the laplace transformweb.eecs.utk.edu/~mjr/ece316/presentationslides/chapter8.pdf · 2011. 1....

The Laplace Transform

1/13/11 M. J. Roberts - All Rights Reserved 2

Generalizing the Fourier Transform

The CTFT expresses a time-domain signal as a linearcombination of complex sinusoids of the form e jωt . In thegeneralization of the CTFT to the Laplace transform, thecomplex sinusoids become complex exponentials of theform est where s can have any complex value. Replacingthe complex sinusoids with complex exponentials leads tothis definition of the Laplace transform.

L x t( )( ) = X s( ) = x t( )e− st dt−∞

∞

∫ x t( ) L← →⎯ X s( )


The variable s is viewed as a generalization of the variable ω ofthe form s = σ + jω . Then, when σ , the real part of s, is zero, theLaplace transform reduces to the CTFT. Using s = σ + jω the Laplace transform is

X s( ) = x t( )e− σ + jω( )t dt−∞

∞

∫= F x t( )e−σ t⎡⎣ ⎤⎦

which is the Fouriertransform of x t( )e−σ t



The extra factor e−σ t is sometimes called a convergence factorbecause, when chosen properly, it makes the integral convergefor some signals for which it would not otherwise converge.For example, strictly speaking, the signal Au t( ) does not havea CTFT because the integral does not converge. But if it ismultiplied by the convergence factor, and the real part of sis chosen appropriately, the CTFT integral will converge.

Au t( )e− jωt dt−∞

∞

∫ = A e− jωt dt0

∞

∫ ← Does not converge

Ae−σ t u t( )e− jωt dt−∞

∞

∫ = A e− σ + jω( )t dt0

∞

∫ ←Converges (if σ > 0)



Complex Exponential Excitation

If a continuous-time LTI system is excited by a complexexponential x t( ) = Aest , where A and s can each be any complexnumber, the system response is also a complex exponential of the same functional form except multiplied by a complex constant. The response is the convolution of the excitation with the impulse response and that is

y t( ) = h τ( )x t −τ( )dτ−∞

∞

∫ = h τ( )Aes t−τ( )dτ−∞

∞

∫ = Aestx t( ) h τ( )e− sτdτ

−∞

∞

∫

The quantity H s( ) = h τ( )e− sτdτ−∞

∞

∫ is called the Laplace transform

of h t( ).


Complex Exponential Excitation

Let x t( ) = 6 + j3( )A

e 3− j2( )s

t = 6.708∠0.4637( )e 3− j2( )t

and let h t( ) = e−4 t u t( ). Then H s( ) = 1s + 4

, σ > −4 and,

in this case, s = 3− j2 = σ + jω with σ = 3 > −4 and ω = −2.

y t( ) = x t( )H s( ) = 6 + j33− j2 + 4

e 3− j2( )t = 0.6793∠0.742( )e 3− j2( )t .

The response is the same functional form as the excitation butmultiplied by a different complex constant. This only happens whenthe excitation is a complex exponential and that is what makescomplex exponentials unique.


Pierre-Simon Laplace

3/23/1749 - 3/2/1827

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 of

y t( ) is Y s( ) = y t( )e−stdt

−∞

∞

∫ = h t( )∗x t( )⎡⎣ ⎤⎦e−stdt−∞

∞

∫

Y s( ) = h τ( )x t −τ( )dτ−∞

∞

∫⎛

⎝⎜⎞

⎠⎟e−stdt

−∞

∞

∫

Y s( ) = h τ( )dτ−∞

∞

∫ x t −τ( )e−stdt−∞

∞

∫



Y s( ) = h τ( )dτ−∞

∞

∫ x t −τ( )e−stdt−∞

∞

∫Let λ = t −τ ⇒ dλ = dt. Then

Y s( ) = h τ( )dτ−∞

∞

∫ x λ( )e−s λ+τ( )dλ−∞

∞

∫ = h τ( )e−sτdτ−∞

∞

∫H s( )

x λ( )e−sλdλ

−∞

∞

∫X s( )

