Laplace transforms

LAPLACE TRANSFORMS
M. Awais Yaqoob
University of Engineering and Technology, Lahore

Definition
 Transforms -- a mathematical conversion
from one way of thinking to another to
make a problem easier to solve
transform
solution
in transform
way of
thinking
inverse
transform
solution
in original
way of
thinking
problem
in original
way of
thinking
2. Transforms

Laplace
transform
solution
in
s domain
inverse
Laplace
transform
solution
in time
domain
problem
in time
domain
• Other transforms
• Fourier
• z-transform
• wavelets
2. Transforms

Laplace transformation
linear
differential
equation
time
domain
solution
Laplace transform
Laplace
transformed
equation
Laplace
solution
time domain
Laplace domain or
complex frequency domain
algebra
inverse Laplace
transform
4. Laplace transforms

Basic Tool For Continuous Time:
Laplace Transform
= = ò¥ -
L[ f (t)] F(s) f (t)e stdt
0
 Convert time-domain functions and operations into
frequency-domain
 f(t) ® F(s) (tÎR, sÎC)
 Linear differential equations (LDE) ® algebraic expression
in Complex plane
 Graphical solution for key LDE characteristics
 Discrete systems use the analogous z-transform

The Complex Plane (review)
Imaginary axis (j)
u = x + jy
Real axis
x
y
fr
-f
r
u y
º º = +
u r u x y
u = x - jy
(complex) conjugate
- y
2 2
1
tan
| | | |
x
Ð ºf = -

Laplace Transforms of Common
Functions
Name f(t) F(s)
Impulse
Step
Ramp
Exponential
Sine
1
1
s
1
s
2
1
s - a
1
2 2
w + s
t
1 0
î í ì
=
0 0
f (t) =1
f (t) = t
=
>
f t
( )
t
f (t) = eat
f (t) = sin(wt)

Laplace Transform Properties
L af t bf t aF s bF s
[ ( ) ( )] ( ) ( )
L d
f t sF s f
dt
± - = úû
( ) ( ) (0 )
é
[ ] [ ]
L f t dt F s
( ) ( ) 1 ( )
f t dt
ò = +
ò
s s
f (t - τ)f (τ dτ =
F s F s
) ( ) ( )
Addition/Scaling
Differentiation
Integratio n
Convolution
Initial value theorem
- f + =
sF s
(0 ) lim ( )
s
®¥
lim ( ) lim ( )
0
0
1 2 1 2
0
1 2 1 2
- f t sF s
t s
t
t
®¥ ®
= ±
=
ù
êë
± = ±
ò
Final value theorem

LAPLACE TRANSFORMS
SIMPLE TRANSFORMATIONS

Transforms (1 of 11)
 Impulse -- d (to)
F(s) =
0
¥
e-st d (to) dt
= e-sto
f(t)
t
d (to)

 Step -- u (to)
F(s) =
0
¥
e-st u (to) dt
= e-sto/s
f(t)
t
1 u (to)

 e-at
F(s) =
0
¥
e-st e-at dt
= 1/(s+a)

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

 Most mathematical handbooks have tables
of Laplace transforms

LAPLACE TRANSFORMS
PARTIAL FRACTION EXPANSION

Definition
 Definition -- Partial fractions are several
fractions whose sum equals a given fraction
 Purpose -- Working with transforms requires
breaking complex fractions into simpler
fractions to allow use of tables of transforms

Partial Fraction Expansions
s 1
B
 Expand into a term for
+
+
A
+
=
+
s s
+ +
s
s
( 2) ( 3) 2 3
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.
( )
A s B s
= + + +
( 3) 2
( 2) ( 3)
1
+
s
( 2) ( 3)
+ +
+ +
s s
s s
A+ B =1 3A+ 2B =1
3
2
1
s
s s s 2
s
1
+
( 2) ( 3)
+
+
= -
+
+ +

Example of Solution of an ODE
d y dy
 ODE w/initial conditions
6 8 2 (0) '(0) 0 2
2
+ + y = y = y =
dt
dt
 Apply Laplace transform
to each term
 Solve for Y(s)
 Apply partial fraction
expansion
 Apply inverse Laplace
transform to each term
s2 Y(s) + 6sY(s) + 8Y(s) = 2 / s
( ) 2
=
s s s
+ +
( 2) ( 4)
Y s
1
s s s
4( 4)
1
= + -
2( 2)
( ) 1
4
+
+
+
Y s
y ( t
) = 1
- e - 2t e -
4t +
4 2 4

