2/5/2015
1
- Systems Described by Ordinary Differential
Equations (ODEs)
- Laplace Transform
- Inverse Laplace Transform. Partial Fraction Expansion
- (Last revised on 2/5/2015)
EE110, S15, Circuits & Systems: Lec03
Prof. Ping Hsu
1
Systems Described by Ordinary Differential Equations
Systems described by algebraic equations do not involve time. Such a system’s output depends on only its input at the instant of time.
2
© 2013 National Technology and Science Press. All rights reserved.
× 3x y=3x 2
4
VoutVin
A dynamic system’s output at a certain instant of time depends on the entire historyof the input. In other words, such a system has some ‘memory effect’. Systems described by differential equations are dynamic systems.
y(t)
x(t)
Dynamicalsystem
x(t) y(t)
2/5/2015
2
Systems Described by Ordinary Differential Equations
- 1st order ODE with constant coefficients- The system is called a first order system
(only 1 reactive element, C or L).
Example: A simple RC circuit:
The output and input signals are related by the simple differential equation (EE98):
(From KVL)
- Note that the resistor R and the voltage source x(t) could be the Thevenin’sequivalent of more elaborate resistive circuit connected across the capacitor C.
3
© 2013 National Technology and Science Press. All rights reserved.
( ) 1 1( ) ( )
dy ty t x t
dt RC RC
Systems Described by Ordinary Differential Equations4
© 2013 National Technology and Science Press. All rights reserved.
For a given input signal x(t), the output signal y(t) can be obtained by solving the first-order differential equation above.
( ) 1 1( ) ( )
dy ty t x t
dt RC RC
systemx(t) y(t)
2/5/2015
3
Given y(0)=0 and x(t)=1 (or unit step u(t)),
Find y(t). Vs
y(t)
+
-
5
/
where is the time constant.
Solution: ( ) (1 ) ( )
ts
RC
y t V e u t
=1=2
=5
=3=4
( ) 1 1( ) ( )
dy ty t x t
dt RC RC The circuit equation:
SECOND-ORDER SYSTEMS (CIRCUITS)
vS vR vC vL 0 (KVL)
Ld 2i
dt 2 Rdi
dt
i
C
dvS
dt
Two reactive elements
22
2
usually written in the normalized form
( ) ( ) 2 ( ) 0
1 undamped natural frequency
= damping ratio2
n n
n
d x t dx tx t
dt dt
LC
R C
L
1 21 2
2 2
1 2
The Characteristic Equation:
2 0;
Two roots, and (natural frequencies),
can be complex. The complementary
solution (i.e., with vs=0), is
( ) , s t s tc
n n
x t K
s
K
s
e
s
s
e
Adapted from Irwin & Nelms, 20116
Two initial conditions are required to determine K1 and K2. In a series RLCcircuit, for example, they may be vC(0-) and iL(0-).
For f(t)=Au(t), the total solution is
1 21 22
( ) s t s t
n
Ax t K e K e
2/5/2015
4
Linear ODE can be solved by using Laplace Transform
Solve ODE
x(t) y(t)Given x(t), find y(t)
Time-Domain (t-domain) approach
x(t) y(t)
Complex Frequency Domain (s-domain) approach
Laplacetransform
Inverse transform
Solve Algeb-raic Eq’n
Don’tSolve ODE
X(s) Y(s)
t-domain
s-domain
algebraic eq’nsare easier to solve than ODEs
a transform is like change of variable
7
Definition of the Laplace Transform
0 σ
ω
Re{s}
Im{s} s
s-plane
X(s) x(t)e st
0
dt, s j ,
's ' is a complex number called
the "complex frequency"
( , ) form the 2 orthogonal axes of a
plane called the "s-plane"
Any given 's ' defines a vector in the s-plane:
Re{s}, Im{s}
are its (x, y) coordinates
0 integral limit allows including initial conditions
Called 1-sided or unilateral Laplace Transform (t ≥ 0-)
8
2/5/2015
5
The Inverse Laplace Transform
- Given an X(s), one is able to recover exactly the same x(t) that was used to compute X(s) with the help of the inverse transform
- {x(t), X(s)} are said to from a unique transform pair.
- Integral evaluation requires knowledge of complex variables theory, avoided here by using table look-up approach instead.
- Notation:
x(t) 1
2 jX(s)est
j
j
ds
9
-1, ( ) ( ), ( ) ( )x t X s X s x t x t X s
Example Transform Pairs: Unit Step Function
x1(t) Au(t T )
0 T
A
t
x1(t) Au(t T ), T 0 is a constant, X1(s) ?
