ee 401 control systems analysis and design a review of the laplace transform wednesday 27 aug 2014...
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The Laplace Transform Wednesday 27 Aug 2014 EE 401: Control Systems Analysis and Design Slide 3 of 19TRANSCRIPT
EE 401 Control Systems Analysis and Design
A Review of The Laplace Transform
Wednesday 27 Aug 2014 EE 401: Control Systems Analysis and Design Slide 1 of 18
The Laplace Transform
Wednesday 27 Aug 2014
• Question: What is the utility of this mathematical tool (the T)?
• Answer: 1. It greatly simplifies the process of solving Linear Time-
Invariant (LTI), homogeneous, Ordinary Differential Equations (ODEs)
2. Provides the basis for a “qualitative” evaluation of linear systems.
The T converts differential equations into algebraic equations
EE 401: Control Systems Analysis and Design Slide 2 of 19
The Laplace Transform
Wednesday 27 Aug 2014
• Definition: The (unilateral) Laplace Transform is defined as
(1)
where is the complex frequency variable with units sec–1. Assumes that (i.e., causal signal)
• Note: Equation (1) is meaningful only if the integral converges, i.e.,
This may be true for only a region of convergence within the imaginary plane.
0
( ) ( ) ( )stf t f t e dt F s
L
0
( ) stf t e dt
EE 401: Control Systems Analysis and Design Slide 3 of 19
Some LT Examples
Wednesday 27 Aug 2014
Example 1: T of an exponential (a > 0);
Similarly,
0
at at ste e e dt
L
For now we will ignore this Region of Convergence.
1ates a
L
( )
0
s a te dt
( )
0
1( )
s a tes a
( ) ( )
0
1( )
s a t s a t
t te e
s a
1s a
Region of Convergence
0 iff Re{s}>a 1
EE 401: Control Systems Analysis and Design Slide 4 of 19
Some LT Examples
Wednesday 27 Aug 2014
Example 2: LT of a constant; . Often denoted by and referred to as the unit step function or Heaviside function
• Note:
0
( ) ( ) stu t u t e dt
L
( ) ( ) aau t a u ts
L L
0
1 stes
1s
EE 401: Control Systems Analysis and Design Slide 5 of 19
Some LT Examples
Wednesday 27 Aug 2014
Example 3: LT of a sinusoidal;
• Also,
0
cos( )2
jbt jbtste ebt e dt
L
2 2sin( ) bbts b
L
Recall Euler’s Formula
cos( )2 cos( ) sin( )
sin( )2
j j
jj j
e e
e je e
j
( ) ( )0 0
1 12( ) 2( )
s jb t s jb te es jb s jb
1 1
2( ) 2( )s jb s jb
( ) ( )2( )( )s jb s jbs jb s jb
2 22
2( )s
s b
2 2s
s b
EE 401: Control Systems Analysis and Design Slide 6 of 19
Some LT Examples
Wednesday 27 Aug 2014
Example 4: LT of a powers of t;
i.e., (LT of a unit ramp)
0
n n stt t e dt
L
21ts
L
u dv
Integration by Parts
( )d uv udv vdudt
( )d uv udv vdudt
21 nn ts
L
1
0 0
1 stn st n et e nt dt
s s
1
0
n stn t e dts
1nn ts
L
1!nns
udv uv vdu
EE 401: Control Systems Analysis and Design Slide 7 of 19
Using MATLAB Directly
Symbolic Toolbox
Using MATLAB MuPAD notepad
MATLAB Symbolic EditorSometimes different syntax
Some LT ExamplesThe MATLAB Symbolic Math Toolbox
Wednesday 27 Aug 2014 EE 401: Control Systems Analysis and Design Slide 8 of 19
Some LT ExamplesThe MATLAB Symbolic Math Toolbox
Wednesday 27 Aug 2014
• Using Mathematica
EE 401: Control Systems Analysis and Design Slide 9 of 19
Some LT Examples
Wednesday 27 Aug 2014
Example 5: LT of a unit impulse;
0
( ) ( ) stt t e dt
L Note: We need to include t = 0
in the integral.
See Handout “Laplace_Transform_Table.pdf” for a table of Laplace Transforms.
