Lecture 3

17
2/5/2015 1 - Systems Described by Ordinary Differential Equations (ODEs) - Laplace Transform - Inverse Laplace Transform. Partial Fraction Expansion - (Last revised on 2/5/2015) EE110, S15, Circuits & Systems: Lec03 Prof. Ping Hsu 1 Systems Described by Ordinary Differential Equations Systems described by algebraic equations do not involve time. Such a system’s output depends on only its input at the instant of time. 2 © 2013 National Technology and Science Press. All rights reserved. × 3 x y=3x 2 4 V out V in A dynamic system’s output at a certain instant of time depends on the entire history of the input. In other words, such a system has some ‘memory effect’. Systems described by differential equations are dynamic systems. y(t) x(t) Dynamical system x(t) y(t)

description

EE 110

Transcript of Lecture 3

Page 1: Lecture 3

2/5/2015

1

- Systems Described by Ordinary Differential

Equations (ODEs)

- Laplace Transform

- Inverse Laplace Transform. Partial Fraction Expansion

- (Last revised on 2/5/2015)

EE110, S15, Circuits & Systems: Lec03

Prof. Ping Hsu

1

Systems Described by Ordinary Differential Equations

Systems described by algebraic equations do not involve time. Such a system’s output depends on only its input at the instant of time.

2

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× 3x y=3x 2

4

VoutVin

A dynamic system’s output at a certain instant of time depends on the entire historyof the input. In other words, such a system has some ‘memory effect’. Systems described by differential equations are dynamic systems.

y(t)

x(t)

Dynamicalsystem

x(t) y(t)

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Systems Described by Ordinary Differential Equations

- 1st order ODE with constant coefficients- The system is called a first order system

(only 1 reactive element, C or L).

Example: A simple RC circuit:

The output and input signals are related by the simple differential equation (EE98):

(From KVL)

- Note that the resistor R and the voltage source x(t) could be the Thevenin’sequivalent of more elaborate resistive circuit connected across the capacitor C.

3

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( ) 1 1( ) ( )

dy ty t x t

dt RC RC

Systems Described by Ordinary Differential Equations4

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For a given input signal x(t), the output signal y(t) can be obtained by solving the first-order differential equation above.

( ) 1 1( ) ( )

dy ty t x t

dt RC RC

systemx(t) y(t)

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Given y(0)=0 and x(t)=1 (or unit step u(t)),

Find y(t). Vs

y(t)

+

-

5

/

where is the time constant.

Solution: ( ) (1 ) ( )

ts

RC

y t V e u t

=1=2

=5

=3=4

( ) 1 1( ) ( )

dy ty t x t

dt RC RC The circuit equation:

SECOND-ORDER SYSTEMS (CIRCUITS)

vS vR vC vL 0 (KVL)

Ld 2i

dt 2 Rdi

dt

i

C

dvS

dt

Two reactive elements

22

2

usually written in the normalized form

( ) ( ) 2 ( ) 0

1 undamped natural frequency

= damping ratio2

n n

n

d x t dx tx t

dt dt

LC

R C

L

1 21 2

2 2

1 2

The Characteristic Equation:

2 0;

Two roots, and (natural frequencies),

can be complex. The complementary

solution (i.e., with vs=0), is

( ) , s t s tc

n n

x t K

s

K

s

e

s

s

e

Adapted from Irwin & Nelms, 20116

Two initial conditions are required to determine K1 and K2. In a series RLCcircuit, for example, they may be vC(0-) and iL(0-).

For f(t)=Au(t), the total solution is

1 21 22

( ) s t s t

n

Ax t K e K e

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Linear ODE can be solved by using Laplace Transform

Solve ODE

x(t) y(t)Given x(t), find y(t)

Time-Domain (t-domain) approach

x(t) y(t)

Complex Frequency Domain (s-domain) approach

Laplacetransform

Inverse transform

Solve Algeb-raic Eq’n

Don’tSolve ODE

X(s) Y(s)

t-domain

s-domain

algebraic eq’nsare easier to solve than ODEs

a transform is like change of variable

7

Definition of the Laplace Transform

0 σ

ω

Re{s}

Im{s} s

s-plane

X(s) x(t)e st

0

dt, s j ,

's ' is a complex number called

the "complex frequency"

( , ) form the 2 orthogonal axes of a

plane called the "s-plane"

Any given 's ' defines a vector in the s-plane:

Re{s}, Im{s}

are its (x, y) coordinates

0 integral limit allows including initial conditions

Called 1-sided or unilateral Laplace Transform (t ≥ 0-)

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The Inverse Laplace Transform

- Given an X(s), one is able to recover exactly the same x(t) that was used to compute X(s) with the help of the inverse transform

- {x(t), X(s)} are said to from a unique transform pair.

