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LAPLACE TRANSFORMS
INTRODUCTION
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Definition Transforms -- a mathematical conversion from
one way of thinking to another to make a problem easier to solve
transformsolution
in transformway of
thinking
inversetransform
solution in original
way of thinking
problem in original
way of thinking
2. Transforms
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Laplace transform
solutionin
s domain
inverse Laplace
transform
solution in timedomain
problem in time domain
• Other transforms• Fourier• z-transform• wavelets
2. Transforms
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All those signals……….A
mpl
itude
time
discrete continuous discrete-time analog signal w(nTs)
Ts
sampling
discrete discrete discrete-time digital signal Cn
111
110
101
100
011
010
001
000 sampling
Time Amplitude Signal
w(t)
continuous continuous continuous-time analog signal w(t)
discrete continuous discrete-time sequence w[n]
n=0 1 2 3 4 5
indexing
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…..and all those transforms
Continuous-time analog signal
w(t)
Continuous-time analog signal
w(t)
Discrete-time analog sequence
w [n]
Discrete-time analog sequence
w [n]
Sample in time,period = Ts
=2f= Ts,scale amplitude by 1/Ts
Sample in frequency, = 2n/N,N = Length of sequence
Continuous Fourier
TransformW(f)
Continuous Fourier
TransformW(f)
f-
dt e w(t) ft2 j-
Discrete Fourier
TransformW(k)
Discrete Fourier
TransformW(k)
10
e [n] w1
0 =n
Nnk2
j-
Nk
N
Discrete-Time Fourier
Transform W(
Discrete-Time Fourier
Transform W(
20
e [n] w- =n
j-
n
LaplaceTransform
W(s)
LaplaceTransform
W(s)
s-
dt e w(t)0
st
z-TransformW(z)
z-TransformW(z)
z
=n
n- z [n] w
z = ej
s = jf
CC
C
C
C D
D
D
C Continuous-variable
Discrete-variable
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Laplace transformation
linear differential equation
timedomainsolution
Laplacetransformed
equation
Laplacesolution
time domain
Laplace domain orcomplex frequency domain
algebra
Laplace transform
inverse Laplace transform
4. Laplace transforms
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Basic Tool For Continuous Time: Laplace Transform
Convert time-domain functions and operations into frequency-domain
f(t) F(s) (tR, sC Linear differential equations (LDE) algebraic expression
in Complex plane Graphical solution for key LDE characteristics Discrete systems use the analogous z-transform
0)()()]([ dtetfsFtf stL
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The Complex Plane (review)
Imaginary axis (j)
Real axis
jyxu
x
y
r
r
jyxu (complex) conjugate
y
22
1
||||
tan
yxuru
x
yu
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Laplace Transforms of Common Functions
Name f(t) F(s)
Impulse
Step
Ramp
Exponential
Sine
1
s
1
2
1
s
as 1
22
1
s
1)( tf
ttf )(
atetf )(
)sin()( ttf
00
01)(
t
ttf
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Laplace Transform Properties
)(lim)(lim
)(lim)0(
)()()
)(1)(
)(
)0()()(
)()()]()([
0
0
2121
0
2121
ssFtf-
ssFf-
sFsFdτ(ττ)f(tf
dttfss
sFdttfL
fssFtfdt
dL
sbFsaFtbftafL
st
s
t
t
theorem valueFinal
theorem valueInitial
nConvolutio
nIntegratio
ationDifferenti
calingAddition/S
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LAPLACE TRANSFORMS
SIMPLE TRANSFORMATIONS
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Transforms (1 of 11)
Impulse -- (to)
F(s) =
0
e-st (to) dt
= e-sto
f(t)
t
(to)
4. Laplace transforms
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Transforms (2 of 11) Step -- u (to)
F(s) =
0
e-st u (to) dt
= e-sto/sf(t)
t
u (to)1
4. Laplace transforms
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Transforms (3 of 11) e-at
F(s) =
0
e-st e-at dt
= 1/(s+a)
4. Laplace transforms
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Transforms (4 of 11)f1(t) f2(t)
a f(t)
eat f(t)
f(t - T)
f(t/a)
F1(s) ± F2(s)
a F(s)
F(s-a)
eTs F(as)
a F(as)
Linearity
Constant multiplication
Complex shift
