23 + T33

–2

T23

–2 1

T23

T13 – 2T23 + T33

T33

–2

T33

T13 + T23 – 2T33

4T13 = 2T13 + T23 + T33

4T23 = T13 + 2T23 + T33

4T33 = T13 + T23 + 2T33

state space analysis - 20.11

• The Eigenvectors can be used to calculate the system response.

Given the following Eigen values and vectors the free


λ1 = 2	λ2 = 1	v1	=	2
λ1 = 2	λ2 = 1	v1	=	3
				3

response is,


v2	=	1
v2	=	1
		1

2e2t + et

( t) = ∑ vieλit =

e2t +

e1t =

i = 1

+ e

• zeros of state space functions can be found using the state matrices.

state space analysis - 20.12

The following relationship locates the zero frequencies in state equations.

s0I – A –B

= 0

Consider the example beginning with state equations,

x = v – 5F

v = – 3x – 4v + 5F

–5

----

x =

1 0

+ 0F

–3 –4

A =

B =

–5

C =

D = 0

1 0

–3 –4

This can be put into the matrix form,

1 0

–

–5

0 1

–3

–4

= 0

1 0

–1

3 s0 + 4 –5

= 0

0( 0)

– ( –1) ( –4.5) + 5( –( s0 + 4) ) = 0

–24.5 = 5s0 s0 = –4.9

There is one zero at -4.9 rad/sec. The order of the polynomial determines the number of zeros. If the polynomial has no ’s’ terms, then there are no zeros.

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