L1 Math Fundamentals

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Math Fundamentals

Ch 2-1 through Ch 2-5 Complex Variables

Fourier Transforms

Laplace Transforms Inverse Laplace Transforms

Partial Fraction Expansion.

Linear TimeInvariant Differential Equations.

Ch 5-1, 5-2

State Space Analysis.

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Complex Variables

Re (s)

Im (s)

s = x + j yFor the spatial relationship,

Where ,

With complex conjugate

2 2 1tanjy

s x jy s e x y x

1j

s x jy

Complex variables follow algebraic rules such that if 1 1 1 2 2 2ands x jy s x jy

1 2 1 2 1 2

1 2 1 2 1 2 1 2 1 2

s s x x j y y

s s x x y y j x y y x

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

j j js s s e s e s s e

1 1 2 2 1 1 2 21 1 1 1 22 2 2

2 2 2 2 2 2 2 2 2 2

x jy x jy x jy x jys x jy s s

s x jy x jy x jy x y s

1

1 2

2

1 1 ( )1

2 22

jj

j

s e sse

s ss e

2 31 1 1, ,j j j

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Useful Exponential and Series Expansions

3 5 7

2 3 4

2 4 6

2 2

1sin , sin2 3! 5! 7!

12! 3! 4!

1

cos , cos 12 2! 4! 6!

1e cos sin , sin

2

1e cos sin , cos

2

cos sin 1

j j

j j

j j j

j j j

e ej

e

e e

j e ej

j e e

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Fourier Series and Transform

In the study of control systems, the principal goal is to designa system such that the response of the system meets the

functional specifications.

Therefore the designer must understand the methods indescribing a function in the time and frequency domains or

s-plane so that the behavior of the system can be analyzed.

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Fourier Series A function in the time domain can be represented by a sum of exponential

functions over the entire interval -


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Fourier Series and Transform

Consider a time function x(t) described by a discrete set of exponential timedomain functions;

where and is a

series of amplitudes.

For a periodic function existing within the time interval [-T/2, T/2], TheFourier transform is given as

Where represents the amplitude of the component of frequency

,

Which can be expanded for a non-periodic function with frequency domaintransform

and inverse time domain transform

( )2 2

ojn t

n

n

T Tx t X e t

2

oT

nX

/ 2

/ 2

1( ) ( ) o

Tjk t

k oT

X k x t e dtT

kX

ok k

( ) ( ) j tF x t e dt

1( ) ( )

2

j tx t F e d

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Fourier Transform

( ) ( )f t F

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Laplace Transform

Since control systems do not operate in the minus infinity

time range, the single sided Fourier Transform with damping

is referred to as the Laplace Transform such that

where s = + j

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0( ) ( )

1( ) ( )

2

st

jst

j

F s f t e dt

f t F s e dsj

The Laplace Transform expresses a causal functionf(t) as a

continuous sum of exponentials of complex frequencies.

Sufficient conditions for existence are the functionf(t) must

be piecewise continuous and of exponential order.

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Existence of the Laplace Transformation

Given a time varying functionf(t), 0 < t < a Laplace

Transform exists such that

With Inverse

For the Laplace Transform to exist, the integral

must be finite. The integral will be finite, provided that

, and real positive finite numbersM and exist

so that , thenf(t) is said to be anexponential function and exists for all and the integral

exists as well.

.

0

dtetfsFtf st

j

j

stdtesFj

tfsF2

11

0

stf t e dt

0

tf t e dt

( ) for all 0tf t Me t

0t

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Fourier and Laplace Transform relationship

=

Complex Fourier Transform

(Bilateral Laplace Transform)

s = +j .f(t) exists in the interval

t

=

Ordinary Fourier Transform

s = j . f(t) exists in the

interval t =

Laplace Transform

s = + j . f(t) exists as a

casual function , t > 0

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Basic Rules of Transformations

Given that

Scaling

Time shifting: For t0 > 0,

Frequency shifting:

Linearity property:

Time differentiation:

Time integration:

Frequency integration:

1f t F s

1 sf at Fa a

00

stf t t F s e

at

f t e F s a

sFasFatfatfathensFtfandsFtf 221122112211

1 2 10 0 0m

m m m mm

d fs F s s f s f f

dt

0

(0)t F s ff t d t

s s

dssFt

tf

s

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Examples

Determine the Laplace transform of the following commandsignals:

Impulse function:

step function:

ramp function:

sinusoidal function:

0, 0( )

1, 0

for tf t

for t

0, 0( )

, 0

for tf t

At for t

0, 0

( )sin , 0

for tf t

A t for t

0

0

0 0

( )( ) 1

for t

tt dt

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Development of Important Properties

Initial Value Theorem: As t 0, the signal in the s- plane is related

as s such that

Which states that it is always possible to determine the initial value

of the time function from its Laplace transform.

