Transcript
Continuous-time Systems/Circuits
Example:
+
-x(t)
R
C y(t)
( )( ) ( )
( ) 1 1( ) ( )
dy tRC y t x t
dtdy t
y t x tdt RC RC
Differential/Integral Equation
+
-
Laplace Transform
Continuous-time Systems/Circuits
Time Domain Analysis (Discussed in the last chapter) Zero-input Response
Zero-state Response
Impulse Response
Convolution
Continuous-time Systems ...contd
Frequency Domain Analysis (This chapter) The Laplace Transform Using Laplace Transform to solve
problems Initial Value Theorem Final Value Theorem Transfer Function Relationship between Impulse response
and Transfer Function
Laplace Transform
Gives us a systematic way for relating time domain behavior of a circuit to its frequency domain behavior
Converts integro-differential equations describing a circuit to a set of algebraic equations
Considers transient behavior of circuits with multiple nodes and meshes, with initial conditions
Using Laplace Transform The idea of using a transform is
similar to using logarithms for multiplication
A = BCLog A = Log B + Log C
A = Antilog (Log B + Log C)
The use of Laplace Transform is universal – mechanical systems, electrical circuits and other systems – as long as the behavior is described by ordinary differential equations.
Laplace Transform
The general definition of Laplace Transform i.e. the Two-sided Transform:
And Inverse Transform:
( ) ( ) stX s x t e dt
1( ) ( )
2
c jst
c j
x t X s e dsj
Laplace Transform
But we shall consider a special case – One-Sided Transform:
We shall use the Table 4.1 (Lathi, page 344) to do inverse transform
0
( ) ( ) stX s x t e dt
Laplace Transform: Definition
t represents time s has the units of 1/t since the exponent must be
dimensionless
This is a one-sided, unilateral Laplace Transform – i.e. it ignores negative values of t.
Again, the definition of Laplace Transform is:
0
( ) ( )
( ) ( )
stL x t x t e dt
X s L x t
0
( ) ( ) ( ) stX s L x t x t e dt
Properties of Laplace Transform Linearity of Laplace Transform:
since
Uniqueness of Laplace Transform:
is a one-to-one relationship
1 2 1 2
( ) ( )
( ) ( ) ( ) ( )
L k x t k X s
L x t x t X s X s
1 2
0
1 2
0 0
1 2
( ) ( )
( ) ( )
( ) ( )
st
st st
x t x t e dt
x t e dt x t e dt
X s X s
( ) ( )x t X s
Properties ...contd Question: Does the integral converge?
We avoid functions like The integral converges for all the cases we shall
consider!
Question: What happens if the function is discontinuous at t=0?We choose the lower limit as 0-.In fact, we shall define the Laplace Transform as:
2
, ,t tt e etc
0
( ) stx t e dt
Properties ...contd
In taking Laplace Transform, it becomes important whether one chooses 0- or 0+
Our choice is always 0- for the lower limit!
0
, 0ate t 0, 0t
x(t)
t 0
ate
x(t)
t
Continuous at t=0 Discontinuous at t=0
Laplace Transform of the Unit Step
0
0 0
( ) ( )
1.
