Transcript Sinusoidal Steady

THE LAPLACE TRANSFORM
IN CIRCUIT ANALYSIS
A Resistor in the s Domain
+
R i
v
v=Ri (Ohm’s Law).
+
V
R I
V(s)=RI(s
An Inductor in the s Domain
Initial current of I0
+
v
L
di
vL
dt
V(s)=L[sI(s)-i(0-)]=sLI(s)-LI0
i
V ( s ) I0
I ( s) 

sL
s
+
sL I
V
+
V
+
LI0
I
sL
I0
s
A Capacitor in the s Domain
Initially charged to V0 volts.

dv
iC
dt
I ( s )  C [ sV ( s )  v (0 )]  sCV ( s )  CV0
V0
 1 
V ( s)  
 I ( s) 
s
 sC 
I
+
v
i
C
1/sC
+
V
1/sC
I
+
CV0
V
+
V0/s
The Natural Response of an RC Circuit
C
+
V0
t=0
+
R
v
i
1/sC
V0/s
+
I
R
+
V
V0
1

I ( s )  RI ( s )
s sC
V0
CV0
R
I ( s) 

RCs  1 s   1 RC 
V0 

i  e u( t )  v  Ri  V0e u(t )
R
t
RC
t
RC
The Step Response of a Parallel Circuit
?
The Step Response of a Parallel Circuit
I dc
V V
sCV  

R sL
s
V
IL 
I dc
C
s   1 RC  s   1 LC 
2
I dc
LC
s[ s   1 RC  s   1 LC ]
2
384  105
IL 
s( s 2  64000 s  16  108 )
384  105
IL 
s( s  32000  j 24000)( s  32000  j 24000)
K1
K2
K 2*
IL 


s
s  32000  j 24000 s  32000  j 24000
384  105
3
K1 

24

10
16  108
384  105
K2 
 20  103 126.870
( 32000  j 24000)( j 48000)
i L ( t )  [24  40e 32000t cos(24000t  126.870 )]u( t )mA
Transient Response of a
Parallel RLC Circuit
Replacing the dc current source with a sinusoidal current source
sI m
i g  I m cos  t A  I g ( s ) 
V ( s) 
 1C  s
s  ( 1 RC ) s  ( 1 LC )
2
 
Im
V ( s) 
C
s 
2
2
I g ( s)
s2
( s   )[ s  ( 1 RC ) s  ( 1 LC )]
2
2
2

Im

LC s
V ( s)
I L ( s) 
 2
sL
( s   2 )[ s 2  ( 1 RC ) s  ( 1 LC )]
I m  24mA,
  40000rad / s
384  105 s
I L ( s)  2
( s  16  108 )( s 2  64000 s  16  108 )
K1
K 1*
K2
K 2*
I L ( s) 



s  j 40000 s  j 40000 s  32000  j 24000 s  32000  j 24000
384  105 ( j 40000)
K1 
 7.5  103  900
( j 80000)(32000  j16000)(32000  j 64000)
384  105 ( 32000  j 24000)
K2 
 12.5  103 900
( 32000  j16000)( 32000  j 64000)( j 48000)
i L ( t )  (15sin 40000t  25e 32000 t sin 24000t )u( t )mA
i Lss  15sin 40000t mA
Mesh Analysis
?
Mesh Analysis (cont.)
336
 (42  8.4 s ) I1  42 I 2
s
0  42 I1  (90  10 s ) I 2
40( s  9)
15
14
1
I1 



s( s  2)( s  12) s s  2 s  12
168
7 8.4
1.4
I2 
 

s( s  2)( s  12) s s  2 s  12
i1  (15  14e
i2  (7  8.4e
2 t
2 t
e
12 t
 1.4e
)u( t ) A,
12 t
336(90)
i1 ( ) 
 15 A,
42(48)
15(42)
i2 (  ) 
7A
90
)u( t ) A.
Thevenin’s Theorem
Use the Thevenin’s theorem to find vc(t)
?
Thevenin’s Theorem (cont.)
(480 / s )(0.002 s )
480
VTh 

4
20  0.002 s
s  10
0.002 s(20) 80( s  7500)
ZTh  60 

4
20  0.002 s
s  10
480 /( s  10 )
IC 
4
5
[80( s  7500) /( s  10 )]  [(2  10 ) / s]
6s
6s
IC  2

