# Aplikasi Transformasi Laplace 21 June 2016 1 ```Aplikasi
Transformasi Laplace
21 June 2016
1
TRANSIENT ANALYSIS
Introduction
• Concerned with circuit behaviour under general conditions
• E.g. Application of voltage pulse or switch closure
Note: AC and DC circuits were special cases that
allowed simplified analysis
• Particularly interested in transitional behaviour between
initial and final conditions
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2
Illustrative case
: Switch closes at t = 0
4Ω
2H
0.5H
+
3Ω
+
6Ω
30V
-
0.05F
0.1F
S
t=0
For t &lt; 0 : vout (t) =18V
For t &gt;&gt; 0 : vout (t) =10V
}
-
DC conditions
For 0 &lt; t &lt; “infinity” : vout (t) = ???
21 June 2016
vout(t)
} Transient conditions
3
Classical Analysis
20H
S
Simple example : Find vout(t)
t=0
+
-
+
+
KVL:
di(t )
vin (t )  20
 4i(t )
dt
vin(t)
10V
-
+
+
4Ω
i(t)
vout(t)
-
-
Boundary conditions :
 0V ; t  0
vin (t )  
10V ; t  0
0.2t
) Amps
PI + CF ( + a lot of work) : i (t )  2.5(1  e
 vout (t )  10(1  e 0.2t ) Volts (t &gt; 0)
21 June 2016
(t &gt; 0)
[Ans]
4
Graph of Solution
12
(Volts)
FV = final value = 10V
10
8
6
vout(t)
4
63% of FV
2
IV = initial value = 0V
0
t
0
2t
3t
10
4t
5t
20
transient
6t
30
7t
8t
40
9t
(secs)
10t
50
• Mathematically takes infinite time for respose to reach final value of 10V:
• Use concept of time constant t : transition 63% complete
• Response is within 1% of FV after 5 time constants (5t)
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5
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6
Laplace transform method
• a generally superior method (avoids differential equations)
Problem in
time domain*
Problem in Laplace domain*
Laplace transformation
* all variables are
functions of t
Solution
Time Domain
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Inverse transformation
* all variables are now functions of
the complex frequency parameter s
Problem solved in Laplace domain
Laplace Domain
7
Laplace Transform

F ( s)   f (t )e  st dt
0
Laplace
domain
Time
domain
• Note: LT only represents time functions in range 0 to infinity
• s is a complex variable
• st is dimensionless

s is complex frequency
• Property of LT : converts DE’s to algebraic equations (easy to solve)
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8
Unit Step Function
u(t)
1
0; t  0
u (t )  
1; t  0
time (sec)
0
1
L{u(t)}=
s
sin(wt)
u(t)
1
time
Useful for representing
sudden changes
0
sin(wt)u(t)
e.g. application of
sinusoid at t = 0
time
0
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9
Unit Impulse Function
Mathematical representation of
short burst of input (lightning,
hammer blow, etc.)
1
DT
approximation
to unit impulse
area = 1
DT
Exact shape unimportant if
duration short relative to effects
time
(sec)
0
Dt
L{d(t)} =
d(t)
1  st
Lim  e dt = 1
Dt 0 Dt
0
representation of
ideal unit impulse
Impuse functions sometimes used to
test system dynamics
0
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time
(sec)
10
Transform Networks
• Networks shifted from time domain to complex frequency domain
• Apply LT to time domain equations
• Circuit elements become functions of s (including sources)
Resistance
v(t )  Ri(t )


 st
[
v
(
t
)

Ri
(
t
)]
e
dt

o

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V ( s)  RI ( s)
+
+
V(s)
-
R
I(s)
-
11
Inductance
di(t )
v(t )  L
dt

 [v(t )  L
o
I(s)
+
+
di(t )  st
]e dt
dt
sL
V(s)





 st 
 st 
V ( s)  L i (t )e
dt    Li (0)  sLI ( s)
o  s  i (t )e


o


Li(0)
-
+
V (s)  sLI (s)  Li (0)
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12
Capacitance
dv(t )
i(t )  C
dt
I(s)
+
+

1
dv(t )  st
[
i
(
t
)

