second-order circuits
DESCRIPTION
SECOND-ORDER CIRCUITS. THE BASIC CIRCUIT EQUATION. Single Loop: Use KVL. Single Node-pair: Use KCL. Differentiating. MODEL FOR RLC PARALLEL. MODEL FOR RLC SERIES. LEARNING BY DOING. THE RESPONSE EQUATION. IF THE FORCING FUNCTION IS A CONSTANT. THE HOMOGENEOUS EQUATION. LEARNING BY DOING. - PowerPoint PPT PresentationTRANSCRIPT
SECOND-ORDER CIRCUITS
THE BASIC CIRCUIT EQUATION
Single Node-pair: Use KCL
Ri Li Ci
0 CLRS iiii
)();()(1
;)(
0
0
tdt
dvCitidxxv
Li
R
tvi CL
t
tLR
SL
t
t
itdt
dvCtidxxv
LR
v )()()(
10
0
Differentiating
dt
di
L
v
dt
dv
Rdt
vdC S 12
2
Single Loop: Use KVL
Rv
Cv
Lv
0 LCRS vvvv
)();()(1
; 0
0
tdt
diLvtvdxxi
CvRiv LC
t
tCR
dt
dv
C
i
dt
diR
dt
idL S2
2
SC
t
t
vtdt
diLtvdxxi
CRi )()()(
10
0
LEARNING BY DOING
LYRESPECTIVE FOR EQUATIONAL DIFFERENTI THE WRITE ),(),( titv
0
00)(
tI
tti
SS
Si
MODEL FOR RLC PARALLEL
dt
di
L
v
dt
dv
Rdt
vdC S 12
2
0;0)( ttdt
diS
01
2
2
L
v
dt
dv
Rdt
vdC
Sv
00
0)(
t
tVtv S
S
MODEL FOR RLC SERIES
dt
dv
C
i
dt
diR
dt
idL S2
2
0;0)( ttdt
dvS
02
2
C
i
dt
diR
dt
idL
THE RESPONSE EQUATION
)()()()( 212
2
tftxatdt
dxat
dt
xd
EQUATION
THE FOR SOLUTIONS THESTUDY WE
solutionary complement
solution particular
:KNOWN
c
p
cp
x
x
txtxtx )()()(
0)()()( 212
2
txatdt
dxat
dt
xdc
cc
SATISIFES
SOLUTIONARY COMPLEMENT THE
IF THE FORCING FUNCTION IS A CONSTANT
solution particular a is 2
)(a
AxAtf p
Axadt
xd
dt
dx
a
Ax p
ppp 22
2
2
0 :PROOF
)()(
)(
2
txa
Atx
Atf
c
FUNCTION FORCING ANY FOR
0)(4)(2)(2
2
txtdt
dxt
dt
xd
LEARNING BY DOING
0)(16)(8)(4 2
2
txtdt
dxt
dt
xd
FREQUENCYNATURAL
ANDRATIO DAMPING EQUATION,
STICCHARACTERI THE DETERMINE
COEFFICIENT OF SECOND DERIVATIVEMUST BE ONE
0)(4)(2)(2
2
txtdt
dxt
dt
xd
2nn2
0422 ss
EQUATION STICCHARACTERI
DAMPING RATIO, NATURAL FREQUENCY
2 n
5.0
THE HOMOGENEOUS EQUATION
0)()()( 212
2
txatdt
dxat
dt
xd
0)()(2)( 22
2
txtdt
dxt
dt
xdnn
FORM NORMALIZED
2
11
22
2
22
a
aa
aa
n
nn
ratio damping
frequency natural (undamped)
n
02 22 nnss
EQUATION STICCHARACTERI
ANALYSIS OF THE HOMOGENEOUS EQUATION
0)()(2)( 22
2
txtdt
dxt
dt
xdnn
FORM NORMALIZED
02
)(22
nn
st
ss
Ketx
iff solution a is
Iff s is solution of the characteristicequation
stst Kesdt
xdsKet
dt
dx 22
2
;)( :PROOF
stnnnn Kesstxt
dt
dxt
dt
xd)2()()(2)( 222
2
2
02 22 nnss
EQUATION STICCHARACTERI
roots)distinct and (real :1 CASE 1tsts eKeKtx 21
21)( roots) conjugate(complex :2 CASE 1
d
nn
js
js
21