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

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

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

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

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

Apply Initial- and Final-Value
Theorems to this Example
 Laplace
transform of the
function.
 Apply final-value
theorem
 Apply initial-value
theorem
( ) 2
=
s s s
+ +
( 2) ( 4)
Y s
lim ( ) 2(0) =
[ ]
1
4
+ +
(0) (0 2) (0 4)
= ®¥ f t t
= ¥ ® f t t
lim ( ) 2( ) 0 =
[ ] 0
¥ ¥+ ¥+
( ) ( 2) ( 4)

LAPLACE TRANSFORMS
SOLUTION PROCESS

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

 Step 1: Put differential equation into
standard form
 D2 y + 2D y + 2y = cos t
 y(0) = 1
 D y(0) = 0

 Step 2: Take the Laplace transform of both
sides
 L{D2 y} + L{2D y} + L{2y} = L{cos t}

 Step 3: Use table of transforms to express
equation in s-domain
 L{D2 y} + L{2D y} + L{2y} = L{cos w 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)

 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)]

 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)
 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

 (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]

 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

LAPLACE TRANSFORMS
TRANSFER FUNCTIONS

Introduction
 Definition -- a transfer function is an
expression that relates the output to the
input in the s-domain
r(t) y(t)
differential
equation
r(s) y(s)
transfer
function
5. Transfer functions

Transfer Function
 Definition
 H(s) = Y(s) / X(s)
X(s) H(s) Y(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

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

Typical block diagram
control
Gc(s)
plant
Gp(s)
feedback
H(s)
pre-filter
G1(s)
post-filter
G2(s)
reference input, R(s)
error, E(s)
plant inputs, U(s)
output, Y(s)
feedback, H(s)Y(s)

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 conditions

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 )
V(s) I(s)
C s / (C L s2 + C R s + 1 )

Block diagram reduction rules
U Y U Y
G1 G2 G1 G2
U + Y
G1
G2
U Y
+ G1 + G2
U + Y
G1
- G1 /(1+G1 G2)
G2
U Y
Series
Parallel
Feedback

Rational Laplace Transforms
F s A s
( ) ( )
B s
( )
n
A ( s ) = a s + ...
+ a s +
a
m
n
1 0
B ( s ) = b s + ...
+ b s +
b
m
1 0
Poles: (So,
' = = ¥
* ( *) 0 ( *) )
Zeroes: (So,
s A s F s
' = =
Poles and zeroes are complex
m
s B s F s
Order of system # poles
= =
=
* ( *) 0 ( *) 0)

First Order System
Reference
Y(s)
R(s)
S E(s)
K
U(s)
Y s
( )
B(s) 1
K
1
1+ sT
K
sT
K sT
R s
+
»
+ +
=
( ) 1 1

First Order System
Impulse
response
Exponential
K
+
Step response Step,
exponential
1/
s T
K
Ramp response Ramp, step,
exponential
1 sT
- KT
s
2 s 1/
T
K
- KT
s
+
- K
s
+
No oscillations (as seen by poles)

Second Order System
Impulse response:
K
Y s
w
N
( )
2 2
Oscillates if poles have non - zero imaginary part (ie,
B
Damping ratio : where
= =
Undamped natural frequency :
B JK
K
J
B
B JK
Js Bs K s s
R s
c
N
c
N N
=
- <
+ +
=
+ +
=
w
x
xw w
2
4 0)
( ) 2
2
2
2

Second Order System: Parameters
Interpreta tion of damping ratio
Undamped oscillation (Re 0, Im 0)
= = ¹
Underdamped (Re Im)
0 :
< x
< ¹ ¹
0 1: 0
1 Overdamped (Re 0,Im 0)
Interpreta tion of undamped natural frequency
gives the frequency of the oscillation
wN
x
x
£ ¹ =
:

Transient Response Characteristics
2
1.75
1.5
1.25
1
0.75
0.5
= maximum overshoot p M
p t s t d t
: Delay until reach 50% of steady state value
: Rise time delay until first reach steady state value
: Time at which peak value is reached
: Settling time =
stays within specified % of steady state
=
d
r
p
s
t
t
t
t
0.5 1 1.5 2 2.5 3
0.25
r t