1
st
stst
L u t u t e dt
ee dt
s
s
0
0
( ) ( )
0 1
st
a stst
a aas
L u t a u t a e dt
edt e dt
se
s
t
1
0
( )u t
t
1
0
( )u t a
a
Laplace Transform of the Unit Step
0
( )
0( )
0
( )
( )1
at at st
s a t
s a t
L e u t e e dt
e dt
e
s a
s a
0
( ) ( )
1
stL t t e dt
0
( )ate u t
x(t)
t
0
( )tx(t)
t
1
Using Step Functions
Step functions can be used for writing analytical expressions for illustrated functions. Consider:
( ) ( ) ( )x t u t a u t b
t
( )x t
1
a b
Using Step Functions ...contd
Subtracting, it is obvious that
( ) 0,1,
u t a t at a
( ) 0,1,
u t b t bt b
( ) ( ) ( )x t u t a u t b
t
1
a
( )u t a
0 t
1
b
( )u t b
0
t
( )x t
1
a b
Using Step Functions ...contd
Consider:
( ) ( ) ( 1) ( 2) ( 1) ( 2)x t t u t u t t u t u t
( ) ( ) 2 2 ( 1) 2 ( 2)x t tu t t u t t u t
t
t
1
210
2
2t
t
1
210
( )x t
Using Step Functions ...contd
Consider:
( ) ( ) ( ) 2 ( ) ( )
( ) ( ) 2 ( )
x t u t a u t b u t b u t c
u t a u t b u t c
Alternative way:
( ) ( ) ( ) ( ) ( )
( ) ( ) 2 ( )
x t u t a u t c u t b u t c
u t a u t b u t c
( ) 2as bs cse e e
sXs s s
t
( )x t
1
a b c
2
A Reminder
cos sin jj e
cos sin jj e
sin2
j je e
j
cos2
j je e
Laplace Transforms of Sinusoids
0
0( ) ( )
0
2 2
(sin ) ( ) sin
2
2
1 1 1
2
st
j t j tst
s j t s j t
L t u t t e dt
e ee dt
j
e edt
j
j s j s j
s
Laplace Transforms of Sinusoids …contd
0
0( ) ( )
0
2 2
(cos ) ( ) cos
2
2
1 1 1
2
st
j t j tst
s j t s j t
L t u t t e dt
e ee dt
e edt
s j s js
s
Laplace Transform of Ramp
t
2
210
( )x t
3
1
( )u tt
0
0 0
2
0
( )
1
1 1
st
st st
st
L t u t t e dt
e et dt
s s
e dts s
Laplace transform of tu(t) :
Inverse Laplace Transform Most functions of interest to us are rational
functions – ratios of polynomials in s.
Roots of P0(s) are the zeros of X(s) Roots of Q0(s) are the poles of X(s)
Assume the degree of the numerator is less than that of the denominator – Otherwise, divide!
Expand into partial fractions and use the table
0
0
( )( )
( )
P sX s
Q s
0
( )( )
( )NewP s
X s PolynomialQ s
Watch out – I have put subscripts so that it does not conflict with later notation
Roots: Real and Distinct
31 296 5 12
( )8 6 8 6
s s kk kX s
s s s s s s
Multiply both sides by s and let s=0
1
321
0 00
96 5 12
8 6 8
2
6
1 0
s ss
s s k sk sk
s s s s
k
Similarly, for k2 and k3 ,
1 32
8 88
2 72
96 5 12 8 8
6 6s ss
k
s s k s k sk
s s s s
Roots: Real and Distinct ...contd
1 23
6 66
3
96 5
48
12 6 6
8 8s ss
s s k s k sk
s
k
s s s
Therefore,
96 5 12 120 72 48
6 8 8 6
s s
s s s s s s
Identify:
Substitute and the two sides become equal
5s
Inverse: 8 6( ) 120 72 48 ( )t tx t e e u t
Laplace Transform
