1 laplace transform. chapter 4 school of computer and communication engineering, unimap pn. nordiana...
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1
Laplace Laplace Transform.Transform.
CHAPTER CHAPTER 44
School of Computer and Communication School of Computer and Communication Engineering, UniMAPEngineering, UniMAP
Pn. Nordiana Mohamad SaaidPn. Nordiana Mohamad Saaid
EKT 230 EKT 230
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4.0 Laplace 4.0 Laplace Transform.Transform.
4.1 Introduction.4.1 Introduction.4.2 The Laplace Transform.4.2 The Laplace Transform.4.3 The Unilateral Transform and 4.3 The Unilateral Transform and
Properties.Properties.4.4 Inversion of the Unilateral.4.4 Inversion of the Unilateral.4.5 Solving Differential Equation with 4.5 Solving Differential Equation with
Initial Conditions.Initial Conditions.4.6 Forced and Natural Responses in 4.6 Forced and Natural Responses in
Unilateral Laplace Transform.Unilateral Laplace Transform.4.7 4.7 Properties of the Bilateral Laplace Properties of the Bilateral Laplace
TransformTransform4.8 Inversion of the Bilateral Laplace 4.8 Inversion of the Bilateral Laplace
Transform.Transform.4.9 The Transfer Function4.9 The Transfer Function4.10 Causality and Stability4.10 Causality and Stability
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4.1 Introduction.4.1 Introduction. In CChapter 3hapter 3 we developed representation of signal
and LTI by using superposition of complex sinusoids.
In this CChapter 4hapter 4 we are considering the continuous-time signal and system representation based on complex exponential signals.
The Laplace transform can be used to analyze a large class of continuous-time problems involving signal that are not absolutely integrable, such as impulse response of an unstable system.
Laplace transform comes in two varietiestwo varieties;(i) Unilateral (one sided); is a tool for solving differential equations with initial condition.
(ii) Bilateral (two sided); offers insight into the nature of system characteristic such as stability, causality, and frequency response.
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4.2 Laplace Transform.4.2 Laplace Transform. Let est be a complex exponential with complex
frequency s = +j. We may write,
The real part of est is an exponential damped cosine And the imaginary part is an exponential damped
sine as shown in Figure 6.1. The real part of s is the exponential damping factor
And the imaginary part of s is the frequency of the
cosine and sine factor, .
.sincos tjetee ttst
5
Figure 4.1: Real and imaginary parts of the complex exponential est, where s = + j.
Cont’d…Cont’d…
6
Apply an input to the form x(t) =est to an LTI system with impulse response h(t). The system output is given by,
Derivation:
We use the input, x(t) =est to obtain
the output, y(t) as:
Thus, transfer function is:
dtxh
txth
txHty
*
dehe
dehty
sst
ts
dehsH s
4.2.1 Eigen Function 4.2.1 Eigen Function Property of eProperty of estst..
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We can write
An eigen function is a signal that passes through the system without being modified except by multiplication by scalar.
The equation below indicates that,- est is the eigenfunction of the LTI
system.- H(s) is the eigen value.
The transfer function expressed in terms of magnitude and phase;
stst esHeHty
sjesHsH
Cont’d…Cont’d…
8
Express complex-value transfer function in Express complex-value transfer function in Rectangular FormRectangular Form
Where |H(s)| and (s) are the magnitude and phase of H(s)
.sincos
)(
jtejHjjtejH
eejHty
js
substitute
eesHty
output; The
tt
jjwtt
stsj
Cont’d…Cont’d…
9
H(s) is the Laplace Transform of h(t) and.. thus the h(t) is the inverse Laplace transform of H(s).
The Laplace transform of x(t) is
The Inverse Laplace Transform of X(s) is
We can express the relationship with the notation
dtetxsX st
4.2.2 Laplace Transform 4.2.2 Laplace Transform Representation.Representation.
j
j
stdsesXj
tx
)(2
1
sXtxL
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The condition for convergence of the Laplace transform is the absolute integrability of x(t)e-t ,
The range of for which the Laplace transform converges is termed the region of convergence (ROC)
Figure 4.2: The Laplace transform applies to more general signals than the Fourier transform does. (a) Signal for which the
Fourier transform does not exist. (b) Attenuating factor associated with Laplace transform.