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)

Effect of pole locations
Oscillations
(higher-freq)
Im(s)
Faster Decay Faster Blowup
Re(s) (e-at) (eat)

Basic Control Actions: u(t)
Proportional control :
Integral control :
Differential control :
( ) ( ) ( )
= =
ò
p p
K
i
K s
u t K e t U s
t
i
( )
u t K e t dt U s
( ) ( ) ( )
= =
( )
e t U s
dt
( ) ( ) ( )
= =
E s
u t K d
s
E s
K
E s
d d
( )
0

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

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:
ö
÷ ÷ø
æ
ç çè
K K
G s U s
( ) ( )
= = + = +
T s
K
s
E s
i
p
I
p
1 1
( )

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

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

LAPLACE TRANSFORMS
LAPLACE APPLICATIONS

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) = 1
s ¥
Example
6. Laplace applications

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) = 0
s 0
Example

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

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

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
x
x
x x x
x
stable unstable
x
jw
s

LAPLACE TRANSFORMS
FREQUENCY RESPONSE

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

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

Process
 The frequency response is computed by
replacing s with j w in the transfer function
Example
f(t) = e -t
F(s) = 1/(s+1)
magnitude in dB
F(j w) = 1/(j w +1)
Magnitude = 1/SQRT(1 + w2)
Magnitude in dB = 20 log10 (magnitude)
Phase = argument = ATAN2(- w, 1)
w

Graphical methods
 Frequency response is a graphical method
 Polar plot -- difficult to construct
 Corner plot -- easy to construct

Constant K
60 dB
40 dB
20 dB
0 dB
-20 dB
-40 dB
-60 dB
+180o
+90o
0o
-90o
-180o
-270o
magnitude
phase
0.1 1 10 100
w, radians/sec
20 log10 K
arg K

Simple pole or zero at origin, 1/ (jw)n
60 dB
40 dB
20 dB
0 dB
-20 dB
-40 dB
-60 dB
+180o
+90o
0o
-90o
-180o
-270o
magnitude
phase
0.1 1 10 100
w, radians/sec
1/ w
1/ w3 1/ w2
1/ w
1/ w2
1/ w3
G(s) = wn
2/(s2 + 2d wns + w n
2)

Simple pole or zero, 1/(1+jw)
60 dB
40 dB
20 dB
0 dB
-20 dB
-40 dB
-60 dB
+180o
+90o
0o
-90o
-180o
-270o
magnitude
phase
0.1 1 10 100
wT

Error in asymptotic approximation
wT
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
84.3

Quadratic pole or zero
60 dB
40 dB
20 dB
0 dB
-20 dB
-40 dB
-60 dB
+180o
+90o
0o
-90o
-180o
-270o
magnitude
phase
0.1 1 10 10wT 0

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.

Derivation of a Transfer Function
M dT = F T + F T - ( F + F ) T
1 1 2 2 1 2  Dynamic model of
dt
CST thermal
mixer
 Apply deviation
variables
 Equation in terms
of deviation
variables.
0 1 1 0 2 2 0 DT = T -T DT = T -T DT =T -T
M dDT = D + D - ( + )D 1 1 2 2 1 2
F T F T F F T
dt

Derivation of a Transfer Function
T s = F T s +
F T s
1 1 2 2 ( ) ( ) ( )
F
1
[ 1 2 ]
G s T s
( ) ( )
1 ( )
M s F F
T s
+ +
= =
 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.
[ M s + F +
F
] 1 2

Poles of the Transfer Function
Indicate the Dynamic Response
( ) 1 s + a s2 + bs + c s -
d
( ) ( ) ( )
C
B
G s
=
Y s A
+
+
=
( ) s +
a
s 2 + bs +
c
s -
d
( ) ( ) ( )
y(t) = A¢e-at + B¢e pt sin(w t) + C¢edt
 For a, b, c, and d positive constants, transfer
function indicates exponential decay, oscillatory
response, and exponential growth, respectively.

Poles on a Complex Plane
Re
Im

Exponential Decay
Re
Im
Time
y

Damped Sinusoidal
Re
Im
Time
y

Exponentially Growing Sinusoidal
Behavior (Unstable)
Re
Im
Time
y

What Kind of Dynamic Behavior?
Re
Im

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.

Laplace transforms

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Laplace transforms

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