Transcript No Slide Title

SECOND-ORDER CIRCUITS
THE BASIC CIRCUIT EQUATION
 vC 
 vR 
iR
iL
Lv
iC
Single Node-pair: Use KCL
 iS  iR  iL  iC  0


Single Loop: Use KVL
 v S  v R  vC  v L  0
1 t
di
v (t )
1t
dv
iR 
; i L   v ( x )dx  i L (t0 ); iC  C (t ) v R  Ri ; vC   i ( x )dx  vC (t0 ); v L  L (t )
C t0
dt
R
L t0
dt
v 1t
dv
  v ( x )dx  i L (t0 )  C (t )  i S
R L t0
dt
Differentiating
d 2v 1 dv v di S
C 2
 
R dt L dt
dt
1 t
di
Ri   i ( x )dx  vC (t0 )  L (t )  v S
C t0
dt
d 2i
di i dv
L 2 R   S
dt C
dt
dt
LEARNING BY DOING
WRITE THE DIFFERENTIAL EQUATION FOR v (t ), i (t ), RESPECTIVE LY

iS
vS

 0 t 0
i S (t )  
I S t  0
di S
(t )  0; t  0
dt
MODEL FOR RLC PARALLEL
d 2v 1 dv v di S
C 2
 
R
dt
L dt
dt
2
C
d v 1 dv v

 0
2
R dt L
dt
V t  0
v S (t )   S
 0 t 0
dv S
(t )  0; t  0
dt
MODEL FOR RLC SERIES
d 2i
di i dv
L 2 R   S
dt C
dt
dt
d 2i
di i
L 2 R  0
dt C
dt
THE RESPONSE EQUATION
WE STUDY THE SOLUTIONS FOR THE
EQUATION
d2x
dx
(
t
)

a
( t )  a2 x ( t )  f ( t )
1
dt
dt 2
KNOWN : x ( t )  x p (t )  xc (t )
xp
particular solution
xc
complement ary solution
THE COMPLEMENTARY SOLUTION
SATISIFES
2
d xc
dxc
(
t
)

a
( t )  a 2 xc ( t )  0
1
2
dt
dt
IF THE FORCING FUNCTION IS A CONSTANT
f (t )  A  x p 
A
is a particular solution
a2
dx p d 2 x p
A
PROOF : x p  

 0  a2 x p  A
a2
dt
dt 2
FOR ANY FORCING FUNCTION f (t )  A
x (t ) 
A
 xc ( t )
a2
THE HOMOGENEOUS EQUATION
2
d x
dx
(
t
)

a
( t )  a2 x ( t )  0
1
2
dt
dt
NORMALIZED FORM
2
d x
dx
2
(
t
)

2

(
t
)


x (t )  0
n
n
2
dt
dt
(undamped) natural frequency
n

damping ratio
CHARACTERI STIC EQUATION
s 2  2n s   n2  0
LEARNING BY DOING
DETERMINE THE CHARACTERI STIC
EQUATION, DAMPING RATIO AND
NATURAL FREQUENCY
d2x
dx
4 2 (t )  8 (t )  16 x (t )  0
dt
dt
COEFFICIENT OF SECOND DERIVATIVE
MUST BE ONE
d2x
dx
(
t
)

2
(t )  4 x (t )  0
2
dt
dt
a2   n2   n  a2
CHARACTERI STIC EQUATION
a
a1  2n    1
2 a2
s2  2s  4  0
DAMPING RATIO, NATURAL FREQUENCY
d2x
dx
(
t
)

2
(t )  4 x (t )  0
dt
dt 2
 n2  n  2
2n

  0.5
ANALYSIS OF THE HOMOGENEOUS EQUATION
NORMALIZED FORM
d2x
dx
2
(
t
)

2

(
t
)


