State-Space Formulation



State-Space Formulation

Review of analysis techniques

• Continuous-time domain:

o Classical solution of differential equations

o Impulse response and convolution

o Fourier transform

o Laplace transform

• Discrete-time domain:

o Classical solution of difference equations

o Impulse response and convolution

o Discrete-time Fourier transform

o Z-transform

State-variable characterization of systems

• It is another time-domain analysis technique for linear systems.

• Advantages:

o Describes all signals in system – not just inputs and outputs

o Handles MIMO systems and time-varying systems readily

o Ideal for computer solutions

• Disadvantage:

o Lots of work even for setting up a simple system

o Knowledge of advanced matrix mathematics required

Definitions

• State: A summary of the past information that affects the system’s future behavior.

• State variables: A minimal set of variables needed to determine the future behavior of the system from the system’s inputs.

• State-space formulation: A mathematical description of the relationships of the input, output, and the state of the system.

Continuous-time systems

• Differential equation:

[pic]

Solving for the highest order derivative:

[pic]

• Block diagram:

• Converting an nth order differential equation to a set of n first-order differential equations: Let xi(t) represent the outputs of the integrators. As a result we can choose them as the state variables of the system.

x1(t) = y(t)

x2(t) = y’(t) = x1‘(t)

x3(t) = y”(t) = x2‘(t)

…

xn(t) = y(n-1)(t) = xn-1‘(t)

Rewriting these equations:

x1‘(t) = x2(t)

x2‘(t) = x3(t)

…

xn-1‘(t) = xn(t)

xn‘(t) = y(n)(t) = b0 u(t) – a0 x1(t) – … – an-1xn(t)

• Matrix formulation:

[pic]

[pic]

• Vector notation:

x’(t) = A x(t) + B u(t)

y(t) = C x(t) + D u(t)

where

o x(t) is the state vector

o y(t) is the output

o u(t) is the input

o A is called the state (dynamic) matrix

o B is called the input matrix

o C is called the output (sensor) matrix

o D is called the feed through matrix

Now, an nth order differential equation has been converted to a set of n first-order differential equations.

• For a multi-input/multi-output system:

x’(t) = A x(t) + B u(t)

y(t) = C x(t) + D u(t)

For a system with n state variable, p inputs and q outputs:

A is n ( n

B is n ( p

C is q ( n

D is q ( p

x(t) is n ( 1

y(t) is q ( 1

u(t) is p ( 1

• The transfer function matrix:

Taking the Laplace transform of the state-variable vector equation we obtain:

s X(s) – x(0) = A X(s) + B U(s)

( X(s) = (s I – A)-1 {x(0) + B U(s)}

and Y(s) = C X(s) + D U(s)

( Y(s) = C (s I – A)-1 {x(0) + B U(s)} + D U(s)

Finally, setting x(0) = 0 gives us the transfer function matrix:

Y(s) = C (s I – A)-1 B U(s) + D U(s) = {C (s I – A)-1 B + D} U(s)

or G(s) = C (s I – A)-1 B + D

• Example 1:

y’’’(t) + y”(t) – 2y’(t) + 3y(t) = u(t)

[pic]

• Example 2: Derivatives of u(t) are involved.

y”(t) + 2y’(t) + 3y(t) = 4u”(t) + 5u’(t) + 6u(t)

Step 1: Rearranging terms so that the derivatives are on the LHS

[y”(t) – 4u”(t)] + [2y’(t) – 5u’(t)] = -3y(t) + 6u(t)

Step 2: Construct the main path

Input state variable state variable output

Step 3: Start with the derivative terms and pass them through the first integrator

Step 4: Remove the non-derivative terms

Step 5: Pass the result through the second integrator

Step 6: Remove the input term

Step 7: Use the input u(t) and feedback from x1(t) and x2(t) to construct x2‘(t)

Step 8: Compute the constants

o K2 = 0 since x2‘(t) has no derivative term.

o K1 = -3 in order to match the -3y(t) term in x2‘(t).

o -3y(t) + 6u(t) = -3 [y(t) – 4u(t)] + K u(t) ( K = -6

Step 9: Use u(t) and feedback from x1(t) and x2(t)to construct x1’(t) = y’(t) – 4u’(t)

Step 10: Compute the constants

o K4 = 1

o K3 = -2 in order to match the -2y(t) term.

o - [2y(t) – 5u(t)] = -2 [y(t) – 4u(t)] + K5 u(t) ( K5 = -3

Step 11: Add feedback path to obtain x2‘(t) (from step 8)

Step 12: Add feedback/through paths to obtain x1‘(t) (from step 10)

Step 13: Add feed through path to obtain y(t) (from step 6)

Step 14: Formulate the state-space equations

x1‘(t) = -2 x1(t) + x2(t) – 3 u(t)

x2‘(t) = -3 x1(t) – 6 u(t)

y(t) = x1(t) + 4 u(t)

[pic]

• Example 3: LRC circuit

[pic]

State variables:

x1(t) – capacitor voltage x2(t) – inductor current

Node voltage at 1: [x1(t) – u(t)] / 6 + x2(t) + x1‘(t) / 30 = 0, or

x1’(t) = -5x1(t) – 30x2(t) + 5u(t)

Inductor I-V equation: 5x2’(t) = x1(t) – u(t), or

x2’(t) = [x1(t) – u(t)] / 5

Rewriting the equations we have:

x1’(t) = (-5)x1(t) + (-30)x2(t) + (5)u(t)

x2’(t) = (1/5)x1(t) + (0) x2(t) + (-1/5)u(t)

y(t) = (1)x1(t) + (0) x2(t) + (0) u(t)

Matrix notation:

[pic]

Block diagram:

• Example 4: (text, p. 26)

• Op Amp implementation:

o Inverter:

[pic]

o Adder:

[pic]

o Integrator:

[pic]

o Example:

[pic]

Rearranging the terms we have:

[pic]

• Constructing y:

[pic]

• Constructing x1:

[pic]

• Constructing x2:

[pic]

• Connecting the stages:

[pic]

• Linearization: Nonlinear systems can be linearized by ignoring the higher order terms in the Taylor series (Text P. 18).