Linear Systems and Signals – Lectures 10Dr. J. K. AggarwalThe University of Texas at Austin
Another Example
2 12( )
1 2 3
1 2 3
sX s
s s sA B C
s s s
Multiply both sides by (s+1), evaluate at s= -1
1
2 12 105
2 3 (1)(2)s
sA
s s
Multiply both sides by (s+2), evaluate at s= -2
2
2 12 88
1 3 ( 1)(1)s
sB
s s
Another Example ...contd
Multiply both sides by (s+3), evaluate at s= -3
3
2 12 63
1 2 ( 2)( 1)s
sC
s s
5 8 3( )
1 2 3X s
s s s
Therefore,
Inverse:
2 3( ) 5 8 3 ( )t t tx t e e e u t
Still Another Example
2
2
2
2
6 7( )
1 2
6 7
3 23 5
13 2
3 51
1 2
s sX s
s s
s s
s ss
s ss
s s
3 5
1 2 1 2
s A B
s s s s
Still Another Example ...contd
1
2
3 5 22
2 1
3 5 11
1 1
s
s
sA
s
sB
s
2 1( ) 1
1 2X s
s s
Therefore,
Inverse: 2( ) ( ) 2 ( )t tx t t e e u t
Roots: Complex and Distinct
2 6 25 3 4 3 4s s s j s j
2
100 3( )
6 6 25
sX s
s s s
31 22
100 3
6 3 4 3 46 6 25
s kk k
s s j s js s s
1
2
3 4
3
12
100 3
6 3 4
100 46 8
3 4 8
6 8
s j
k
sk
s s j
jj
j j
k j
Roots: Complex and Distinct …contd
2
100 3 12 6 8 6 8
6 3 4 3 46 6 25
12 10 53.13 10 53.13
6 3 4 3 4
o o
s j j
s s j s js s s
s s j s j
Substituting k1, k2, k3 :
Inverse: 3 46 53.13 53.13 (3 4)( ) 12 10 10 ( )
o oj tt j j j tx t e e e e e u t
Simplifying,
6 3( ) 12 20 cos 4 53.13 ( )t t ox t e e t u t
Generalization
1
( ) ( )
( ) ( )
2cos ( )
*
( )j j t j j t
t j t t
t t u t
k kL
s j s j
k e e k e e u t
k e e e
k e
For complex root pairs,
jwhere k k e
1 *2 cos ( )tk k
L k e t u ts j s j
Another Example
2
2
6 100 4150( )
14 625
s sX s
s s
2
1
16 400( ) 6
14 625*
67 24 7 24
*( )
7 24 7 24
sX s
s sk k
s j s j
k kLet F s
s j s j
Complete the problem!
7 24 7 24
16 400 16 400*
7 24 7 24s j s j
s sk k
s j s j
Repeated Roots
31 2 42 2
180 30
5 35 3 3
s kk k k
s s ss s s s
1
2
120
225
k
k
...as before
For k3, multiply by (s+3)2 and evaluate at s=-3
2 2
1 23 4 3
3 3 3
3
180 30 3 33
5 5
810
ss s s
s k s k sk k s
s s s s
k
Repeated Roots ...contd
Multiply by (s+3)2, differentiate with respect to s and evaluate at s=-3
4
3
422
3
180 30
5
5 1 30 2 5180 105
5
s
s
sdk
ds s s
s s s sk
s s
Now, for k4 :
Therefore, 2 2
180 30 120 225 810 105
5 35 3 3
s
s s ss s s s
1 5 3 32
180 30120 225 810 105 ( )
5 3t t ts
L e te e u ts s s
Repeated Roots ...contd
Another method:
31 2 4
2 2
180 30
5 35 3 3
s kk k k
s s ss s s s
4
4
4
4
4
1 0, 5, 3
180 31 120 225 810
1 6 16 1 6 16 4
5580 120 96 225 16 810 6 6 4
5580 11520 3600 4860 24
2520 24
105
Let s
k
k
k
k
k
Laplace Transform
Linear Systems and Signals – Lectures 11Dr. J. K. AggarwalThe University of Texas at Austin
Properties of Laplace Transform
Again, Definition:
Uniqueness:
Linearity :
0
( ) ( ) stX s x t e dt
( ) ( )x t X s
1 1
2 2