Y s( ) = H s( )X s( )H s( ) is called the transfer function.

y t( ) = x t( )∗h t( ) L← →⎯ Y s( ) = H s( )X s( )



Let x t( ) = u t( ) and let h t( ) = e−4t u t( ). Find y t( ).y t( ) = x t( )∗h t( ) = x τ( )h t −τ( )dτ

−∞

∞

∫ = u τ( )e−4 t−τ( ) u t −τ( )dτ−∞

∞

∫

y t( ) = e−4 t−τ( ) dτ0

t

∫ = e−4t e4τ dτ0

t

∫ = e−4t e4t −14

= 1− e−4t

4 , t > 0

0 , t < 0

⎧

⎨⎪

⎩⎪

y t( ) = 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 + 4x t( ) = 1/ 4( ) 1− e−4t( )u t( )1/13/11 M. J. Roberts - All Rights Reserved 10


Cascade-Connected Systems

If two systems are cascade connected the transfer function ofthe overall system is the product of the transfer functions of thetwo individual systems.

A very common form of transfer function is a ratio of two polynomials in s,

H s( ) = Y s( )X s( ) =

bksk

k=0

N∑aks

kk=0

N∑= bNs

N + bN−1sN−1 ++ b1s + b0

aNsN + aN−1s

N−1 ++ a1s + a0

Direct Form II Realization



The transfer function can be conceived as the product of twotransfer functions,

H1 s( ) = Y1 s( )X s( ) =

1aNs

N + aN−1sN−1 ++ a1s + a0

and

H2 s( ) = Y s( )Y1 s( ) = bNs

N + bN−1sN−1 ++ b1s + b0



From

H1 s( ) = Y1 s( )X s( ) =

1aNs

N + aN−1sN−1 ++ a1s + a0

we get

X s( ) = aNsN + aN−1s

N−1 ++ a1s + a0⎡⎣ ⎤⎦Y1 s( )orX s( ) = aNsN Y1 s( ) + aN−1s

N−1 Y1 s( ) ++ a1sY1 s( ) + a0 Y1 s( )Rearranging

sN Y1 s( ) = 1aN

X s( )− aN−1sN−1 Y1 s( ) ++ a1sY1 s( ) + a0 Y1 s( )⎡⎣ ⎤⎦{ }


1/13/11 M. J. Roberts - All Rights Reserved 17 1/13/11 M. J. Roberts - All Rights Reserved 17


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

H s( ) = s − 5s2 + 3s + 7

Inverse Laplace Transform

There is an inversion integral

y t( ) = 1j2π

Y s( )estdsσ − j∞

σ + j∞

∫ , s = σ + jω

for finding y t( ) from Y s( ), but it is rarely used in practice.

Usually inverse Laplace transforms are found by using tablesof standard functions and the properties of the Laplace transform.


Existence of the Laplace Transform

Time Limited Signals

If x t( ) = 0 for t < t0 and t > t1 it is a time limited signal. If

x t( ) is also bounded for all t, the Laplace transform integral

converges and the Laplace transform exists for all s.



Let x t( ) = rect t( ) = u t +1/ 2( )− u t −1/ 2( ).X s( ) = rect t( )e−st dt

−∞

∞

∫ = e−st dt−1/2

1/2

∫ = e−s/2 − es/2

−s= es/2 − e−s/2

s , All s



Right- and Left-Sided Signals

Right-Sided Left-Sided



Right- and Left-Sided Exponentials

Right-Sided Left-Sided x t( ) = eαt u t − t0( ) , α ∈ x t( ) = eβt u t0 − t( ) , β ∈



Right-Sided Exponential

x t( ) = eα t u t − t0( ) , α ∈

X s( ) = eα te− stdtt0

∞

∫ = e α −σ( )te− jω tdtt0

∞

∫If Re s( ) = σ >α the asymptotic

behavior of e α −σ( )te− jω t as t→∞ is to approach zero and the Laplace transform integral converges.