tAtAetx ddt sincos)( 21
tjttjst dndn eeee )( :HINTtjte dd
tj d sincos
roots) equal and (real 1 :3 CASE ns
tnetBBtx 21)(
)022()02( 22 nnn
st
sANDss
te
iff solution is :HINT
tsts eKeKtx 2121)(
frequency noscillatio dampedd
*12)( KKtx real
tj
d
deKtxjs
KK )(1
*12 Re2)(
2/)( 211 jAAK ASSUME
1
0)()(
2
222
2222
nn
nnn
nnn
s
s
s
(modes of the system)
factor damping
LEARNING EXTENSIONS
0)(4)(4)(2
2
txtdt
dxt
dt
xd
0442 ss
EQUATION STICCHARACTERI
0)2(044 22 sss
242 nn 142 n
system 3) (case damped critically a is this
t
st
etBBtx
etBBtx2
21
21
)()(
)()(
DETERMINE THE GENERAL FORM OF THE SOLUTION
0)(16)(8)(4 2
2
txtdt
dxt
dt
xd
0)(4)(2)(2
2
txtdt
dxt
dt
xd
Divide by coefficient of second derivative
242 nn 5.022 n
system 2) (case dunderdampe
325.0121;1 2 ndn
tAtAetx
tAtAetxt
ddt
3sin3cos)(
sincos)(
21
21
3103)1(42 22 jssss
Roots are real and equal Roots are complex conjugated
LEARNING EXTENSIONS
FCHLR
RLC
2,2,1 WITH CIRCUIT PARALLEL
01
2
2
L
v
dt
dv
Rdt
vdC 02
2
C
i
dt
diR
dt
idL
042
1
02
2
2
2
2
2
v
dt
dv
dt
vd
v
dt
dv
dt
vd
016
3)4
1(
4
1
222 s
ss
2
1
4
1;2
1 nn
tAtAetv
t
c 4
3sin
4
3cos)( 21
4
4
3
4
11
2
11 2 nd
FFFCHLR
RLC
2,1,5.0,1;2 WITH CIRCUIT SERIES
HOMOGENEOUS EQUATION
022
2
C
i
dt
di
dt
id
valuesreplace& L/:
CC nn 22;1
C=0.5 underdampedC=1.0 critically dampedC=2.0 overdamped
Classify the responses forthe given values of C
Form of the solution
4
1
C
44 ntdiscrimina
THE NETWORK RESPONSE
DETERMINING THE CONSTANTS
Atxtdt
dxt
dt
xdnn )()(2)( 2
2
2
FORM NORMALIZED
tsts
n
eKeKA
tx 21212)(
212)0( KKA
xn
2211)0( KsKsdt
dx
tAtAeA
tx ddt
n
n
sincos)( 212
12)0( AA
xn
21)0( AAdt
dxdn
t
n
netBBA
tx
212)(
12)0( BA
xn
21)0( BBdt
dxn
LEARNING EXAMPLE FCHLR 51,5,2 VvAi CL 4)0(,1)0(
t
L dt
dvCidxxv
LR
v
0
0)0()(1
011
2
2
vLCdt
dv
RCdt
vd
015.22 ss
EQUATION STICCHARACTERI
5.1;1 n
2
5.15.2
2
4)5.2(5.2 2
s
tt eKeKtv 5.02
21)(
To determine the constants we need
)0();0( dt
dvv
Vvvv CC 4)0()0()0( 0t KCL AT
0)0()0()0(
dt
dvCi
R
vL
C
5)5/1(
)1(
)5/1(2
4)0(
dt
dv
2;255.02
421
21
21
KKKK
KK
0;22)( 5.02 teetv tt
RiLi Ci
0 CLR iii
STEP 1 MODEL
STEP 2
STEP 3ROOTS
STEP 4FORM OFSOLUTION
STEP 5: FIND CONSTANTS
)0(),0( LC iv FIND GIVEN NOT IF
ANALYZECIRCUIT ATt=0+
%script6p7.m%plots the response in Example 6.7%v(t)=2exp(-2t)+2exp(-0.5t); t>0t=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')
USING MATLAB TO VISUALIZE THE RESPONSE
LEARNING EXAMPLE FCHLR 04.0,1,6 VvAi CL 4)0(;4)0(