C
]e dt

dt
o
sC
V(s)
+





 st 
 st 
I ( s)  C v(t )e o  s  v(t )e dt   Cv(0)  sCV ( s)


o


v(0)
s
-
-
1
v(0)
V ( s) 
I ( s) 
sC
s
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13
Summary
Resistance
I(s)
+
+
V(s)
+
R
-
-
Inductance
I(s)
+
Capacitance
+
sL
V(s)
-
I(s)
V(s)
+
1
-
sC
+
Li(0)
-
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+
-
-
v(0)
s
14
Example 1
R1= 6W
+
• Solve for vout(t)
• Sketch response
• Specify time constant
+
S
-
+
t=0
i (t)
i1(t)
10V
-
R= 4W
+
vout(t)
C= 0.5F -
Solution :
• Capacitor voltage at t = 0: vc(0) = 4V
• Switch OPEN in transform network ( t &gt; 0 )
• Voltage source = 10/s in transform network
• Capacitance has impedance 1/sC = 2/s
• In series with voltage vc(0)/s = 4/s
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15
Solution (ctd.)
6
o/c
+
10/s
R= 4W
-
KVL:
LT Tables :
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4
s  0.5

2/s
Vout(s)
-
Transform network
[Ans]
L-1  s 1 a  = e-at

+
4/s
-
2
4
 4I ( s)  I ( s)   0
s
s
1
 I ( s)  
s  0.5
Vout ( s)  4  I ( s) 
+
I(s) +
 vout (t )  4e 0.5t [Ans]
16
Solution (ctd.)
5
(Volts)
• Can be drawn freehand
4
3
vout(t)
transition 63% complete
2
1
0
37% t
2t
3t
4t
2
4
6
8
0
constant
5t
(secs)
10
transient
• t &lt; 0 : vout = 4V
• t &gt; 0 : Capacitor discharges to FV of 0V
• Time constant of decay : t = 1/0.5 = 2 sec
• Transition 63% complete after t secs ( 99+ % complete after 5t = 10 secs)
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17
Example 2
2Ω
5H
+
• Find vout(t)
• Sketch response
+
20V
3/5 W
-
• Verify IV and FV
t=0
S
3Ω
vout(t)
-
Solution
• Inductor current at t &lt; 0 and at t = 0 : iL(0) = 4A
• Switch CLOSED in transform network ( t &gt; 0 )
• Voltage source = 20/s in transform network
• Inductance has impedance sL = 5s
• In series with voltage source LiL(0) = 20
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18
Solution (ctd.)
5s
2
-
20
+
+
+
20/s
3/5 W
3Ω
Vout(s)
-
Exercise : Solve network for Vout(s) [ without superposition ]
Solution obtained in notes :
Vout ( s) 
36  156 s 4
8
 
s(9  13s) s s  9
13
Invert LT :
21 June 2016
vout (t )  4  8e
9t
13
Volts [t  0]
[Ans]
19
Solution (ctd.)
INITIAL
VALUE
10V
Sketch :
9t
13
4  8e
12V
vout(t)
Volts
transition 63% complete
8V
6V
37%
FINAL
VALUE
FINAL VALUE = 4V
• Sketch starts at IV (t = 0)
• 63% transition at t = t
2V
0V
•Asymptotic on FV
 36  156 s  156
IV  Lim s  
 12V

s
 s(9  13s)  13
0
2
t = 1.44 sec
4
6
8
(secs)
10
5t
 36  156 s  36
FV  Lim s  
 4V

s0
 s(9  13s)  9
i.e. no need to invert LT to find correct IV and FV (also found by circuit inspection)
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20
Inversion of LTs
• Time consuming
E.g.
Vout ( s ) 
40( s  5)
s ( s 2  4 s  20)
• Not in LT Tables
10
10 s