Left-Sided Exponential

x t( ) = eβt u t0 − t( ) , β ∈

X s( ) = eβte− stdt−∞

t0

∫ = e β−σ( )te− jωtdt−∞

t0

∫If σ < β the asymptotic behavior of

e β−σ( )te− jωt as t→−∞ is to approach zero and the Laplace transformintegral converges.



The two conditions σ >α and σ < β define the region ofconvergence (ROC) for the Laplace transform of right- and left-sided signals.



Any right-sided signal that grows no faster than an exponential in 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( ) + xl t( ) where xr t( ) is the right-sided part andxl t( ) is the left-sided part and if xr t( ) < Kre

α t and xl t( ) < Kleβt

and α and β are as small as possible, then the Laplace-transform integral converges and the Laplace transform exists for α < σ < β. Therefore if α < β the ROC is the region α < β. If α > β, there is no ROC and the Laplace transform does not exist.



Laplace Transform Pairs

The Laplace transform of g1 t( ) = Aeαt u t( ) is

G1 s( ) = Aeαt u t( )e− stdt−∞

∞

∫ = A e− s−α( )tdt0

∞

∫ = A e α−σ( )te− jωtdt0

∞

∫ = As −α

This function has a pole at s = α and the ROC is the region to the right of that point. The Laplace transform of g2 t( ) = Aeβt u −t( ) is

G2 s( ) = Aeβt u −t( )e− stdt−∞

∞

∫ = A e β−s( )tdt−∞

0

∫ = A e β−σ( )te− jωtdt−∞

0

∫ = − As − β

This function has a pole at s = β and the ROC is the region to the left of that point.


Region of Convergence

The following two Laplace transform pairs illustrate the importance of the region of convergence.

e−αt u t( ) L← →⎯ 1s +α

, σ > −α

− e−αt u −t( ) L← →⎯ 1s +α

, σ < −α

The two time-domain functions are different but the algebraic expressions for their Laplace transforms are the same. Only the ROC’s are different.

Region of Convergence

δ t( ) L← →⎯ 1 , All σ

u t( ) L← →⎯ 1/ s , σ > 0 − u −t( ) L← →⎯ 1/ s , σ < 0

ramp t( ) = t u t( ) L← →⎯ 1/ s2 , σ > 0 ramp −t( ) = −t u −t( ) L← →⎯ 1/ s2 , σ < 0

e−αt u t( ) L← →⎯ 1/ s +α( ) , σ > −α − e−αt u −t( ) L← →⎯ 1/ s +α( ) , σ < −α

e−αt sin ω0t( )u t( ) L← →⎯ω0

s +α( )2+ω0

2, σ > −α − e−αt sin ω0t( )u −t( ) L← →⎯

ω0

s +α( )2+ω0

2, σ < −α

e−αt cos ω0t( )u t( ) L← →⎯ s +α

s +α( )2+ω0

2, σ > −α − e−αt cos ω0t( )u −t( ) L← →⎯ s +α

s +α( )2+ω0

2, σ < −α

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← →⎯ 1s +1

, σ > −1

e2t u −t( ) L← →⎯ − 1s − 2

, σ < 2

e− t u t( ) + e2t u −t( ) L← →⎯ 1s +1

− 1s − 2

, −1<σ < 2



Find the inverse Laplace transform of

X s( ) = 4s + 3

− 10s − 6

, − 3 <σ < 6

The ROC tells us that 4s + 3

must inverse transform into a

right-sided signal and that 10s − 6

must inverse transform into

a left-sided signal. x t( ) = 4e−3t u t( ) +10e6t u −t( )




X s( ) = 4s + 3

− 10s − 6

, σ > 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( )




X s( ) = 4s + 3

− 10s − 6

, σ < −3

The ROC tells us that both terms must inverse transform into a left-sided signal. x t( ) = −4e−3t u −t( ) +10e6t u −t( )


MATLAB System Objects

A MATLAB system object is a special kind of variable inMATLAB that contains all the information about a system.It can be created with the tf command whose syntax is sys = tf(num,den)where num is a vector of numerator coefficients of powers of s, denis a vector of denominator coefficients of powers of s, both in descending

order and sys is the system object.