(c) The modified signal x(t)e -t is absolutely integrable for > 1.
4.2.3 Convergence.4.2.3 Convergence.
dtetx t
11
4.2.4 The s-Plane.4.2.4 The s-Plane. It is convenience to represent the complex
frequency s graphically in terms of the s-plane. (i) the horizontal axis represents the real part of s
(exponential damping factor ).
(ii) The vertical axis represents the imaginary part of s (sinusoidal frequency )
In s-plane, =0 correspond to imaginary axis. Fourier transfrom is given by the Laplace
transform evaluated along the imaginary axis.
0| sXjX
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The j-axis divides the s-plane in half. (i) The region to the left of the j-axis is termed the left half of the s-plane.(ii) The region to the right of the j-axis is termed the right half of the s-plane.
The real part of s is negative in the left half of the s-plane and positive in the right half of the s -plane..
Figure 4.3: The s-plane. The horizontal axis is Re{s}and the vertical axis is Im{s}. Zeros are depicted at s = –1 and s = –4
2j, and poles are depicted at s = –3, s = 2 3j, and s = 4.
Cont’d…Cont’d…
13
Zeros. The ck are the root of the numerator polynomial and are termed the zeros of X(s). Location of zeros are denoted as “o”.
Poles. The dk are the root of the denominator polynomial and are termed the poles of X(s). Location of poles are denoted as “x”.
The Laplace transform does not uniquely correspond to a signal x(t) if the ROC is not specified.
Two different signal may have identical Laplace Transform, but different ROC. Below is the example.
Figure 4.4a Figure 4.4b
Figure 4.4a. The ROC for x(t) = eatu(t) is depicted by the shaded region. A pole is located at s = a.
Figure 4.4b. The ROC for y(t) = –eatu(–t) is depicted by the shaded region. A pole is located at s = a.
4.2.5 Poles and Zeros.4.2.5 Poles and Zeros.
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Example 4.1:Example 4.1: Laplace Transform of a Laplace Transform of a Causal Exponential Signal.Causal Exponential Signal.
Determine the Laplace transform of x(t)=eatu(t).
Solution:Solution:
Step 1Step 1:: Find the Laplace transform. Find the Laplace transform.
To evaluate e-(s-a)t, Substitute s= + j
0
)(
0
)(
1 tas
tas
st
eas
dte
dtetxsX
0
)(1 tajweaj
sX
15
If > 0, then e-(s-a)t goes to zero as t approach infinity, while as t is zero, e-(s-a)t goes to minus one, hence
*The Laplace transform does not exist for =<a because the integral does not converge.
*The ROC is at >a, the shaded region of the s-plane in figure below. The pole is at s=a.
.
.)Re(,1
),10(1
asas
aaj
sX
Figure 4.5: The ROC for x(t) = eatu(t) is depicted by the shaded region. A pole is located at s = a.
Cont’d…Cont’d…
16
4.3 The Unilateral Laplace 4.3 The Unilateral Laplace Transform and Properties.Transform and Properties. The Unilateral Laplace Transform of a signal
x(t) is defined by
The lower limit of implies that we do include discontinuities and impulses that occur at t = 0 in the integral. H(s) depends on x(t)for t >= 0.
The relationship between X(s) and x(t) as
The unilateral and bilateral Laplace transforms are equivalent for signals that are zero for time t<0.
dtetxsX st
0
sXtx uL
0
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Properties of Unilateral Laplace Transform.Properties of Unilateral Laplace Transform.ScalingScaling
Linearity, Linearity,
For a>0Time ShiftTime Shift
for all t such that x(t - )u(t) = x(t - )u(t - ) A shift in in time correspond to multiplication
of the Laplace transform by the complex exponential e-s.
sbYsaXtbytax uL
a
sX
aatx uL 1
sXetx sLu
Cont’d…Cont’d…
18
s-Domain Shifts-Domain Shift
Multiplication by a complex exponential in time introduces a shift in complex frequency s into the Laplace transform.
Figure 4.6: Time shifts for which the unilateral Laplace transform time-shift property does not apply. (a) A nonzero portion of x(t) that occurs at times t 0 is
shifted to times t < 0. (b) A nonzero portion x(t) that occurs at times t < 0 is shifted to times t 0.
00 ssXtxe uLts
Cont’d…Cont’d…
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ConvolutionConvolution..