t
CvdxxiC
tdt
diLtRi
0
0)0()(1
)()(
0)(1
)(2
2
tiLC
tdt
di
L
R
dt
id
0)(25)(62
2
titdt
di
dt
id
02562 ss :Eq. Ch.6.062
5252
n
nn
432
100366js :roots d
)4sin4cos()( 213 tAtAeti t
Aii L 4)0()0(
)0( dt
di COMPUTE TO )()( t
dt
diLtvL
)0()0()0( CvRidt
diL 20)0(
dt
di
41 A
)4cos44sin4()(3)( 213 tAtAetit
dt
di t
24)4(320:0@ 22 AAt
0];)[4sin24cos4()( 3 tAtteti t
Rv Lv
Cv0 CLR vvv
t
CC dxxiC
vtdt
diLtRitv
0
)(1
)0()()()(
0];)[4sin224cos4()( 3 tVttetv tC
NO SWITCHING OR DISCONTINUITY AT t=0. USE t=0 OR t=0+
model
Form:
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 timetau=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 HLFCRR 2,1,8,10 21 AiVv LC 5.0)0(,1)0(
KVL
0)()()( 1 tvtiRtdt
diL
KCL
)()(
)(2
tdt
dvC
R
tvti
0)()()(
)(1
212
2
2
tvt
dt
dvC
R
tvR
dt
vdCt
dt
dv
RL
0)()(1
)(2
211
22
2
tv
LCR
RRt
dt
dv
L
R
CRt
dt
vd
0)(9)(6)(2
2
tvtdt
dvt
dt
vd 0962 ss :Eq. Ch.
162,3 nn
22 )3(096 sss :Eq. Ch.
tBBetv t21
3)(
Vvv c 1)0()0(
)0()0(
)0()0(
0
2 dt
dvC
R
vii
t
L
KCL AT
3)0( dt
dv
11)0( Bv
63)0(3)0( 22 BBvdt
dv
0;61)( 3 ttetv t
NO SWITCHING OR DISCONTINUITYAT t=0. USE t=0 OR 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 0),( tti FIND
Once the switch opens the circuit is RLCseries
0)()0()(2)(30
t
C dxxivtdt
diti
0)(2
1)(
2
3)(2
2
titdt
dit
dt
id
5.0,1
05.05.12
s
ss
:roots
:Ch.Eq.
0;)( 221 teKeKtit
t
To find initial conditions use steady stateanalysis for t<0
dt
diCvC
AiL 2)0(
VvC 0)0(
0)0( dt
di
21
21
2
10
2
KK
KK
0;42)( 2 teetit
t
Cv
Rv
Lv
And analyze circuit at t=0+
0tat KCL Ai 2)0(
=0=2
LEARNING EXTENSION 0),(0 ttv FIND
For t>0 the circuit is RLC series
)(2)(0 titv
)(ti
KVL
0)(2)0()(3/2
1)(
2
1
0
tivdxxitdt
diC
t
0)(3)(4)(2
2
titdt
dit
dt
id
3,1
0342
s
ss
:roots
: Eq. Ch.
0;)( 321 teKeKti tt
To find initial conditions we use steadystate analysis for t<0
AiL 2)0(
0)0(Cv
And analyze circuit at t=0+
Ai 2)0( 0)0()0( dt
diCvC
030)0(
20)0(
21
21
KKdt
di
KKi
3
1
1
2
K
K
0;32)(
0;3)(3
0
3
teetv
teetitt
tt
LEARNING EXTENSION 0);(),( 00 ttvti DETERMINE )(12)(18)( 00 Vtitv
KVL
012)(18)(2)0()(36/1
14
0
titdt
divdxxi
t
C
0)(18)(9)(2
2
titdt
dit
dt
id
6,3
01892
s
ss
:roots
: Eq. Ch.
0;)( 62
310 teKeKti tt
0)0( Cv AiL 5.0)0(
Analysis at t=0+
)(5.0)0()0(0 Aii L
)0()0()0( 0 dt
diL
dt
diLv L
L )0(Lv
0)0(Cv
012)0(18)0(4 LL iv
210
210
5.0)0(
632/17)0(
KKi
KKdt
di
Steady state t<0
)0(Li
)0(CvV24
17)0( Lv
0;6
14
6
11)( 63
0 teeti tt
6
14;6
1121 KK
SecondOrder