- Integral evaluation requires knowledge of complex variables theory, avoided here by using table look-up approach instead.

- Notation:

x(t) 1

2 jX(s)est

j

j

ds

9

-1, ( ) ( ), ( ) ( )x t X s X s x t x t X s

Example Transform Pairs: Unit Step Function

x1(t) Au(t T )

0 T

A

t

x1(t) Au(t T ), T 0 is a constant, X1(s) ?

0

0

( ) sTAAu t T e

s T 0

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Example Transform Pairs: Rectangular Pulse

h- x2(t) is the rectangular signal shown. X2(s) = ?

Integrals are linear operation:

u(t T1) u(t T2 ) 1

s(e sT1 e sT2 )

0

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Example Transform Pairs: Delta Function

- x3(t) is the Dirac delta function shown. X3(s) = ?

0

(t T ) 1 e sT

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Example Laplace Transform Pairs

Re{a} 0

Re{a}

e( ja)t 0, t , only if

Called the region of convergence, or, ROC

x1(t) eatu(t)

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Exponential & Sinusoidal Signals

Example Laplace Transform Pairs14

Euler’s formula:

cos( ) sin( ) cos( )2

cos( ) sin( ) sin( )2

jx jxjx

jx jxjx

e ee x j x x

e ee x j x x

j

2

1 1cos( )

2 2 2

1 1 1 1

2 2

j tj tj t j te e

t e e

s

s j s j s

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Some Basic Laplace Transform Pairs

Extractions from Table 3-2

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Poles & Zeros of X(s) (= L(x(t)))16

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In general, X(s) is a rational function of the form

11 1 0

11 1 0

1 2 1

1 2 1

( )( )

( )

( )( ) ( )( )

( )( ) ( )( )

m mm m

n nn n

m m

n n

b s b s b s bN sX s

D s a s a s a s a

A s z s z s z s z

s p s p s p s p

The zeros of X(s) are the values of s that render N(s)=0 (which are also the roots of N(s)).

The poles of X(s) are the roots of its denominator D(s).

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Inverse Laplace Transform using Partial Fraction Expansion

17

3 2

2

5 30 49 10( )

( 2)( 4 5)

s s sX s

s s s s

Partial Fraction Expansion

2

1 4 8( )

2 4 5X s

s s s s

L-1 (X(s)) = ? Not in the Table!

x(t) = L-1 (X(s)) = L−1( )+ L-1 ( L−1( )

2 2( ) ( ) 4 ( ) 8 sin( ) ( )t tx t u t e u t e t u t

Table-look-up

Inverse Laplace Transform using Partial Fraction Expansion

18

11 1 0

11 1 0

( )m m

m mn n

n n

b s b s b s bX s

a s a s a s a

When apply the PFE method, X(s) must be in a ‘strictly proper’ form.

If n>m, X(s) is called “strictly proper”. (If n=m, it is call “proper”.)

3 2

2

5 30 49 10( )

( 2)( 4 5)

s s sX s

s s s s

2

2

1( )

4 5

sX s

s s

Strictly proper(n=4, m=2)

NOT strictly proper(n=2, m=2)

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If the given X(s) is not strictly proper, use polynomial division :

2 3 2

3 2

2

2

6 8

2 1 6 4 8 6

6 12 6

8 2 6

8 16 8

18 14

s

s s s s s

s s s

s s

s s

s

2

18 14( ) 6 8

2 1

sX s s

s s

3 2

2

6 4 8 6( )

2 1

s s sX s

s s

Inverse Laplace Transform using Partial Fraction Expansion

where L-1 (6s) = 6’(t)L-1 (8) = 8(t)

then apply PFE to the last term before table look-up.

Example

Partial Fraction Expansion: 1) Case of Distinct Real Poles

A2 (s 1) X(s)s1

s2 4s 3

s(s 3)s1

4

A1, A2 and A3 are called the residuals at poles s = 0, -1, and -3, respectively

requires knowing the poles of H(s)

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2 3

23

23 3

( 3) ( 3) ( 3)( 3) ( )

1 3

( 3) ( 3)

1

( 3) ( 3)

1 3

s A s A s As X s

s s s

s A s AA

s s

s A s AA A

s s s

2

2

2

4 3( 3) ( ) ( 3)

( 1)( 3)

4 3

( 1)

4 34

( 1) 3

s ss X s s

s s s

s s

s s

s s

s s s

This is why the method work!

