Transcript EEE 302 Electrical Networks II Dr. Keith E. Holbert Summer 2001 Lecture 14 Solving Differential Equations • Laplace transform approach automatically includes initial conditions in the.

EEE 302
Electrical Networks II
Dr. Keith E. Holbert
Summer 2001
Lecture 14
1
Solving Differential Equations
• Laplace transform approach automatically includes
initial conditions in the solution
L  d x(t )   s X(s)  x(0)
 dt 
 d 2 y (t ) 
L  2   s 2 Y(s)  s y(0)  y' (0)
 dt 
• Exercise: For zero initial conditions, solve
d 2 y(t )
d y (t )
 11
 30 y (t )  4 u (t )
2
dt
dt
Lecture 14
2
Inverse Laplace Transform
• Consider that F(s) is a ratio of polynomial
expressions
N ( s)
F ( s) 
D ( s)
• The roots of the denominator, D(s) are called the
poles
– Poles really determine the response and stability of the
system
• The roots of the numerator, N(s), are called the zeros
Lecture 14
3
Inverse Laplace Transform
• We will use partial fractions expansion with the
method of residues to determine the inverse Laplace
transform
• Three possible cases (need proper rational, i.e., n>m)
(1) simple poles (real and unequal)
(2) simple complex roots (conjugate pair)
(3) repeated roots of same value
Lecture 14
4
Simple Poles
• Simple poles are placed in a partial fractions expansion
K 0 s  z1 s  z m 
Kn
K1
K2
F ( s) 



s  p1 s  p2 s  pn  s  p1 s  p2
s  pn
• The constants, Ki, can be found from (use method of residues)
K i  ( s  pi ) F ( s) s  p
i
• Finally, tabulated Laplace transform pairs are used to invert
expression, but this is a nice form since the solution is
f (t )  K1 e  p1 t  K 2 e  p2 t    K n e  pn t
Lecture 14
5
Class Examples
• Extension Exercise E13.9
• Extension Exercise E13.10
Lecture 14
6
Complex Conjugate Poles
• Complex poles result in a Laplace transform of the form
K1 
K1   
K1
K1*
F ( s) 

 


s  (  j ) s  (  j )
s  (  j ) s  (  j )
• The K1 can be found using the same method as for simple
poles
K 1  ( s    j ) F ( s ) s    j
WARNING: the "positive" pole of the form -+j MUST be
the one that is used
• The corresponding time domain function is
f (t )  2 K1 e  t cos t     
Lecture 14
7
Class Example
• Extension Exercise E13.11
Lecture 14
8
Repeated Poles
• When F(s) has a pole of multiplicity r, then F(s) is
written as
F ( s) 
P1 (s)
Q 1 (s) s  p1 
r
K11
K12
K1r




2
r
s  p1 s  p1 
s  p1 
• Where the time domain function is then
f (t )  K11 e  p1t  K12 t e  p1t
t r 1  p1t
   K1r
e 
r  1!
• That is we get the usual exponential multiplied by t's
Lecture 14
9
Repeated Poles (cont’d.)
• The K1j terms are evaluated from
K1 j

1
d r j
r



s

p
F ( s)
1
r j
r  j ! ds

s   p1
• This actually simplifies nicely until you reach s³
terms, that is for a double root (s+p1)²
K12  s  p1  F (s)
2
s   p1

d
s  p1 2 F(s)
K11 
ds

s   p1
• Thus K12 is found just like for simple roots
• Note this reverse order of solving for the K values
Lecture 14
10
Class Examples
• Extension Exercise E13.12
• Extension Exercise E13.13
Lecture 14
11