1 2 1 2
( ) ( )
( ) ( )
( ) ( ) ( ) ( )where X s x t
X s x t
X s X s L x t x t
Shifting in Time We have already noticed
In general,
t
1
0
( )u t
t
1
a
( )u t a
0
1s
1 ases
00( ) ( )
( ) ( )stx t t
x t X s
X s e
Shifting in Time …contd
It is better to carry u(t)
00 0
( ) ( ) ( )
( ) ( ) ( ) st
x t u t X s
x t t u t t X s e
0
0
0
0
0
0 0 0 0
0
0
;
( ) ( )
( )
st
s t
t
st s
st
Let t t dt d
L x t t u t t x t t u t t e dt
x u e d
e u x e d
e X s
Example 1 Exercise E 4.3, Lathi, page 363
slope = -2( ) 2
0 66
x t t cc
c
( ) ( ) ( 2) 2 6 ( 2) ( 3)
( ) ( ) ( 2) 2 ( 2) 6 ( 2)
2 ( 3) 6 ( 3)
x t t u t u t t u t u t
x t t u t t u t t u t u t
t u t u t
t
2
210
( )x t
3
Expanding the expression,
Example 1 ...contd
2 32 2 2
1 3 2( ) s sX s e e
s s s
( ) ( ) ( 2) 2 ( 2) 6 ( 2)
2 ( 3) 6 ( 3)
x t t u t t u t t u t u t
t u t u t
= t u(t) - (t-2) u(t-2) -2 u(t-2)
- 2 (t - 2) u (t - 2) + 2 u (t - 2)
+ 2 (t - 3) u (t - 3)
= t u(t) – 3(t-2) u (t - 2)+ 2 (t - 3) u (t - 3)
Example 2
Linear segments with breakpoints at 1,3 and 4 seconds Slope of line at t=0 is 2; (0,0) Slope of line at t=1 is -2; (2,0) Slope of line at t=3 is 2; (4,0)
t
2
210
( )x t
3 4
22 42 8
ttt
( )x t
Example 2 ...contd
( ) 2 ( ) ( 1)2 4 ( 1) ( 3)
2 8 ( 3) ( 4)
x t t u t u tt u t u t
t u t u t
2 ( ) 4 1 1 4 3 32 4 4
t u t t u t t u tt u t
3 4
2 2 2 2
1( ) 2 4 4 2
s s se e eX s
s s s s
Laplace Transform:
Collecting terms
Example 2 ...contd
Another way of looking at the expression for x(t):
t
2
210
( )x t
3 4
2 ( )t u t 4 3 3t u t
4 1 1t u t 2 4 4t u t
Example 3
( ) 10sin ( ) ( 2)
10sin ( ) 10sin ( 2)
10sin ( ) 10sin ( 2) ( 2)
x t t u t u t
t u t t u t
t u t t u t
10
0 1 2 t
( )x t
22 2 2 2
( ) 10 10 sX s es s
Laplace Transform:
Frequency Shifting
Proof:
Examples:
00
( ) ( )
( ) ( )s t
x t X s
x t e X s s
0 0
0
0
0
0
( ) ( )
( )
( )
s t s t st
s s t
L x t e x t e e dt
x t e dt
X s s
2 2
2 2
cos ( )
cos ( )( )
at
sbt u t
s bs a
e bt u ts a b
Time Differentiation Property
( ) ( )
( ) (0 )
x t X s
dxsX s x
dt
0
00
( )
( ) ( )
(0 ) ( ) ( ) 0
st
st st
s
dx dx tL e dt
dt dt
x t e s x t e dt
x sX s x e
Example: Time Differentiation Property
2H
+-3u(t)
i(t)4
( )2 4 ( ) 3 ( )
32 ( ) (0 ) 4 ( )
di ti t u t
dt
sI s i I s s
Taking LT:
(0 ) 5i A Initial conditions:
32 ( ) 10 4 ( )
3( ) 2 4 10
s I s I ss
I s s s
Solving,
Example ...contd1.5 5 1.5 1 1 5
( )2( 2) 2 2 2
0.75 4.25( )
2
I ssss s s s
I ss s
Inverse:
Supposing the input was ,3 ( ) ( )u t t
( )2 4 ( ) 3 ( ) ( )
32 ( ) (0 ) 4 ( ) 1
3 3 11( ) 2 4 10
1.5 5.5( )
( 2) 2
di ti t u t t
dt
s I s i I ss
s sI s s
s s
I ss s s
2( ) 0.75 ( ) 4.25 ( )ti t u t e u t
…Complete the problem!