Real shift
Scaling
4. Laplace transforms
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Transforms (5 of 11) Most mathematical handbooks have tables
of Laplace transforms
4. Laplace transforms
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Table of Laplace Transforms
Definition of Laplace transform
0
dttfetfL st )()}({
)}({)( 1 sFLtf )}({)( tfLsF
1
nt 1
!ns
n
ateas
1
tsin 22 s
tcos 22 s
s
s
1
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Translation and Derivative Table for Laplace Transforms
First translation and derivative theorems
)(tfeat )( asF
)(tft n )()1( sFds
dn
nn
)(' tf )0()( fssF
)('' tf )0(')0()(2 fsfsFs
)(''' tf )0('')0(')0()( 23 fsffssFs
sFLtf 1)( )(tfLsF
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Unit step and Dirac delta function
Unit step and Dirac delta function
)( atu se as
)()( atutf )}({ atfLe as
)()( atuatf )(sFe as
)( at ase
sFLtf 1)( )(tfLsF
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Convolution theorem
Convolution theorem
sFLtf 1)( )(tfLsF
t
dtgftgtf
0
)()()()( )()( sGsF
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LAPLACE TRANSFORMS
SOLUTION PROCESS
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Solution process (1 of 8)
Any nonhomogeneous linear differential equation with constant coefficients can be solved with the following procedure, which reduces the solution to algebra
4. Laplace transforms
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Solution process (2 of 8)
Step 1: Put differential equation into standard form
D2 y + 2D y + 2y = cos t y(0) = 1 D y(0) = 0
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Solution process (3 of 8)
Step 2: Take the Laplace transform of both sides
L{D2 y} + L{2D y} + L{2y} = L{cos t}
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Solution process (4 of 8)
Step 3: Use table of transforms to express equation in s-domain
L{D2 y} + L{2D y} + L{2y} = L{cos t}
L{D2 y} = s2 Y(s) - sy(0) - D y(0) L{2D y} = 2[ s Y(s) - y(0)] L{2y} = 2 Y(s) L{cos t} = s/(s2 + 1)
s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) = s /(s2 + 1)
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Solution process (5 of 8) Step 4: Solve for Y(s)
s2 Y(s) - s + 2s Y(s) - 2 + 2 Y(s) = s/(s2 + 1) (s2 + 2s + 2) Y(s) = s/(s2 + 1) + s + 2
Y(s) = [s/(s2 + 1) + s + 2]/ (s2 + 2s + 2) = (s3 + 2 s2 + 2s + 2)/[(s2 + 1) (s2 + 2s + 2)]
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Solution process (6 of 8) Step 5: Expand equation into format covered by
table Y(s) = (s3 + 2 s2 + 2s + 2)/[(s2 + 1) (s2 + 2s + 2)] = (As + B)/ (s2 + 1) + (Cs + E)/ (s2 + 2s + 2) (A+C)s3 + (2A + B + E) s2 + (2A + 2B + C)s + (2B
+E) Equate similar terms
1 = A + C 2 = 2A + B + E 2 = 2A + 2B + C 2 = 2B + E A = 0.2, B = 0.4, C = 0.8, E = 1.2
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Solution process (7 of 8)
(0.2s + 0.4)/ (s2 + 1) = 0.2 s/ (s2 + 1) + 0.4 / (s2 + 1) (0.8s + 1.2)/ (s2 + 2s + 2) = 0.8 (s+1)/[(s+1)2 + 1] + 0.4/ [(s+1)2 + 1]
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Solution process (8 of 8) Step 6: Use table to convert s-domain to
time domain 0.2 s/ (s2 + 1) becomes 0.2 cos t 0.4 / (s2 + 1) becomes 0.4 sin t 0.8 (s+1)/[(s+1)2 + 1] becomes 0.8 e-t cos t 0.4/ [(s+1)2 + 1] becomes 0.4 e-t sin t
y(t) = 0.2 cos t + 0.4 sin t + 0.8 e-t cos t + 0.4 e-
t sin t
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LAPLACE TRANSFORMS
PARTIAL FRACTION EXPANSION
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Definition Definition -- Partial fractions are several
fractions whose sum equals a given fraction Example --
(11x - 1)/(x2 - 1) = 6/(x+1) + 5/(x-1) = [6(x-1) +5(x+1)]/[(x+1)(x-1))] =(11x - 1)/(x2 - 1)
Purpose -- Working with transforms requires breaking complex fractions into simpler fractions to allow use of tables of transforms
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Partial Fraction Expansions
32)3()2(
1
s
B
s
A
ss
s Expand into a term for each factor in the denominator.