x (t )  0
n
n
2
dt
dt
x (t )  Ke
st
is a solution iff
s 2  2n s   n2  0
Iff s is solution of the characteristic
equation
CASE 1 :   1 (real and distinct roots)
x (t )  K1e s1t  K 2e s2t
CASE 2 :   1 (complex conjugate roots)
x (t )  K1e s1t  K 2e s2t
s  n  j n 1   2
x (t ) real  K 2  K1*
s    j d
d  damped oscillatio n frequency
  damping factor
x (t )  e t  A1 cosd t  A2 sin d t 
2
dx
st d x
PROOF :
(t )  sKe ;
 s 2 Ke st
2
dt
HINT : e st  e (n  jd ) t  e nt e  jd t
dt
d2x
dx
e  jd t  cosd t  j sin d t
2
2
2
st
(t )  2n (t )   n x (t )  ( s  2n s   n ) Ke
dt
dt 2
ASSUME K1  ( A1  jA2 ) / 2
CHARACTERI STIC EQUATION
K 2  K1* 
 x (t )  2 Re K1e (  j d ) t
2
2

s  2n s   n  0
s    j d 
CASE 3 :   1 (real and equal roots)
( s   n ) 2  ( n2   2 n2 )  0
s  n
2 2
2
s   n    n   n
x (t )  B1  B2t e  nt
s   n   n  2  1
HINT : te st is solution iff

(modes of the system)
( s 2  2n s   n2  0) AND (2 s  2n  0)

LEARNING EXTENSIONS
DETERMINE THE GENERAL FORM OF THE SOLUTION
d2x
dx
(
t
)

4
(t )  4 x (t )  0
dt
dt 2
CHARACTERI STIC EQUATION
s  4s  4  0
2
 n2  4   n  2 2n  4    1
s 2  4s  4  0  ( s  2)2  0
Roots are real and equal
this is a critically damped (case 3) system
x (t )  ( B1  B2 t )e st
x (t )  ( B1  B2 t )e 2 t
d2x
dx
4 2 (t )  8 (t )  16 x (t )  0
dt
dt
Divide by coefficient of second derivative
d2x
dx
(
t
)

2
(t )  4 x (t )  0
dt
dt 2
 n2  4   n  2
2n  2    0.5
s 2  2s  4  ( s  1)2  3  0  s  1  j 3
Roots are complex conjugate
underdampe d (case 2) system
d
  n  1; d  n 1   2  2 1  0.25  3
x (t )  e t  A1 cos d t  A2 sin  d t 
x (t )  e t A1 cos 3t  A2 sin 3t 
Form of the solution
LEARNING EXTENSIONS
RLC PARALLEL CIRCUIT WITH
RLC SERIES CIRCUIT WITH
R  2; L  1H , C  0.5F ,1F ,2 F
R  1, L  2 H , C  2 F
Classify the responses for
the given values of C
HOMOGENEOUS EQUATION
d 2v 1 dv v
C 2 
 0
R dt L
dt
d 2v dv v
2 2 
 0 2 s 1
1
3
dt 2
s    ( s  )2   0
dt
2 4
4
16
2
d v 1 dv v

  0  n  1 ;n  1    1
2
2 dt 4
dt
2
4
2
1

4
1
1
3
d   n 1   2  1  
2
4 4
t
 
3
3 
vc (t )  e 4  A1 cos t  A2 sin
t 
4
4 

d 2i
di i
L 2  R   0 : / L & replace values
dt C
dt
d 2i
di i

2
 0
2
dt C
dt
1
n 
; 2n  2    C
C
C=0.5
C=1.0
C=2.0
underdamped
critically damped
overdamped
discrimina nt  4 
4
C
THE NETWORK RESPONSE
DETERMINING THE CONSTANTS
NORMALIZED FORM
x (t ) 
A
 n2
 K1e s1t
x (0 ) 
A
 n2
d2x
dx
2
(
t
)

2

(
t
)


n
n x (t )  A
2
dt
dt
A
 n t
s2 t
 A1 cos d t  A2 sin  d t 
x
(
t
)


e
 K 2e
2
n
 K1  K 2
x (0  ) 
A
 n2
 A1
dx
(0)  n A1  d A2
dt
dx
(0)  s1K1  s2 K 2
dt
x (t ) 
A
 n2
 B1  B2 t e  n t
x (0  ) 
A
 n2
 B1
dx
(0)  n B1  B2
dt
LEARNING EXAMPLE
R  2, L  5 H , C  1 F
5
iL
iR
iL (0)  1A, vC (0)  4V
iC
iR  iL  iC  0
v 1t
dv
  v ( x )dx  i L (0)  C  0
R L0
dt
STEP 1
d 2v
1 dv
1


