filter analysis3

8/12/2019 Filter Analysis3

1/13

1

The Laplace Transform

M. J. Roberts - All Rights Reserved.

Edited by Dr. Robert Akl

1

Generalizing the Fourier Transform

The CTFT expresses a time-domain signal as a linear

combination of complex sinusoidsof the form ej!t

. In thegeneralization of the CTFT to the Laplace transform, the

complex sinusoids become complex exponentialsof the

form est

where scan have any complex value. Replacing

the complex sinusoids with complex exponentials leads to

this definition of the Laplace transform.

L x t( )( ) = X s( ) = x t( )e" stdt"#

#

$

x t( ) L% &' X s( )

M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 2

Generalizing the Fourier Transform

The variable sis viewed as a generalization of the variable !of

the form s = "+ j!. Then, when ",the real part of s, is zero, the

Laplace transform reduces to the CTFT. Using s = "+ j!the

Laplace transform is

X s( ) = x t( )e# " +j!( )t dt#$

$

%

= F x t( )e#"t&' ()which is the Fourier

transform of x t( )e#"t


Generalizing the Fourier TransformThe extra factor e!

"t is sometimes called a convergence factor

because, when chosen properly, it makes the integral converge

for some signals for which it would not otherwise converge.

For example, strictly speaking, the signal A u t( ) does not have

a CTFT because the integral does not converge. But if it is

multiplied by the convergence factor, and the real part of s

is chosen appropriately, the CTFT integral will converge.

A u t( )e!j#t dt!$

$

% = A e!j#t dt

0

$

% &Does not converge

Ae!"t u t( )e!j#t dt

!$

$

% =A e! " +j#( )t

dt0

$

% &Converges (if "> 0)


Complex Exponential Excitation


Complex Exponential Excitation



2/13

2

Pierre-Simon Laplace

3/23/1749 - 3/2/1827M. J. Roberts - All Rights Reserved. Edited by Dr. Robert Akl 7

The Transfer Function

Let x t( ) be the excitation and let y t( ) be the response of a

system with impulse response h t( ). The Laplace transform ofy t( ) is

Y s( ) = y t( )e!stdt!"

"

# = h t( )$x t( )%& '(e!stdt!"

"

#

Y s( ) = h )( )x t!)( )d)!"

"

#*

+,

-

./e

!stdt

!"

"

#

Y s( ) = h )( )d)!"

"

# x t!)( )e!stdt!"

"

#


The Transfer Function

Let x t( ) = u t( ) and let h t( ) =e!4tu t( ). Find yt( ).

yt( ) = x t( )"h t( ) = x #( )h t! #( )d#!$

$

% = u #( )e!4t!#( )

u t! #( )d#!$

$

%

yt( ) =e!4t!#( )

d#

0

t

% =e!4t

e4#

d#

0

t

% =e!4te

4t!1

4=

1!e!4t

4 , t> 0

0 , t< 0

&

'(

)(

yt( ) = 1/ 4( ) 1!e!4t( )u t( )

X s( ) = 1/ s, H s( ) =1

s+ 4*Y s( ) =

1

s+

1

s+ 4=

1/ 4

s!

1/ 4

s + 4

x t( ) = 1/ 4( ) 1!e!4t( )u t( )


Cascade-Connected Systems

If two systems are cascade connected the transfer function of

the overall system is the product of the transfer functions of the

two individual systems.


Direct Form II Realization





3/13

3








A system is defined by !!y t( )+ 3 !y t( )+ 7y t( ) = !x t( )" 5 x t( ).

H s( ) =s " 5

s2+ 3s + 7


Inverse Laplace Transform

There is an inversion integral

y t( ) =1

j2!Y s( )estds

"#j$

"+j$

% , s = "+ j&

for finding y t( ) from Y s( ), but it is rarely used in practice.Usually inverse Laplace transforms are found by using tables

of standard functions and the properties of the Laplace transform.


Existence of the Laplace

TransformTime Limited Signals

If x t( ) = 0 for t< t0 and t> t1it is a time limitedsignal. If

x t( ) is also bounded for all t, the Laplace transform integralconverges and the Laplace transform exists for alls.



