Background Review Elementary functions Complex numbers Common test input signals Differential equations Laplace transform –Examples –properties –Inverse

Background Review• Elementary functions• Complex numbers• Common test input signals• Differential equations• Laplace transform

– Examples– properties– Inverse transform– Partial fraction expantion

• Matlab

Elementary functions

0

1lnln ln

Exponential Function

, /

( ) , 1

Logrithmic Function ln( )

ln ln ln

ln ln ln

ln ln , ln(1) 0

1

x

x y x y x y x y

x y xy

a

x x x

e

e e e e e e

e e e

x

(xy) x y

x x yy

(x ) a x

e x, e ex

The most beautiful equation

• It contains the 5 most important numbers: 0, 1, i, , e.

• It contains the 3 most important operations: +, *, and exponential.

• It contains equal sign for equations

* 1 0ie

2 2

Even Function: ,

Odd Function: ,

sin sin cos cos

sin cos 1

sin sin cos cos sin

sin sin cos cos sin

sin 2 2sin cos

cos c

f( x) f(x) x

f( x) f(x) x

( x) x, ( x) x

x x

(x y) x y x y

(x y) x y x y

x x x

(x y)

os cos sin sin

cos cos cos sin sin

x y x y

(x y) x y x y



x x,

xx,

x

xx,

x

xx

x x) (π

x x) (π

-x)π

()π

( xx

-x)π

()π

( x-x

x)(x

x)(x

xxxxx

sin

1csc

cos

1sec

sin

coscot

cos

sintan

coscos

sinsin2

sin2

sincos

2cos

2cossin

2cos12

1sin

2cos12

1cos

1cos2sin21sincos2cos

2

2

2222


2sin

2sin2coscos

2cos

2cos2coscos

2cos

2sin2sinsin

22then Substitute

sinsin2

1cossin

coscos2

1coscos

coscos2

1sinsin

vuvuvu

vuvuvu

vuvuvu

vu,y

vuxx-y, y, vxu

y)](xy)(x[yx

y)](xy)(x[yx

y)](xy)(x[yx


phase, δ magnitudeC

δ)(xBAxBxA

B

Aδ, BAδ, then CCδ, BCA

xBxAxδCxδCδ)(xC

δ)(xBAxBxA

A

Bδ, BACδ, Cδ, BCA

xBxAxδCxδCδ)(xC

xB x A

::

sinsincos

tancossin

sincoscossinsincossin

Also

cossincos

tanthen sincos

sincossinsincoscoscos

sincos gSimplifyin

22

22

22

22


• F(t)=3sin t +4cos t

• F(t)=Asin(3t-)=Acos sin3t –Asin cos3t

• Acos =3

• Asin =-4

• A2=25, A=5

• tan =-

• F(t)=5sin(3t+)

Complex Numbers

• X2+1=0 x=i where i2=-1

• X2+4=0, then x=2i, or 2j

• If z1=x1+iy1, z2=x2+iy2

• Then z1+ z2= (x1+ x2)+i(y1 + y2)

• z1 z2=(x1+iy1)(x2+iy2)=(x1x2 -y1y2) +i(x1y2

+x2y1)

22

22

211222

22

2121

2222

2211

22

11

2

1

))((

))((

yx

yxyxi

yx

yyxxz

iyxiyx

iyxiyx

iyx

iyx

z

zz

Polar form of Complex Numbers• z=x+iy, let’s put x=rcos, y= rsin• Then z = r(cos+i sin = r cisr• Absolute value (modulus) r2=x2+y2

• Argument= tan-1(y/x)

• Example z=1+i

,...2,1,0,24

arg

2

nnz

z

Euler Formula

• z=x+iy

• ez =ex+iy= ex eiy= ex (cos y+i sin y)

• eix =cos x+i sin x = cis x

• | eix | = sqrt(cos2 x+ sin2 x) = 1

• z=r(cos+i sinr ei

• Find e1+i

• Find e-3i

In Matlab>> z1=1+2*i

z1 = 1.0000 + 2.0000i

>> z2=3+i*5

z2 = 3.0000 + 5.0000i

>> z3=z1+z2

z3 = 4.0000 + 7.0000i

>> z4=z1*z2

z4 = -7.0000 +11.0000i

>> z5=z1/z2

z5 = 0.3824 + 0.0294i

>> r1=abs(z1)

r1 = 2.2361

>> theta1=angle(z1)

theta1 = 1.1071

>> theta1=angle(z1)*180/pi

theta1 = 63.4349

>> real(z1)

ans = 1

>> imag(z1)

ans = 2

Poles and zeros• Pole of G(s) is a value of s near which the

value of G goes to infinity• Zero of G(s) is a value of s near which the

value of G goes to zero.

