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[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PERegistered Electrical & Mechanical Engineer
Engineering 43
2nd OrderRLC
Circuits
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Cap & Ind Physics Summary
Under Steady-State (DC) Conditions• Caps act as OPEN Circuits• Inds act as SHORT Circuits
Under Transient (time-varying) Conditions• Cap VOLTAGE can NOT Change Instantly–Resists Changes in Voltage Across it
• Ind CURRENT can NOT change Instantly–Resists Changes in Curring Thru it
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ReCall RLC VI Relationships
Resistor Capacitor
01
0
C
t
CC vduuiC
tv tiRtv RR dt
tdiLtv L
L
dt
tdvCti C
C tvGti RR 01
0
L
t
LL iduuvL
ti
Inductor
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Transient Response
The VI Relations can be Combined with KCL and/or KVL to solve for the Transient (time-varying) Response of RL, RC, and RLC circuits Kirchoff’s Current Law
The sum of all Currents entering any Circuit-Node is equal to Zero
Kirchoff’s Voltage Law
The sum of all the Voltage-Drops around any Closed Circuit-Loop is equal to Zero
N
kk ti
1
0
N
kk tv
1
0
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Second Order Circuits Single Node-Pair
Ri Li Ci
0 CLRS iiii
)();()(1
;)(
0
0
tdt
dvCitidxxv
Li
R
tvi CL
t
tLR
SL
t
t
itdt
dvCtidxxv
LR
v )()()(1
0
0
• By KCL
Rv
Cv
Lv
• By KVL0 LCRS vvvv
)();()(1
; 0
0
tdt
diLvtvdxxi
CvRiv LC
t
tCR
SC
t
t
vtdt
diLtvdxxi
CRi )()()(
10
0
Single Loop
Differentiatingdt
di
L
v
dt
dv
Rdt
vdC S 1
2
2
dt
dv
C
i
dt
diR
dt
idL S2
2
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Second Order Circuits Single Node-Pair
Ri Li Ci
• By KCL Obtained
Rv
Cv
Lv
• By KVL Obtained
Single Loop
dt
di
L
v
dt
dv
Rdt
vdC S 1
2
2
dt
dv
C
i
dt
diR
dt
idL S2
2
Make CoEfficient of 2nd Order Term = 1
1∙(2nd Order Term)dt
di
CLC
v
dt
dv
RCdt
vd S112
2
dt
dv
Li
LCdt
di
L
R
dt
id S112
2
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ODE for iL(t) in SNP
Single-Node Ckt
By KCL
Note That Use Ohm & Cap
Laws
Recall v-i Relation for Inductors
Sub Out vL in above
Ri Li Ci
SCLR iiii
tvtv L
SL
LL i
dt
dvCi
R
v
dt
diLv L
L
SL
LL i
dt
diL
dt
dCi
dt
diL
R
1
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ODE Derivation Alternative
Take Derivative and ReArrange
Make CoEff of 2nd Order Term = 1
Use Similar Method to vC(t) for Single LOOP Circuit
Then
SL
LL i
dt
diL
dt
dCi
dt
diL
R
1
SLLL iidt
di
R
L
dt
idLC 2
2
LC
ii
LCdt
di
RCdt
id SL
LL 11
2
2
C
SLCR
ii
vvvv
and
dt
dvCi
vdt
diLvRi
CC
SC
CC
but
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
ODE for vC(t) in SNP
Single LOOP Ckt
Importantly
Thus the KVL eqn
Cleaning Up
Rv
Cv
LvS
CC
C
SC
CC
vdt
dvC
dt
dLv
dt
dvCR
vdt
diLvRi
dt
dvCi C
C
LC
vv
LCdt
dv
L
R
dt
vd
vdt
vdLCv
dt
dvRC
SC
CC
SC
CC
12
2
2
2
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Illustration Write The Differential Eqn for v(t) & i(t) Respectively
0
00)(
tI
tti
SS
dt
di
L
v
dt
dv
Rdt
vdC S
12
2
0;0)( ttdt
diS
00
0)(
t
tVtv S
S
dt
dv
C
i
dt
diR
dt
idL S
2
2
0;0)( ttdt
dvS
Si
Sv
The Forcing Function
Parallel RLC Model
In This Case
So 01
2
2
L
v
dt
dv
Rdt
vdC
The Forcing Function
Series RLC Model
