The Laplace Transform

(1)

مود هسلج

The Laplace Transform

wlg

(2)

The Laplace Transform

دنکیم لقتنم سناکرف هزوح هب ار نامز هزوح رد عبات هک یطخ هتسویپ عبات کی سلاپلا لیدبت











0

) ( )

( )]

(

[ f t F s f t e dt

L

^st

سلاپلا لیدبت سوکعم



^







 

j

ts

ds e s j F

t f s

F L







( )

2 ) 1 ( )]

(

1

[

*notes

Eq A

Eq B

هحفص زا یا هدودحم هب ای و ییارگمه هیحان دشابیم ارگمه نآ رد قوف لارگتنا هکs

دنمانیمROC .

دحاو هلپ عبات سلاپلا لیدبت

*notes

|

₀

0

1 1 )]

(

[

^ ^







 



^st

e

^st

dt s

e t

u L

t s u

L 1

)]

(

[ 

(3)

Pictorially, the unit impulse appears as follows:

0 t0

f(t) (t – t₀)

Mathematically:

(t – t₀) = 0 t



⁰

*note

0 1

) (

0

 



^









dt t t

t

The Laplace transform of a unit impulse:

An important property of the unit impulse is a sifting or sampling property. The following is an important.



²

        

1 0 1 0 2

2 0 1 0

0

0 ,

) ) (

( ) (

t

t t t t

dt f t t t

f 

(4)

The Laplace Transform

The Laplace transform of a unit impulse:

In particular, if we let f(t) = (t) and take the Laplace

1 )

( )]

(

[

⁰

0



^

 t estdt e s

t

L  

Building transform pairs:

e

L(



^ ^ ^

  

^at

u t  e

^at

e

^st

dt 

^e

e

^s ^a ^t

dt e

L

0

) ( 0

)]

( [

a s a

s t e

u e L

st

at

 



 

^

 

1 )

)] ( (

[ |

₀

a t s

u e

^at

 



1

) (

A transform pair

(5)







 0

)]

(

[ tu t te dt

L

^st

 

 







0 0

|

0

vdu

uv

udv

^{u = t}_{dv = e}_-st_dt

2

) 1

( t s

tu 

A transform

pair

2 2 0

1 1

2 1

2 ) )] (

[cos(

w s

s

jw s jw s

dt e e

wt e

L ^st

jwt jwt

 



 





 

 

  ^

 



2

) 2

( )

cos( s w

t s u

wt   A transform

pair

(6)

The Laplace Transform

Time Shift

 



 







 

























0 0

)

( ( )

) (

, . ,

0 ,

,

) ( )]

( ) ( [

dx e x f e dx e

x f

So x

t as and x

a t As

a x t and dt dx then a t x Let

e a t f a

t u a t f L

sx as

a x s

a

st

) ( )]

( ) (

[ f t a u t a e F s

L   

^^as

The Laplace Transform

Frequency Shift









  







0

) ( 0

) (

)]

( [

)]

( [

a s F dt e

t f

dt e t f e t

f e L

t a s

st at

at

) (

)]

(

[ e f t F s a

L

^^at

 

(7)

Example: Using Frequency Shift Find the L[e^-atcos(wt)]

In this case, f(t) = cos(wt) so,

2 2 2

2

) (

) ) (

( ) (

w a

s

a a s

s F and

w s s s F



 



 

2

2 ( )

) (

) )] (

cos(

[ s a w

a wt s

e L ^at



 



The Laplace Transform

Time Integration:

The property is:

st st

t

st t

se v dt e dv and

dt t f du dx x f u Let

parts by

Integrate

dt e dx x f dt

t f L











 



 



 







 



, 1

) ( ,

) (

: ) ( )

(

0

0 0 0

(8)

The Laplace Transform

Time Integration:

Making these substitutions and carrying out The integration shows that

) 1 (

) (

0 0

s sF

dt e t s f dt t f

L ^st







 



^



^



^

Time Differentiation:

If the L[f(t)] = F(s), we want to show:

) 0 ( )

( ) ]

[ ( sF s f

dt t

L df  

Integrate by parts:

) ( ),

) ( ( ,

t f v so t df dt dt

t dv df

and dt se du

e

u

^st ^st







^ ^

*note

(9)

Making the previous substitutions gives,

 









 













 



0 0 0

) ( )

0 ( 0

) ( )

(

|

dt e t f s f

dt se t f e

t dt f

L df

st st st

So we have shown:

) 0 ( ) ) (

( sF s f

dt t

L df  



The Laplace Transform

We can extend the previous to show;

) 0 ( .