Convolution in time domain corresponds to multiplication of Laplace transform. This property apply when x(t)=0 and y(t) = 0 for t < 0.
Differentiation in the s-Domain.Differentiation in the s-Domain.
Differentiation in the s-domain corresponds to multiplication by -t in the time domain.
.* sYsXtytx uL
.sXds
dttx uL
Cont’d…Cont’d…
20
Differentiation in the Time Domain.Differentiation in the Time Domain.
Initial and Final Value Theorem.Initial and Final Value Theorem.
The initial value theorem allow us to determine the initial value, x(0-), and the final value, x(infinityof x(t) directly from X(s).
The initial value theorem does NOT apply to rational functions X(s) in which the order of the numerator polynomial is greater than or equal to that of the denominator polynomial.
0xssXtxdt
duL
.0
xssXLim
s
Cont’d…Cont’d…
21
Example 4.2:Example 4.2: Applying Properties.Applying Properties.Find the unilateral Laplace Transform of x(t)=(-e3tu(t))*(tu(t)).
Solution:Solution:Find the Unilateral Laplace Transform.
And
Apply s-domain differentiation property,
Use the convolution property,
.
)3(
1)(3
s
tue uLt
2
1)(
1)(
sttu
stu
u
u
L
L
1 )3(
1)())((*)(()(
23
ss
sXttutuetx uLt
22
We can determine the inverse Laplace transforms using one-to-one relationship between the signal and its unilateral Laplace transform.Appendix D1 consists of the table of Laplace Transform.
X(s) is the sum of simple terms,
Using the residue method, solve for a system linear equation.
Then sum the Inverse Laplace transform of each term.
k
kLtd
k ds
AtueA uk
.!1
1
nk
Ltdn
ds
Atue
n
Atuk
4.4 Inversion of the 4.4 Inversion of the Unilateral Laplace Unilateral Laplace Transform.Transform.
N
kk
k
ds
AsX
1)(
~
23
Example 4.3:Example 4.3: Inversion by Partial-Inversion by Partial-Fraction Expansion.Fraction Expansion.
Find the Inverse Laplace Transform of
Solution:Solution:
Step 1Step 1:: Use the partial fraction expansion of Use the partial fraction expansion of XX((ss) ) to writeto write
Solving the A, B and C by the method of residues
2)2)(1(
43
ss
ssX
2)2()2()1(
s
C
s
B
s
AsX
)1()2(
)1(
)1()2(
)2)(1(
)2)(1(
)2(
)2)(1(
)43(222
2
2
ss
sC
ss
ssB
ss
sA
ss
s
24.2
3)1(3)1(4
)2(int,
1
0
)1(
.1
1
);2()3(
)3(424
)2(334
)1(0
,,
)24()34()(
)1()23()44(
)1()1)(2()2()43(
2
22
2
C
C
oAandBSubstitute
A
BA
From
B
B
CBA
CBA
BA
tcoefficiencompareso
CBAsCBAsBA
sCssBssA
sCssBsAs
Cont’d…Cont’d…
25
A=1, B=-1 and C=2
Step 2Step 2:: Construct the Inverse Laplace transform from Construct the Inverse Laplace transform from the above partial-fraction term above.the above partial-fraction term above.
- The pole of the 1st term is at s = -1, so
- The pole of the 2nd term is at s = -2, so
-The double pole of the 3rd term is at s = -2, so
Step 3: Combining the terms.
.
)1(
1)(
stue uLt
2)2(
2
)2(
1
)1(
1
ssssX
)2(
1)(2
stue uLt
22
)2(
2)(2
stute uLt
).(2)()()( 22 tutetuetuetx ttt
Cont’d…Cont’d…
26
Example 4.4:Example 4.4: Inversion An Improper Rational Laplace Inversion An Improper Rational Laplace Transform.Transform.
Find the Inverse Laplace Transform of
Solution:Solution:
Step 1Step 1:: Use the long division to espress X( Use the long division to espress X(ss) as sum of ) as sum of rational polynomial function.rational polynomial function.
We can write,
43
104922
23
ss
ssssX
23
1293
10123
862
__________32
1049243
2
2
23
232
s
ss
ss
sss
s
sssss
43
2332
2
ss
sssX
27
Use partial fraction to expand the rational function,
Step 2Step 2:: Construct the Inverse Laplace transform Construct the Inverse Laplace transform from the above partial-fraction term above. Refer from the above partial-fraction term above. Refer to the Laplace transform Table.to the Laplace transform Table.