MATLAB System Objects

For example, the transfer function

H1 s( ) = s2 + 4s5 + 4s4 + 7s3 +15s2 + 31s + 75

can be created by the commands»num = [1 0 4] ; den = [1 4 7 15 31 75] ;

»H1 = tf(num,den) ;

»H1

Transfer function:

s ^ 2 + 4

----------------------------------------

s ^ 5 + 4 s ^ 4 + 7 s ^ 3 + 15 s ^ 2 + 31 s + 75


Partial-Fraction Expansion

The inverse Laplace transform can always be found (in principle at least) by using the inversion integral. But that is rare in engineering practice. The most common type of Laplace-transform expression is a ratio of polynomials in s,

G s( ) = bM sM + bM −1s

M −1 ++ b1s + b0

sN + aN−1sN−1 +a1s + a0

The denominator can be factored, putting it into the form,

G s( ) = bM sM + bM −1s

M −1 ++ b1s + b0

s − p1( ) s − p2( ) s − pN( )



For now, assume that there are no repeated poles and that N > M , making the fraction proper in s. Then it is possible to write the expression in the partial fraction form,

G s( ) = K1

s − p1

+ K2

s − p2

++ KN

s − pNwhere

bM sM + bM −1s

M −1 +b1s + b0

s − p1( ) s − p2( ) s − pN( ) = K1

s − p1

+ K2

s − p2

++ KN

s − pNThe K’s can be found be any convenient method.



Multiply both sides by s − p1

s − p1( )bM sM + bM −1s

M −1 ++ b1s + b0

s − p1( ) s − p2( ) s − pN( ) =K1 + s − p1( ) K2

s − p2

+

+ s − p1( ) KN

s − pN

⎡

⎣

⎢⎢⎢⎢

⎤

⎦

⎥⎥⎥⎥

K1 =bM p1

M + bM p1M −1 ++ b1p1 + b0

p1 − p2( ) p1 − pN( )All the K’s can be found by the same method and the inverse Laplace transform is then found by table look-up.


Partial-Fraction Expansion H s( ) = 10s

s + 4( ) s + 9( ) =K1

s + 4+ K2

s + 9 , σ > −4

K1 = s + 4( ) 10ss + 4( ) s + 9( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥s=−4

= 10ss + 9

⎡⎣⎢

⎤⎦⎥s=−4

= −405

= −8

K2 = s + 9( ) 10ss + 4( ) s + 9( )

⎡

⎣⎢⎢

⎤

⎦⎥⎥s=−9

= 10ss + 4

⎡⎣⎢

⎤⎦⎥s=−9

= −90−5

= 18

H s( ) = −8s + 4

+ 18s + 9

= −8s − 72 +18s + 72s + 4( ) s + 9( ) = 10s

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

↓ ↓ ↓ ↓ ↓

h t( ) = −8e−4 t +18e−9t( )u t( )



If the expression has a repeated pole of the form,

G s( ) = bM sM + bM −1s

M −1 ++ b1s + b0

s − p1( )2 s − p3( ) s − pN( )the partial fraction expansion is of the form,

G s( ) = K12

s − p1( )2 +K11

s − p1

+ K3

s − p3

++ KN

s − pNand K12 can be found using the same method as before.But K11 cannot be found using the same method.



Instead K11 can be found by using the more general formula

Kqk =1

m − k( )!dm−k

dsm−k s − pq( )m H s( )⎡⎣

⎤⎦s→ pq

, k = 1,2,,m

where m is the order of the qth pole, which applies to repeated poles of any order.

