ENGR-43_Lec-04b_2nd_Order_Ckts

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Engineering 43
nd
2
Order
RLC Circuits
Bruce Mayer, PE
Registered Electrical & Mechanical Engineer
[email protected]
Engineering-43: Engineering Circuit Analysis
1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
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
Engineering-43: Engineering Circuit Analysis
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
ReCall RLC VI Relationships
Resistor
Capacitor
vR t   R  iR t 
iR t   G  vR t 
dv t 
iC t   C C
dt
Engineering-43: Engineering Circuit Analysis
3
t
1
vC t    iC u du vC 0
C0
Inductor
vL t   L
t
diL t 
dt
1
iL t    vL u du  iL 0
L0
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Transient Response
 The VI Relations can be Combined with KCL
and/or KVL to solve for the Transient (timevarying) Response of RL, RC, and RLC
circuits
 Kirchoff’s Current Law
 The sum of all Currents
entering any Circuit-Node
is equal to Zero
N
 i t   0
k 1
k
Engineering-43: Engineering Circuit Analysis
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 Kirchoff’s Voltage Law
 The sum of all the VoltageDrops around any Closed
Circuit-Loop is equal to Zero
N
 v t   0
k 1
k
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Second Order Circuits
 Single Node-Pair
 Single Loop
 vC 
 vR 
iR
iL
vL
iC
 v S  v R  vC  v L  0
 iS  iR  iL  iC  0
v (t )
1t
dv
iR 
; i L   v ( x )dx  i L (t0 ); iC  C (t )
R
L t0
dt
t
v 1
dv
  v ( x )dx  i L (t0 )  C (t )  i S
R L t0
dt
d 2v 1 dv v di S
C 2
 
R dt L dt
dt
Engineering-43: Engineering Circuit Analysis

• By KVL
• By KCL
5

1 t
di
v R  Ri ; vC   i ( x )dx  vC (t0 ); v L  L (t )
C t0
dt
1 t
di
Ri   i ( x )dx  vC (t0 )  L (t )  v S
C t0
dt
Differentiating
d 2i
di i dv
L 2 R   S
dt C
dt
dt
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Second Order Circuits
 Single Node-Pair
 Single Loop
 vC 
 vR 
iR
iL

vL
iC
• By KCL Obtained

• By KVL Obtained
d 2v 1 dv v di S
C 2
 
R dt L dt
dt
d 2i
di i dv
L 2 R   S
dt C
dt
dt
 Make CoEfficient of 2nd Order Term = 1
d 2v 1 dv v
1 diS



dt 2 RC dt LC C dt
1∙(2nd Order Term)
Engineering-43: Engineering Circuit Analysis
6
d 2i R di
1
1 dvS


i

dt 2 L dt LC
L dt
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
ODE for iL(t) in SNP
 Single-Node Ckt
iR
iL
iC
vL
dvL
 iL  C
 iS
R
dt
 Recall v-i
diL
Relation for vL  L
dt
Inductors
 Sub Out vL in above
 By i  i  i  i
1  diL 
d  diL 
R
L
C
S
KCL
L
  iL  C  L
  iS
R  dt 
dt  dt 
 Note That vt   vL t 
 Use Ohm & Cap
Laws
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
ODE Derivation Alternative
 Use Similar Method
1  diL 
d  diL 
L
  iL  C  L
  iS
to find 𝑣𝐶 𝑡 for
R  dt 
dt  dt 
 Take Derivative and
ReArrange
d 2iL L diL
LC

 iL  iS
2
dt  R dt
 Make CoEff of 2nd
Order Term = 1
iS
d 2 iL
1 diL
1


iL 
2
LC
dt  RC dt LC
Engineering-43: Engineering Circuit Analysis
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Single LOOP Circuit
vR  vC  vL  vS
and i  iC
 Then
diC
RiC  vC  L
 vS
dt
dvC
but iC  C
dt
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
ODE for 𝒗𝑪 𝒕 in SNP
 Single LOOP Ckt
 vC 
 vR 

vL

diC
RiC  vC  L
 vS
dt

d  dvC
 dvC 
R C
  vC  L  C
dt  dt
 dt 

  vS

 Cleaning Up
 Importantly
dvC
iC  C
dt
 Thus the KVL eqn
Engineering-43: Engineering Circuit Analysis
9
dvC
d 2vC
RC
 vC  LC 2  vS
dt
dt
d 2 vC R dvC
vS
1


vC 
2
dt
L dt LC
LC
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Illustration
 Write The Differential Eqn for v(t) & i(t) Respectively

vS
iS

 The Forcing Function
0 t 0
iS (t )  
I S t  0
V t  0
vS (t )   S
0 t 0
 Parallel RLC Model
d 2v 1 dv v diS
C 2
 
dt R dt L dt
 In This Case
diS
(t )  0; t  0
dt
2
d
 So C v  1 dv  v  0
2
dt
R dt L
Engineering-43: Engineering Circuit Analysis
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 The Forcing Function
 Series RLC Model
d 2i
di i dv
L 2 R   S
dt
dt C
dt
 In This Case
dvS
(t )  0; t  0
dt
2
d
 So L i  R di  i  0
dt 2
dt
C
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
2nd Order Response Equation
 Need Solutions to the
2nd Order ODE
d 2x
dx
1 2 (t )  a1 (t )  a2 x(t )  f (t )
dt
dt
 As Before The Solution
Should to this Linear
Eqn Takes This form
x(t )  x p (t )  xc (t )
 Where
• xp  Particular Solution
• xc  Complementary
Solution
Engineering-43: Engineering Circuit Analysis
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 If the Forcing Fcn is a
Constant, A, Then
Discern a Particular Soln
A
f (t )  A  x p 
a2
 Verify xp
dx p d 2 x p
A
xp 