• Must decompose into partial fractions Vout ( s) 
s s 2  4s  20
• More work on 2nd term :
• Apply LT Theorem :
• Finally :
21 June 2016
Vout ( s) 
10 10( s  2)  20

s ( s  2) 2  4 2
L1 F ( s   )  e t L1 F ( s )
vout (t )  10  10e 2t cos(4t )  5e 2t sin( 4t )
21
Graphical result
• Still need to draw graph (more work):
20V
Contains vital info:
• FV = 10V
• Takes 2.5 secs to settle
c. 30%
10V
10(1 - e-2t)
• Involves decaying oscn
63%
• Period circa 1.6 sec
• Overshoot c. 30%
0V
0
t = 0.5
1.0
1.5
2.0 (sec) 2.5
GOOD NEWS : Can get basic info without inverting LT
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22
Pole-zero diagrams
( First step in sketching transient results )
• Applies to rational functions e.g. Vout ( s) 
• General form :
• Factorize :
40( s  5)
s ( s 2  4s  20)

40 s  200
s 3  4s 2  20 s
P( s ) bm s m  bm1s m1  b1s  b0

 H (s)
n
n

1
Q( s ) an s  an 1s
   a1s  a0
( s  z1 )(s  z 2 ) (s  z m )
P( s )
 H.
 H ( s)
Q( s )
( s  p1 )(s  p 2 ) ( s  p n )
z1, z2, … zm are the m roots of P(s) – called ZEROS
p1, p2, … pm are the n roots of Q(s) – called POLES
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23
Illustration
P( s )
2s 2  20 s  50

Q( s) s 4  6s 3  18s 2  40 s
P( s )
2( s  5) 2

Q( s) s( s  4)(s  1  3 j )(s  1  3 j )
Function has poles at :
s = 0 , s = - 4 , s = -1 – 3j , s = - 1 + 3j
Function has zeros at :
s = - 5 , s = - 5 ( i.e. double zero @ s = - 5 )
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24
Im
Illustration (ctd.)
j3
Mark location of poles and zeros in
s-plane as shown
j2
poles
zeros
j1
Re
multiplier = 2
-6
-5
-4
-3
s-plane
-2
-1
1
2
-j1
-j2
pole-zero diagram of :
P( s )
2( s  5) 2

Q( s) s( s  4)(s  1  3 j )(s  1  3 j )
21 June 2016
-j3
25
Exercises
Draw pole-zero maps of the following
functions :
s 1
F (s) 
s( s  2)
F ( s) 
21 June 2016
s 2  2s  2
2
s ( s  3)(s  4s  8)
F ( s) 
3s( s  1)
( s  2)(s 2  2s  2)
F ( s) 
5s  5
s 3  4s 2  4s
26
Exercises (ctd.)
Find functions of s corresponding to the following
pole-zero diagrams :
s-plane
Im
Im
j6
j3
j4
j2
s-plane
j1
j2
Re
-12 -10
-8
-6
-4
-2
1
Re
-6
-5
-4
multiplier = 100
-j6
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-2
-1
1
2
-j1
-j2
-j4
-3
multiplier = 20
-j2
-j3
27
Sketching time functions
Im
j4
Illustrative function and its pole-zero map :
j3
Vout ( s ) 
60( s  4)
s-plane
j2
2
s ( s  6)(s  2 s  17 )
j1
Re
-6
-5
-4
-3
-2
-1
1
2
-j1
If we used partial fractions, function would
decompose into following form :
A
B
Cs  D
Vout (s)  

s s  6 s 2  2s  17
21 June 2016
-j2
-j3
-j4
28
Sketching time functions
Im
j4
Re-write as :
Vout ( s ) 
E ( s  1)  2 F
A
B