In This Case
So 02
2
C
i
dt
diR
dt
idL
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
2nd Order Response Equation Need Solutions to the
2nd Order ODE
As Before The Solution Should Take This form
If the Forcing Fcn is a Constant, A, Then Discern a Particular Soln
Verify xp
)()()()(1 212
2
tftxatdt
dxat
dt
xd
)()()( txtxtx cp Where
• xp Particular Solution
• xc Complementary Solution
)(2a
AxAtf p
Aa
Aaxa
dt
xd
dt
dx
a
Ax
p
ppp
222
2
2
2
0
For Any const Forcing Fcn, f(t) = A
)()(2
txa
Atx c
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
The Complementary Solution The Complementary
Solution Satisfies the HOMOGENOUS Eqn
Nomenclature• α Damping Coefficient• Damping Ratio• 0 Undamped (or
Resonant) Frequency
Need xc So That the “0th”, 1st & 2nd Derivatives Have the same form so they will CANCEL in the Homogeneous Eqn
Look for Solution of the form
ReWrite in Std form
0)()()( 212
2
txatdt
dxat
dt
xd
0)()(2)( 202
2
txtdt
dxt
dt
xd
Where• a1 2α = 20• a2 02
stKetx )(
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complementary Solution cont Sub Assumed Solution
(x = Kest) into the Homogenous Eqn
A value for “s” That SATISFIES the CHARACTERISTIC Eqn ensures that Kest is a SOLUTION to the Homogeneous Eqn
Units Analysis
Canceling Kest
The Above is Called the Characteristic Equation
1-
222
S
Unitless Also
A/SA/S;
[A] Amps)()(Let
s
ste
dtiddtdi
titx
st
02 20
2 ststst KesKeKes
0)()(2)( 202
2
txtdt
dxt
dt
xd
02 20
2 ss
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complementary Solution cont.2 Recall Homog. Eqn. Short Example: Given
Homogenous Eqn Determine• Characteristic Eqn• Damping Ratio, • Natural frequency, 0
Given Homog. Eqn
Discern Units after Canceling Amps
0)()(2)( 202
2
titdt
dit
dt
id
UnitLess1
11)(
1
secradians/1
1)(1
/11
1
002
10
220
202
1
SS
dtdtt
dt
SS
Stdt
SSsst
SameUnits
SameUnits
SameUnits
0)(16)(8)(4
2
2
txtdt
dxt
dt
xd
Coefficient of 2nd Order Term MUST be 1
0)(4)(2)(2
2
txtdt
dxt
dt
xd
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complementary Solution cont.3 Example Cont. Before Moving On,
Verify that Kest is a Solution To The Homogenous Eqn
Then
K=0 is the TRIVIAL Solution• We need More
042
0)(4)(2)(
2
2
2
ss
txtdt
dxt
dt
xd
UnitLess)(5.02
1
1222
2/2
44
00
10
20
20
SS
stst Kesdt
xdsKe
dt
dx 22
2
;
0)2(
or
0)()(2)(
20
2
202
2
stKess
txtdt
dxt
dt
xd
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complementary Solution cont.4 If Kest is a Solution Then
Need Solve By Completing
the Square
• The CHARACTERISTIC Equation
Solve For s by One of• Quadratic Eqn• Completing The Square• Factoring (if we’re
REALLY Lucky)
The Solution for s Generates 3 Cases1. >1
2. <1
3. =1
02 20
2 ss
1
1
0)()(
2002,1
20
202,1
20
2
220
2
s
s
s
s
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Aside: Completing the Square
Start with: ReArrange: Add Zero → 0 = y−y:
ReArrange: Grouping• The First Group is a PERFECT Square
ReWriting:
02 20
2 ss
02 20
2 ss
02 20
222 ss
02 220
22 ss
02 220
22 ss
0220
2 s
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx18
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Initial Conditions Summarize the TOTAL solution for f(t) = const, A