Higher Order Derivatives
Similarly for nth order derivative
( )( )
( ) ( ) (0 )
dx tg t
dt
G s s X s x
2
2
( ) ( )dg t d x t
dt dt
2
( )( ) (0 )
( ) (0 ) '(0 )
dg tL sG s g
dt
s X s s x x
Properties: Integration
0 0 0
0 00
( ) ( )
( ) ( )
( )
t tst
tst st
L x d x d e dt
e e
X s
s
x d x t dts s
uv v du
Scale Change
1( )
( ) ( )s
x a
x t X
t
s
Xa a
for a>0
0
0
0
( ) (
1
)
( )
1( )
st
s
a
s
a
L x at x at e dt
de x
a
e x da
sX
a a
Convolution
Time Convolution:
Frequency Convolution:
1 2 1 2
1 1 2 2( ) ( ) , ( ) ( )
( )* ( ) ( ) ( )
x t X s x t X s
x t x t X s X s
If
then
convolution product
1 2 1 2
1( ) ( ) ( )* ( )
2x t x t X s X s
j
convolutionproduct
Review
Definition:
Step function:
Delta function:
Exponential function:
0
( ) ( ) stL x t x t e dt
1( )L u t
s
( ) 1L t
1( )atL e u t
s a
Review ...contd
Differentiation:
Integration:
Time Shifting:
Frequency Shifting:
( )( ) (0 )
d x tL s X s x
dt
0
( )( )
t X sL x t dt
s
( ) ( ) ( )asL x t a u t a e X s
( ) ( )atL e x t X s a
Zero-Input/Zero-State Response Consider the system
2H
+-3u(t)
i(t)4
( )2 4 ( ) 3 ( )
( ) 32 ( ) ( )
23 1
( ) (0 ) 2 ( )2
(0 ) 3 1( )
2 2 ( 2)
di ti t u t
dtdi t
or i t u tdt
s I s i I ss
iI s
s s s
Taking LT:
Part due to initial conditions : Zero-Input Response
Part due to input : Zero-State Response
Laplace Transform
Linear Systems and Signals – Lectures 12-??Dr. J. K. AggarwalThe University of Texas at Austin
Transfer Functions Consider the system
If the initial conditions are zero,
i.e.
and x(t) is causal so that
and
Then,
11 1
10 1 1
( ) ( ) ( ) ( )
( )
( )
N NN N
N NN N
Q D y t P D x t
D a D a D a y t
b D b D b D b x t
1(0 ) (0 ) (0 ) 0Ny yy
1(0 ) (0 ) (0 ) 0Nx xx ( ) ( ) ( ) ( )y t Y s x t X s
( )( ) ( )
( )
P sY s X s
Q s
Transfer Functions ...contd
A Transfer Function is a ratio of Laplace Transform of to the Laplace Transform of
( ) ( )y t Y s( ) ( )x t X s
( )LT of Output
H sLT of Input
When the initial conditions are all zero
Transfer Functions: An Example
2
( )2 4 ( ) ( )
( ) 12 ( ) ( )
21
( ) 2 ( ) ( )
1( ) 2( )( )
2
1( ) ( )
2
2
t
di ti t x t
dtdi t
P sH s
Q s
i t x tdt
s I s I s X s
h t e u
s
t
Taking LT:
Transfer function:
In time-domain:
Transfer Functions ...contd If x(t) is the input and y(t) is the output
(assuming zero initial conditions)then,
Also, if the input isthen the output is
( ) ( ) ( )( )
( )( )
Y s H s X sP s
H sQ s
( ) 1t ( ) ( )y t h t
0
( ) ( )
( ) ( ) st
L h t H s
H s h t e dt
1 1 ( )
( ) ( )( )
P sh t L H s L
Q s
Laplace Transform & Circuit Analysis
The Laplace Transform provides a systematic method for analyzing circuits.
It converts integro-differential equations into algebraic equations.
In addition, it takes into account the initial conditions.
One may write differential equations or go directly to the algebraic equations. The latter is the desirable route.