Recombine RHS
Equate terms in s and constant terms. Solve.
Each term is in a form so that inverse Laplace transforms can be applied.
)3()2(
2)3(
)3()2(
1
ss
sBsA
ss
s
3
2
2
1
)3()2(
1
ssss
s
1 BA 123 BA
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Example of Solution of an ODE
0)0(')0(2862
2
yyydt
dy
dt
yd ODE w/initial conditions
Apply Laplace transform to each term
Solve for Y(s)
Apply partial fraction expansion
Apply inverse Laplace transform to each term
ssYsYssYs /2)(8)(6)(2
)4()2(
2)(
ssssY
)4(4
1
)2(2
1
4
1)(
sss
sY
424
1)(
42 tt eety
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Different terms of 1st degree To separate a fraction into partial fractions
when its denominator can be divided into different terms of first degree, assume an unknown numerator for each fraction
Example -- (11x-1)/(X2 - 1) = A/(x+1) + B/(x-1) = [A(x-1) +B(x+1)]/[(x+1)(x-1))] A+B=11 -A+B=-1 A=6, B=5
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Repeated terms of 1st degree (1 of 2) When the factors of the denominator are of
the first degree but some are repeated, assume unknown numerators for each factor
If a term is present twice, make the fractions the corresponding term and its second power
If a term is present three times, make the fractions the term and its second and third powers
3. Partial fractions
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Repeated terms of 1st degree (2 of 2) Example --
(x2+3x+4)/(x+1)3= A/(x+1) + B/(x+1)2 + C/(x+1)3
x2+3x+4 = A(x+1)2 + B(x+1) + C = Ax2 + (2A+B)x + (A+B+C) A=1 2A+B = 3 A+B+C = 4 A=1, B=1, C=2
3. Partial fractions
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Different quadratic terms When there is a quadratic term, assume a
numerator of the form Ax + B Example --
1/[(x+1) (x2 + x + 2)] = A/(x+1) + (Bx +C)/ (x2 + x + 2)
1 = A (x2 + x + 2) + Bx(x+1) + C(x+1) 1 = (A+B) x2 + (A+B+C)x +(2A+C) A+B=0 A+B+C=0 2A+C=1 A=0.5, B=-0.5, C=0
3. Partial fractions
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Repeated quadratic terms Example --
1/[(x+1) (x2 + x + 2)2] = A/(x+1) + (Bx +C)/ (x2 + x + 2) + (Dx +E)/ (x2 + x + 2)2
1 = A(x2 + x + 2)2 + Bx(x+1) (x2 + x + 2) + C(x+1) (x2 + x + 2) + Dx(x+1) + E(x+1)
A+B=0 2A+2B+C=0 5A+3B+2C+D=0 4A+2B+3C+D+E=0 4A+2C+E=1 A=0.25, B=-0.25, C=0, D=-0.5, E=0
3. Partial fractions
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Apply Initial- and Final-Value Theorems to this Example Laplace
transform of the function.