A r-th order pole p:

( )( ) ; 0,lim r

s p

G s s p R R

A zero z:

( ) 0lims z

G s

Poles and zeros in Matlab>> s=tf(‘s’)

Transfer function: s

>> G=exp(-2*s)/s/(s+1)

Transfer function: 1exp(-2*s) * ----------- s^2 + s

>> pole(G)

ans = 0, -1

>> zero(G)

ans = Empty matrix: 0-by-1

2

2( )

2

seG s

s

Test waveforms

used in control

systems

1st order differential equations• y’ + a y = 0; y(0)=C, and zero input• Solution: y(t) = Ce-at

• y’ + a y = (t); y(0)=0, input = unit impulse• Unit impulse response: h(t) = e-at

• y’ + a y = f(t); y(0)=C, non zero input• Total response: y(t) = zero input response +

zero state response = Ce-at + h(t) * f(t)

• Higher order LODE: use Laplace

Laplace Transform

• Definition and examples

ssF

es

es

es

dteLsF

tutf

FLtf

dttfefLsF

sstst

st

1)(

1111)1()(

1)()(

Example

Lapalce inverse );()(

)()()(

0

00

1

0

Unit Step Function u(t)

Laplace Transform

aseL

ssF

se

s

dtedteeeLsF

etf

at

ts

tststt

t

1)(

3

1)(

3

1

3

1

)()(

)(

Example

0

)3(

0

)3(

0

33

3

1lexponentia )(

1 impulseunit )(

1stepunit

t

astu

seat

___;)tan(sin______sincos 2 ji)b

a(xxbxa

The single most important thing to remember is that whenever there is feedback, one should worry about __________

Name:____________

Laplace Transform

320

00

2

0

22

20

20

000

2122

211

)(

111

11)(

)(

ssstdte

s

dttes

ets

dttetL

se

sdte

s

dtes

tes

tdtetL

ttf

st

ststst

stst

ststst

Laplace Transform

)(1

)'(1

)(1

a

ss'Let

)(1

)(

)()(

zatLet ?)}({about what ),()}({ If

0

'

0

0

a

sF

asF

adzzfe

a

dzzfea

atfL

dtatfeatfL

atfLsFtfL

zs

a

zs

st

Laplace Transform

220

20

2

2

02

2

2

0

2

02

0

000

sin

sin)1(

sin

sin)1

(coscos

cos)1

(sin1

sin)(sin

sin)(

stdte

stdte

s

tdtess

tdtess

tes

tdtes

tdtes

tes

tdtetL

ttf

st

st

st

ststst

ststst

Laplace Transform

)(sin)(cos)sin(cos)(

11

11

)(

cos)( ,sin)(

222222

0

)(

0

)(

0

tiLtLtitLeL

s

i

s

s

s

is

is

is

isis

ise

is

dtedteeeL

ttfttf

ti

tis

tististti

Laplace transform table

Laplace transform theorems

Laplace Transform

220

220

20

2

2

0

)(2

2

2

0

2)(

0

)(2

0

)(

0

)(

0

)(

0

)(cos

)(sin

)(sin)

)(1(

sin)()(

sin)1

(cos)(

cos

cos)1

(sin1

sin)sin(

as

astdte

astdte

astdte

as

tdteasas

tdteasas

teas

tdteas

tdteas

teas

tdteeteL

st

st

st

tas

tastas

tas

tastas

atstat

Laplace Transform

?)52

23(

sin2cos3)1

2

1

3()

1

23(

)3sin3

13(cos3sin

3

13cos

)3)2(

1

3)2(

2()

3)2(

3(

3sin3

1)

3

3

3

1()

3

1()

9

1(

21

22221

221

222

22221

221

221

221

21

ss

sL

Find

ttss

sL

s

sL

ttetete

ss

sL

s

sL

ts

Ls

Ls

L

ttt

Laplace Transform

)0(')0()(

)]0()([)0(')(')0('

)(')()(')(")}("{

)0()()'(

)0()()'(

)()0()()0(

)()()()(')}('{

2

0

00

0

0

00

0

fsfsFs

fssFsfdttfesf

dttfestfedttfetfL

yssYyL

xssXxL

ssFfdttfesf

dttfestfedttfetfL

st

ststst

st

ststst

Laplace Transform

• y”+9y=0, y(0)=0, y’(0)=2

• L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2

• L(y)=Y(s)

• (s2+9)Y(s)=2

• Y(s)=2/ (s2+9)

• y(t)=(2/3) sin 3t

Matlab

F=2/(s^2+9)

F =

2/(s^2+9)

>> f=ilaplace(F)

f =

2/9*9^(1/2)*sin(9^(1/2)*t)

>> simplify(f)

ans =

2/3*sin(3*t)

Laplace Transform

• y”+2y’+5y=0, y(0)=2, y’(0)=-4

• L(y”)=s2Y(s)-sy(0)-y’(0)= s2Y(s)-2s+4

• L(y’)=sY(s)-y(0)=sY(s)-2

• L(y)=Y(s)

• (s2+2s+5)Y(s)=2s

• Y(s)=2s/ (s2+2s+5)=2(s+1)/[(s+1)2+22]-2/[(s+1)2+22]