0
0
( ) sTAAu t T e
s T 0
10
© 2013 National Technology and Science Press. All rights reserved.
2/5/2015
6
Example Transform Pairs: Rectangular Pulse
h- x2(t) is the rectangular signal shown. X2(s) = ?
Integrals are linear operation:
u(t T1) u(t T2 ) 1
s(e sT1 e sT2 )
0
11
© 2013 National Technology and Science Press. All rights reserved.
Example Transform Pairs: Delta Function
- x3(t) is the Dirac delta function shown. X3(s) = ?
0
(t T ) 1 e sT
12
© 2013 National Technology and Science Press. All rights reserved.
2/5/2015
7
Example Laplace Transform Pairs
Re{a} 0
Re{a}
e( ja)t 0, t , only if
Called the region of convergence, or, ROC
x1(t) eatu(t)
13
© 2013 National Technology and Science Press. All rights reserved.
Exponential & Sinusoidal Signals
Example Laplace Transform Pairs14
Euler’s formula:
cos( ) sin( ) cos( )2
cos( ) sin( ) sin( )2
jx jxjx
jx jxjx
e ee x j x x
e ee x j x x
j
2
1 1cos( )
2 2 2
1 1 1 1
2 2
j tj tj t j te e
t e e
s
s j s j s
2/5/2015
8
Some Basic Laplace Transform Pairs
Extractions from Table 3-2
15
© 2013 National Technology and Science Press. All rights reserved.
Poles & Zeros of X(s) (= L(x(t)))16
© 2013 National Technology and Science Press. All rights reserved.
In general, X(s) is a rational function of the form
11 1 0
11 1 0
1 2 1
1 2 1
( )( )
( )
( )( ) ( )( )
( )( ) ( )( )
m mm m
n nn n
m m
n n
b s b s b s bN sX s
D s a s a s a s a
A s z s z s z s z
s p s p s p s p
The zeros of X(s) are the values of s that render N(s)=0 (which are also the roots of N(s)).
The poles of X(s) are the roots of its denominator D(s).
2/5/2015
9
Inverse Laplace Transform using Partial Fraction Expansion
17
3 2
2
5 30 49 10( )
( 2)( 4 5)
s s sX s
s s s s
Partial Fraction Expansion
2
1 4 8( )
2 4 5X s
s s s s
L-1 (X(s)) = ? Not in the Table!
x(t) = L-1 (X(s)) = L−1( )+ L-1 ( L−1( )
2 2( ) ( ) 4 ( ) 8 sin( ) ( )t tx t u t e u t e t u t
Table-look-up
Inverse Laplace Transform using Partial Fraction Expansion
18
11 1 0
11 1 0
( )m m
m mn n
n n
b s b s b s bX s
a s a s a s a
When apply the PFE method, X(s) must be in a ‘strictly proper’ form.
If n>m, X(s) is called “strictly proper”. (If n=m, it is call “proper”.)
3 2
2
5 30 49 10( )
( 2)( 4 5)
s s sX s
s s s s
2
2
1( )
4 5
sX s
s s
Strictly proper(n=4, m=2)
NOT strictly proper(n=2, m=2)
2/5/2015
10
19
If the given X(s) is not strictly proper, use polynomial division :
2 3 2
3 2
2
2
6 8
2 1 6 4 8 6
6 12 6
8 2 6
8 16 8
18 14
s
s s s s s
s s s
s s
s s
s
2
18 14( ) 6 8
2 1
sX s s
s s
3 2
2
6 4 8 6( )
2 1
s s sX s
s s
Inverse Laplace Transform using Partial Fraction Expansion
where L-1 (6s) = 6’(t)L-1 (8) = 8(t)
then apply PFE to the last term before table look-up.
Example
Partial Fraction Expansion: 1) Case of Distinct Real Poles
A2 (s 1) X(s)s1
s2 4s 3
s(s 3)s1
4
A1, A2 and A3 are called the residuals at poles s = 0, -1, and -3, respectively
requires knowing the poles of H(s)
20
© 2013 National Technology and Science Press. All rights reserved.
2/5/2015
11
© 2013 National Technology and Science Press. All rights reserved.