v

0
MODEL
dt 2 RC dt LC
CHARACTERI STIC EQUATION STEP 2
s 2  2.5 s  1  0  n  1;   1.5
 2.5  (2.5)2  4  2.5  1.5 STEP 3
s

ROOTS
2
2
v (t )  K1e 2 t  K 2e 0.5t
STEP 4
FORM OF
SOLUTION
STEP 5: FIND CONSTANTS
To determine the constants we need
dv
(0 )
dt
IF NOT GIVEN FIND vC (0), iL (0)
v (0)  vC (0)  vC (0)  4V
ANALYZE
KCL AT t  0 
CIRCUIT AT
vC (0)
dv
t=0+
 i L (0  )  C ( 0  )  0
R
dt
dv
4
(1)
(0 )  

 5
dt
2(1/ 5) (1 / 5)
K1  K 2  4 
  K1  2; K 2  2
 2 K1  0.5 K 2  5
v (0);
v (t )  2e 2t  2e 0.5t ; t  0
USING MATLAB TO VISUALIZE THE RESPONSE
%script6p7.m
%plots the response in Example 6.7
%v(t)=2exp(-2t)+2exp(-0.5t); t>0
t=linspace(0,20,1000);
v=2*exp(-2*t)+2*exp(-0.5*t);
plot(t,v,'mo'), grid, xlabel('time(sec)'), ylabel('V(Volts)')
title('RESPONSE OF OVERDAMPED PARALLEL RLC CIRCUIT')
LEARNING EXAMPLE

vR

R  6, L  1H , C  0.04F
 vL 

v R  v L  vC  0
vC
di
1t
Ri (t )  L (t )   i ( x )dx  vC (0)  0
dt
C0
iL (0)  4 A; vC (0)  4V

TO COMPUTE
NO SWITCHING OR
DISCONTINUITY AT t=0.
USE t=0 OR t=0+
di
( 0 )
dt
v L (t )  L
di
(t )
dt
di
di
d 2 i R di
1
L
(
0
)


Ri
(
0
)

v
(
0
)

(0)  20
C
model

(
t
)

i
(
t
)

0
dt
dt
LC
dt 2 L dt
di
d 2i
di
(t )  3i (t )  e 3t (4 A1 sin 4t  4 A2 cos 4t )
 6 (t )  25i (t )  0
dt
dt
dt 2
2
 n  25   n  5 @ t  0 :  20  3  (4)  4 A2  A2  2
2
Ch. Eq. : s  6s  25  0
2n  6    0.6
i (t )  e 3t (4 cos 4t  2 sin 4t )[ A]; t  0
 6  36  100
roots : s 
 3  j 4  d
di
1t
vC (t )   Ri (t )  L (t )  vC (0)   i ( x )dx
2
dt
C0
3t
i
(
t
)

e
(
A
cos
4
t

A
sin
4
t
)
Form:
1
2
vC (t )  e 3t (4 cos 4t  22 sin 4t )[V ]; t  0
i (0)  i L (0)  4 A  A1  4
USING MATLAB TO VISUALIZE THE RESPONSE
%script6p8.m
%displays the function i(t)=exp(-3t)(4cos(4t)-2sin(4t))
% and the function vc(t)=exp(-3t)(-4cos(4t)+22sin(4t))
% use a simle algorithm to estimate display time
tau=1/3;
tend=10*tau;
t=linspace(0,tend,350);
it=exp(-3*t).*(4*cos(4*t)-2*sin(4*t));
vc=exp(-3*t).*(-4*cos(4*t)+22*sin(4*t));
plot(t,it,'ro',t,vc,'bd'),grid,xlabel('Time(s)'),ylabel('Voltage/Current')
title('CURRENT AND CAPACITOR VOLTAGE')
legend('CURRENT(A)','CAPACITOR VOLTAGE(V)')
LEARNING EXAMPLE
R1  10, R2  8, C  1F , L  2 H vC (0)  1V , iL (0)  0.5 A
Ch. Eq. : s 2  6s  9  0  ( s  3)2
3t
n  3, 2n  6    1 v (t )  e B1  B2t 
KVL
v (0)  vc (0)  1V
KCL
NO SWITCHING OR DISCONTINUITY
AT t=0. USE t=0 OR t=0+
KCL AT t  0 
dv
v (0)
dv
 (0)  3
i (0)  i L (0) 
 C (0)
dt
R2
dt
v (t )
dv
di
 C (t )
L (t )  R1i (t )  v (t )  0 i (t ) 
R2
dt
dt
 1 dv
 v (t )
d 2v 
dv 
L
(t )  C 2   R1 
 C (t )   v (t )  0
dt 
dt 
 R2
 R2 dt
 1
d 2v
R1  dv
R1  R2