6
2
s  10000 s  25  10
( s  5000)
4
30000
6
IC 

2
s  5000
( s  5000)
ic ( t )  ( 30000te
5000 t
 6e
5000 t
)u( t ) A
1
2  10
6s
12  10
Vc 
IC 

2
sC
s ( s  5000)
( s  5000)2
5
vc ( t )  12  10 te
5
5000 t
5
u( t )V
MUTUAL INDUCTANCE EXAMPLE
i2(t)=?
60

i1 (0 ) 
 5 A, i2 (0 )  0
12

?
MUTUAL INDUCTANCE EXAMPLE
Using the T-equivalent of the
inductors, and s-domain
equivalent gives the following circuit
(3  2 s ) I1  2 sI 2  10
2 sI1  (12  8 s ) I 2  10
2.5
1.25 1.25
I2 


( s  1)( s  3) s  1 s  3
i2 ( t )  1.25(e
t
e
3 t
)u( t ) A
THE TRANSFER FUNCTION
The transfer function is defined as the ratio of the Laplace
transform of the output to the Laplace transform of the input
when all the initial conditions are zero.
Y ( s)
H ( s) 
X ( s)
Y(s) is the Laplace transform of the output,
X(s) is the Laplace transform of the input.
THE TRANSFER FUNCTION
(cont.)
I ( s)
1
H1 ( s ) 

Vg ( s ) R  sL  1/ sC
sC
 2
s LC  RCs  1
V ( s)
1
H 2 ( s) 
 2
Vg ( s ) s LC  RCs  1
EXAMPLE
Find the transfer function V0/Vg and
determine the poles and zeros of H(s).
?
EXAMPLE
V0  Vg
V0
V0 s

 6 0
1000
250  0.05 s 10
1000( s  5000)
V0  2
V
6 g
s  6000 s  25  10
V0
1000( s  5000)
H ( s) 
 2
Vg s  6000 s  25  106
p1  3000  j 4000,
p2  3000  j 4000
z1  5000
Assume that vg(t)=50tu(t). Find v0(t). Identify the
transient and steady-state components of v0(t).
1000( s  5000)
50
V0 ( s )  H ( s )Vg ( s )  2
6
2
( s  6000 s  25  10 ) s
K1
K 1*
K2 K3


 2 
s  3000  j 4000 s  3000  j 4000 s
s
K1  5 5  10
4
79.7 , K 2  10, K 3  4  10
0
4 3000 t
v0  [10 5  10 e
4
cos(4000t  79.7 )
 10t  4  10 ]u( t )V
0
4
The transient component is generated by the poles of
the transfer function and it is:
4 3000 t
10 5  10 e
cos(4000t  79.7 )
0
The steady-state components are generated by the poles
of the driving function (input):
4
(10t  4  10 )u( t )
Time Invariant Systems
If the input delayed by a seconds, then
L x ( t  a )u( t  a )  e
Y ( s )  H ( s ) X ( s )e
1
y( t )  L
 as
X ( s)
 as
Y ( s )  y(t  a )u(t  a )
Therefore, delaying the input by a seconds simply delays the
response function by a seconds. A circuit that exhibits this
characteristic is said to be time invariant.
Impulse Response
If a unit impulse source drives the circuit, the response of the
circuit equals the inverse transform of the transfer function.
x( t )   ( t )  X ( s )  1
Y ( s)  H ( s)
1
y( t )  L
 H ( s )  h( t )
Note that this is also the natural response of the circuit
because the application of an impulsive source is equivalent
to instantaneously storing energy in the circuit.
CONVOLUTION INTEGRAL
x(t)
y(t)
x(t)
a
b
t
y(t)
N
a
t
b
Circuit N is linear with no initial stored energy. If we know the
form of x(t), then how is y(t) described? To answer this
question, we need to know something about N. Suppose we know
the impulse response of the system.
y(t )  h(t )
x(t )   (t )
(1)
t
x(t)
N
y(t)
t
Instead of applying the unit impulse at t=0, let us suppose that it is
applied at t=λ. The only change in the output is a time delay.
 (t   )
h(t   )
N
Next, suppose that the unit impulse has some strength other than
unity. Let the strength be equal to the value of x(t) when t= λ. Since
the circuit is linear, the response should be multiplied by the same
constant x(λ)
x( ) (t   )
N
x( )h(t   )
Now let us sum this latest input over all possible values of λ and
use the result as a forcing function for N. From the linearity, the
response is the sum of the responses resulting from the use of all
possible values of λ


