Preliminaries-Signals and Systems-Lecture 39 Slides-Electrical and Computer Engineering, Slides for Signals and Systems Theory
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Preliminaries-Signals and Systems-Lecture 39 Slides-Electrical and Computer Engineering, Slides for Signals and Systems Theory

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Preliminaries, Forced Equation, Output Equation, Solution Via the Laplace Transform, Discrete-time Systems, Solutions Using the Z-transform, Joseph Picone, Signals and Systems, Electrical and Computer Engineering, Missis...
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PowerPoint Presentation

ECE 8443 – Pattern RecognitionECE 3163 – Signals and Systems

Objectives: Useful Matrix Properties

Time-Domain Solutions

Relationship to Convolution

Laplace Transform Solution

Resources: SHS: State Equation Solutions

MIT: State Equation Solutions

LECTURE 39: SOLUTIONS OF THE STATE EQUATIONS

Audio:URL:

ECE 3163: Lecture 39, Slide 1

Recall our state equations:

To solve these equations, we will need a few mathematical tools. First:

where I is an NxN identity matrix. Ak is simply AxAx…A.

For any real numbers t and :

Further, setting  = -t:

Next:

We can use these results to show that the solution to

is:

Preliminaries

)()()(

)()()(

ttxt

ttt

DvCy

BvAxx





 !3!2

3322 tt te t

AA AI

A

 AAA eee tt  )(

I AAA   ttt eee )( 

t

t

e tt

t

t t

tt t

dt

d e

dt

d

A

A

A AA

AIA

A AA

AA AI

 

  

 

 

  

 



!3!2

!2!3!2

3322

23 2

3322

)()( tt Axx 

0),0()(  tet txx A

ECE 3163: Lecture 39, Slide 2

If:

is referred to as the state-transition matrix.

We can apply these results to the state equations:

Note that:

Integrating both sides:

Solutions to the Forced Equation

teA

      )()0()0()0()( tee dt

d e

dt

d t

dt

d ttt AxxAxxx

AAA 

0),0()(  tet txx A

  )()()(

)()()(

)()()(

tette

ttt

ttt

tt BvAxx

BvAxx

BvAxx

AA  





   

  )()(

)()()()()()()(

tete dt

d

ttetetete dt

d tete

dt

d

tt

tttttt

Bvx

AxxxAxxxx

AA

AAAAAA





 

  

  

  0,)()0()(

)()0()(

0

0









tdeet

dete

t

tt

t

t





Bvxx

Bvxx

AA

AA Generalization of our

convolution integral

ECE 3163: Lecture 39, Slide 3

Recall:

Using the definition of the unit impulse:

Recall our convolution integral for a single-input single-output system:

Equating terms:

Solution to the Output Equation

 

  0),()()0(

0),()()0()()()(

0

0



 

  

 





ttdee

ttdeettxt

t

tt

t

tt

DvBvCxC

DvBvxCDvCy

AA

AA





 

    

 

0,)()()0()( 0

   tdteet

t

t

t

t

t

zs

zi    

y

A

y

A vDBvCxCy 

    

 teth

dvthdvtve

t

t t

t



DBC

DBC

A

A



  

)(

)()()()( 0 0

  0,)()()(*)( 0

  tdvthtvthty t

zs 

The impulse response can be computed

directly from the coefficient matrices.

ECE 3163: Lecture 39, Slide 4

Solution via the Laplace Transform

Recall our state equations:

Using the Laplace transform on the first equation:

Comparing this to:

reveals that:

Continuing with the output equation:

For zero initial conditions:

)()()(

)()()(

ttxt

ttt

DvCy

BvAxx





 

    )()0()(

)()0()(

)()()0()(

11 ssss

sss

ssss

BVAIxAIX

BVxXAI

BVAXxX

 





  0,)()0()( 0

   tdeet

t

tt  Bvxx AA

  11   AIA se t L

     )()0( )()()(

)()()(

11 ssxs

sss

ttxt

VDBAICAI

DVCXY

DvCy









  DBAICHXHY  1)(where)()()( sssss

The transfer function

can be computed

directly from the

system parameters.

ECE 3163: Lecture 39, Slide 5

The discrete-time equivalents of the state equations are:

The process is completely analogous to CT systems, with one exception:

these equations are easy to implement and solve using difference equations

For example, note that if v[n] = 0 for n ≥ 0,

and An is the state-transition matrix for the discrete-time case.

The output equation can be derived following the procedure for the CT case:

The impulse response is given by:

Discrete-Time Systems

           nnxn

nnn

DvCy

BvAxx



1

      1,0 1

0

1   

 nin n

i

inn BvAxAx

   0xAx nn

      1,][0

][

1

0

1

][

  

 nnin

n

n

i

in

n

n

zs

zi    

y

y

DvBvCAxCAy

    

 

 1,

0, 1 n

n nh

n BCA

D

ECE 3163: Lecture 39, Slide 6

Just as we solved the CT equations with the Laplace transform, we can solve

the DT case with the z-transform:

Note that historically the CT state equations were solved numerically using a

discrete-time approximation for the derivatives (see Section 11.6), and this

provided a bridge between the CT and DT solutions.

Example:

Solutions Using the z-Transform

     

   

     

       

  

)(

)(

11

11

)(]0[)(

)(]0[)(

)()(]0[)(

1

z

z

zs

zzzzz

nnxn

zzzzz

zzzzz

nnn

Y

H

VDBAICxAICY

DvCy

BVAIxAIX

BVAXxX

BvAxx















][][][][

][][

][][]1[

][][]1[

][][][]1[

2312

21

323

212

2121

nvnxnxny

nxny

nvnxnx

nvnxnx

nvnvnxnx









ECE 3163: Lecture 39, Slide 7

Summary

Introduced some useful properties of matrices that allowed us to solve the

state equations.

Demonstrated a solution to the forced equation.

Discussed how this is a generalization of the single-input single-output

analysis approach previously studied (e.g., convolution).

Applied the Laplace transform to solve the state equations using algebra.

Derived an expression for the transfer function.

Extended the state equations to the discrete-time case.

Demonstrated a DT signal flow graph implementation of the DT solution.

Next: We will work some examples involving circuit analysis using the state

variable approach.

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