If the expression is not a proper fraction in s the partial-fraction method will not work. But it is always possible to synthetically divide the numerator by the denominator until the remainder is a proper fraction and then apply partial-fraction expansion.



H s( ) = 10ss + 4( )2 s + 9( )

= K12

s + 4( )2 +K11

s + 4+ K2

s + 9 , σ > 4

↑ Repeated Pole

K12 = s + 4( )2 10ss + 4( )2 s + 9( )

⎡

⎣

⎢⎢

⎤

⎦

⎥⎥s=−4

= −405

= −8

Using

Kqk =1

m − k( )!dm−k

dsm−k s − pq( )m H s( )⎡⎣

⎤⎦s→ pq

, k = 1,2,,m

K11 =1

2 −1( )!d 2−1

ds2−1 s + 4( )2 H s( )⎡⎣ ⎤⎦s→−4= dds

10ss + 9

⎡⎣⎢

⎤⎦⎥s→−4


Partial-Fraction Expansion H s( ) = 10s2

s + 4( ) s + 9( ) , σ > −4 ← Improper in s

H s( ) = 10s2

s2 +13s + 36 , σ > −4

Synthetic Division→ s2 +13s + 36 10s2

10s2 +130s + 360−130s − 360

10

H s( ) = 10 − 130s + 360s + 4( ) s + 9( ) = 10 − −32

s + 4+ 162s + 9

⎡⎣⎢

⎤⎦⎥

, σ > −4

h t( ) = 10δ t( )− 162e−9t − 32e−4 t⎡⎣ ⎤⎦u t( )


Method 1

G s( ) = ss − 3( ) s2 − 4s + 5( ) , σ < 2

G s( ) = ss − 3( ) s − 2 + j( ) s − 2 − j( ) , σ < 2

G s( ) = 3 / 2s − 3

−3+ j( ) / 4s − 2 + j

−3− j( ) / 4s − 2 − j

, σ < 2

g t( ) = − 32e3t + 3+ j

4e 2− j( )t + 3− j

4e 2+ j( )t⎛

⎝⎜⎞⎠⎟ u −t( )

Inverse Laplace Transform Example


g t( ) = − 32e3t + 3+ j

4e 2− j( )t + 3− j

4e 2+ 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 intothe form

g t( ) = 3 / 2( ) e2t cos t( ) + 1 / 3( )sin t( )⎡⎣ ⎤⎦ − e3t{ }u −t( )

which has only real values.



Method 2

G s( ) = ss − 3( ) s2 − 4s + 5( ) , σ < 2

G s( ) = ss − 3( ) s − 2 + j( ) s − 2 − j( ) , σ < 2

G s( ) = 3 / 2s − 3

−3+ j( ) / 4s − 2 + j

−3− j( ) / 4s − 2 − j

, σ < 2

Getting a common denominator and simplifying

G s( ) = 3 / 2s − 3

−14

6s −10s2 − 4s + 5

=3 / 2s − 3

−64

s − 5 / 3s − 2( )2 +1

, σ < 2



Method 2

G s( ) = 3 / 2s − 3

− 64

s − 5 / 3s − 2( )2 +1

, σ < 2

The denominator of the second term has the form of the Laplacetransform of a damped cosine or damped sine but the numeratoris not yet in the correct form. But by adding and subtracting thecorrect expression from that term and factoring we can put it intothe form

G s( ) = 3 / 2s − 3

− 32

s − 2s − 2( )2 +1

+ 1 / 3s − 2( )2 +1

⎡

⎣⎢

⎤

⎦⎥ , σ < 2



Method 2

G s( ) = 3 / 2s − 3

− 32

s − 2s − 2( )2 +1

+ 1 / 3s − 2( )2 +1

⎡

⎣⎢

⎤

⎦⎥ , σ < 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.