• y(t)= e-t(2cos 2t –sin 2t)

Matlab

>> F=2*s/(s^2+2*s+5)

F =

2*s/(s^2+2*s+5)

>> f=ilaplace(F)

f =

2*exp(-t)*cos(2*t)-exp(-t)*sin(2*t)

Laplace transform

• Y”-2 y’-3 y=0, y(0)= 1, y’(0)= 7

• Y”+2 y’-8 y=0, y(0)= 1, y’(0)= 8

• Y”+2 y’-3 y=0, y(0)= 0, y’(0)= 4

• 4Y”+4 y’-3 y=0, y(0)= 8, y’(0)= 0

• Y”+2 y’+ y=0, y(0)= 1, y’(0)= -2

• Y”+4 y=0, y(0)= 1, y’(0)= 1

Y”+2 y’+ y=0, y(0)= 1, y’(0)= -2

>> A=[0 1;-1 -2]; B=[0;1]; C=[1 0]; D=0;>> x0=[1;-2];>> t=sym('t');>> y=C*expm(A*t)*x0 y = exp(-t)-t*exp(-t)

Y”+2 y’+ y=f(t)=u(t), y(0)= 2, y’(0)= 3

Partial Fraction

10

3

3

)2()2(

)3(

1

2sput and 2,-sby multiply

6

1

32)3)(2(

1

0sput and s,by multiply

32)3)(2(

1

6

1)(

y(t)? is What , 6

1)(

222

000

23

23

Bs

sC

s

sAB

ss

s

As

Cs

s

BsA

ss

s

s

C

s

B

s

A

sss

s

sss

ssY

sss

ssY

sss

sss

Partial Fraction

tt

sss

eety

ssssss

ssY

C

s

sA

s

sBC

ss

s

32

23

333

15

2

10

3

6

1)(

3

1

15

2

2

1

10

31

6

1

6

1)(

15

2

)3(

)2(

)3(

)2(

1

3sput and 3,sby multiply

Partial fraction; repeated factor

2)23)(3(

124137)(A

?Abut before, as obtained becan D C, B, ,A123

)(

)(

)(

)23)(3(

124137)(

3

142)1(31

-1(0)y' 1,y(0) ,42'3"

0

2

234

02

12

122

22

234

22

3

ss

t

sss

sssssQ

s

D

s

C

s

B

s

A

s

Aty

sH

sG

ssss

sssssY

ssYsYsYs

etyyy

Partial fraction; repeated factor

3)('A

0sput and s, with ateDifferenti

123)23)(3(

124137

obtained becan A 0,sput and ,sby Multiply

123

)23)(3(

124137)(

01

222

122

234

22

122

22

234

ssQ

s

Ds

s

Cs

s

BssAA

sss

ssss

s

D

s

C

s

B

s

A

s

A

ssss

sssssY

But No FUN

Partial fraction; exercise

22 )5

1 )4

44

1411 )3

9

99 )2

23 )1

2

2

23

3

2

2

ss

s

s

ssss

sss

ss

ss

s

>> [r p k]=residue(n,d)

r =

1

2

p =

1

0

k =

[]

>> d=[1 -1 0]

d =

1 -1 0

>> n=[3 -2]

n =

3 -21/(s-1) + 2/s

Matlab

>> n=[1 9 -9]

n =

1 9 -9

>> d=[1 0 -9 0]

d =

1 0 -9 0


r =

1.5000

-1.5000

1.0000

p =

3

-3

0

k =

[]

1.5/(s-3)-1.5/(s+3)+1/s

Matlab

>> n=[11 -14]

n =

11 -14

>> d=[1 -1 -4 4]

d =

1 -1 -4 4


r =

2.0000

-3.0000

1.0000

p =

2.0000

-2.0000

1.0000

k =

[]

2/(s-2)-3/(s+2)+1/(s-1)

Matlab

>> b=[1 2 1]

b =

1 2 1

>> a=[1 0]

a =

1 0

>> [r p k]=residue(a,b)

r =

1

-1

p =

-1

-1

k =

[]

1/(s+1)-1/(s+1)2

Matlab

>> Y=(s^4-7*s^3+13*s^2+4*s-12)/s^2/(s-3)/(s^2-3*s+2) Transfer function:s^4 - 7 s^3 + 13 s^2 + 4 s - 12------------------------------------ s^5 - 6 s^4 + 11 s^3 - 6 s^2>> [n,d]=tfdata(Y,'v')n = 0 1 -7 13 4 -12d = 1 -6 11 -6 0 0>> [r,p,k]=residue(n,d)r = 0.5000 -2.0000 -0.5000 3.0000 2.0000p = 3.0000 2.0000 1.0000 0 0k = [ ]

2

0.5 2 0.5 3 2

3 2 1Y

s s s s s

Documents

Background Review Elementary functions Complex numbers Common test input signals Differential equations Laplace transform –Examples –properties –Inverse