4/13

4


Transform

Let x t( ) = rect t( ) = u t+ 1/ 2( )! u t!1/ 2( ).

X s( ) = rect t( )e!stdt!"

"

# = e!st

dt

!1/2

1/2

# =e!s/2 ! es/2

!s=

es/2 ! e!s/2

s, Alls



TransformRight- and Left-Sided Signals

Right-Sided Left-Sided



TransformRight- and Left-Sided Exponentials

Right-Sided Left-Sidedx t( ) = e!t u t" t0( ) , !#! x t( ) = e!t u t0 " t( ) , !#!



Transform

Right-Sided Exponentialx t( ) =e!t u t" t0( ) , !#!

X s( ) = e!te" stdtt0

$

% = e! "&( )t

e" j'tdtt0

$

%

If Re s( ) = & >!the asymptotic

behavior of e! "&( )t

e" j't

as t($

is to approach zero and the Laplace

transform integral converges.



Transform

Left-Sided Exponentialx t( ) =e!t u t0 " t( ) , !#!

X s( ) = e!te"stdt"$

t0

% = e!"&( )t

e" j't

dt"$

t0

%

If &< !the asymptotic behavior of

e!"&( )t

e" j't as t("$is to approach

zero and the Laplace transform

integral converges.



TransformThe two conditions !> "and !< #define the region of

convergence(ROC) for the Laplace transform of right- and

left-sided signals.



5/13

5


Transform

Any right-sided signal that grows no faster than an exponentialin positive time and any left-sided signal that grows no faster

than an exponential in negative time has a Laplace transform.

If x t( ) = xr t( ) + x

l t( ) where x

r t( ) is the right-sided part and

xl t( ) is the left-sided part and if x

r t( ) < K

re!t and x

l t( ) < K

le"t

and !and "are as small as possible, then the Laplace-transform

integral converges and the Laplace transform exists for !< # &' & e

&'tsin (0t( )u &t( ) L" #$

(0

s+'( )2

+(0

2, %< &'

e&'tcos (

0t( )u t( ) L" #$

s+'

s+'( )2

+(0

2

, %> &' & e&'tcos (

0t( )u &t( ) L" #$

s+'

s+'( )2

+(0

2

, %< &'

Some of the most common Laplace transform pairs(There is more extensive table in the book.)


Laplace Transform Example

Find the Laplace transform of

x t( ) = e!t u t( )+ e2t u !t( )

e!t u t( ) L" #$ 1

s +1, %> !1

e2t u !t( ) L" #$ !

1

s ! 2 , %< 2

e!t u t( )+ e2t u !t( ) L" #$

1

s +1!

1

s ! 2 , ! 1


6/13

6


Find the inverse Laplace transform of

X s( ) =4

s + 3!

10

s ! 6, " > 6

The ROC tells us that both terms must inverse transform into a

right-sided signal.

x t( ) = 4e!3t u t( )!10e6t u t( )



Find the inverse Laplace transform of

X s( ) =4

s + 3!

10

s ! 6, "


7/13

7

Partial-Fraction Expansion


Partial-Fraction ExpansionH s( ) =

10s

s + 4( ) s + 9( )=

K1

s + 4+

K2

s + 9 , ! >"4

K1= s + 4( )

10s

s + 4( ) s + 9( )

#

$%%

&

'((s="4

=10s

s + 9

#

$%&

'(s="4=

"40

5="8

K2= s + 9( )

10s

s + 4( ) s + 9( )

#

$%%

&

'((s="9

=10s

s + 4

#

$%&

'(s="9=

"90

"5=18

H s( ) ="8

s + 4+

18

s + 9=

"8s" 72+18s + 72

s + 4( ) s + 9( ) =

10s

s + 4( ) s + 9( ) . Check.