0
st
te
1
EE 401: Control Systems Analysis and Design Slide 10 of 19
Some Properties of the LT
Wednesday 27 Aug 2014
• Linearity:
Example:
0
( ) ( ) sta f t a f t e dt
L
0
( ) sta f t e dt
( )a f t L
0
( ) ( ) ( ) ( ) stf t g t f t g t e dt
L
0 0
( ) ( )st stf t e dt g t e dt
( ) ( )f t g t L L
3 ( ) 2 3 ( ) 2t tt e t e L L L
13 21s
3 5
1ss
EE 401: Control Systems Analysis and Design Slide 11 of 19
Some Properties of the LT
Wednesday 27 Aug 2014
• Differentiation:
'
0
( ) ( ) std f t f t e dtdt
L
00
( ) ( )st stf t e s f t e dt
( ) (0)sF s f
du v
F(s)
EE 401: Control Systems Analysis and Design Slide 12 of 19
𝑠Input Output
+-
( )F s ( ) ( ) (0)Y s sF s f
𝑑𝑑𝑡
Input Output
( )f t ( ) '( )y t f t
(0)f
Some Properties of the LT
Wednesday 27 Aug 2014
• Integration:
0 0 0
( ) ( )t t
stf d f d e dt
L
0 00
1 1( ) ( )t
st stf d e f t e dts s
1 ( )F ss
u dv
(.)Input Output
( )f t
EE 401: Control Systems Analysis and Design Slide 13 of 19
1𝑠
Input Output
( ) ( )dy t f ( )F s1( ) ( )Y s F ss
Some Properties of the LT
Wednesday 27 Aug 2014
• Convolution:
0
( ) ( ) ( ) ( )t
f g t d F s G s
L
𝑔∗Input Output
( )f t ( )y t f g
EE 401: Control Systems Analysis and Design Slide 14 of 19
𝐺 (𝑠 )Input Output
( ) ( ) ( )Y s F s G s( )F s
(t)* ( ) ( ) ( )g f t F s G sL
Inverting the Laplace Transform
Wednesday 27 Aug 2014
• Inverting the Laplace Transform:
• Use the tables instead!!
1 1( ) ( ) ( )2π
jst
j
f t F s F s e dsj
L
DO NOT USE THIS FORMULA
EE 401: Control Systems Analysis and Design Slide 15 of 19
Inverting the Laplace TransformExample #1:
Wednesday 27 Aug 2014
Solve the first order ODE (ordinary differential equation)
Take the LT of the equation.
5 ( ) 10 ( ) 3 ( ), with (0) 1y t y t u t y
35 ( ) (0) 10 ( )sY s y Y ss
35( 2) ( ) 5s Y ss
1 0.6( )2 ( 2)
Y ss s s
1 2( ) 0.32 ( 2)
Y ss s s
2 2( ) 0.3(1 ), 0t ty t e e t
1s a
( )a
s s a
= 1
Unforced Response(due to initial conditions)
Forced Response(due to input 3u(t))
EE 401: Control Systems Analysis and Design Slide 16 of 19
Inverting the Laplace TransformExample #2:
Wednesday 27 Aug 2014
Solve the second order ODE (ordinary differential equation)
Take the LT of the equation.
( ) 3 ( ) 2 ( ) 5 ( ), with (0) 1, (0) 2y t y t y t u t y y
2 5( ) (0) (0) 3 ( ) (0) 2 ( )s Y s sy y sY s y Y ss
2 5( 3 2) ( ) (0) (0) 3 (0)s s Y s sy y ys
2( 1)( 2) ( ) 5 2 3s s s Y s s s s 2 5( )
( 1)( 2)s sY s
s s s
1 2
A B Cs s s
5 / 2 5 3 / 21 2s s s
25 3( ) 5 , 0
2 2t ty t e e t
EE 401: Control Systems Analysis and Design Slide 17 of 19
Inverting the Laplace TransformExample #3:
Wednesday 27 Aug 2014
Partial fraction expansion for the case of complex roots
2
3( )2 5
Y ss s s
2 2 5A Bs Cs s s
3 3 3( ) cos(2 ) sin(2 ), 05 5 10
t ty t e t e t t
3( 1 2 )( 1 2 )s s j s j
2
3 63 / 5 5 5
2 5
s
s s s
20
3 352 5 s
As s
233 2 5
5s s Bs C s
2 23 65 5s s Bs Cs
2 2
3 / 5 3 25 1 2
ss s
2 22 2
3 / 5 3 1 1 25 21 2 1 2
ss s s
cos(2 )te t L
sin(2 )te t L
EE 401: Control Systems Analysis and Design Slide 18 of 19
Inverting the Laplace TransformExample #3:
Wednesday 27 Aug 2014
Partial fraction for the case of complex rootsThis result can be further simplified:
3 3 3( ) cos(2 ) sin(2 )5 5 10
t ty t e t e t
3 3 1cos(2 ) sin(2 )5 5 2
te t t
cos( )cos( ) sin( )sin( )r a b r a b
cos( )r a b
221 1 / 2r
1/ 2arctan1
b
22 13 3 1 1/ 2 cos(2 tan (0.5))5 5
te t
0.6 0.6708 cos(2 26.6 )te t
EE 401: Control Systems Analysis and Design Slide 19 of 19