Method 3 When we have a pair of poles p2 and p3 that are complex conjugates

we can convert the form G s( ) = As − 3

+ K2

s − p2

+ K3

s − p3

into the

form G s( ) = As − 3

+s K2 + K3( )− K3p2 − K2 p3

s2 − p1 + p2( )s + p1p2

= As − 3

+ Bs +Cs2 − p1 + p2( )s + p1p2

In this example we can find the constants A, B and C by realizing that

G s( ) = ss − 3( ) s2 − 4s + 5( ) ≡

As − 3

+ Bs +Cs2 − 4s + 5

, σ < 2

is not just an equation, it is an identity. That means it must be anequality for any value of s.



Inverse Laplace Transform Example Method 3 A can be found as before to be 3 / 2. Letting s = 0, the

identity becomes 0 ≡ − 3 / 23

+ C5

and C = 5 / 2. Then, letting

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

G s( ) = 3 / 2s − 3

+−3 / 2( )s + 5 / 2s2 − 4s + 5

, σ < 2

or

G s( ) = 3 / 2s − 3

− 32

s − 5 / 3s2 − 4s + 5

, σ < 2

This is the same as a result in Method 2 and the rest of the solutionis 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 residue that can be very helpful inpartial fraction expansion. Its syntax is [r,p,k] = residue(b,a)where b is a vector of coefficients of descending powers of s in the numerator of the expression and a is a vector of coefficients of descending powers of s in the denominator of the expression, r is a vector of residues, p is a vector of finite pole locations and k is 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 α g t( ) + β h t( ) L← →⎯ αG s( ) + βH 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 , (s is in ROC if s − s0 is in ROCG )


Time Scaling g at( ) L← →⎯ 1 / a( )G s / a( ) ROC = ROCG scaled by a (s is in ROC if s / a is in ROCG )

Time Differentiation ddt

g t( ) L← →⎯ sG s( ) ROC ⊇ ROCG

s-Domain Differentiation − t g t( ) L← →⎯ dds

G s( ) ROC = ROCG



Convolution in Time g t( )∗h t( ) L← →⎯ G s( )H s( ) ROC ⊇ ROCG ∩ROCH

Time Integration g τ( )dτ−∞

t

∫ L← →⎯ G s( ) / s

ROC ⊇ ROCG ∩ σ > 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: lim

t→∞g t( ) = lim

s→0sG s( ) if lim

t→∞g t( ) exists



Final Value Theorem limt→∞

g t( ) = lims→0

sG s( )This theorem only applies if the limit lim

t→∞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 ω0t( ) L← →⎯ X s( ) = ss2 +ω0

2

lims→0

sX s( ) = lims→0

s2

s2 +ω02 = 0

but limt→∞

cos ω0t( ) does not exist.



Final Value Theorem


The final value theorem applies to a function G s( ) if all thepoles of sG s( ) lie in the open left half of the s plane. Be sureto notice that this does not say that all the poles of G s( ) must

lie in the open left half of the s plane. G s( ) could have a singlepole 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, σ > 0.Then, using the time-shifting property u t − a( ) L← →⎯ e−as / s, σ > 0.

Using the linearity property u t( )− u t − a( ) L← →⎯ 1− e−as( ) / s, σ > 0.Using the time-scaling property

u 2t( )− u 2t − a( ) L← →⎯ 12

1− e−as

s⎡

⎣⎢

⎤

⎦⎥s→s /2

= 1− e−as /2

s, σ > 0


Use of Laplace Transform Properties

Use the s-domain differentiation property and u t( ) L← →⎯ 1 / s , σ > 0 to find the inverse Laplace transform of 1 / s2 . The s-domain

differentiation property is − t g t( ) L← →⎯ dds

G s( )( ). Then

−t u t( ) L← →⎯ dds

1s

⎛⎝⎜

⎞⎠⎟ = − 1

s2 . Then using the linearity property

t u t( ) L← →⎯ 1s2 .