Final Value Theorem: The final value of a signal as t is related

to the value of the Laplace transform of the function as s 0, such

that

0lim ( ) (0 ) lim ( )

stf t f sF s

0lim ( ) lim ( )t s

f t sF s

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Laplace transforms of common signals encountered in

control engineering

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Laplace Transform Examples

23 0( )

0 0

t te e t

f t

t

( ) cosatf t e t

Determine the Laplace Transform for the following signals;

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Inverse Laplace transform by Partial

Fraction expansion

Given a rational function

By factoring the polynomials the function can be expressed as

the product of the factors such as

Where F(s) is the transfer function representing any physical

system . When , ziis referred to as the zeros of the

system andpiare the poles of the system. For the case that

the poles are either real or complex, but distinct, the transferfunction can be written in terms of partial fractions as

1 21 2

1 21 2

( )( )

( )

n nn

n n nn

b s b s bY SF s

U S s a s a s a

1

1

( )( )

( )

m

ii

n

ii

s zF s K

s p

m n

31 2

1 2 3

( ) n

n

CC C CF s

s p s p s p s p

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Fraction expansion

The set of constants Ci are determined by multiplying bothsides of the above equation by the corresponding pole and

solving for the constant. For example to determine C1 the

following method would be applied

And evaluate the expression at s = p1with results

For the case of repeated roots consider the function with

three repeated roots, with partial fraction expansion given by

31 2

1 1 1 1 11 2 3

( )n

n

CC C C

s p F s s p s p s p s ps p s p s p s p

1

1 1 ( ) s pC s p F s

31 2 4

2 31 21 1

( ) n

n

CC C C C F s

s p s p s ps p s p

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Fraction expansion

Set s = p1, and C3 follows as

To determine the remaining constants related to the repeated

root, differentiate equation with respect to the Laplace

variable s.

Then

In general

1

3

3 1 ( )s p

C s p F s

33 1

1 1 1 2( ) 2n

n

C s pd ds p F s C s p C

ds ds s p

1

3

2 1 ( )s p

dC s p F s

ds

1

1

1( ) , 0, , 1

!

ik

k i i

s p

dC s p F s i k

i ds

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Examples

2

3( )

3 2

sF s

s s

2

3( )

( 1)( 2)

sF s

s s

21( )

( 1)F s

s s s

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Transfer function of spring-damper-mass

system

Given the differential equation for a typical spring (spring constant k)

damper (damping constant c) system with mass m; from Newtons law, the

equation can be written as

Where u(t) is a forcing function on the system. The Laplace Transform for

the system equation (Output/Input) is written as:

In general, define the natural frequency as and

damping ratio

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tukyycym

kcsmssUsY

2

1

n

k

m

2 n

c

m

2

2 2 2

1 1

2

n

n n

Y s

U s kms cs k s s

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Matlab Example

For a unit step function input

num=[2 10];

den=[1 2 10];

sys=tf(num,den)

t=0:0.01:8;

step(sys,t); %U(s)=1/s

title('Response');

xlabel('time');ylabel('Displacement');0 1 2 3 4 5 6 7 8

0

0.5

1

1.5Response

time (sec )

Displacement

2( ) 2 10( )( ) 2 10

Y s sH sU s s s

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Solution of an Ordinary Differential

Equation

2 0,y y y (0) 1 and (0) 0y y

2 35 6 3 ,t ty y y e e (0) 0 and (0) 0.y y

22 1( )3 3

t ty t e e

( ) 3 ( ) 2 ( ) 2 ( ) ( ),y t y t y t u t u t 0 0 0(0) , (0) , and (0)y y y y u u

With unit impulse input

1 0( )

0 0

tt

t2( ) 3t ty t e e

2 3 2 31 2

1( ) 3

30

t t t t y t c e c e te e

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State Space Representation Example

For the ODE

Define the state variables

In state variable form:

Given the matrix form as

2 3 0y y y

1

2

2

x y

x y

x y

2 2 1

1 2 1 1

2 1 2 2 2

2 3 0

0 1

3 2 3 2

x x x

x x x x

x x x x x

= u

= u

x Ax + B

Cx+Dy

A is the state matrix .