.
).(2)()(3)(2)( 4)1( tuetuetttx tt
4
2
1
132
ssssX
Cont’d…Cont’d…
28
Primary application of unilateral Laplace transform in system analysis, solving differential equations with nonzero initial condition.
Refer to the example.
4.5 Solving Differential 4.5 Solving Differential Equation with Initial Equation with Initial Condition.Condition.
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Example 4.5:Example 4.5: RC Circuit Analysis (Initial RC Circuit Analysis (Initial condition)condition)
Use the Laplace transform to find the voltage across Use the Laplace transform to find the voltage across the capacitor , the capacitor , yy((tt), for the RC circuit shown in Figure ), for the RC circuit shown in Figure 4.7 in response to the applied voltage 4.7 in response to the applied voltage xx((tt)=(3/5)e)=(3/5)e-2t-2tuu((tt) ) and the and the initial condition initial condition yy(0-) = -2.(0-) = -2.
Solution: Solution:
Step 1Step 1:: Derive differential equation from the Derive differential equation from the circuit.circuit.
KVL around the loop.
Figure 4.7: RC circuit for Examples 6.4 and 6.10. Note that RC = 1/5.
txtytydt
d
(1), into substitute
smFKRC where
txRC
tyRC
tydt
d
RCbydividetxtytydt
dCR
tytydt
dCRtx
55
2.0200*1
)1(11
0
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Step 2Step 2:: Get the unilateral Laplace Transform. Get the unilateral Laplace Transform.
Apply the properties of differential in time domain,
Step 3Step 3:: Substitute Unilateral Laplace Transform Substitute Unilateral Laplace Transform of x(t) into Y(s).of x(t) into Y(s).
Laplace transform of the applied voltage, x(t);
Given initial condition, substitute y(0-)=-2 into (2),And substitute (3) into (2),
)2()0()(55
1)(
);(
)(5)(5)0()(
ysXs
sY
sY of terms in equation the Rearrange
sXsYyssY
)3(2
1
5
3)()(
5
)(3)(
2
ssXtx
tuetx
uL
t
)5(
2
)5)(2(
3)(
sss
sY
0xssXtxdt
duL
Cont’d…Cont’d…
31
Step 4Step 4:: Expand Y(s) into partial fraction. Expand Y(s) into partial fraction.
Step 5Step 5: : Take Inverse Unilateral Laplace Take Inverse Unilateral Laplace Transform of Y(s).Transform of Y(s).
by referring to the Laplace transform table.
.
5
3
2
1)(
5
2
5
1
2
1)(
sssY
ssssY
).(3)()( 52 tuetuety tt
Cont’d…Cont’d…
QUIZ 1Use the basic Laplace transforms & Laplace transform
properties in Appendix D to determine the following signals:
a)
b)
c)
32
)()2cos(*)()( tutttutx
)1(*)1()( 2 tuetutx t
)()cos()( tutedt
dttx t
QUIZ 2 Use the method of partial fractions to find the time
signals corresponding to the following Laplace transforms:
a)
b)
C)
33
23
3)(
2
ss
ssX
12
12)(
2
ss
ssX
sss
ssX
23
45)(
23
4.6 Forced and Natural 4.6 Forced and Natural Responses in Unilateral Responses in Unilateral Laplace Transform.Laplace Transform. Primary application of the unilateral Laplace
transform is to solve differential equations with non-zero initial conditions.
The initial conditions are incorporated into the solution as the values of the signal and its derivatives that occur at time zero in the differentiation property.
General form of differentiation property:-
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6.19 Eq. xstxdt
ds
txdt
dstx
dt
dsXStx
dt
d
n
t
n
tn
n
tn
nnL
n
nu
).()(...
)()()()(
01
0
2
02
2
01
1
4.6 Forced and Natural 4.6 Forced and Natural Responses in Unilateral Responses in Unilateral Laplace Transform.Laplace Transform. The forced response of a system represents the
component of the response associated entirely with the input, denoted as . This response represents the output when the initial conditions are zero.
The natural response represents the component of the output due entirely to the initial conditions, denoted as . This response represents the system output when the input is zero.