Apply final-value theorem
Apply initial-value theorem
)4()2(
2)(
ssssY
4
1
)40()20()0(
)0(2)(lim
tft
0)4()2()(
)(2)(lim 0
tft
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LAPLACE TRANSFORMS
TRANSFER FUNCTIONS
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Introduction Definition -- a transfer function is an
expression that relates the output to the input in the s-domain
differentialequation
r(t) y(t)
transferfunction
r(s) y(s)
5. Transfer functions
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Transfer Function
Definition H(s) = Y(s) / X(s)
Relates the output of a linear system (or component) to its input
Describes how a linear system responds to an impulse
All linear operations allowed Scaling, addition, multiplication
H(s)X(s) Y(s)
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Block Diagrams
Pictorially expresses flows and relationships between elements in system
Blocks may recursively be systems Rules
Cascaded (non-loading) elements: convolution Summation and difference elements
Can simplify
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Typical block diagram
controlGc(s)
plantGp(s)
feedbackH(s)
pre-filterG1(s)
post-filterG2(s)
reference input, R(s)
error, E(s)
plant inputs, U(s)
output, Y(s)
feedback, H(s)Y(s)
5. Transfer functions
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Example
v(t)
R
C
L
v(t) = R I(t) + 1/C I(t) dt + L di(t)/dt
V(s) = [R I(s) + 1/(C s) I(s) + s L I(s)]
Note: Ignore initial conditions5. Transfer functions
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Block diagram and transfer function V(s)
= (R + 1/(C s) + s L ) I(s) = (C L s2 + C R s + 1 )/(C s) I(s)
I(s)/V(s) = C s / (C L s2 + C R s + 1 )
C s / (C L s2 + C R s + 1 )V(s) I(s)
5. Transfer functions
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Block diagram reduction rules
G1 G2 G1 G2
U Y U Y
G1
G2
U Y+
+ G1 + G2
U Y
G1
G2
U Y+
- G1 /(1+G1 G2)U Y
Series
Parallel
Feedback
5. Transfer functions
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Rational Laplace Transforms
m
sFsAs
sFsBs
bsbsbsB
asasasA
sB
sAsF
mm
nn
poles # system ofOrder
complex are zeroes and Poles
(So, :Zeroes
(So, :Poles
)0*)(0*)(*
)*)(0*)(*
...)(
...)(
)(
)()(
01
01
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First Order System
Reference
)(sY)(sR
)(sE
1)(sB
)(sUsT1
1K
sT
K
sTK
K
sR
sY
11)(
)(
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First Order System
Impulse response
Exponential
Step response Step, exponential
Ramp response Ramp, step, exponential
1 sT
K
/1
2 Ts
KT-
s
KT-
s
K
/1
Ts
K-
s
K
No oscillations (as seen by poles)
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Second Order System
:frequency natural Undamped
where :ratio Damping
(ie,part imaginary zero-non have poles if Oscillates
:response Impulse
J
K
JKBB
B
JKB
ssKBsJs
K
sR
sY
N
cc
NN
N
2
)04
2)(
)(
2
22
2
2
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Second Order System: Parameters
noscillatio the offrequency the gives
frequency natural undamped of tionInterpreta
0)Im0,(Re Overdamped 1
Im) (Re dUnderdampe
0)Im 0,(Re noscillatio Undamped
ratio damping of tionInterpreta
N
:
0:10
:0
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Transient Response Characteristics
statesteady of % specified within stays time Settling :
reached is valuepeak whichat Time :
valuestatesteady reachfirst untildelay time Rise :
valuestatesteady of 50% reach untilDelay :
s
p
r
d
t
t
t
t
0.5 1 1.5 2 2.5 3
0.25
0.5
0.75
1
1.25
1.5
1.75
2
rt
overshoot maximumpM
pt stdt
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Transient Response
Estimates the shape of the curve based on the foregoing points on the x and y axis
Typically applied to the following inputs Impulse Step Ramp Quadratic (Parabola)
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Effect of pole locations
Faster Decay Faster Blowup
Oscillations(higher-freq)
Im(s)
Re(s)(e-at) (eat)
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Basic Control Actions: u(t)
:control alDifferenti
:control Integral
:control alProportion
sKsE
sUte
dt
dKtu
s
K
sE
sUdtteKtu
KsE
sUteKtu
dd
it
i
pp
)(
)()()(
)(
)()()(