Circuit Elements in s-Domain
Resistorv R i
R is constant
Taking Laplace Transform,
( )( )
( ) ( )
where V L v tI L i t
V s R I sor V R I
I(s)R
+
-
V(s)
iR
+
-
v
( ) ( )v and i are really v t and i t
Circuit Elements in s-Domain ..contd
Inductordi
v Ldt
Taking Laplace Transform,
0
( ) ( ) (0 )V s sL I s L i
or V sLI LI
-
I
+
V-+
sL
LI0
i
+
-
I0
Lv
0
0IVI
s I V L I
s
L
s L
Rewriting,
i
+
-
Lv
When initial condition I0 is zeroI
sL 0Is
I0: initial current
Circuit Elements in s-Domain ..contd
Capacitordv
i Cdt
Taking Laplace Transform,
0
( ) ( ) (0 )I s s CV s CV
or I sCV CV
0
0VIV
s V I CV
s
C
s C
Rewriting,
i+
-CV0
+
-
V
I
+-
1sC
0Vs
0CV1sC
+
-V
When initial condition V0 is zero
I+
-
V 1sC
V0: initial voltage
Ohm’s Law for s-Domain
If no “initial” energy is stored in a capacitor or inductor,
Kirchoff’s Laws
where V is voltage transformI is current transform
V Z I
Z is the s domain impedance1, ,R s L sC -All rules as before for combining, etc
0 0I V
Ohm’s Law for s-Domain ...contd
I(s)R
+
-
V(s)
iR
+
-
v
i
+
-
I0
Lv
I
sL 0Is
( ) ( )
v R i
V s R I s
0
0
( )
( ) ( )( )
( )
di tv L
dtV s s L I s L I
IV sI s
s L s
I
+
V-+
sL
LI0
-I0: initial current
Ohm’s Law for s-Domain ...contd
i+
-
CV0
+
-
V
I
+-
1sC
0Vs
V0: initial voltage
0
0
( ) ( )
( )( )
dvi C
dtI s s CV s CV
VI sV s
s C s
0CV1sC
+
-V
I
Example 1
Equivalent circuit:0
00
0
1( ) ( ) 0
( )11
( ) ( )tRC
VI s R I s
s s CV
CV RI ssRC s RC
Vi t e u t
R
C R
+
-
t=0+
-
V0 ˆ( )i t
ˆ( )I s
ˆ( )V sR
+
-
+
-0V
s
+-
1sC
0VIV
s C s
( )I s
( )V s
ˆ 0v v for t
Example 1 ...contd
ˆ( )I s
ˆ( )V sR
+
-
+
-0V
s
+-
1sC
0VIV
s C s
( )I s
( )V s
0
1
0
0
0
0
1 ˆ ˆ( ) ( )
1ˆ( )
1
1
ˆ( ) ( )tRC
VI s R I s
s sCV
I s RsC sCV
s RCV R
s RC
Vi t e u t
R
Example 1 ...contd
ˆ( )I s
ˆ( )V sR
+
-
+
-0V
s
+-
1sC
0VIV
s C s
( )I s
( )V s
0 0
0 0
0
0
1( )
1
1
1
1
( ) ( )tRC
V V RV s
s s C s RC
V V
s s s RC
V
s RC
v t V e u t
Notice ˆ( ) ( )I s I s
Example 2 For the circuit below, switch is in position
‘a’ for a long time. At t=0, switch is changed to position ‘b’
1 2
1 1 2 2
1 2
1 2
60
0.5
2
V V
C V C V
V V
V V
60 V +-
t=0
a b10 k
3k1V
