Equations and Laplace Transform

ةبيط ةعماج

EE 242 – Signals and Systems

Lecture 12: Systems represented by Linear Constant-Coefficient Differential and Difference

Equations and Laplace Transform

Omar Siddiqui

Department of Electrical Engineering College of Engineering

Taiba University Madinah

Email:[email protected]

Constant-Coefficient Differential and Difference Equations Differential Equations System

k M k

k k k

N k

k

k

dt

t x b d

dt t y

a d ( ) ( )

0

0





 



Nth order

) (t ) y

(t x

Example of First Order System

) ( )

) ( (

0 0

1

a y t b x t

dt t

a dy  

Difference Equations System

] [n ] y

[n x

] [

] [

0 0

k n x b k

n y a

M

k

k N

k

k

   



 

Example of First Order System

] [ ]

[ ]

1

[

_{0 0}

1

y n a y n b x n

a   

Nth order

Example of a Linear Differential Equation from Circuits

System Equation

It relates the output of a system to its inputs

) ( )

( )

( t Ri t y t

x  

Example of a series RC Circuit

The capacitor voltage relates to current in the following manner:

dt C dv t

i ( ) 

^c

dt t C dy

t

i ( )

) ( 



) ) (

) (

( y t

dt t RC dy

t

x  



) ( )

) (

( y t x t

dt t

RC dy  



The Solution to the Differential Equation

RC Example

The solution to this differential equation has two parts:

) ( )

( )

( t y t y t

y 

_h



_p

) (t y

_h

Homogenous solution Or Zero Input

response

Particular solution Or Zero State

response

The Homogenous solution or the Zero Input response of the system, obtained by setting x(t) = 0

) (t

y

_p

The Particular solution or the Zero State response of the system, obtained by setting the initial conditions to zero

) ( )

) (

( y t x t

dt t

RC dy  

0 )

) (

(  t y  dt

t

RC dy

The Homogenous or Zero input Solution

Let us solve the differential equation by first writing the characteristic equation Characteristic equation : Obtained by replacing D by  in the differential

equation and leaving y(t):

0 )

) (

(  t y  dt

t RC dy

Let

Homogenous equation

dt d  D

0 )

( 1 )

(  

 y t

D RC

RC 1 ( ) 0

 



 

  

 y t

D RC

1 0

 



 

  

  RC

) ( )

( t Ae u t

y

^RC

t h







Solution of the zero state response y

_p

(t) depends on specific x(t)

The Solution of RC Circuit Using FT

System Equation: We start with the System Equation

) ( )

) (

( y t x t

dt t

RC dy  

Fourier Transform: Take the FT of both sides

) ( )

( )

(   

 Y Y X

RCj  

Find H(  ):

RC j

X H Y





 

 

 1

1 )

( ) ) (

(

) ( )

( ) 1

(  j  RC Y   X 



Solution to any input is given by Y (  )  X (  ) H (  )

In time domain y ( t )  x ( t )  h ( t )  

^1

[ Y (  )]

X(t) =cos

_

t : Comparison between the FT Method and the Phasor Method

 

t t

x( )  cos ₀

) ( ) ( )

(  X  H 

Y 

) (

) (

)

(      

₀

    

₀

X

RC H j

 

  1 ) 1 (

RC j 

 1

 1

 Y (  )  ^{ (   }

₀

^{)   (   }

₀

⁾ 

Applying the sifting property of the delta functions:

) (

) ( )

( )

(     

₀

 f 

₀

   

₀

f

RC j

₀

0

1

) (











 

 Y (  )

RC j

₀

0

1

) (











 

FT Method Phasor Method

(only for sinusoidal inputs)

c

c

IZ

V Y (  )  

c c

Z Z R

V

 

C j

C

R j

⁰

0

1 1

1

 





Multiplying num and den by j

_

C

) ( 

 Y

RC j

₀

1

1



 

V  1 0

o

Both of the method give steady state

solution without taking into account the

initial conditions

RC j

₀

0

1

) (











 

) ( 

Y j

₀

RC

0

1

) (











 

FT Method Phasor Method

) (  Y

RC j

₀

1

1



  )]

( [ )

( t

¹

Y 

y  

^

(Taking IFT) y ( t )  Re[ Y (  ) e

^j0t

]

 ) (t

y 2 ( 1

₀

)

0

RC j

e

^{j t}





 2 ( 1

₀

)

0

RC j

e

^{j t}





 

^

Using e

^j0t

 ^{2     }

⁰

^

) 1

)(

1 ( 2

) 1

( )

1 (

0 0

0 0

0 0

RC j

RC j

e RC j

e RC

j ^{j t j t}







 ^{ }









  ^

) 1

)(

1 (

2 _{0 0}

0 0

0 0

0 0

RC j

RC j

RCe j

e RCe

j

e^{j t j t j t j t}









^{  }











  ^{ }

) 1

)(

1 ( 2

sin 2

cos ) 2

(

0 0

0 0

0

RC j

RC j

t RC

t t

y  











 