Partial Fraction Expansion: 1) Case of Distinct Real Poles

Partial Fraction Expansion: 2) Case of Repeated Real Poles

Example:

22

2

4

4 6( )

( 1)( 2)( 3)

s sX s

s s s

Where A, B, and C4 can be found using the same method as in the distinct pole case.

1 2 3 42 3 4

( )1 2 3 ( 3) ( 3) ( 3)

A B C C C CX s

s s s s s s

The pole at -3 is a repeated pole of multiplicity of 4. In this case, the PFE will produce a total of 6 terms (2 from the distinct poles at -1 and at -2 and 4 from the 4 poles at -3.

1

2

44 3

( 1) ( )

B ( 2) ( )

( 3) ( )

s

s

s

A s X s

s X s

C s X s

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Partial Fraction Expansion: 2) Case of Repeated Real Poles

23

1 2 3 42 3 4

( )1 2 3 ( 3) ( 3) ( 3)

A B C C C CX s

s s s s s s

43

3

24

2 2

3

34

1 3

3

( 3) ( )

( 3)1

2!

1

3

( )

( 3) ( )!

s

s

s

dC s X s

ds

dC s X s

ds

dC s X s

ds

where ! ( 1)( 2) .... 3 2 1n n n n

Partial Fraction Expansion: 2) Case of Repeated Real Poles

24

2

311

2

322

23

3 23

2 23

2

8( 4 6)( 1) ( ) = = 3

( 2)( 3)

8( 4 6)B ( 2) ( ) = = -16

( 1)( 3)

8( 4 6)C ( 3) ( ) = = 12

( 1)( 2)

8( 4 6) 8(C ( 3) ( ) =

( 1)( 2)

ss

ss

ss

s sA s X s

s s

s ss X s

s s

s ss X s

s s

d d s s ss X s

ds ds s s

2

3

2 3 23

1 2 3

3

8 10)10

(( 1)( 2))

16( 12 30 22)C ( 3) (

1 1

2 2) 13

(( 1)( 2))

s

s

s

s s

d s s ss X s

ds s s

2

1 2 33 2 3

8 4 6( )

( 1)( 2)( 3) 1 2 3 ( 3) ( 3)

s s A B C C CX s

s s s s s s s s

Example:

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Partial Fraction Expansion: 2) Case of Repeated Real Poles

25

2

3 2 3

8 4 6 3 16 13 10 12( )

( 1)( 2)( 3) 1 2 3 ( 3) ( 3)

s sX s

s s s s s s s s

Example:

1 1

1 1

- -2

-3 -32

3 16 3 ( ) 16 ( )

1 213 10

13 ( ) 10 ( )3 ( 3)

12

( 3

t t

t t

e u t e u ts s

e u t te u ts s

s

L L

L L

1 2 -33 3

- -2 -3 -3 2 -3

2=6 6 ( )

) ( 3)

( ) 3 16 13 10 6 ( )

t

t t t t t

t e u ts

x t e e e te t e u t

L

4 4 4 4 4 44 1 2 3 4

2 3 4

4 43 2

3 2 3 4

4 43 2

3 2 3 4

( 3) ( 3) ( 3) ( 3) ( 3) ( 3)( 3) ( )

1 2 3 3 3 3

( 3) ( 3) 3 3 3

1 2

( 3) ( 3)3 3 3

1 2

s A s B s C s C s C s Cs X s

s s s s s s

s A s Bs C s C s C C

s s

d s A s Bs C s C s C C

ds s s

4 43 2

3 2 3

4 43 2

3 2 3

4

all these terms contai

( 3) ( 3)3 3

n (s+3) after differenti

31 2

( 3) ( 3)3 3

1 2

( 3) ( 3

1

ation!

d s A s Bs C s C s C

ds s s

d s A s Bs C s C C

ds s s

d s A s

ds s

4

3 2

3 2 3 3

)3 3

23

Bs C s C C C

ss

26

Why does the method work (consider how we found C3) ?

C4 is by itself !

C3 is by itself !