The Unilateral Laplace Transform

In most practical signal and system analysis using the Laplacetransform a modified form of the transform, called the unilateralLaplace transform, is used. The unilateral Laplace transform is

defined by G s( ) = g t( )e− st dt0−

∞

∫ . The only difference between

this version and the previous definition is the change of the lowerintegration limit from − ∞ to 0−. With this definition, all the Laplace transforms of causal functions are the same as beforewith the same ROC, the region of the s plane to the right of allthe finite poles.



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 avoid ambiguity 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.



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 isunnecessary. The inverse Laplace transform is unchanged. It is

g t( ) = 1j2π

G s( )e+ stdsσ − j∞

σ + j∞

∫



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 Derivative ddt

g t( ) L← →⎯ sG s( ) − g 0−( )

Nth Time Derivative dN

dt Ng t( )( ) L← →⎯ sN G s( ) − sN −n dn−1

dt n−1 g t( )( )⎡

⎣⎢

⎤

⎦⎥t=0−n=1

N

∑

Time Integration g τ( )dτ0−

t

∫ L← →⎯ G s( ) / s



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 thesame reason scaling in time must only be done with positive scalingcoefficients so that time is not reversed producing an anti-causal function.The derivative property must now take into account the initial valueof the function at time t = 0− and the integral property applies only tofunctional behavior after time t = 0. Since the unilateral and bilateralLaplace transforms are the same for causal functions, the bilateral tableof transform pairs can be used for causal functions.



The Laplace transform was developed for the solution of differentialequations and the unilateral form is especially well suited for solvingdifferential equations with initial conditions. For example,

d2

dt 2 x t( )⎡⎣ ⎤⎦ + 7 ddt

x t( )⎡⎣ ⎤⎦ +12x t( ) = 0

with initial conditions x 0−( ) = 2 and ddt

x t( )( )t=0− = −4.

Laplace transforming both sides of the equation, using the new derivative property for unilateral Laplace transforms,

s2 X s( ) − sx 0−( ) − ddt

x t( )( )t=0− + 7 sX s( ) − x 0−( )⎡⎣ ⎤⎦ +12 X s( ) = 0



Solving for X s( )

X s( ) =sx 0−( )

=2+ 7 x 0−( )

=2+ ddt

x t( )( )t=0−

=−4

s2 + 7s +12

or X s( ) = 2s +10s2 + 7s +12

=4

s + 3−

2s + 4

. The inverse transform yields

x t( ) = 4e−3t − 2e−4 t( )u t( ). This solution solves the differential equationwith the given initial conditions.


Pole-Zero Diagrams and Frequency Response

If the transfer function of a stable system is H(s), the frequency response is H(jω ). The most common type of transfer functionis of the form,

H s( ) = A s − z1( ) s − z2( ) s − zM( )s − p1( ) s − p2( ) s − pN( )

Therefore H(jω ) is

H jω( ) = A jω − z1( ) jω − z2( ) jω − zM( )jω − p1( ) jω − p2( ) jω − pN( )


Let H s( ) = 3ss + 3

.

H jω( ) = 3 jωjω + 3

The numerator jω and thedenominator jω + 3 can be conceived as vectors in the s plane.

H jω( ) = 3 jωjω + 3

H jω( ) = 3=0 +jω − jω + 3( )



limω→0+

H jω( ) = limω→0+

3 jωjω + 3

= 0limω→0−

H jω( ) = limω→0−

3 jωjω + 3

= 0

limω→+∞

H jω( ) = limω→+∞

3 jωjω + 3

= 3limω→−∞

H jω( ) = limω→−∞

3 jωjω + 3

= 3



limω→0+H jω( ) = π

2− 0 = π

2 limω→0−H jω( ) = − π

2− 0 = − π

2

limω→−∞H jω( ) = − π

2− − π

2⎛⎝⎜

⎞⎠⎟ = 0

limω→+∞H jω( ) = π

2− π2= 0


the laplace transformweb.eecs.utk.edu/~mjr/ece316/presentationslides/chapter8.pdf · 2011. 1....

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