s s  6 ( s  1) 2  4 2
j3
s-plane
j2
Corresponding time function is of form :
vout (t )  A  Be
6t
 Ge
G  E2  F 2
1t
j1
Re
cos(4t   )
-6
-5
  tan 1 ( F E )
-4
-3
-2
-1
1
-j1
-j2
Thus response has three components :
1st is due to pole at s = 0
2nd is due to pole at s = - 6
3rd is due to conjugate pair at s = - 1 &plusmn; j4
21 June 2016
-j3
-j4
29
2
Sketching time functions
Im
vout (t )  A  Be 6t  Ge 1t cos(4t   )
j4
j3
• Pole @ origin gives constant term
s-plane
• Pole @ s = - 6 gives term decaying as e- 6t
j2
j1
• Pole pair gives oscillatory term
( frequency 4 rad/s decaying as e- 1t )
Re
-6
-5
-4
-3
-2
-1
1
-j1
Can find (constant) FV without inverting LT :
vout ()  Lim sVout ( s)  2V
s0
Also
vout (0)  Lim sVout ( s)  0V
-j2
-j3
-j4
s
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30
2
Sketching time functions
Im
j44
FV theorem takes care of pole @ origin
j3
vout (t )  A  Be 6t  Ge 1t cos(4t   )
Fast decay
j2
Slow decay
j1
Re
FV thm.
-6
-5
-4
-3
-2
-11
1
-j1
Dominant term
-j2
Decaying oscillation
• Period of oscn = T = 2p/4 = 1.6 sec
Dominant poles
-j3
-j4
• Decay time const. t = 1/1 = 1 sec
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31
2
Im
Sketching time functions
j4
j3
Approximating sketch
j2
ignore
j1
Re
4V
T = 1.6 sec
-6
-5
-4
-3
-2
-1
1
-j1
upper boundary
-j2
FV = 2V
-j3
63%
0V 0
t=1
T/2
21 June 2016
lower boundary
-j4
2
3
4
(sec)
5
3T/2
32
2
Guidelines for sketching
1. Write down LT in factored form
2. Draw pole-zero map
3. Draw time axis
4. Determine IV (dot on vertical axis)
5. Determine FV
6. Calculate t
7. Place dot 63% between IV and FV
8. Sketch exponential curve
FV
95%
63%
IV
time
0
21 June 2016
t
3t
5t
33
Case 1
: Dominant pole on real axis
• Stop at step 8
Im
dominant pole
• Response non-oscillatory
• FV = 0 if NO pole @ origin
Re
-
FV
s-plane
63%
95%
IV
time
0
21 June 2016
t  1/
3t
5t
34
Case 2
: Dominant conjugate poles
Im
9. Draw mirror image of
exponential
jb
10. T = 2p/b (insert markers)
@ T/2 etc. as shown.
Sketch decaying sinusoid
thro’ markers convergent
on FV
Re
-
T/2
3T/2
-jb
T
FV
95%
63%
IV
time
0
21 June 2016
t  1/
3t
5t
35
Examination-type question
In given circuit, S has been closed for a long time before being opened at t=0.
(a) Determine values for iL(0) and vC(0)
(b) Draw transform network describing circuit behaviour for t &gt; 0.
(c) Determine expression for I(s), and draw its pole-zero map.
(d) Use IV theorem to find initial value of i(t) and FV theorem to find final value of i(t).
(e) Sketch nature if i(t) for t &gt; 0.
S
2Ω
t=0
+
2Ω
-
2H
8V
21 June 2016
i(t)
2 F
3
36
t&lt; 0:
iL(0) = 2A [Ans]
2Ω
t&lt;0
2Ω
+
8V
o/c
2A
-
2H
vC(0) = 4V [Ans]
2
Transform
network :
[Ans]
4V
S
Solution
+
8/s
-
s/c
o/c
+
0
2
2s
-
+
-
I(s)
-
3/2s
+
+
4/s
4
+
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37
Solution (ctd.)
2
o/c
+
0
+
2
8/s
2s
-
+
I(s)
3/2s
+
+
-
-
-
4/s
4
-
+
Circuit analysis :
I ( s) 
4
4
s
2  2s 
21 June 2016
3
2s

2( s  1)
2
s s
3
4

2( s  1)
2
( s  1)  0.707
2
[Ans]
38
Solution (ctd.)
I ( s) 
 2( s  1) 


i(0)  Lim s 
  2A
2
3
s   s  s  
4

 2( s  1) 


i()  Lim s 
  0A
2
3
s0  s  s  
4

2( s  1)
j0.707
( s  1) 2  0.707 2
-0.5
-1
[Ans]
-j0.707
[Ans]
[Ans]
t
IV=2A
t
63%
t
1
 2 sec
0.5
2p
T
 9 sec
0.707
sec
FV=0
0
2
6
4
T/2
8
10
T
[Ans]
21 June 2016
39
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