. .

) 0 ( ' )

0 ( )

) ( (

) 0 ( '' ) 0 ( ' ) 0 ( )

) ( (

) 0 ( ' ) 0 ( ) ) (

(

) 1 (

2 1

2 3

3 3

2 2

2







 











 











 





n

n n

n

f

f s f

s s F dt s

t L df

case general

f sf

f s s F dt s

t L df

f sf

s F dt s

t L df

(10)

Transform Pairs:

______

__________

) ( )

(t Fs

f

f(t) F(s)

1 2

! 1

1 ) 1

(

1 )

(







n n

st

s t n

t s

a e s

t s u

 t

Transform Pairs:

f(t) F(s)

 

2 2

1 2

) cos(

) sin(

) (

! 1

w s wt s

w s wt w

a s e n

t

a s te

n at



 



(11)

Transform Pairs:

f(t) F(s)

2 2

sin ) cos

cos(

cos ) sin

sin(

) ) (

cos(

) ) (

sin(

w s

w wt s

w s

w wt s

w a

s

a wt s

e

w a

s wt w

e

at at



 



 







 



  Yes !

The Laplace Transform

Common Transform Properties:

f(t) F(s)

) 1 ( )

(

) ) (

(

) 0 ( .

. . ) 0 ( ' )

0 ( )

) ( (

) ( )

(

) ( [ 0

), ( ) (

) ( 0

), ( ) (

0

1 0 2

1 0 0

0 0 0

s sF d

f

ds s t dF

tf

f f s f

s f s s F s dt

t f d

a s F t

f e

t t f L e t

t t u t f

s F e t

t t u t t f

t

n n

at

os t

















(12)

نآ سوکعم و سلاپلا لیدبت یارب بلتم روتسد

Example Use Matlab to find the transform of t

te^⁴

The following is written in italic to indicate Matlab code

syms t,s

laplace(t*exp(-4*t),t,s) ans =

1/(s+4)^2

سوکعم سلاپلا لیدبت یارب بلتم روتسد

Example Use Matlab to find the inverse transform of

19 . 12 . )

18 6 )(

3 (

) 6 ) (

( 2 prob

s s s

F   

 

syms s t

ilaplace(s*(s+6)/((s+3)*(s^2+6*s+18))) ans =

-exp(-3*t)+2*exp(-3*t)*cos(3*t)

(13)

The Laplace Transform

Theorem: نامز هزوح رد هیلوا رادقم هیضق

If the function f(t) and its first derivative are Laplace transformable and f(t) Has the Laplace transform F(s), and the exists, thenlimsF(s)

0

) 0 ( ) ( lim ) ( lim









 t s

f t f s

sF

The utility of this theorem lies in not having to take the inverse of F(s) in order to find out the initial condition in the time domain. This is particularly useful in circuits and systems.



 s

Initial Value Theorem

The Laplace Transform

Initial Value Theorem:

Example: Given;

2 2 5 ) 1 (

) 2 ) (

(  

  s s s F

Find f(0)

1 ) 26 ( 2

lim 2

25 1 2 lim 2

5 ) 1 (

) 2 lim (

) ( lim ) 0 (

2 2

2

2 2 2

2

 



 















 



 



s s

s s s

s s s s

s s

s s s

sF

f s s s



 s

(14)

Theorem: نامز هزوح رد ییاهن رادقم هیضق

If the function f(t) and its first derivative are Laplace transformable and f(t) has the Laplace transform F(s), then

) ( ) ( lim ) (

limsF s  f t  f 

0

s t

Again, the utility of this theorem lies in not having to take the inverse of F(s) in order to find out the final value of f(t) in the time domain.