2V-+
-+
1C
2C1
2
0.51.0
C FC F
1
2
0
0
40
20
,Before the switchV V
V V
Example 2 ...contdi
3k10 40V V
20 20V V-+
1C
2C
-+
1( )v t
2 ( )v t
1
2
1 21 2
1 1 0
2 2 0
,
( ) ( )
( ) ( )
dv dvC i C i
dt dtC sV s v I s
C sV s v I s
1 2( ) ( ) ( )V s V s R I s
1 2
1 2
0 0
1 21 2
0 0
1 2
1 1
V VI IV V
s C s s C s
V VI I R
s s s C s C
Example 2 ...contd
6 3
63
3 3
3
60 11 10 3 10 ( )
0.560
3 103 10
60 1
3 10 100.02
10
II
s s
sI
s
s
s
1000( ) 0.02 ( )ti t e u t A
-+
-+
R
1
1s C
2
1s C
10Vs
20Vs
( )i t
Example 2 ...contd
3
1
60
1 3
3
3
10
3
1
0.02 10
102
40 101
10
40
1
( ) 40
0
( )t
VV
ss s
s s
s
v t e u t
-+
-+
R
1
1s C
2
1s C
10Vs
20Vs
( )i t
Example 3
i
+ -t=0
-+ 42
+
-
8.4 H
336 ( )V u t
336 ( )
3368.4 ( ) (0 ) 42 ( )
336( ) 42 8.4
diu t L i R
dt
s I s i I ss
I s ss
336 40
( )42 8.4 (5 )
I ss s s s
Let (0 ) 0,i
Example 3 ...contd
40( )
( 5) 5
A BI s
s s s s
0
5
408
540
8
s
s
As
Bs
5
40 8 8
( 5) 5
( ) 8 ( ) 8 ( )t
s s s s
i t u t e u t A
Example 3 ...contd
Suppose the current 0(0 )i I
0
0
0
5 50
3368.4 ( ) (0 ) 42 ( )
3368.4 ( ) 42 8.4
8.4336( )
(42 8.4 ) 42 8.4
40
( 5) 5
( ) 8 ( ) 8 ( ) ( )t t
s I s i I ss
I I s ss
II s
s s s
I
s s s
i t u t e u t I e u t
Notice initial current changes
Example 4
1
2
(0 ) 0(0 ) 0
i Ampi Amp
Given:
1 2
1 2
336 (42 8.4 ) 42
0 42 (90 10 )
s I IsI s I
Solving for I1 and I2 :
i2i1
t=0
-+ 42
8.4 H
336 ( )V u t 48
10 H
1( )I s2 ( )I s
1 2
2 12 2 121 2
40 9 168
( 2)( 12) ( 2)( 12)
15 14 ( ) 7 8.4 1.4 ( )t t t t
sI I
s s s s s s
i e e u t i e e u t
Example 4 ...contd Writing these equations
Are these the currents that should be there?
1 1 2
1 2 2 2
3368.4 42
0 42 10 48
I s I Is
I I s I I
1
2
1
2
(0 ) 0
(0 ) 0
( ) 15
( ) 7
i
i
i A
i A
Example 4A
1
2
(0 ) 5(0 ) 3
i Ampi Amp
Given:
1 2
1 2
336 (42 8.4 ) 8.4(5) 42
0 42 (90 10 ) 10(3)
s I IsI s I
Solving for I1 and I2 :
i2i1
t=0
-+ 42
8.4 H
336 ( )V u t 48
10 H
1( )I s2 ( )I s
2 2
1 2
2 12 2 121 2
5 20 72 168 36 3
( 2)( 12) ( 2)( 12)
15 9 ( ) 7 5.4 1.4 ( )t t t t
s s s sI I
s s s s s s
i e e u t i e e u t
Example 4A ...contd Writing these equations
Are these the currents that should be there?