Find K1 and K2 From INITIAL CONDITIONS x(0) AND (this is important) [dx/dt]t=0; e.g.;
2021
21)()()( AeKeKtxtxtx tstspcTOT
22110
2021
2021
)0()(
and
)0(
then
)( 21
KsKsdt
dx
dt
tdx
AKKx
AeKeKtx
t
tsts
Two Eqns in Two Unknowns
Must Somehow find a NUMBER for 0
)(
tdt
tdx
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx19
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Case 1: >1 → OVERdamped The Damped Natural Frequencies, s1 and s2,
are REAL and UNequal The Natural Response Described by the Relation
The TOTAL Natural Response is thus a Decaying Exponential plus a Constant
tstsc eKeKtx 21
21)(
20
1
2
1
1
200
200
)()(
AeKeK
txtxtx
tt
pcTOT
020
20
202,1 and as s
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx20
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Case 2: <1 → UNDERdamped
Then The UnderDamped UnForced (Natural) Response Equation
Where• n Damped natural
Oscillation Frequency• α Damping Coefficient
Since <1 The Characteristic Eqn Yields COMPLEX Roots as Complex Conjugates• So with j=(-1)
n
n
jjs
jjs
2
002
2001
1
1
tAtAetx nnt
c sincos 21
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx21
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
UnderDamped Eqn Development Start w/ Soln to
Homogeneous Eqn
From Appendix-A; The Euler Identity
Since K1 & K2 are Arbitrary Constants, Replace with NEW Arbitrary Constants
tstsc eKeKtx 21
21)(
tjte nntj n sincos
Then
tKKjtKKe
tjKtKtjKtKe
eeKeeK
eKeKtx
nnt
nnnnt
tjttjt
tjtjc
ndn
nn
sincos
sincossincos
)(
2121
2211
21
)(2
)(1
212
211
KKjA
KKA
Sub A1 & A2 to Obtain
tAtAetx nnt
c sincos)( 21
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx22
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
UnderDamped IC’s Find Under Damped
Constants A1 & A2
Given “Zero Order” IC
With xp = D (const) then at t=0 for total solution
Now dx/dt at any t
0)0( Xx
For 1st-Order IC
DXA
DAAe
Xx
nn
01
210
0
0sin0cos
)0(
10Xdtdx
t
121
21
120
21
12
)0(
0sin
0cos)0(
sin
cos
AAXdt
dx
AA
AAe
dt
dx
tAA
tAAe
dt
dx
n
nd
nn
nn
nnt
Arrive at Two Eqns in Two Unknowns• But MUST have a
Number for X1
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx23
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Case 3: =1 →CRITICALLY damped
The Natural Response is a Decaying Exponential against The Sum of a CONSTANT and a LINEAR Term
Find Constants from Initial Conditions and TOTAL response
The Damped Natural Frequencies, s1 and s2, are REAL and EQUAL
The Natural Response Described by Relation
tttc etBBteBeBtx 2121
00)(
EXERCISE• VERIFY that the Above
IS a solution to the Homogenous Equation
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx24
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example: Case Analyses
Recast To Std Form Characteristic Eqn
Determine The General Form Of The Solution
Factor The Char. Eqn
0)(4)(4)()1(2
2
txtdt
dxt
dt
xd
0442 ss
0)2(044 22 sss
24 020
1 & 2422 0
t
st
etBBtx
etBBtx2
21
21
)()(
)()(
Real, Equal Roots → Critically Damped (C3)
0)(16)(8)(4)2(2
2
txtdt
dxt
dt
xd
0)(4)(2)(2
2
txtdt
dxt
dt
xd
Then The Undamped Frequency and Damping Ratio
10
20 24 S
122
5.022 0
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx25
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example: Case Analyses cont.