)(
)()()(
0
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Effect of Control Actions
Proportional Action Adjustable gain (amplifier)
Integral Action Eliminates bias (steady-state error) Can cause oscillations
Derivative Action (“rate control”) Effective in transient periods Provides faster response (higher sensitivity) Never used alone
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Basic Controllers
Proportional control is often used by itself Integral and differential control are typically
used in combination with at least proportional control
eg, Proportional Integral (PI) controller:
sTK
s
KK
sE
sUsG
ip
Ip
11
)(
)()(
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Summary of Basic Control
Proportional control Multiply e(t) by a constant
PI control Multiply e(t) and its integral by separate constants Avoids bias for step
PD control Multiply e(t) and its derivative by separate constants Adjust more rapidly to changes
PID control Multiply e(t), its derivative and its integral by separate constants Reduce bias and react quickly
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Root-locus Analysis
Based on characteristic eqn of closed-loop transfer function
Plot location of roots of this eqn Same as poles of closed-loop transfer function Parameter (gain) varied from 0 to
Multiple parameters are ok Vary one-by-one Plot a root “contour” (usually for 2-3 params)
Quickly get approximate results Range of parameters that gives desired response
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LAPLACE TRANSFORMS
LAPLACE APPLICATIONS
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Initial value In the initial value of f(t) as t approaches 0
is given by
f(0 ) = Lim s F(s)s
f(t) = e -t
F(s) = 1/(s+1)
f(0 ) = Lim s /(s+1) = 1s
Example
6. Laplace applications
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Final value In the final value of f(t) as t approaches
is given by
f(0 ) = Lim s F(s)s 0
f(t) = e -t
F(s) = 1/(s+1)
f(0 ) = Lim s /(s+1) = 0s 0
Example
6. Laplace applications
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Apply Initial- and Final-Value Theorems to this Example Laplace
transform of the function.
Apply final-value theorem
Apply initial-value theorem
)4()2(
2)(
ssssY
4
1
)40()20()0(
)0(2)(lim
tft
0)4()2()(
)(2)(lim 0
tft
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Poles The poles of a Laplace function are the
values of s that make the Laplace function evaluate to infinity. They are therefore the roots of the denominator polynomial
10 (s + 2)/[(s + 1)(s + 3)] has a pole at s = -1 and a pole at s = -3
Complex poles always appear in complex-conjugate pairs
The transient response of system is determined by the location of poles
6. Laplace applications
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Zeros The zeros of a Laplace function are the
values of s that make the Laplace function evaluate to zero. They are therefore the zeros of the numerator polynomial
10 (s + 2)/[(s + 1)(s + 3)] has a zero at s = -2
Complex zeros always appear in complex-conjugate pairs
6. Laplace applications
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Stability A system is stable if bounded inputs produce bounded
outputs The complex s-plane is divided into two regions: the
stable region, which is the left half of the plane, and the unstable region, which is the right half of the s-plane
s-plane
stable unstable
x
x
xx x
x
x
j
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LAPLACE TRANSFORMS
FREQUENCY RESPONSE
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Introduction Many problems can be thought of in the
time domain, and solutions can be developed accordingly.
Other problems are more easily thought of in the frequency domain.
A technique for thinking in the frequency domain is to express the system in terms of a frequency response
7. Frequency response
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Definition The response of the system to a sinusoidal
signal. The output of the system at each frequency is the result of driving the system with a sinusoid of unit amplitude at that frequency.