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Partial Fraction Expansion: 3) Distinct Complex Poles (method #1)

Example: 27

2 11

6 3 4 41/ 4

21 2 4 20

BB

B CC

B C

2

2 2 2 2 2

1

2

2

2

2

22 6 21 (4)( )

( 1)( 4 20) 1 ( 2) 4 ( 2) 4

where ( 1) ( ) 1

(1)( 4 20) ( 1)( 2) 4( 1)( )

( 1)( 4 20)

( 1) (3 4 4) (2 4 20)

( 1)( 4 20)

s

ss sX s

s s s s s s

A s X s

s s B s s C sX s

s

B

s s

B s B C s B C

A C

s s s

Comparing the coefficients:

2 20

2 20

Table:

cos( ) ( )( )

sin( ) ( )( )

ato

atoo

s ae t u t

s a

e t u ts a

Partial Fraction Expansion: 3) Distinct Complex Poles (Method #1)

28

2 21( ) cos(4 ) sin(4 ) ( )

4t t tx t e e t e t u t

where 1

1

22 2

22 2

11

( 1/ 4

2( ) cos(4 ) ( )

( 2) 4

4sin(4 ) ( )

( 2)

) 1

4 4

L t

L t

se t u t

s

e t u ts

2

2 2 2 2 2

22 6 21 (4)( )

( 1)( 4 20) 1 ( 2) 4 ( 2) 4

11 ( 1/ 4)ss sX s

s s s s s s

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Partial Fraction Expansion: 3) Distinct Complex Poles (Method #1)

29

2 2

2 2 2

2

1( ) cos(4 ) sin(4 ) ( )

4

1 (1 4) cos 4 atan2 1 4,1 ( )

1.031 cos 4 0.245 ( )

t t t

t t

t t

x t e e t e t u t

e e t u t

e e t u t

Trigonometric identity: cos( ) sin( ) cos( )X t tMt Y 2 2where and atan2(Y,X)M X Y

-1

-1

-1

0tan ( / ),

0, 0tan ( / ),

0, 0tan ( / ),where atan2(,) is defined as: atan2( , )

0, 0/ 2,

0, 0/ 2,

0, 0undefined,

xy x

y xy x

y xy xy x

y x

y x

y x

What is atan2(y,x)?30

“atan2(y,x) is an arc-tangent function that takes both the x and y component of a 2-D vector and returns the angle of the vector.

The regular tan-1(y/x) gives the right answer only if the vector is in the 1st

and the 4th quadrant.

Example:

For x = -1, y = -1, if you use the function tan-1(y/x), you get

tan-1(-1/(-1)) = tan-1(1) =/4=45o.

which is wrong. The function atan2(y,x) will give you the correct answer:

atan2(-1,-1) = - /4=-135o.

(-1,-1)

-135o

Page 16: Lecture 3

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Partial Fraction Expansion: 3) Distinct Complex Poles (Method #2)

31

2

2

2

2 42 4

2 6 21 1( )

( 1)( 4 20) 1 ( 2 4) ( 2 4)

where

2 6 21( 2 4) ( )

( 1)( 2 4)

1 1

2 8

s js j

s s A jB A jBX s

s s s s s j s j

s sA jB s j X s

s s j

j

1 cos( ) sin( ) ( )2 2atA jB A jBL e A bt B bt u t

s a jb s a jb

2 21( ) cos(4 ) sin(4 ) ( )

4t t tx t e e t e t u t

(Pole determines the oscillation frequency () and exponent (a)).

Partial Fraction Expansion: 3) Distinct Complex Poles (Method #3)

32

2

2

2

2 4

0.245

2 6 21 1( )

( 1)( 4 20) 1 ( 2 4) ( 2 4)

2 6 21( 2 4) ( )

( 1)( 2 4)

1 1 0.515

2 8

s j

j

s s Ae AeX s

s s s s s j s j

s sAe s j X s

s s j

j e

2( ) ( ) 2(0.515) cos(4 0.245) ( )t tx t e u t e t u t

1 ( ) ( )

( ) ( )

( ) ( )

( ) 2 cos( ) ( )

j jj a jb t j a jb t

at j bt j bt at

Ae AeL Ae e u t Ae e u t

s a jb s a jb

Ae e e u t Ae bt u t

0.245 radian (obtained by atan2) 0.515A

Its pole has positive

imaginary part.

. ., (-2 4) i e j

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Summary Results for 4 Cases

n! n(n 1)(n 2) .... 4 32 1

33

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