Then the Solution The Roots are Complex and Unequal → an Underdamped (Case 2) System• Find the Damped
Parameters
For Char. Eqn Complete the Square
31
3)1(
03)1(
31242
2
2
22
js
s
s
ssss
1220
120
10
0
0
314
325.0121
122
2122
24
S
S
S
n
n
tAtAetx
tAtAetxt
nnt
3sin3cos)(
sincos)(
21
21
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx26
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
UnderDamped Parallel RLC Exmpl Find Damping Ratio
and Undamped Natural Frequency given• R =1 Ω• L = 2 H• C = 2 F
The Homogeneous Eqn from KCL (1-node Pair)
Or, In Std From
01
2
2
L
v
dt
dv
Rdt
vdC
021
12
2
2
v
dt
dv
dt
vd
042
12
2
v
dt
dv
dt
vd
Recognize Parameters
2
1
2
12;
2
1
4
100
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx27
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Parallel RLC Example cont Then: Damping Factor,
Damped Frequency
4
3
4
11
2
11 2
0 n
4
10
Then The Response Equation
tAtAetv
t
c 4
3sin
4
3cos)( 21
4
If: v(0)=10 V, and dv(0)/dt = 0 V/S, Then Find:
31010 21 AA
Plot on Next Slide
ttetv
t
c 4
3sin
3
1
4
3cos10)( 4
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx28
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Underdamped Parallel RLC Circuit Response
-10.0
-7.5
-5.0
-2.5
0.0
2.5
5.0
7.5
10.0
0 2 4 6 8 10 12 14 16 18 20
Time (s)
vO (
V)
Envelope (top)
v(t) (V)
Envelope (bot)
file = Engr44_Lec_06-1_Last_example_Fall03..xls
ttetv
t
c 4
3sin
3
1
4
3cosV10)( 4
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx29
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Determine Constants Using ICs Standardized form of
the ODE Including the FORCING FCN “A”
Case-1 → OverDamped
Case-2 → UnderDamped
Atxtdt
dxt
dt
xd )()(2)( 2
02
2
tsts eKeKA
tx 21212
0
)(
2211)0( KsKsdt
dx
2120
)0( KKA
x
tAtAeA
tx nnt
sincos)( 212
0
120
)0( AA
x
21)0( AAdt
dxn
Case-3 → Crit. Damping
tetBBA
tx
212
0
)(
120
)0( BA
x
21)0( BBdt
dx
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx30
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
KEY to 2nd Order → [dx/dt]t=0+
Most Confusion in 2nd Order Ckts comes in the from of the First-Derivative IC
If x = iL, Then Find vL
MUST Find at t=0+ vL or iC
Note that THESE Quantities CAN Change Instantaneously• iC (but NOT vC)
• vL (but NOT iL)
LvX
vdt
diL
L
Lt
L
0or
0
1
0
If x = vC, Then Find iC
10Xdtdx
t
CiX
idt
dvC
C
Ct
C
0or
0
1
0
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx31
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[dx/dt]t=0+ → Find iC(0+) & vL(0+)
If this is needed
Then Find a CAP and determine the Current through it
If this is needed
Then Find an IND and determine the Voltage through it
0tdt
dv
C
i
dt
dv
t
0
0
0tdt
di
L
v
dt
di
t
0
0
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx32
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard → Find vO(t)
V24
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx33
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx34
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx35
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx36
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx37
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx38
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx39
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example For The Given 2nd Order
Ckt Find for t>0• io(t), vo(t)
From Ckt Diagram Recognize by Ohm’s Law
KVL at t>0 The Char Eqn & Roots
V24
)(12)(18)( 00 Vtitv
012)(18)(2)0()(36/1
14
0
tit
dt
divdxxi
t
C
0)(36)(18)(22
2
titdt
dit
dt
id
6,3 :roots REAL
630
0189 : Eq. Ch. 2
s
ss
ss
0;)( 62
31 teKeKti tt
o
Taking d(KVL)/dt → ODE The Solution Model
KVL
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx40
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example cont Steady State for t<0
The Analysis at t = 0+
Then Find The Constants from ICs
Then di0/dt by vL = LdiL/dt Solving for K1 and K2
)0(Li
)0(CvV24
0)0( Cv AVV
iL 5.0186
1224)0(
)(5.0)0()0( Aii Lo
)0()0()0( dt
diL
dt
diLv oL
L
012)0(18)0(4 LL iv
VAvL 175.0181240
KVL at t=0+ (vc(0+) = 0)
HVdt
dio 2/17)0(
21
21
A 5.0)0(
63S
A
2
17)0(
KKi
KKdt
di
o
o
A6
14;A
6
1121 KK
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx41
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Numerical Example cont.2 Return to the ODE
Yields Char. Eqn Roots
V24
63 21 ss
0)(18)(9)(2
2
titdt
dit
dt
id
Write Soln for i0
0;6
14
6
11)( 63 teeti tt
o
And Recall io & vo reln
)(12)(18)( Vtitv oo
0;124233)( 63 tVeetv tto
0;)( 62
31 teKeKti tt
o
630
0189 2
ss
ss
So Finally
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx42
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
General Ckt Solution Strategy
Apply KCL or KVL depending on Nature of ckt (single: node-pair? loop?)