2 2 2

0 0

0

1

sin ) cos

( R C

t RC

t t y



o









 



 ) (t

y 



 





 

RC j

e

^{j t}

1

0

Re 1

0





 

   ^ _ ^

 







 

 j RC j RC

RC j

t e y

t j

0 0

0

1 1

Re 1 )

(

0









  

  ^{  } _ ^

 









 

 j RC j RC

RC j

t j

t t y

0 0

0 0

0

1 1

1 sin

Re cos )

(  







0



 











 

 _{2 2 2}

0

0 0

0 0

0 0

1

sin cos

sin Re cos

)

( R C

t RC

t RC

j t j

t t

y 













2 2 2 0

0 0

0

1

sin ) cos

( R C

t RC

t t

y 









 



X(t) =cos

_

t : Comparison between the FT Method and the

Phasor Method



 Y (  )

) (t u



RC Circuit with Other sources

) ( )

( t u t

x 

t )

(t x

1

 



 ^

j 1

) ( ) ( )

(  X  H 

Y 

 

 





 j  RC 1

 1

 Y (  )   _



 



   

 j

1



⁰



Re ),

( 

 u t a

e ^at _a¹_j

  

^_^





 j RC j 1 

1 



 





 

RC j





1

) (



 Y (  )

RC j

B j

A



   1

  ^{j j RC} 

RC j

B j

A







  

 

1 1 1

 ^{1 }  ^{  1}

 A j  RC j  B

Put j=-1/RC ^{ B   RC} Put j=0 ^{ A  1}



 Y()

 j

1



FT Pairs

) (t y

Partial Fractions

) ( )

( t u t

x 

 ) ( 

X    

 ^ j

1

RC H j

 

  1 ) 1 (

  



 

 Y (  )

RC j RC j    

1

1     

  







 y (t )

 j RC 

 1 / 1

) (t

u e

^RC

u (t )

 t

  y (t )  _ ^{u (t )}



 



 1  e

^RCt

Note: Such problems cannot be solved with the Phasor method

(sifting property)

t )

(t y

1

Limitations of FT: Initial Value Problems and sources with non-converging FT

t )

(t x

1

) ( )

( t tu t

x  y (t )

) (t tu

FT method does not work for non converging sources and initial value problems

For example, consider the source x(t) = tu(t). Its Fourier Transform does not converge because it does not satisfy the Direchlet condition

 Or if sources have converging FT like cost but there is an initial value on capacitor, FT method or phasor methods cannot be used

 Some of these types of problems can be solved with Laplace Transforms



 







dt t x ) (

0

t

cos  ^V

⁰

^{y (t )}

Initial Value Problem

Initial Voltage on capacitor = V

0

Non-converging X()

) ( ) ( )

(  X  H 

Y 

Laplace Transform









 x t e



dt X (  ) ( )

^jt



 



d e

X t

x

^



^{j t}





 ( )

2 ) 1 (

Inverse Fourier Transform

Fourier Transform Laplace Transform









 x t e



dt s

X ( ) ( )

^st

Inverse Laplace Transform

ds e s j X

t

x 

^{ st}







^





^{( )}

2 ) 1 (



 j

s  

Relation between Laplace and Fourier Transform













 X ( j  ) x ( t ) e

^jt

dt

- Fourier Transform can be obtained from Laplace Transform by putting  = 0









 x t e



dt s

X ( ) ( )

^st

s    j 

Assume  = 0

















 X (  j  ) x ( t ) e

^{( j ) t}

dt

- Laplace Transform can be obtained from Fourier Transform by writing the LT in the following form















 j x t e dt

X (   ) ( )

^{( j ) t }















 X (  j  ) x ( t ) e

^t

e

^jt

dt

 ^{x t e}

^t



j

X     

^

 ( ) ( )

Hence Laplace transform is the Fourier Transform of x ( t ) e

^t

Examples of Laplace Transform









 x t e



dt s

X ( ) ( )

^st















 X ( s ) e

^at

u ( t ) e

^st

dt















 X ( s ) e

^{(s a)t}

u ( t ) dt













0

)

)

(

( s e dt

X

^{s a t}

) (

0 ) (

a s e

^{s a t}



 

 



) (

1 0

a s 



 

a s s

X  

 1

) (

From Laplace Integral From Fourier integral









 x t e



dt s

X ( ) ( )

^st

s    j 

















 X (  j  ) x ( t ) e

^t

e

^jt

dt



















 X (  j  ) e

^at

u ( t ) e

^t

e

^jt

dt

 



















 X (  j  ) e

^{(a )t}

u ( t ) e

^jt

dt

) (t u e

^at

 j a 

 1 FT Pair







j t a

u e

^{a t}



 