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)()( sY f
)()( sY n
Example 4.6: Finding forced and Example 4.6: Finding forced and natural responses.natural responses.Use the unilateral Laplace transform to determine
the output of a system represented by the differential equation,
in response to the input . Assume that the initial conditions on the system are;
Solution.Using differentiation property in Eq. 6.19 and
taking Laplace transform of both sides of the differential equation,
36
)(6)()(6)(5)(2
2
txtxdt
dtyty
dt
dty
dt
d
)()( tutx
2)(1)0(0
tty
dt
d and y
)(6)()(6)0(5)(5)0()()(0
2 sXssXsYyssYsytydt
dsYs
t
Contd…Contd…
Rearranging the terms;
Solving for Y(s), we get;
The first term is associated with the forced response , . The second term corresponds to the natural response, .
Using the transform of input, and the initial conditions, we obtain;
and37
65
)0(5)()0(
65
)(6)(
2
0
2
ss
ytydtd
sy
ss
sXssY
t
)(6)0(5)0()()(650
2 sXsysytydt
dsYss
t
)()( sY f
)()( sY n
ssX /1)(
)3)(2(
6)()(
sss
ssY f
)3)(2(
7)()(
ss
ssY n
Next, taking the inverse Laplace transforms of and , we obtain;
and
Hence, the output of the system is;
38
Contd…Contd…
)()( sY f )()( sY n
)()(2)()( 32)( tuetuetuty ttf
)(4)(5)( 32)( tuetuety ttn
)()()( )()( tytyty nf
a) Forced response of the system, y(f)(t). (b) Natural response of the system, y(n)(t). (c) System output.
39
Contd…Contd…
40
The Bilateral Laplace Transform is suitable to the problems involving non-causal signals and system.
The properties of linearity, scaling, s-domain shift, convolution and differentiation in the s-domain is identical for the bilateral and unilateral LT, the operations associated by these properties may change the ROC.
Example; a linearity property.
ROC of the sum of the signals is an intersection of the individual ROCs.
yxL
yL
xL
RRROCwithsbYsaXtbytax
then
RROCwithsYty
RROCwithsXtx
.
.
4.7 Properties of 4.7 Properties of Bilateral Laplace Bilateral Laplace Transform.Transform.
41
Time ShiftTime Shift
The bilateral Laplace Transform is evaluated over both positive and negative values of time. ROC is unchanged by a time shift.
Differentiation in the Time Domain.Differentiation in the Time Domain.
Differentiation in time domain corresponds to multiplication by s.
sXetx sL
,xL
RleastatROCwithssXtxdt
d
Cont’d…Cont’d…
42
Integration with Respect to Time.Integration with Respect to Time.
Integration corresponds to division by s Pole is at s=0, we are integrating to the
right, hence the ROC must lies to the right of s=0.
.0Re,
sRROCwiths
sXdx x
Lt
Cont’d…Cont’d…
43
4.8 Inversion of Bilateral 4.8 Inversion of Bilateral Laplace Transform.Laplace Transform. The inversion of Bilateral Laplace transforms are
expressed as a ratio of polynomial in s. Compare to the unilateral, in the bilateral Laplace
transform we must use the ROC to determine the unique inverse transform in bilateral case.
sidedrightdsROCwithds
AtueA k
k
kLtdk
k
)Re(,)(
sided-left dsROCwith ds
AtueA k
k
kLtdk
k
)Re(,)(
44
Example 4.7:Example 4.7: Inverting a Proper Rational Laplace Inverting a Proper Rational Laplace Transform.Transform.
Find the Inverse bilateral Laplace Transform of
With ROC -1<Re(s)<1.
Solution:Solution:
Step 1Step 1:: Use the partial fraction expansion of Use the partial fraction expansion of XX((ss) to write) to write
Solving the A, B and C by the method of residues
)2)(1)(1(
75
sss
ssX
)2()1()1(
s
C
s
B
s
AsX
)2(
1
)1(
2
)1(
1
ssssX
45
Step 2Step 2:: Construct the Inverse Laplace transform Construct the Inverse Laplace transform from the above partial-fraction term above.from the above partial-fraction term above.
- The pole of the 1st term is at s = -1, the ROC lies to the right of this pole, choose the right-sided inverse Laplace Transform.