21
2 3
23
23 3
( 3) ( 3) ( 3)( 3) ( )
1 3
( 3) ( 3)
1
( 3) ( 3)
1 3
s A s A s As X s
s s s
s A s AA
s s
s A s AA A
s s s
2
2
2
4 3( 3) ( ) ( 3)
( 1)( 3)
4 3
( 1)
4 34
( 1) 3
s ss X s s
s s s
s s
s s
s s
s s s
This is why the method work!
Partial Fraction Expansion: 1) Case of Distinct Real Poles
Partial Fraction Expansion: 2) Case of Repeated Real Poles
Example:
22
2
4
4 6( )
( 1)( 2)( 3)
s sX s
s s s
Where A, B, and C4 can be found using the same method as in the distinct pole case.
1 2 3 42 3 4
( )1 2 3 ( 3) ( 3) ( 3)
A B C C C CX s
s s s s s s
The pole at -3 is a repeated pole of multiplicity of 4. In this case, the PFE will produce a total of 6 terms (2 from the distinct poles at -1 and at -2 and 4 from the 4 poles at -3.
1
2
44 3
( 1) ( )
B ( 2) ( )
( 3) ( )
s
s
s
A s X s
s X s
C s X s
2/5/2015
12
Partial Fraction Expansion: 2) Case of Repeated Real Poles
23
1 2 3 42 3 4
( )1 2 3 ( 3) ( 3) ( 3)
A B C C C CX s
s s s s s s
43
3
24
2 2
3
34
1 3
3
( 3) ( )
( 3)1
2!
1
3
( )
( 3) ( )!
s
s
s
dC s X s
ds
dC s X s
ds
dC s X s
ds
where ! ( 1)( 2) .... 3 2 1n n n n
Partial Fraction Expansion: 2) Case of Repeated Real Poles
24
2
311
2
322
23
3 23
2 23
2
8( 4 6)( 1) ( ) = = 3
( 2)( 3)
8( 4 6)B ( 2) ( ) = = -16
( 1)( 3)
8( 4 6)C ( 3) ( ) = = 12
( 1)( 2)
8( 4 6) 8(C ( 3) ( ) =
( 1)( 2)
ss
ss
ss
s sA s X s
s s
s ss X s
s s
s ss X s
s s
d d s s ss X s
ds ds s s
2
3
2 3 23
1 2 3
3
8 10)10
(( 1)( 2))
16( 12 30 22)C ( 3) (
1 1
2 2) 13
(( 1)( 2))
s
s
s
s s
d s s ss X s
ds s s
2
1 2 33 2 3
8 4 6( )
( 1)( 2)( 3) 1 2 3 ( 3) ( 3)
s s A B C C CX s
s s s s s s s s
Example:
2/5/2015
13
Partial Fraction Expansion: 2) Case of Repeated Real Poles
25
2
3 2 3
8 4 6 3 16 13 10 12( )
( 1)( 2)( 3) 1 2 3 ( 3) ( 3)
s sX s
s s s s s s s s
Example:
1 1
1 1
- -2
-3 -32
3 16 3 ( ) 16 ( )
1 213 10
13 ( ) 10 ( )3 ( 3)
12
( 3
t t
t t
e u t e u ts s
e u t te u ts s
s
L L
L L
1 2 -33 3
- -2 -3 -3 2 -3
2=6 6 ( )
) ( 3)
( ) 3 16 13 10 6 ( )
t
t t t t t
t e u ts
x t e e e te t e u t
L
4 4 4 4 4 44 1 2 3 4
2 3 4
4 43 2
3 2 3 4
4 43 2
3 2 3 4
( 3) ( 3) ( 3) ( 3) ( 3) ( 3)( 3) ( )
1 2 3 3 3 3
( 3) ( 3) 3 3 3
1 2
( 3) ( 3)3 3 3
1 2
s A s B s C s C s C s Cs X s
s s s s s s
s A s Bs C s C s C C
s s
d s A s Bs C s C s C C
ds s s
4 43 2
3 2 3
4 43 2
3 2 3
4
all these terms contai
( 3) ( 3)3 3
n (s+3) after differenti
31 2
( 3) ( 3)3 3
1 2
( 3) ( 3
1
ation!
d s A s Bs C s C s C
ds s s
d s A s Bs C s C C
ds s s
d s A s
ds s
4
3 2
3 2 3 3
)3 3
23
Bs C s C C C
ss
26
Why does the method work (consider how we found C3) ?
C4 is by itself !
C3 is by itself !