The frequency response has both amplitude and phase
7. Frequency response
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Process The frequency response is computed by
replacing s with j in the transfer function
f(t) = e -t
F(s) = 1/(s+1)
Example
F(j ) = 1/(j +1)
Magnitude = 1/SQRT(1 + 2)
Magnitude in dB = 20 log10 (magnitude)
Phase = argument = ATAN2(- , 1)
magnitude in dB
7. Frequency response
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Graphical methods Frequency response is a graphical method Polar plot -- difficult to construct Corner plot -- easy to construct
7. Frequency response
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Constant K
+180o
+90o
0o
-270o
-180o
-90o
60 dB40 dB20 dB 0 dB
-20 dB-40 dB-60 dB
magnitude
phase
0.1 1 10 100, radians/sec
20 log10 K
arg K
7. Frequency response
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Simple pole or zero at origin, 1/ (j)n
+180o
+90o
0o
-270o
-180o
-90o
60 dB40 dB20 dB 0 dB
-20 dB-40 dB-60 dB
magnitude
phase
0.1 1 10 100, radians/sec
1/
1/ 21/ 3
1/ 1/ 2
1/ 3
G(s) = n2/(s2 + 2 ns + n
2)
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Simple pole or zero, 1/(1+j)
+180o
+90o
0o
-270o
-180o
-90o
60 dB40 dB20 dB 0 dB
-20 dB-40 dB-60 dB
magnitude
phase
0.1 1 10 100T
7. Frequency response
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Error in asymptotic approximationT
0.01
0.1
0.5
0.76
1.0
1.31
1.73
2.0
5.0
10.0
dB
0
0.043
1
2
3
4.3
6.0
7.0
14.2
20.3
arg (deg)
0.5
5.7
26.6
37.4
45.0
52.7
60.0
63.4
78.7
84.37. Frequency response
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Quadratic pole or zero
+180o
+90o
0o
-270o
-180o
-90o
60 dB40 dB20 dB 0 dB
-20 dB-40 dB-60 dB
magnitude
phase
0.1 1 10 100T
7. Frequency response
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Transfer Functions
Defined as G(s) = Y(s)/U(s) Represents a normalized model of a process,
i.e., can be used with any input. Y(s) and U(s) are both written in deviation
variable form. The form of the transfer function indicates the
dynamic behavior of the process.
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Derivation of a Transfer Function
TFFTFTFdt
dTM )( 212211 Dynamic model of
CST thermal mixer
Apply deviation variables
Equation in terms of deviation variables.
0220110 TTTTTTTTT
TFFTFTFdt
TdM
)( 212211
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Derivation of a Transfer Function
21
1
1 )(
)()(
FFsM
F
sT
sTsG
Apply Laplace transform to each term considering that only inlet and outlet temperatures change.
Determine the transfer function for the effect of inlet temperature changes on the outlet temperature.
Note that the response is first order.
21
2211 )()()(
FFsM
sTFsTFsT
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Poles of the Transfer Function Indicate the Dynamic Response
For a, b, c, and d positive constants, transfer function indicates exponential decay, oscillatory response, and exponential growth, respectively.
)()()()(
2 ds
C
cbss
B
as
AsY
dtptat eCteBeAty )sin()(
)()()(
1)(
2 dscbssassG
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Poles on a Complex Plane
Re
Im
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Exponential Decay
Re
Im
Time
y
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Damped Sinusoidal
Re
Im
Time
y
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Exponentially Growing Sinusoidal Behavior (Unstable)
Re
Im
Time
y
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What Kind of Dynamic Behavior?
Re
Im
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Unstable Behavior
If the output of a process grows without bound for a bounded input, the process is referred to a unstable.
If the real portion of any pole of a transfer function is positive, the process corresponding to the transfer function is unstable.
If any pole is located in the right half plane, the process is unstable.