^{ }

) (

) 1

)

(

(

a

j 

 

) (

1



 a

s s

X  

 1

) (

Example 9.1: Find the Laplace Transform of e

^at

u (t )

Region of Convergence (ROC)















 X ( s ) e

^at

u ( t ) e

^st

dt

^













 X ( s ) e

^{(s a)t}

u ( t ) dt

) ) (

(

⁰

) (

a s e s

X

t a s



 



 





 j s  

For a Laplace Transform, we have to identify the region of Convergence (ROC)

) (

0 ) (

a s

e

^{j a t}



 

 



  

) (

0 ) (

a s

e

^{a j t}



 

 



  





plane s

 0



  a

For convergence    Re[ s ]   a

t

) ( )

( t e u t x

^st

a

 

ROC

 a

) ( )

( t e u t x

^st

a

 

t

a

 

a s 

 1 )

(

0 )

(

a s

e

^{a t j t}



 

 



  

Converging

Non-Converging

Region of Convergence (ROC)















 X ( s ) e

^at

u ( t ) e

^st

dt

) ) (

(

⁰

) (

a s e s

X

t a s



 



 



Two x(t) may have same Laplace but different ROC

) (

0 ) (

a s

e

^{a j t}



 

 



  

t

) ( )

( t e u t x

^st

 0



  a For convergence

a

 





plane s

ROC

a

a s 

 1

Example 9.2 Example 9.1 x ( t )  e

^at

u ( t )

a s  

 Re[ ]

) ( )

( t e u t x  

^at





















 X ( s ) e

^at

u ( t ) e

^st

dt 













0

dt e

e

^{at st}









 0

dt e

e

^{at st}

) ) (

(

) 0 (

a s e s

X

t a s



 



^





) (

) 0 (

a s

e

^{a j t}

 

^





  

 0

 a if 

a s 

 1





plane s

ROC

a

a

s ]  

Re[

Region of Convergence (ROC)

Combination of two functions

Example 9.3 x ( t )  e

^2t

u ( t )  e

^t

u ( t )



 )

2

( t u e

^t

2 1



s e

^t

u (t ) 

1 1

 2 s

]

Re[ s   Re[ s ]   1

1 1 2

) 1

(  

 

 X s s s   ^{2 2}  ³  ¹ 

 

s s

s ²   ²  ^{3 / 2}  ¹ 

 

s s

s





 2  1

1 ]

Re[ s   Plot of poles and zeroes and ROC

2

 3

plane

s

Laplace Transform Examples

Example 9.4

) ( 3 cos )

( )

( t e

²

u t e t u t

x 

^{ t}



^t

( )

2 ) 1 2 (

) 1

(

^{3 3}

2

u t e e u t e e u t

e

^{ t}



^{t j t}



^{t  j t}



We know the Laplace of the first term:

2 ) 1

2

(

 



t s u

e

^t

Re[ s ]   2

Using the following general Integral for the second and third terms:















 X

₂

( s ) e

^at

cos btu ( t ) e

^st

dt

) 2 (

) 1 2 (

) 1 ( )

( t e

²

u t e

^{(1 3)}

u t e

^{(1 3)}

u t x 

^{ t}



^{t  j t}



^{t j}







^{   }

   





0

) (

0

)

(

dt e dt

e

^{s a jb t s a jb t}

jb a

s jb a

s   



  1 / 2 1 / 2

a s ]   Re[

3 1

2 / 1 3

1 2 / ) 1

2

(

j s

j s s

X   



  3

, 1 

 b a

For

) (

2 1 )

( 2

1

^{( )} ₀ ^{( )} ₀

jb a

s e jb

a s

e

^{s a jb t s a jb t}





 



 



 





 



 



 

 



 







1 ]

Re[ s  

Laplace Transform Examples

Example 9.4 (continued) Combining

2 ) 1

(   s s

X

3 1

2 / 1 3

1 2 / 1

j s

j

s   



  Re[ s ]   1

2 ) 1

(  

 X s s    

 ^{1 1 3 3}  ^{0 . 5 1 1 3}  ³

5 . 0

j s

j s

j s

j s

















 

2 ) 1

(  

 X s s

 ^{1 . 1 5}  ^{0 ( . 5 3 )}

²

^{0 . 5 1 . 5}

5 . 0 5

. 0

j s

j s

j s













 

2 ) 1

(  

 X s s

9 2

1 1

2

  

 

s s

s

2 1

 

s 2 10

1

2

 

 

s s

s

 ) (s

X   ^{  }

 ^{2 2}   ¹⁰

²

^{2 10 1}  ²

2



















s s

s

s s

s s

 ^{2 2}   ¹⁰

²

^{2 3 10}  ²

2 2















 

s s

s

s s

s s



 X (s )

 ^{2 2}  

²

^{5 2 12 10} 

2











s s

s

s

s Re[ s ]   1

Laplace Transform Examples

Example 9.4 (continued)

 ) (s

X  ^{2 2}  

²

^{5 2 12 10} 

2











s s

s

s

s Re[ s ]   1

Poles and zeros and ROCs: Write the result in the form:

Factorizing the numerator

) (

) ) (

( D s

s s N

X 

 ) (s

N 



 



   6

2 2 s

²

5 s Applying quadratic formula to find roots of N (s )

2

) 6 )(

1 ( 4 5

. 2 5

.

