State equationlecture

ME 381

Mechanical and Aerospace Control Systems

Dr. Robert G. Landers

State Equation Solution

Unforced Response 2State Equation Solution Dr. Robert G. Landers

( ) ( )t t=x Ax

( ) ( )0tt e= Ax x

( )tt

d ee

dt=

AAA

( ) ( ) ( ) ( )0 0t tt t e e= ⇒ =A Ax Ax A x A x

( ) ( ) ( )0t t=x Φ x

The matrix Φ(t) is called the state transition matrix. The state transition matrix transforms the initial conditions.

The state equation for an unforced dynamic system is

Assume the solution is

The derivative of eAt with respect to time is

Checking the solution

Letting Φ(t) = eAt, the solution is written as

Properties of State Transition Matrices 3

( ) 00 e= =AΦ I

( ) ( ) ( ) ( ) ( )1 1 1t tt e e t t t− − −−= = = − → = −⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦

A AΦ Φ Φ Φ

( ) ( ) ( ) ( ) ( ) ( )1 2 1 21 2 1 2 2 1

t t t tt t e e e t t t t++ = = = =A A AΦ Φ Φ Φ Φ

( ) [ ] ( )n tt t t n tt e e e e e nt+ + += = = =⎡ ⎤⎣ ⎦A A AA A A AΦ Φ

State Equation Solution Dr. Robert G. Landers

State transition matrix evaluated at t = 0

Inverse of the state transition matrix

State transition matrix evaluated at t1 + t2

State transition matrix raised to a power

Forced Response 4

( ) ( ) ( )t t t= +x Ax Bu

( ) ( ) ( )t t t− =x Ax Bu

( ) ( ) ( ) ( )t t tde t t e t e tdt

− − −⎡ ⎤− = =⎡ ⎤⎣ ⎦ ⎣ ⎦A A Ax Ax x Bu

( ) ( )t

te e d

λλ λ

λ ττ

λ λ λ=− −

=⎡ ⎤ =⎣ ⎦ ∫A Ax Bu


The state equation for a forced dynamic system is

Rearranging

Multiply by e–At

Integrating from τ to t

Forced Response 5

( ) ( ) ( ) ( ) ( )free response

forced response

tt tt e e dτ λ

τ

τ λ λ− −= + ∫A Ax x Bu

( ) ( ) ( ) ( ) ( ) ( )t

t tt t e e dτ λ

τ

τ λ λ− −= = + ∫A Ay Cx C x C Bu

If A = A(t), B = B(t), and C = C(t), these results do not hold and the state transition matrix must, in general, be calculated numerically.


Solving for x(t)

Solving for y(t)

( ) ( ) ( )t

te t e e dτ λ

τ

τ λ λ− − −− = ∫A A Ax x BuEvaluating the left hand side

Example 1 6The dynamic equation and schematic of an RL circuit are shown below. Calculate the current if the voltage source is a constant V and the initial current in the circuit is I0.

( ) ( ) ( )1RI t I t V tL L

= − +


Example 1 7

( ) ( ) ( ) ( ) 1tR Rt tL LI t e I e Vd

Lτ λ

τ

τ λ− − − −

= + ∫

( ) ( )0

0

tR Rt tL LL VI t I e e

R L

λλ

λ

=− − −

=

⎡ ⎤= + ⎢ ⎥

⎣ ⎦

( ) 0 1R Rt tL LVI t I e e

R− −⎡ ⎤

= + −⎢ ⎥⎣ ⎦


Letting τ = 0 and evaluating the integral

Evaluating the limits of the integral

The state matrices are A = –R/L and B = 1/L. The state I(t) is

Laplace Transform 8( ) ( ) ( )

( ) ( ) ( ) ( )0

t t t

s s s s−

= +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦⇓

− = +

x Ax Bu

X x AX BU

L L

( ) ( ) ( ) ( )0s s s−− = +I A X x BU

( ) ( ) ( ) ( ) ( )1 10s s s s− −−= − + −X I A x I A BU


Taking the Laplace transform of the state equation

Rearranging

Solving for X(s)

φ = (sI–A)–1 is known as the resolvant. The state transition matrix may be found by taking the inverse Laplace Transform of φ element by element.

The Laplace transform of eAt is ( ) ( )1 1at te s a e s− −⎡ ⎤ ⎡ ⎤= − → = −⎣ ⎦ ⎣ ⎦

A I AL L

Example 2 9

Determine the state transition matrix for the matrix A by using Laplace Transform techniques

0 16 5

⎡ ⎤= ⎢ ⎥− −⎣ ⎦

A


Example 2 10

Evaluating sI–A1

6 5s

ss−⎡ ⎤

− = ⎢ ⎥+⎣ ⎦I A

( )2 21

2 2

5 15 6 5 66

5 6 5 6

ss s s ss

ss s s s

−

+⎡ ⎤⎢ ⎥+ + + +− = ⎢ ⎥

−⎢ ⎥⎢ ⎥+ + + +⎣ ⎦

I A


Taking the inverse of sI–A

Example 2 11

( ){ }2 3 2 3

112 3 2 3

3 26 6 2 3

t t t tt

t t t t

e e e ee s

e e e e

− − − −−−

− − − −

⎡ ⎤− −= − = ⎢ ⎥− + − +⎣ ⎦

A I AL

( ) 1

3 2 1 12 3 2 3

6 6 2 32 3 2 3

s s s ss

s s s s

−

⎡ ⎤− −⎢ ⎥+ + + +− = ⎢ ⎥⎢ ⎥− + − +⎢ ⎥+ + + +⎣ ⎦

I A


Applying partial fraction expansion

The state transition matrix is found by taking the inverse Laplace transform of the inverse of sI–A