x( ) (t   )d 
N

 x( )h(t   )d
From the sifting property of the unit impulse, we see that the input is
simply x(t)
X(t)
N

 x( )h(t   )d
Our question is now answered. When x(t) is known, and h(t), the
unit impulse response of N is known, the response is expressed by

y(t )   x( )h(t   )d 
This important relation is known as the convolution integral. It
is often abbreviated by means of
y(t )  x(t )* h(t )
Where the asterisk is read “convolved with”. If we let z=t-λ,
then dλ=-dz, and the expression for y(t) becomes


y(t )    x(t  z )h( z )dz   x(t  z )h( z )dz


y(t )  x(t )* h(t )   x( z )h(t  z )dz   x(t  z )h( z )dz


y(t )  x(t )* h(t )   x( z )h(t  z )dz   x(t  z )h( z )dz
Convolution and Realizable Systems
For a physically realizable system, the response of the system
cannot begin before the forcing function is applied. Since h(t)
is the response of the system when the unit impulse is applied at t=0,
h(t) cannot exist for t<0. It follows that, in the second integral, the
integrand is zero when z<0; in the first integral, the integrand is zero
when (t-z) is negative, or when z>t. Therefore, for realizable
systems the convolution integral becomes
t

y(t )  x(t )* h(t )   x( z )h(t  z )dz  0 x(t  z )h( z )dz
EXAMPLE
x(t)
1
h(t)
t
h( t )  2e u( t )
y(t)
t
x( t )  u( t )  u( t  1)

y( t )  x( t ) * h( t )  0 x( t  z )h( z )dz

z
 0 [u( t  z )  u( t  z  1)][2e u( z )]dz
Graphical Method of Convolution
Since h(z) does not exist prior to t=0 and vi(t-z) does not exist
for z>t, product of these functions has nonzero values only in
the interval of 0<z<t for the case shown where t<1.
y(t )  0 2e  z dz  2(1  e  t )
t
0 t 1
When t>1, the nonzero values for the product are obtained in
the interval (t-1)<z<t.
t
z
y(t )  t 1 2e dz  2(e  1)e
t
t 1
EXAMPLE
Apply a unit-step function, x(t)=u(t), as the input to a system whose
impulse response is h(t) and determine the corresponding output
y(t)=x(t)*h(t).
?
h(t)=u(t)-2u(t-1)+u(t-2),
When t<0, there is no overlap and y(t)=0 for t<0
For 0<t<1, the curves overlap from z=0 to z=t and product is 1.
Thus,
t
y(t )  0 1dz  t
0 t 1
When 1<t<2, h(t-z) has slid far enough to the right to bring under
the step function that part of the negative square extending from
0 to z=t-1. Thus,
t 1
t
y(t )  0 1dz  t 11dz
t 1
 z 0

t
z t 1
 2  t,
1 t  2
Finally, when t>2, h(t-z) has slid far enough to the right so that it
lies entirely to the right of z=0
t 1
t
y(t )  t  2 1dz  t 11dz  0
t2
Convolution and the Laplace Transform
Let F1(s) and F2(s) be the Laplace transforms of f1(t) and f2(t),
respectively. Now, consider the laplace transform of f1(t)*f2(t),
L f1 ( t ) * f 2 ( t )  L


f ( ) f 2 (t
 1
  )d 

Since we are dealing with the time functions that do not exist
prior to t=0-, the lower limit can be changed to 0
  st
e f1( ) f2 (t
0
L f1 (t )* f2 (t )  0 [

  )dt ]d 
f1(λ) does not depend on t, and it can be moved outside the inner
integral

  st
e f 2 ( t   )dt ]d 
0
  s( x   )
e
f 2 ( x )dx ]d 
0
L f1 ( t ) * f 2 ( t )  0 f1 ( )[ 


 0 f1 ( )[ 


 0 f1 ( )e
 s

  sx
e f 2 ( x )dx ]d 
0
[

 0 f1 ( )e  s [ F2 ( s )]d 


 F2 ( s )0 f1 ( )e  s d 

 F1 ( s )  F2 ( s )
STEADY-STATE SINUSOIDAL RESPONSE
If the input of a circuit is a sinusoidal function
x(t )  Acos(t   )
x ( t )  A cos  t cos  A sin  t sin 
( A cos ) s ( A sin )
X ( s)  2
 2
2
2
s 
s 
A(s cos   sin )
=
s2   2
A(s cos   sin )
Y ( s)  H ( s)
2
2
s 
The partial fraction expansion of Y(s) is
K1
K1*
Y ( s) 

  terms generated by the poles of H(s)
s  j s  j
If the poles of H(s) lie in the left half of the s plane, the corresponding
time-domain terms approach zero as t increases and they do not
contribute to the steady-state response. Thus only the first two terms
determine the steady-state response.
H ( s ) A( s cos   sin  )
K1 
s  j
s  j
H ( j ) A( j cos   sin )