) ) ) ) )

h t( ) = "8e"4 t +18e"9t( )u t( )







H s( ) =10s

s + 4( )2s + 9( )

=K12

s + 4( )2 +

K11

s + 4+

K2

s + 9 , ! > 4

"

Repeated Pole

K12 = s + 4( )2 10s

s +4( )

2s +

9( )

#

$

%%

&

'

((s=)4

=)40

5=)8

Using

Kqk=1

m )k( )!d

m)k

dsm)k

s ) pq( )m

H s( )#$

&'s*pq

, k= 1,2,!,m

K11 =1

2 )1( )!

d2)1

ds2)1 s + 4( )

2H s( )#$

&'s*)4

=d

ds

10s

s + 9

#

$%&

'(s*)4





8/13

8


H s( ) =10s

2

s + 4( ) s + 9( ) , !> "4# Improper in s

H s( ) =10s

2

s2+13s + 36

, ! >"4

Synthetic Division$ s 2 +13s + 36 10s2

10s2+130s + 360

"130s" 360

10

H s( ) =10 "130s + 360

s + 4( ) s + 9( )= 10 "

"32

s + 4+

162

s + 9

%

&'(

)* , !> "4

h t( ) = 10+ t( )" 162e"9t " 32e"4 t%& ()u t( )


Method 1

G s( ) =s

s! 3( ) s2 ! 4s + 5( ) , " < 2

G s( ) =s

s! 3( ) s!2 + j( ) s!2 ! j( ) , "< 2

G s( ) =3 / 2

s! 3!

3+ j( ) / 4

s!2 + j!

3! j( ) / 4s!2 ! j

, " < 2

g t( ) = !3

2e

3t+

3+ j

4e

2!j( )t+

3! j4

e2+j( )t#

$% &

'(u !t( )

Inverse Laplace Transform Example


g t( ) = !3

2e

3t+

3+ j

4e

2!j( )t+

3! j4

e2+j( )t"

#$ %

&'u ! t( )

This looks like a function of time that is complex-valued. But,

with the use of some trigonometric identities it can be put into

the form

g t( ) = 3 / 2( ) e2t cos t( )+ 1 / 3( )sin t( )() *+!e3t{ }u ! t( )

which has only real values.



Method 2

G s( ) =s

s ! 3( ) s2 ! 4s + 5( ) , " < 2

G s( ) =s

s ! 3( ) s ! 2 + j( ) s ! 2 ! j( ) , " < 2

G s( ) =3 / 2

s ! 3!

3+ j( ) / 4

s ! 2 + j!

3! j( ) / 4

s ! 2 ! j , " < 2

Getting a common denominator and simplifying

G s( ) =3 / 2

s ! 3!

1

4

6s !10

s2! 4s + 5

=3 / 2

s ! 3!

6

4

s ! 5 / 3

s ! 2( )2+1

, " < 2



Method 2

G s( ) =3 / 2

s! 3!

6

4

s! 5 / 3

s! 2( )2+1

, "< 2

The denominator of the second term has the form of the Laplace

transform of a damped cosine or damped sine but the numerator

is not yet in the correct form. But by adding and subtracting thecorrect expression from that term and factoring we can put it into

the form

G s( ) =3 / 2

s! 3!

3

2

s! 2

s! 2( )2+1

+1 / 3

s! 2( )2+1

#

$%

&

'( , " < 2



Method 2

G s( ) =3 / 2

s! 3!

3

2

s! 2

s! 2( )2+1

+1 / 3

s! 2( )2+1

"

#$

%

&' , (< 2

This can now be directly inverse Laplace transformed into

g t( ) = 3 / 2( ) e2t cos t( )+ 1 / 3( )sin t( )"# %&!e3t{ }u !t( )

which is the same as the previous result.




9/13

9

Method 3

When we have a pair of poles p2 and p3 that are complex conjugates

we can convert the form G s( ) =A

s ! 3+

K2

s ! p2+

K3

s ! p3into the

form G s( ) =A

s ! 3+

s K2 +

K3( )! K3p2 ! K2p3

s2! p1 + p2( ) s + p1p2

=A

s ! 3+

Bs + C

s2! p1 + p2( )s +p1p2

In this example we can find the constants A, Band Cby realizing that

G s( ) =s

s ! 3( ) s2 ! 4s + 5( )"

A

s ! 3+

Bs + C

s2! 4s + 5

, #< 2

is not just an equation, it is an identity. That means it must be an

equality for any value of s.