(
t
)


(
t
)

v (t )  0


2
R2 LC
dt
 R2C L  dt
d 2v
dv
Ch. Eq. : s 2  6s  9  0
(t )  6 (t )  9v (t )  0
dt
dt 2
v (0)  1  B1
dv
(0)  3v (0)  B2  3  B2  6
dt
v (t )  e 3t 1  6t ; t  0
USING MATLAB TO VISUALIZE RESPONSE
%script6p9.m
%displays the function v(t)=exp(-3t)(1+6t)
tau=1/3;
tend=ceil(10*tau);
t=linspace(0,tend,400);
vt=exp(-3*t).*(1+6*t);
plot(t,vt,'rx'),grid, xlabel('Time(s)'), ylabel('Voltage(V)')
title('CAPACITOR VOLTAGE')
LEARNING EXTENSION FIND i (t ), t  0
d 2i
3 di
1
(
t
)

(
t
)

i (t )  0
2
2 dt
2
dt
Ch.Eq. : s 2  1.5s  0.5  0
roots : s  1,0.5

t
2;
t 0


vL



vC


vC (0)  0V
di
vC  C
dt

di
(0 )  0
dt
=0
Once the switch opens the circuit is RLC
series
t
di
3i (t )  2 (t )  vC (0)   i ( x )dx  0
dt
0
i L (0)  2 A


i (t )  K1e  K 2e

To find initial conditions use steady state
analysis for t<0

vR
t
i (0 )  2 A
And analyze
circuit at t=0+
=2
KCL at t  0 
2  K1  K 2
1
0   K1  K 2
2
i (t )  2e t  4e

t
2;
t 0
LEARNING EXTENSION FIND v0 (t ), t  0
To find initial conditions we use steady
state analysis for t<0

vC (0)  0
KVL
iL (0)  2 A

i (t )

v0 (t )  2i (t )
For t>0 the circuit is RLC series
1 di
1 t
(t ) 
i ( x )dx  vC (0)  2i (t )  0
2 dt
2 / 3 0
2
d i
di
(
t
)

4
(t )  3i (t )  0
2
dt
dt
Ch. Eq. : s 2  4 s  3  0
roots : s  1,3
i (t )  K1e t  K 2e 3t ; t  0
And analyze circuit at t=0+
i (0)  2 A
vC (0)  C
i (0 )  0  K1  K 2  2
di
(0 )  0
dt
K2  1
di
( 0  )  0   K1  3 K 2  0
dt
K1  3
 i (t )  3e t  e 3t ; t  0


v0 (t )  2  3e t  e 3t ; t  0
LEARNING EXTENSION DETERMINE i0 (t ), v0 (t ); t  0
v0 (t )  18i0 (t )  12(V )
Steady state t<0




KVL
24V
2
d i
di
(
t
)

9
(t )  18i (t )  0
dt
dt 2
Ch. Eq. : s 2  9 s  18  0
roots : s  3,6
i0 (t )  K1e 3t  K 2e 6t ; t  0
i0 (t )  
11 3t 14 6 t
e  e ; t 0
6
6
 v L (0) 

vC ( 0  )  0


v C ( 0)
24V
1 t
di
4
i
(
x
)
dx

v
(
0
)

2
(t )  18i (t )  12  0
C

1/ 36 0
dt
iL (0)

vC (0)  0
iL (0)  0.5 A
Analysis at t=0+
i0 (0)  iL (0)  0.5( A)
di
di
v L (0  )  L L ( 0  )  L 0 (0  )
dt
dt
 4  v L (0)  18iL (0)  12  0 v L (0)  17
di0
(0 )  17 / 2  3 K1  6 K 2
dt
i0 (0 )  0.5  K1  K 2 Second
11
14
Order
K1   ; K 2 
6
6