1 1 2
1 2 2 2
3368.4 8.4(5) 42
0 42 10 10(3) 48
I s I Is
I I s I I
1
2
1
2
(0 ) 5
(0 ) 3
( ) 15
( ) 7
i
i
i A
i A
Initial and Final Value Theorems
Again, the definition:
One can examine the limiting behavior of x(t) by observing the behavior of sX(s) at and at 0
These results are called Initial Value and Final Value Theorems
0
( ) ( ) ( ) stX s L x t x t e dt
Initial Value Theorem
Taking limit
Since is independent of s, subtracting
0
( ) (0 ) stdx dxL s X s x e dt
dt dt
00
0
0 0
lim ( ) (0 )
lim
lim
(0 ) (0 )
s
st
s
st
s
sX s x
dxe dt
dt
dx dxe dt e dt
dt dt
x x
,s
(0 )x (0 ),x
0lim ( ) (0 ) lim ( )s t
s X s x x t
Final Value Theorem
This is useful only if as exists
0 00
lim ( ) (0 ) lim st
s s
dxs X s x e dt
dt
00
0
lim
lim ( ) (0 )
st
s
t
dxRHS e dt
dt
dx
x t x
0
0
lim ( ) (0 ) lim ( ) (0 )
lim ( ) lim ( )s t
s t
sX s x x t x
sX s x t
Thus,
, ( )t x
Initial and Final Value Theorems
Initial Value Theorem:
Final Value Theorem:
Example:
0lim ( ) lim ( )
stx t sX s
0lim ( ) lim ( )t s
x t sX s
0
0
1( )
1lim ( ) 1 lim 1
1lim ( ) 1 lim 1
st
t s
L u t s
x t s s
x t s s
Initial & Final Value Theorems: Example
1 6 896 5 12
120 48 72 ( )8 6
t ts sL e e u t
s s s
5 1296 1 1
lim ( ) 968 6
1 1
0 , ( ) 120 48 72 96
s
s ssX s
s s
As t x t
0
96 5 12lim ( ) 120
8 6, ( ) 120
ssX s
As t x t
In earlier example,
For steady state, is the current correct?
( ) 8
(0 ) 0
i A
i
0
40 40( )
( 5) 5
lim ( ) 8
lim ( ) 0s
s
s I s ss s s
s I s A
s I s
Example 3A
+ -
-+ 42
+
-
8.4 H
336 ( )V u t
1 2
0
0
336 ( )
3368.4 ( ) (0 ) 8.4 ( ) 42 ( )
3368.4 ( ) 16.8 42
8.4336( )
16.8 42 16.8 42
di diu t L L i R
dt dt
s I s i s I s I ss
I I s ss
II s
s s s
8.4 H
I0-+
Example 3A ...contd
2.5 2.50
0
0
120 2 ( )2.5 2.5
120 1 1 2 ( )2.5
18 ( )
2.5 2.5
8 ( )2t tu t e I e i t
II s
s s s
II s
s s s
00
00 0
00 , 8.4
0
, 0
1( )
2, 16.8
2
8.4
8.4
i I i
i I for bo
At t Fl
th
u I
I
x I
IAt t Flux
Therefore,
Example 5
iL
t=0
dcI C R L
ILdcI R sL1sC
242562525
dcI mAC nFRL mH
2 1
L
dc
dc
VI
sLIV V
s CVsR s L
I CV
ss
RC LC
Example 5 ...contd
2
5
2 8
5
1 2 2
1
384 10
( 64000 16 10 )384 10
( 32000 24000)( 32000 24000)*
dcL
I LCI
ss s
RC LC
s s s
s s j s jk k ks s a jb s a jb
3 31 224 10 ; 24 10 126.87k k
Solving,
3200024 40 cos 24000 126.87 ( )tLi e t u t mA
Example 6
iL
t=0
gi C R L
ILgI R sL1sC
cos24
40000 /2562525
g m
m
i I tI mA
rad sC nFRL mH
2 2
2 2
2
2 2 2
/
1
mg
m
m
s II
ss IV V
sCVR s L s
s I CV
ss s
RC LC
Example 6 ...contd
2 2 2
5
2 8 2 8
1 1
2 2
/
1
384 10
16 10 64000 16 10
*
40000 40000*
32000 24000 32000 24000
mL
s I CVI
ssL s sRC LC
s
s s s
k k
s j s jk k
s j s j
3 31 27.5 10 90 ; 12.5 10 90k k
3200015sin 40000 25 sin 24000 ( )tLi t e t u t mA
Solving,
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