Discrete-time systems

• Difference equation:

y(k+n) + an-1 y(k+n-1) + … + a1 y(k+1) + a0 y(k) = b0 u(k)

Solving for the highest order delay:

y(k+n) = -an-1 y(k+n-1) – … – a0 y(k)+ b0 u(k)

• Block diagram:

+

• Converting an nth order difference equation to a set of n first-order difference equations: Let xi(k) represent the outputs of the unit delay. As a result we can choose them as the state variables of the system.

x1(k) = y(k)

x2(k) = y(k+1) = x1(k+1)

x3(k) = y(k+2) = x2(k+1)

…

…

…

xn-1(k) = y(k+n-2) = xn-2(k+1)

xn(k) = y(k+n-1) = xn-1(k+1)

Rewriting these equations:

x1(k+1) = x2(k)

x2(k+1) = x3(k)

…

…

…

xn-1(k) = xn(k)

xn(k) = y(k+n) = b0 u(k) – a0 x1(k) – … – an-1xn(k)

• Matrix formulation:

[pic]

[pic]

• Vector notation:

x(k+1) = A x(k) + B u(k)

y(k) = C x(k) + D u(k)

where

o x(k) is the state vector

o y(k) is the output

o u(k) is the input

o A is called the state (dynamic) matrix

o B is called the input matrix

o C is called the output (sensor) matrix

o D is called the feed through matrix

Now, an nth order difference equation has been converted to a set of n first-order difference equations.

• For a multi-input/multi-output system:

x(k+1) = A x(k) + B u(k)

y(k) = C x(k) + D u(k)

For a system with n state variable, p inputs and q outputs:

A is n ( n x(k) is n ( 1

B is n ( p y(k) is q ( 1

C is q ( n u(k) is p ( 1

D is q ( p

• The transfer function matrix:

Taking the z-transform of the state-variable vector equation we obtain:

G(z) = C (z I – A)-1 B + D

• Example:

y(k+2)+ 0.5 y(k+1) – 0.3 y(k) = 0.4 u(k)

[pic]

MIMO Example

For a multi-input/multi-output system:

x’(t) = A x(t) + B u(t)

y(t) = C x(t) + D u(t)

Example:

p’’’ + 2 p’ + 3 q – 4 r’ = 5 α(t)

p + 8 q’ +9 q = 10 α(t) – 11 β(t)

p’’ – 6 r’’ + 7 r = 0

Let: x1 = p x2 = p’ x3 = p’’

x4 = q x5 = r x6 = r’

u1 = α(t) u2 = β(t)

Substituting and rearranging the equations:

x3’ = -2 x2 –3 x4 + 4 x6 + 5 u1

x4’ = -(1/8) x1 – (9/8) x4 + (10/8) u1 – (11/8) u2

x6’ = (1/6) x3 + (7/6) x5

In matrix notation:

[pic]

-----------------------

(

(

(

(

y(t)

b0

an-1

an-2

a0

y’(t)

y(n-1)(t)

y(n-2)(t)

y(n)(t)

u(t)

y(t)

x1(t)

-1

2

-3

(

(

(

(

y”(t)

x3(t)

y’’’(t)

(

y’(t)

x2(t)

1

u(t)

x2‘(t) x2(t) x1‘(t) x1(t) y(t)

-3

-3

-2

-3

(

(

(

-6

(

u(t)

(

x2‘(t)= [y”(t) – 4u”(t)]

+ [2y’(t) – 5u’(t)]

x2(t) = [y’(t) – 4u’(t)]

+ [2y(t) – 5u(t)]

x2(t) = [y’(t) – 4u’(t)]

+ [2y(t) – 5u(t)]

K1

u(t)

(

-6

(

-[2y(t) – 5u(t)]

x1‘(t) = y’(t) – 4u’(t)

x1‘(t)= y’(t) – 4u’(t)

x1(t) = y(t) – 4u(t)

(

(

x1(t) = y(t) – 4u(t)

y(t)

4u(t)

(

x2‘(t) = -3y(t) + 6u(t)

(

x1(t) = y(t) – 4u(t)

(

K2

x2(t) = [y’(t) – 4u’(t)]

+ [2y(t) – 5u(t)]

u(t)

K

K3

x1(t) = y(t) – 4u(t)

-3

(

x2‘(t) x2(t) x1‘(t) x1(t) y(t)

(

y’(t) – 4u’(t)

K5

u(t)

(

(

u(t)

x2‘(t) x2(t) x1‘(t) x1(t) y(t)

-6

x2‘(t) x2(t) x1‘(t) x1(t) y(t)

(

(

(

(

u(t)

K

(

(

(

(

(

-2

-3

4

x2(t) = [y’(t) – 4u’(t)]

+ [2y(t) – 5u(t)]

K4

note: this is the feed through term.

note: we will obtain this term in step 9.

5

x2‘(t) x2(t) x1‘(t) x1(t) = y(t)

-1/5

(

(

(

an-2

u(t)

(

-5

1/5

-30

a0

(

y(k+1)

b0

u(k)

z-1

y(k+n-1)

y(k+n)

y(k+n-2)

z-1

z-1

y(k)

an-1

+

y(k+2)

y(k)

x1(k)

y(k+1)

x2(k)

z-1

z-1

0.4

(

u(k)

-0.5

0.3

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download