Convert between VI using• Ohm’s Law • Cap Law • Ind Law
00
1tvdxxi
Cv
dt
dvCi
c
t
t cc
cc
Rvi
Riv
RR
RR
00
1tidxxv
Li
dt
diLv
L
t
t LL
LL
Solve Resulting Ckt Analytical-Model using Any & All MATH Methods
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx43
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
2nd Order ODE SuperSUMMARY-1
Find ANY Particular Solution to the ODE, xp (often a CONSTANT)
Homogenize ODE → set RHS = 0 Assume xc = Kest; Sub into ODE
Find Characteristic Eqn for xc a 2nd order Polynomial
dt
di
L
v
dt
dv
Rdt
vdC S
12
2
dt
dv
C
i
dt
diR
dt
idL S
2
2
Differentiating
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx44
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
2nd Order ODE SuperSUMMARY-2
Find Roots to Char Eqn Using Quadratic Formula (or Sq-Completion)
Examine Nature of Roots to Reveal form of the Eqn for the Complementary Solution:• Real & Unequal Roots → xc = Decaying
Constants• Real & Equal Roots → xc = Decaying Line
• Complex Roots → xc = Decaying Sinusoid
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
2nd Order ODE SuperSUMMARY-3
Then the TOTAL Solution: x = xc + xp
All TOTAL Solutions for x(t) include 2 UnKnown Constants
Use the Two INITIAL Conditions to generate two Eqns for the 2 unknowns
Solve for the 2 Unknowns to Complete the Solution Process
pnnt
pst
ptsts
xtAtAex
xbmtex
xeKeKx
sincos 21
2121
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
All Done for Today
2nd OrderIC is
Critical!
0Case
Series
0L
t
L vdt
diL
0Case
Parallel
0C
t
C idt
dvC
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx47
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complete the Square -1 Consider the General
2nd Order Polynomial• a.k.a; the Quadratic Eqn
• Where a, b, c are CONSTANTS
Solve This Eqn for x by Completing the Square
First; isolate the Terms involving x
Next, Divide by “a” to give the second order term the coefficient of 1
Now add to both Sides of the eqn a “quadratic supplement” of (b/2a)2
02 cbxax
cbxax 2
a
cx
a
bx 2
a
cababx
a
bx
222
22
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complete the Square -2 Now the Left-Hand-Side
(LHS) is a PERFECT Square
Solve for x; but first let
Use the Perfect Sq Expression
Finally Find the Roots of the Quadratic Eqn
a
c
a
b
a
bx
a
cababx
a
bx
22
222
22
22
Fab
DRHSa
cab
2
2
2
DFx
a
c
a
b
a
bx
2
22
or
22
DFxx
DFx
DFx
21
2
,
or
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx49
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Derive Quadratic Eqn -1 Start with the PERFECT
SQUARE Expression
Take the Square Root of Both Sides
Combine Terms inside the Radical over a Common Denom
a
c
a
b
a
bx
22
22
a
c
a
b
a
bx
2
22 2
2
2
2
2
2
4
4
2
4
4
42
42
a
acb
a
bx
aa
ac
a
b
a
bx
a
c
a
b
a
bx
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx50
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Derive Quadratic Eqn -2 Note that Denom is,
itself, a PERFECT SQ
Next, Isolate x
But this the Renowned QUADRATIC FORMULA
Note That it was DERIVED by COMPLETING theSQUARE
a
acb
a
bx
a
acb
a
bx
2
4
2
4
4
22
2
2
a
acb
a
bx
2
4
2
2
a
acbbx
2
42
Now Combine over Common Denom
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Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Complete the Square