2 

²



  s

1 . 2 25

.

1 j

s   

 

 ^{1 3}  ^{1 3} 

) 2 (

6 2 / 5 2

²

j s

j s

s

s s













 

  

 ^{1 3}  ^{1 3} 

) 2 (

1 . 2 25

. 1 1

. 2 25

. 1 ) 2

( s s j s j

j s

j s s

X     







 



Re Im

 2  1

plane s

3



Laplace Transform Examples

Example 9.5

) (

) ) (

( D s

s s N

X 

2 Re

1 



) 3 (

) 1 3 (

) 4 ( )

( t t e u t e

²

u t

x   

^t



^t

1 1 3 ) 4

3 ( 4

 





^

t s u e

^t









 t e



dt s

X

₃

( )  ( )

^st

2 1 3 ) 1

3 ( 1

₂

 



 e

^t

u t s

0

e

^s

  ROC is the whole s plane 1 1

]

Re[ s   Re[ s ]  2

2 1 3 1 1 1 3 1 4 )

(  

 



 X s s s

2 1 3 1 1 1 3 1 4 )

(  

 



 X s s s

 ¹  ² 

3

1 8

4 6 6

3 ) 3

(

2

















 

 s s

s s

s s

s s

X ^{3 3}  ¹  ^{6 2 3} 

2







 

s s

s

s ^{3 3}  ^{( 1}  ^{2 2 1 )} 

2







 

s s

s s

 ^{( 1}  ^{1 ) 2} 

) (

2





 

 s s

s s X

 1

2 ] Re[ s 

Observation: If ROC does not include

j axis, FT does not converge

Properties of ROC

Property 1: The ROC of X(s) consists of strips parallel to the j-axis in the s-plane

Property 2: For rational Laplace transform, the ROC does not contain any poles

t

) (t x

a s ]   Re[





plane s

ROC

a

) ( )

( t e u t x 

^at

Example: ROC:

Pole is outside ROC

Properties of ROC

Property 3: If x(t) is of finite duration and is absolutely integrable, the ROC is the entire s-plane

T t e

t

x ( ) 

^at

0   Example 9.6:

A finite duration signal otherwise

 0

1

0 T



^{ }





T

st

at

e dt

e s

X

0

) (

t T a s

a s s e

X

0 ) (

) ) (

(   



^{ }

 ^e

^{s a T}



a s s

X 1 1

^{( )}

)

( 

^{ }

 



e

^at

Seems like there is a pole at s = -a which contradicts the property.

0

) 0

( a 

Lets put s = 0 to confirm X

Properties of ROC

From the L’ Hospital’s rule







( )

lim X s

a s



 lim



X ( s )

a s

T a

X  

 ( )

Hence there is no pole at s = a and the ROC is the entire s-plane Example 9.6 (continued):

 ^e

^{s a T}



a s s

X 1 1

^{( )}

)

( 

^{ }

 

A finite duration signal

1

0 T

e

^ at

 

 ^{s a} 

ds d ds e

d

_{s a T}

a

s





^{ }





)

1

(

lim

1

) ( lim 0

)

(

T

e

^{s a T}

a s















plane s

Properties of ROC

Property 4: If x(t) is a right sided signal, and if the line Re[s] = 

₀

is in the ROC, then all values of s for which Re[s] > 

₀

will also be in the ROC

Property 5: If x(t) is a left sided signal, and if the line Re[s] = 

₀

is in the ROC, then all values of s for which Re[s] < 

₀

will also be in the ROC

Property 6: If x(t) is a two sided signal, and if the line Re[s] = 

₀

is in the ROC,

then ROC will consist of a strip that includes the line Re[s] = 

₀

Properties of ROC

Explanation of the Properties 4,5,and 6 Right-sided

Left-sided

Two sided

Properties of ROC

Property 7: If Laplace transform of x(t) is rational, the ROC is bounded by poles or extends to infinity. No poles are contained in ROC

Property 8: If Laplace transform of x(t) is rational, then if x(t) is right sided, the ROC is in the right side of the rightmost pole. If x(t) is left-sided the ROC is left to the leftmost pole

Right-sided Left-sided

Explanation

Table of Laplace Transforms

Table of Properties of Laplace Transform

Reading from the book

 Chapter 9: Page 654 - 670