2 j
H ( j ) A(cos  j sin ) 1

 H ( j ) Ae j
2
2
H ( j )  H ( j ) e j ( )
A
j[  ( )]
K1  H ( j ) e
2
yss ( t )  A H ( j ) cos[ t     ( )]
EXAMPLE
If the input is 120 cos(5000t+300)V, find the steady-state expression for v0
1000( s  5000)
H ( s)  2
s  6000 s  25  106
1000(5000  j 5000)
H ( j 5000) 
6
6
25  10  j 5000(6000)  25  10
1  j1
2


 450
j6
6
v0 ss
120 2

cos(5000t  300  450 )
6
=20 2 cos(5000t  150 )V
THE IMPULSE FUNCTION
IN CIRCUIT ANALYSIS
The capacitor is charged to an initial voltage
V0 at the time the switch is closed. Find the
expression for i(t) as R
0
I
V0
s
R  ( 1 sC )  ( 1 sC )
1
2

V0
s(
R
1
RCe )
C1C 2
 V0  t / RC
i(t )   e
Ce 
 R
C1  C2
e

 u( t )

As R decreases, the initial current (V0/R)
increases and the time constant (RCe)
decreases. Apparently i is approaching an
impulse function as R approaches zero.
The total area under the i versus t curve represents the total charge
transferred to C2 after the switch is closed.
 V0
Area=q  0

R
e  t / RC dt  V0Ce
e
Thus, as R approaches zero, the current approaches an impulse
strength V0Ce.
i  V0Ce (t )
Series Inductor Circuit
Find v0. Note that opening the switch forces
an instantaneous change in the current L2.
i1 (0 )  10 A,
i2 (0 )  0
V0  [(100 s )  30]
V0

0
2 s  15
3 s  10
40( s  7.5) 12( s  7.5)
V0 

s( s  5)
s5
60
10
V0 
 12 
s
s5
v0 ( t )  12 ( t )  (60  10e 5 t )u( t )V
Does this solution make sense? To answer this question, first let
us determine the expression for the current.
(100 s )  30 4
2
I
 
5 s  25
s s5
i ( t )  (4  2e 5t )u( t ) A
Before the switch is opened, current through L1 is 10A and in L2 is 0 A,
after the switch is opened both currents are 6A. Then the current in L1
changes instantaneously from 10 A to 6 A, while the current in L2 changes
instantaneously from 0 to 6 A. How can we verify that these instantaneous
jumps in the inductor current make sense in terms of the physical behavior
of the circuit?
Switching operation places two inductors in series. Any impulsive voltage
appearing across the 3H inductor must be balanced by an impulsive
voltage across the 2H inductor. Faraday’s law states that the induced
voltage is proportional to the change in flux linkage (v  d  )
dt
before switching
  L1i1  L2i2  3(10)  2(0)  30 Wb-turns
After switching
  ( L1  L2 )i (0 )  5i (0 )
30
i (0 ) 
 6A
5

Thus the solution agrees with the principle of the conservation
of flux linkage.
Impulsive Sources
When the voltage source is applied, the initial
energy in the inductor is zero; therefore the initial
current is zero. There is no voltage drop across R,
so the impulsive source appears directly across L
V0
1 t

i  0 V0 ( x )dx  i (0 ) 
A
L
L

Thus, in an infinitesimal
moment, the impulsive voltage
source has stored
2
1  V0 
1 V02
w
L  
J
2  L
2 L
Current in the circuit decays to
zero in accordance with the
natural response of the circuit
V0  (
i e
L
R
L
)t
u( t )
EXAMPLE
Find i(t) and v0(t) for t>0
I
50   100 s   30
25  5 s
4
12


s5 s
i ( t )  (12e 5 t  4)u( t ) A
60
60

V0  (15  2 s ) I  32 
s5 s
v0 ( t )  32 ( t )  (60e 5 t  60)u( t )V