This is particularly useful in circuits and systems.

Final Value Theorem

The Laplace Transform

Final Value Theorem:

Example: Given:



^s^s



^note ^F ^s ^te ^t

s

F ( ) ^tcos3

3 ) 2 (

3 ) 2 ) (

( ₂ ₂ ¹ ²

2 2

 





  ^

Find f()^.



⁽⁽ ²²⁾⁾ ³³



⁰

lim )

( lim )

( 2 2

2

2 





 





s s s s

sF f

0 s

(15)

State variable technique

Linear System

) m (

u y(p)

) n ( x













 ) t ( x

) t ( x

) t ( x ) n ( x

n i 2 1

State vector













 ) t ( u

) t ( u

) t ( u ) m ( u

m i 2 1

Input vector

















 ) t ( y

) t ( y

) t ( y ) p ( y

p i 2 1

Output vector

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

) t ( u ) t ( B ) t ( x ) t ( dt A x dx









 

Multivariable linear system (MIMO) Representation of x^ AxBu



^x^^dt



x ^x

Bu +



A +

(16)

State representation of a linear system

D





A B

x u



x

Bu Ax x^  

C



^y

Du Cx

y 

A= System Matrix(n,n) B= Input Matrix (n,m) x= State Vector (n,1) u= Input Vector (m,1)

C= Output Matrix (r,n)

D= Direct Transmission Matrix (r,m) y= Output Vector (r,1)

Transition matrix x^  AxBu

Assuming that the system is continuous and linear that A and B are time-invariant and

Using Laplace transform

) s ( BU ) s ( AX ) 0 ( x ) s (

sX   

) s ( BU ) 0 ( x ) s ( X ) A sI

(   

)]

s ( BU ) 0 ( x [ ) A sI ( ) s (

X   ^¹ 

Taking the inverse Laplace transform of resolvent matrix

] ) A sI [(

L ) t

(  ^¹  ^¹



Transition matrix

(17)

Transition matrix

The state vector will take the following form (convolution)



^



 ^t

0

d ) ( Bu ) t ( )

0 ( x ) t ( ) t (

x     

Or more generally



^





 ^t

t 0

0

d ) ( Bu ) t ( )

0 ( x ) t t ( ) t (

x     

The output vector will take the following form Assuming that C and D are time-invariant

) t ( Du ) t ( Cx ) t (

y  

) t (

u R L

C

i(t) v(t) State variable technique

dt Cdv i

dt v Ldi Ri u







Basic circuit equation

Can be arranged

C i dt dv

Lu v 1 L i 1 L u R dt di





 



In matrix form



 



















 



 



















 

















2 1

1 1

1 1 1

2

u u R R

R R R

R v

i C 0

1 L

R

dt dv dt di

(18)

  _



 



 



 



 

v 1 i v 0

C i

 y



 



















 



 



















 

















2 1

1 1

1 1 1

2

u u ) R R ( L

R ) R R ( L

R v

i C 0

1 L

R

dt dv dt di









































 ^

) s ( Ld

R ) s ( d

s ) s ( Cd

1

) s ( Ld

1 )

s ( sd

1

L s R C

1 L

s 1

LC 1 L s sR ] 1 A sI [

2 1

1

C s 1

L 1 L s R ] A sI [





















 



Solution: resolvent matrix

LC 1 L s sR ) s (

d  ² 

b] s

1 a s [ 1 a b

1 ) b s )(

a s (

1 )

s ( d

1

 



 



 

b ] e a [ e a b dt 1 ) t ( f

] be ae a[ b

1 dt df

] e e a[ b ) 1 ) s ( d ( 1 L ) t ( f

bt at

bt at 1





 





 



 





) b s )(

a s LC (

1 L s sR ) s (

d  ²    

(19)

بلتم رد سلاپلا لیدبت (

لاثم )