Transfer Function 12

( )0− =x 0

( ) ( ) ( )1s s s−= −X I A BU

( ) ( ) ( ) ( ) ( ) ( )t t t s s s= + → = +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦y Cx Du Y CX DUL L

( ) ( ) ( )1s s s−⎡ ⎤= − +⎣ ⎦Y C I A B D U


Let the state initial conditions be zero

Taking the Laplace transform of the state equation and solving for X(s)

Taking the Laplace transform of the output equation

Solving for Y(s)

Transfer Function 13

The transfer function matrix for a general multiple input, multiple output system having p outputs and q inputs is

( )

( ) ( )( ) ( )

( ) ( )

11 1

21 1

1

q

q

p pq

H s H sH s H s

s

H s H s

⎡ ⎤⎢ ⎥⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

H


( ) ( ) 1s s −= − +H C I A B DThe transfer function matrix is

Transformation of State Variables 14

( ) ( ) ( ) ( )t t t t= ⇒ =z Tx z Tx

Sometimes it is desirable to redefine the states for greater physical understanding. A linear transformation between states is given by

Assuming T is invertible and constant, x(t) = T–1z(t). Recall the general state space representation is

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

( ) ( ) ( ) ( ) ( ) ( )1 1 1t t t t t t− − −= + = +T z AT z Bu y CT z Du

( ) ( ) ( ) ( ) ( ) ( )1 1t t t t t t− −= + = +z TAT z TBu y CT z Du


Substituting for x(t) and its derivative with respect to time

Left multiplying the state equation by T


The new system is also linear

A_bar is similar to A and A_bar = TAT–1 is a similarity transform

The transfer function remains the same

1 1− −= = = =A TAT B TB C CT D D

( ) ( ) 11s s−−− + = − +C I A B D C I A B D


( ) ( ) ( ) ( ) ( ) ( )t t t t t t= + = +z Az Bu y Cz Du

The transformed state space representation is

where


( )1 1 1 1s s s− − − −− = − = −I TAT TT TAT T I A T

The characteristic polynomial does not change

Determinate of the product of matrices is the product of the determinates of the individual matrices. Therefore

( ) 1 1 1s s s s− − −− = − = − = −T I A T T I A T T T I A I A

1s s−− = −I TAT I A

Therefore, the characteristic polynomial does not change and the eigenvalues remain the same.



The eigenvectors of A and A_bar are related via the transformation matrix

1−= → =Tv v v T v

1s s−= → =Av v TAT v v

1s s s s−= ⇒ = = ⇒ =TAv v Av T v v Av v


Using the relationship between v and v_bar

Therefore, if v is an eigenvector of A, then Tv is an eigenvector of A_bar

If v_bar is an eigenvector of A_bar

Example 3 18

For the system below, calculate the output when the system is unforced and the initial conditions are x1(0) = 2 and x2(0) = 0, when the system is subjected to a unit impulse and the initial conditions are zero, when the system is subjected to a unit step and the initial conditions are zero, and when the system is subjected to a step with magnitude 0.5 and the initial conditions are x1(0) = 2 and x2(0) = 0. For each case, plot the response and compare to the Matlab solution.

[ ]0 1 01 0 0

6 5 1⎡ ⎤ ⎡ ⎤

= = = =⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎣ ⎦A B C D


Example 3 19

2 3 2 33 2t t t t te e e e e− − − −⎡ ⎤= − −⎣ ⎦AC

( ) ( ) ( ) ( )0

0t

tty t e e u dλ λ λ−= + ∫ AAC x C BThe general output response is

( ) ( )2 310 3 2 0t t te e e x− −⎡ ⎤= −⎣ ⎦

AC xThe unforced output is


The term CeAt is

The state transition matrix is2 3 2 3

2 3 2 3

3 26 6 2 3

t t t tt

t t t t

e e e ee

e e e e

− − − −

− − − −

⎡ ⎤− −= ⎢ ⎥− + − +⎣ ⎦

A

Example 3 20

For a unit impulse, u(λ) = 0 except at λ = 0 and the magnitude of u(0) is such that the integral of u(λ) is one.

( ) ( ) ( )0

2 3

0 0

tt t t te u d e u d e eλ λ λ λ λ− − −= = −∫ ∫A AC B C B

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 2 3 2 303 2

1t t t t t t te e e e e e eλ λ λ λ λ λ λ− − − − − − − − − − − − −⎡ ⎤⎡ ⎤= − − = −⎢ ⎥⎣ ⎦ ⎣ ⎦

AC B

The output response to a unit impulse with zero initial conditions is

( ) ( ) ( ) 2 3

0

tt t ty t e u d e eλ λ λ− − −= = −∫ AC B


The term CeA(t–λ)B is

Example 3 21

( ) ( )2 3 2 31

1 1 13 2 0 0.56 2 3

t t t ty t e e x e e− − − −⎡ ⎤⎡ ⎤= − + − +⎣ ⎦ ⎢ ⎥⎣ ⎦

The output response to a step input with magnitude M = 0.5 is

( ) ( ) ( ) ( ) ( ) ( ) ( )2 3 2 3

00 0

2 3 2 3

1 12 3

1 1 1 1 1 1 12 3 2 3 6 2 3

tt tt t t t t

t t t t

e u d e e M d M e e

M e e M e e

λ λ λ λ λλ λ λ− − − − − − − − −

− − − −

⎡ ⎤⎡ ⎤= − = −⎢ ⎥⎣ ⎦ ⎣ ⎦

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎡ ⎤= − − − = − +⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥⎝ ⎠ ⎝ ⎠ ⎣ ⎦⎣ ⎦

∫ ∫AC B

The analytical solutions are now compared to the Matlab numerical solutions.


When the input is a step with a magnitude M, the forced output response is

State equationlecture

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