Inverse Laplace Transform Example Method 3

Acan be found as before to be 3/ 2. Letting s = 0, the

identity becomes 0 ! "3 / 2

3+C

5and C= 5 / 2. Then, letting

s = 1, and solving we get B = " 3/ 2. Now

G s( ) =3 / 2

s " 3+

"3 / 2( )s + 5 / 2s

2" 4s + 5

, #< 2

or

G s( ) =3 / 2

s " 3"

3

2

s " 5 / 3

s2

" 4s + 5 , #< 2

This is the same as a result in Method 2 and the rest of the solution

is also the same. The advantage of this method is that all the

numbers are real.


Use of MATLAB in Partial Fraction

Expansion

MATLAB has a function residuethat can be very helpful in

partial fraction expansion. Its syntax is[r,p,k] = residue(b,a)

where bis a vector of coefficients of descending powers of s

in the numerator of the expression and ais a vector of coefficients

of descending powers of sin the denominator of the expression,

ris a vector of residues, pis a vector of finite pole locations and

kis a vector of so-called direct terms which result when the

degree of the numerator is equal to or greater than the degree

of the denominator. For our purposes, residues are simply the

numerators in the partial-fraction expansion.


Laplace Transform Properties

Let g t( ) and h t( ) form the transform pairs, g t( ) L! "# G s( )

and h t( ) L! "# H s( ) with ROC's, ROCG and ROCH respectively.

Linearity $g t( )+ %h t( ) L! "# $G s( )+ %H s( )

ROC & ROCG ' ROCH

Time Shifting g t ( t0( )L

! "# G s( )e(st0

ROC = ROCG

s-Domain Shift es0t g t( ) L! "# G s ( s0( )

ROC = ROCG shifted by s0,

(sis in ROC if s ( s0 is in ROCG )


Time Scaling g at( ) L! "# 1/ a( )G s /a( ) ROC = ROCG scaled by a

(sis in ROC if s /ais in ROCG )

Time Differentiationd

dtg t( ) L! "# sG s( )

ROC $ ROCG

s-Domain Differentiation % tg t( ) L! "#d

dsG s( )

ROC= ROCG



Convolution in Time g t( )!h t( ) L" #$ G s( )H s( )

ROC % ROCG & ROCH

Time Integration g '( )d'()

t

*L" #$ G s( ) /s

ROC % ROCG & + > 0( )

If g t( ) = 0 , t< 0 and there are no impulses or higher-order

singularities at t= 0 then

Initial Value Theorem: g 0+( ) = lims#)

sG s( )

Final Value Theorem: limt#)

g t( ) = lims#0

sG s( ) if limt#)

g t( ) exists




10/13

10

FinalValueTheorem limt!"

g t( ) = lims!

0

sG s( )

This theorem only applies if the limit limt!"

g t( ) actually exists.

It is possible for the limit lims!0

sG s( ) to exist even though the

limit limt!"

g t( ) does not exist. For example

x t( ) = cos #0t( )L

$ !% X s( ) =s

s2+#0

2

lims!0

s X s( ) = lims!0

s2

s2+#0

2 = 0

but limt!"

cos #0t( ) does not exist.



Final Value TheoremLaplace Transform Properties

The final value theorem applies to a function G s( ) if all the

poles of sG s( ) lie in the open left half of the splane. Be sure

to notice that this does not say that all the poles of G s( ) must

lie in the open left half of the splane. G s( ) could have a single

pole at s = 0 and the final value theorem would still apply.


Use of Laplace Transform

Properties

Find the Laplace transforms of x t( ) = u t( ) ! u t! a( ) and

x 2t( ) = u 2t( ) ! u 2t! a( ). From the table u t( ) L" #$ 1 /s, %> 0.

Then, using the time-shifting property u t! a( ) L" #$ e!as / s, % > 0.

Using the linearity property u t( ) ! u t! a( ) L" #$ 1! e!as( ) /s, %> 0.Using the time-scaling property

u 2t( ) ! u 2t! a( ) L" #$1

2

1! e!as

s

&

'(

)

*+

s#s/2

=1! e!as/2

s

, %> 0


Use of Laplace Transform

Properties

Use the s-domain differentiation property and

u t( ) L! "# 1 /s , $ > 0

to find the inverse Laplace transform of 1/ s2. The s-domain

differentiation property is %tg t( ) L! "# d

dsG s( )( ). Then

%tu t( ) L! "# d

ds

1

s

&'(

)*+= %

1

s2

. Then using the linearity property

tu t( ) L! "#1

s2

.