سناکرف هزوح رد یطخ یاهمتسیس

• یلع یاهمتسیس :

یلع ییاه متسیس

ز طرش اهنآ هبرض خساپ هک دنتسه ری

دنک هدروآرب ار :

h(t)=0 for t<0

• یلع یاه متسیس ییارگمه هیحان

روحم تسار تمس رد هشیمه

jw دنراد رارق

رادیاپ یاه متسیس :

ئارا یفلتخم فیراعت یرادیاپ یارب هدش ه

تسا . فیصوت نامز هزوح رد هک یطخ متسیس کی یارب

پ ینحنم ریز حطس ندوب دودحم یانعم هب یرادیاپ دوشیم خسا

ینعی تسا هبرض :

کی هب سناکرف هزوح رد هدش فیصوت یاه متسیس یارب زا ی

دشاب دناوتیم ریز ینعم ود :

ییارگمه هیحان وزج یموهوم روحم و دشاب یلع متسیس 1. دشاب . لاقتنا عبات جرخم یاه هشیر مامت 2. (

متسیس یاهبطق )

یارب

دنشاب یموهوم روحم پچ تمس رد یلع یاه متسیس .

(20)

کبدیف و یزاوم ،یرس یاهمتسیس

clear all

% system and subsystems g1(s)=s/(s(s+1)) and g2(s)=1/(s^2+2s+2) s=tf('s');

g1=s/(s*(s+1));

g2=1/(s^2+2*s+1);

%%%%%%%%%%%

g_seri=series(g1,g2);

%%%%%%%%%%%

g_feed=feedback(g1,g2);

%%%%%%%%%%%%%%

g_paral=parallel(g1,g2);

%%%%%%%%%%%%%%

figure(1) impulse(g_feed) figure(2) step(g_feed)

0 2 4 6 8 10

-0.2 0 0.2 0.4 0.6 0.8 1

1.2 Impulse Response

Time (seconds)

Amplitude

0 2 4 6 8 10

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7

0.8 Step Response

Time (seconds)

Amplitude

لاقتنا عبات کی یاهرفص و اهبطق

• ار لاقتنا عبات کی تروص یاه هشیر

« رفص » ار جرخم یاه هشیر و

« بطق

» .دنمانیم

• تسب هقلح متسیس کی یرادیاپ یارب مزلا ه

هتسب هقلح یاهبطق یاه هشیر لحم تسا روحم پچ تمس رد دنشابjw

.

K 𝐻(𝑠)

𝐻 𝑠 =𝑠(𝑠+1)(𝑠+5)¹ and k=20 and k=50 𝑌(𝑠)

𝑅(𝑠)= 𝑘

𝑠³+ 6𝑠²+ 5𝑠 + 𝑘 اب هتسب هقلح یاهبطق لحم k=20

k=50 و

>> roots([1 6 5 20]) ans =

-5.7362 + 0.0000i -0.1319 + 1.8626i -0.1319 - 1.8626i

>> roots([1 6 5 50]) ans =

-6.4314 + 0.0000i 0.2157 + 2.7799i 0.2157 - 2.7799i

>>

𝐻 𝑠 = ¹

𝑠(𝑠+1)(𝑠+5) and k=10 𝑌(𝑠)

𝑅(𝑠)= 𝑘 𝑠²+ 𝑠 + 𝑘 ءازاب هتسب هقلح یاهبطق لحم ابk=10

بلتم روتسد

roots([1 1 10]) ans =

-0.5000 + 3.1225i -0.5000 - 3.1225i

دنرادیاپ و پچ تمس یگمه .

0 2 4 6 8 10 12 14

-2 -1 0 1 2

3 Impulse Response

Time (seconds)

Amplitude

0 5 10 15 20 25 30 35 40 45

-1.5 -1 -0.5 0 0.5 1 1.5

2 Impulse Response

Time (seconds)

Amplitude

k=20

0 50 100 150 200 250 300

-3 -2 -1 0 1 2

3x 10²⁸ Impulse Response

Time (seconds)

Amplitude

k=50