B is the input matrix .

C is the output matrix .

u is the input vector .

y is the output vector. D is the transmission matrix

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Block Diagram of State Space Model

A

B

D

C

+

+

y(t)

+

u(t) dt( )x t ( )x t

= u

= u

x Ax + B

Cx+Dy

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State Space Representation Example

With state variables:

The state formulation becomes

1 1 1 1

2 2 2 2

1

2

0 1,

3 2

1 0

x x x x

x x x x

x

x

x x

y

1

2

2

x y

x y

x y

= u

= u

x Ax + B

Cx+Dy

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2 3 0y y y

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MatLab Representation

Applying initial conditions such

that x(0) = 0.1m,

A= [0 1;-3 -2];C = [1 0];

x0 = [0.1 ; 0];

sys = ss(A,[],C,[]);

initial(sys,x0);

title('response to initial displacement x(0)=0.1');

ylabel('Displacement');

xlabel('time');

0 1 2 3 4 5 6-0.02

0

0.02

0.04

0.06

0.08

0.1response to initial displacement x(0)=0.1

time (s ec)

Displa

cement

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System Velocity

x1 is the displacementx2is the velocity

The output vectors become

Matlab code

A= [0 1;-3 -2];Cy = [1 0]; Cv=[0 1]; %Displacement x0 = [0.1 ; 0]

sysd = ss(A,[],Cy,[]); %Cy represents Displacement

sysv= ss(A[],Cv,[]); %Cv represents Velocity

initial(sysd,sysv,x0);

title('response to initial displacement x(0)=0.1');

ylabel('Amplitude')

xlabel('time')

1

2

2

x y

x y

x y

0 1 2 3 4 5 6-0.1

-0.08

-0.06

-0.04

-0.02

0

0.02

0.04

0.06

0.08

0.1response to initial displacement x(0)=0.1

time (s ec)

Amplitude

Displacement

Velocity

1 12 2

1 0 , 0 1x x

y vx x

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= u

= u

x Ax + B

Cx+Dy

MIE 312 M. Brown


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System Response with forcing function

For the following system equation

A= [0 1;-3 -2]; B=[0;10]; D=[0;0]

C= [1 0;0 1] %[Displacement; Velocity]

sys = ss(A,B,C,D);

t=0:0.01:5;

U = sin(5*t);

[Y,T,X] = lsim(sys,U,t);

x1=[1 0 ]*X'; %Displacement

x2=[ 0 1 ]*X'; %Velocityplot(t,x1,t,x2)

title('Response to Forcing Function 10sin(5t)');

ylabel('Amplitude');xlabel('time')

2 3 10sin5y y y t

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5-3

-2

-1

0

1

2

3

Response to Forcing Function 10sin(5t)

Amplitude

time

Velocity

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DC motor Transfer Function

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e

t

Ke

iKT

KVRidt

diL

KibJ

( ) I( )

I( ) ( ) ( )

s Js b s K s

Ls R s V s Ks s

2)()(

KRLsbJs

K

sV

sW

Torque Equation:

System Equations:

Transfer Function:

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DC Motor State Space Formulation

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KVRidt

diL

KibJ

1

b KiJ J

di K R i V

dt L L L

1

2

2

3

3

x

x

x

x idi

xdt

1 2

2 2 3

3 2 31

x x

b Kx x x

J J

K Rx x x VL L L

ux Ax B

+ uy x DC

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2 DOF State Space

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1 1 3 2

2 1 4 2

2 1 4 2

x y x y

x y x y

x y x y


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Runge-Kutta numerical soln

An alternate method to the Controls Tool Box is the Runge-Kutta Algorithm

ODE45 in Matlab follows. ODE45 is designed to solve first order ordinarydifferential equations. Open an M-file and save as any name you want. In

this case the file is named mike.m

function dx = mike(t,x)

dx = zeros(2,1); % designates a column vector

dx(1) = x(2) ;

dx(2) = -3*x(1) - 2*x(2) ;

In the Matlab command window or generate an m-file, enter the following

x0 = [0.1 ; 0]; %initial displacement in radians

[T,X] = ode45(@mike,[0 6],x0); % @mike is the call to the M-file and [0 6]

% is the time span.

plot(T,X(:,1)); title('response to initial displacement x(0)= 0.1')

ylabel('Displacement'); xlabel('time') End of lecture

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L1 Math Fundamentals