The Unilateral Laplace

TransformIn most practical signal and system analysis using the Laplace

transform a modified form of the transform, called the unilateral

Laplace transform, is used. The unilateral Laplace transform is

defined by G s( ) = g t( )e!st dt0!

"

# . The only difference betweenthis version and the previous definition is the change of the lower

integration limit from ! "to 0 !. With this definition, all the

Laplace transforms of causal functions are the same as before

with the same ROC, the region of the splane to the right of all

the finite poles.



Transform

The unilateral Laplace transform integral excludes negative time.

If a function has non-zero behavior in negative time its unilateral

and bilateral transforms will be different. Also functions with the

same positive time behavior but different negative time behavior

will have the same unilateral Laplace transform. Therefore, to avoidambiguity and confusion, the unilateral Laplace transform should

only be used in analysis of causal signals and systems. This is a

limitation but in most practical analysis this limitation is not significant

and the unilateral Laplace transform actually has advantages.



11/13

11


Transform

The main advantage of the unilateral Laplace transform is that

the ROC is simpler than for the bilateral Laplace transform and,

in most practical analysis, involved consideration of the ROC is

unnecessary. The inverse Laplace transform is unchanged. It is

g t( ) =1

j2!G s( )e+stds

"#j$

"+j$

%



Transform

Some of the properties of the unilateral Laplace transform aredifferent from the bilateral Laplace transform.

Time-Shifting g t! t0( )L" #$ G s( )e!st0 , t0 > 0

Time Scaling g at( ) L" #$ 1 / a( )G s /a( ) , a > 0

First Time Derivatived

dtg t( ) L" #$ sG s( )! g 0!( )

Nth Time Derivatived

N

dtN

g t( )( ) L" #$ sN G s( )! sN!n d

n!1

dtn!1

g t( )( )%

&'

(

)*t=0!n=1

N

+

Time Integration g ,( )d,0!

t

-L" #$ G s( ) /s



Transform

The time shifting property applies only for shifts to the right because

a shift to the left could cause a signal to become non-causal. For the

same reason scaling in time must only be done with positive scaling

coefficients so that time is not reversed producing an anti-causal function.

The derivative property must now take into account the initial value

of the function at time t = 0! and the integral property applies only to

functional behavior after time t = 0. Since the unilateral and bilateral

Laplace transforms are the same for causal functions, the bilateral table

of transform pairs can be used for causal functions.



TransformThe Laplace transform was developed for the solution of differential

equations and the unilateral form is especially well suited for solving

differential equations with initial conditions. For example,

d

2

dt2

x t( )!" #$ + 7d

dtx t( )!" #$ +12x t( ) = 0

with initial conditions x 0%( ) = 2 and d

dtx t( )( )

t=0% =%4.

Laplace transforming both sides of the equation, using the new

derivative property for unilateral Laplace transforms,

s2 X s( )%s x 0%( )%

d

dtx t( )( )

t=0% + 7 s X s( )% x 0

%( )!" #$ +12 X s( ) = 0



Transform


Pole-Zero Diagrams and

Frequency Response



12/13

12


Frequency Response



Frequency Response

lim!"0

+ H j!( ) = lim!"0+ 3 j!

j!+ 3=0lim

!"0# H j!( ) = lim!"0# 3

j!

j!+ 3=0

lim!"+#

H j!( ) = lim!"+#

3 j!

j!+ 3= 3lim

!"#$

H j!( ) = lim!"#$

3 j!

j!+ 3= 3



Frequency Response

lim!"0+!H j!( ) =#

2$ 0 =

#

2lim!"0#!H j!( ) = #$

2# 0 = #

$

2

lim!"#$!H j!( ) =#%

2# #

%

2

&'(

)*+ = 0 lim!"+#!H j!( ) =

$

2%$

2=0



Frequency Response



Frequency Response



Frequency Response



13/13

13


Frequency Response



Frequency Response



Frequency Response



Frequency Response



Frequency Response


filter analysis3

Documents