- The pole of the 2nd term is at s = 1, the ROC is to the left of the pole, choose the left-sided inverse Laplace Transform.
-The pole of the 3rd term is at s = -2, the ROC is to the right of the pole, choose the right-sided inverse Laplace Transform.
)1(
1)(
stue Lt
)1(
2)(2
stue Lt
)2(
1)(2
stue Lt
Cont’d…Cont’d…
46
Step 3Step 3:: Combining the terms. Combining the terms.
Combining this three terms we obtain,
Figure 4.12 : Poles and ROC for Example 6.17.Figure 4.12 : Poles and ROC for Example 6.17.
).()(2)()( 2 tuetuetuetx ttt
Cont’d…Cont’d…
47
4.9 Transfer Function.4.9 Transfer Function. The transfer function of an LTI system is defined as
the Laplace transform of the impulse response. Take the bilateral Laplace transform of both sides
of the equation and use the convolution properties, it results in;
Rearrange the above equation will result in the ratio of Laplace transform of the output signal to the Laplace transform of the input signal. (Note: X(s) is non-zero)
)()()( sXsHsY
)(
)()(
sX
sYsH
48
4.9.1 Transfer Function 4.9.1 Transfer Function and Differential-Equation and Differential-Equation System Description.System Description.Given a differential equation.
Step 1Step 1:: Substitute Substitute yy((tt) = ) = eeststHH((ss) into the ) into the equation.equation.y(t) = estH(s), substitute to the above equation result in,
Step 2Step 2:: Solve for H(s). Solve for H(s).H(s) is a ratio of polynomial and s is termed as rational transfer function.
)()(00
txdt
dbty
dt
da
M
kk
k
k
N
kk
k
k
stM
kk
k
kst
N
kk
k
k edt
dbsHe
dt
da
00
N
k
kk
M
k
kk
sa
sbsH
0
0
49
Example 4.8:Example 4.8: Find the Transfer Function.Find the Transfer Function.Find the transfer function of the LTI system described by the
differential equation below,
Solution:Solution:Step 1Step 1:: eestst is an eigenfunction of LTI system. If input, is an eigenfunction of LTI system. If input,
x(t)=x(t)=eestst
Then, Then, yy((tt) = ) = eeststHH((ss). Hence, substitute into the equation..). Hence, substitute into the equation..
Step 2Step 2:: Solve for Solve for H(s)H(s)..
ststststst eedt
dsHesHe
dt
dsHe
dt
d32)(2)(3)(
2
2
)(32232
2
txtxdt
dtyty
dt
dty
dt
d
ststststst eedt
dee
dt
de
dt
dsH 3223)(
2
2
stkstk
k
esedt
d
50
stkstk
k
esedt
d
ststst
stst
eedtd
edtd
eedtd
sH
23
32)(
2
2
23
32)(
2
ss
ssH
Hence, the transfer function is,
.
Cont’d…Cont’d…
51
4.10 Causality and 4.10 Causality and StabilityStability
4.10.1 Causality : Impulse response of a causal system is zero for t <
0. A system pole at s = dk in the left half plane
[Re(dk) < 0] contributes an exponentially decaying term to the impulse response (See Figure 4.13 (a)).
A pole in the right half plane [Re(dk) > 0] contributes an increasing exponential term to the impulse response (See Figure 4.13(b)).
52
Figure 4.13: The relationship between the locations of poles and the impulse response in a causal system. (a) A pole in the left half of the s-plane corresponds to an exponentially decaying impulse response. (b) A pole in the right half of the s-plane corresponds to an exponentially increasing impulse response. The system is unstable in this case.
53
4.10.2 Stability : If the system is stable, the impulse response is
absolutely integrable. A pole of the system transfer function that locates
in the right half plane contributes a left sided decaying exponential term to the impulse response (See Figure 4.14 (a)).
A pole in the left half plane contributes a right-sided decaying exponential term to the impulse response (See Figure 4.14 (b)).
54
Figure 4.14: The relationship between the locations of poles and the impulse response in a stable system. (a) A pole in the left half of the s-plane corresponds to a right-sided impulse response. (b) A pole in the right half of the s-plane corresponds to an left-sided impulse response. In this case, the system is noncausal.
55
A system that is both stable and causal must have a transfer function with all of its poles in the left half of the s-plane, as shown here.
56
Reference Table.Reference Table.