2/5/2015
14
Partial Fraction Expansion: 3) Distinct Complex Poles (method #1)
Example: 27
2 11
6 3 4 41/ 4
21 2 4 20
BB
B CC
B C
2
2 2 2 2 2
1
2
2
2
2
22 6 21 (4)( )
( 1)( 4 20) 1 ( 2) 4 ( 2) 4
where ( 1) ( ) 1
(1)( 4 20) ( 1)( 2) 4( 1)( )
( 1)( 4 20)
( 1) (3 4 4) (2 4 20)
( 1)( 4 20)
s
ss sX s
s s s s s s
A s X s
s s B s s C sX s
s
B
s s
B s B C s B C
A C
s s s
Comparing the coefficients:
2 20
2 20
Table:
cos( ) ( )( )
sin( ) ( )( )
ato
atoo
s ae t u t
s a
e t u ts a
Partial Fraction Expansion: 3) Distinct Complex Poles (Method #1)
28
2 21( ) cos(4 ) sin(4 ) ( )
4t t tx t e e t e t u t
where 1
1
22 2
22 2
11
( 1/ 4
2( ) cos(4 ) ( )
( 2) 4
4sin(4 ) ( )
( 2)
) 1
4 4
L t
L t
se t u t
s
e t u ts
2
2 2 2 2 2
22 6 21 (4)( )
( 1)( 4 20) 1 ( 2) 4 ( 2) 4
11 ( 1/ 4)ss sX s
s s s s s s
2/5/2015
15
Partial Fraction Expansion: 3) Distinct Complex Poles (Method #1)
29
2 2
2 2 2
2
1( ) cos(4 ) sin(4 ) ( )
4
1 (1 4) cos 4 atan2 1 4,1 ( )
1.031 cos 4 0.245 ( )
t t t
t t
t t
x t e e t e t u t
e e t u t
e e t u t
Trigonometric identity: cos( ) sin( ) cos( )X t tMt Y 2 2where and atan2(Y,X)M X Y
-1
-1
-1
0tan ( / ),
0, 0tan ( / ),
0, 0tan ( / ),where atan2(,) is defined as: atan2( , )
0, 0/ 2,
0, 0/ 2,
0, 0undefined,
xy x
y xy x
y xy xy x
y x
y x
y x
What is atan2(y,x)?30
“atan2(y,x) is an arc-tangent function that takes both the x and y component of a 2-D vector and returns the angle of the vector.
The regular tan-1(y/x) gives the right answer only if the vector is in the 1st
and the 4th quadrant.
Example:
For x = -1, y = -1, if you use the function tan-1(y/x), you get
tan-1(-1/(-1)) = tan-1(1) =/4=45o.
which is wrong. The function atan2(y,x) will give you the correct answer:
atan2(-1,-1) = - /4=-135o.
(-1,-1)
-135o
2/5/2015
16
Partial Fraction Expansion: 3) Distinct Complex Poles (Method #2)
31
2
2
2
2 42 4
2 6 21 1( )
( 1)( 4 20) 1 ( 2 4) ( 2 4)
where
2 6 21( 2 4) ( )
( 1)( 2 4)
1 1
2 8
s js j
s s A jB A jBX s
s s s s s j s j
s sA jB s j X s
s s j
j
1 cos( ) sin( ) ( )2 2atA jB A jBL e A bt B bt u t
s a jb s a jb
2 21( ) cos(4 ) sin(4 ) ( )
4t t tx t e e t e t u t
(Pole determines the oscillation frequency () and exponent (a)).
Partial Fraction Expansion: 3) Distinct Complex Poles (Method #3)
32
2
2
2
2 4
0.245
2 6 21 1( )
( 1)( 4 20) 1 ( 2 4) ( 2 4)
2 6 21( 2 4) ( )
( 1)( 2 4)
1 1 0.515
2 8
s j
j
s s Ae AeX s
s s s s s j s j
s sAe s j X s
s s j
j e
2( ) ( ) 2(0.515) cos(4 0.245) ( )t tx t e u t e t u t
1 ( ) ( )
( ) ( )
( ) ( )
( ) 2 cos( ) ( )
j jj a jb t j a jb t
at j bt j bt at
Ae AeL Ae e u t Ae e u t
s a jb s a jb
Ae e e u t Ae bt u t
0.245 radian (obtained by atan2) 0.515A
Its pole has positive
imaginary part.
. ., (-2 4) i e j
2/5/2015
17
Summary Results for 4 Cases
n! n(n 1)(n 2) .... 4 32 1
33
© 2013 National Technology and Science Press. All rights reserved.
Top Related