0
2
a2
dt
dt
 a2 x p  a2
A
A
a2
 For Any const Forcing
Fcn, f(t) = A
x(t ) 
A
 xc (t )
a2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx

The Complementary Solution
 The Complementary
Solution Satisfies the
HOMOGENOUS Eqn
d 2x
dx
(t )  a1 (t )  a2 x(t )  0
2
dt
dt
 ReWrite in Std form
d 2x
dx
2
(
t
)

2

(
t
)


0 x(t )  0
2
dt
dt
 Where
• a1  2α = 20
• a2  02
Engineering-43: Engineering Circuit Analysis
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 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
st
form x(t )  Ke
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Complementary Solution cont
 Sub Assumed Solution
(x = Kest) into the
Homogenous Eqn
d 2x
dx
2
(
t
)

2

(
t
)


0 x(t )  0
2
dt
dt
s 2 Ke st  2sKe st  02 Ke st  0
 Units Analysis
 Canceling Kest
Let x(t )  i (t )  Amps [A]
s  2s    0
2
2
0
 The Above is Called the
Characteristic Equation
Engineering-43: Engineering Circuit Analysis
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 A value for “s” That
SATISFIES the
CHARACTERISTIC
Eqn ensures that Kest is
a SOLUTION to the
Homogeneous Eqn
di dt  A/S; d 2i dt   A/S2
2
Also e st  st  Unitless
s  S1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Complementary Solution cont.2
 Recall Homog. Eqn.
d 2i
di
2
(
t
)

2

(
t
)


0
0 i (t )  0
2
dt
dt
 Discern Units after
Canceling Amps
SameUnits
st 
1  s  1 / S  S 1
1
SameUnits
2
2
2
(
t
)









1
S
0
0
dt 2
 0  1 / S  S 1 radians sec 
1
1
1
SameUnits
(
t
)









0
dt 2
dt
0 dt
  
1
1
S S
 UnitLess
Engineering-43: Engineering Circuit Analysis
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 Short Example: Given
Homogenous Eqn
Determine
• Characteristic Eqn
• Damping Ratio, 
• Natural frequency, 0
 Given Homog. Eqn
d 2x
dx
4 2 (t )  8 (t )  16 x(t )  0
dt
dt
 Coefficient of 2nd Order
Term MUST be 1
d 2x
dx
(t )  2 (t )  4 x(t )  0
2
dt
dt
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Complementary Solution cont.3
 Example Cont.
d 2x
dx
(t )  2 (t )  4 x(t )  0
2
dt
dt
s 2  2s  4  0
 Then
d 2x
dx
2
(
t
)

2

(
t
)


0 x (t )  0
dt 2
dt
or
02  4  02  4
 0  2 / S  2S 1
20  2  2   
1
0
1
    0.5 ( UnitLess)
2
Engineering-43: Engineering Circuit Analysis
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 Before Moving On,
Verify that Kest is a
Solution To The
Homogenous Eqn
2
dx
d
x
2
st
 sKe st ;

s
Ke
dt
dt 2
( s 2  2s  02 ) Ke st  0
 K=0 is the TRIVIAL
Solution
• We need More
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Complementary Solution cont.4
 If Kest is a Solution
Then Need
 Solve By Completing
the Square
( s   ) 2  (02   2 )  0
s  2s    0
2
2
0
s     2  02 
• The CHARACTERISTIC
Equation
s1, 2    0  2 02  1
s1, 2  0  0  2  1
 Solve For 𝑠 by One of
• Quadratic Eqn
• Completing The Square
• Factoring (if we’re
REALLY Lucky)
Engineering-43: Engineering Circuit Analysis
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
The Solution for s
Generates 3 Cases
1. >1
2. <1
3. =1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Aside: Completing the Square
 Start with: s 2  2s  02  0
 ReArrange: s 2  2s 
02  0
 Add Zero → 0 = y−y:


s 2  2s   2   2  02  0
 ReArrange: s 2  2s   2  02   2  0
2
2
2
2



 0
s

2

s






 Grouping
0
• The First Group is a PERFECT Square
 ReWriting: s   2  02   2   0
Engineering-43: Engineering Circuit Analysis
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Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Initial Conditions
 Summarize the TOTAL solution for f(t) = const, A
xTOT (t )  xc (t )  x p (t )  K1e s1t  K 2e s2t  A 02
 Find K1 and K2 From INITIAL CONDITIONS x(0) AND
(this is important) [dx/dt]t=0; e.g.;
x(t )  K1e s1t  K 2 e s2t  A 02
then
x(0)  K1  K 2  A 02
and
dx(t )
dt t 0
 Must Somehow
find a
dx(t )
dx(0)

 s1 K1  s2 K 2 NUMBER dt
t 0
dt
for
Engineering-43: Engineering Circuit Analysis
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Two Eqns in
Two Unknowns
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Case 1: >1 → OVERdamped
 The Damped Natural Frequencies, s1 and s2,
are REAL and UNequal
 The Natural Response Described by the Relation
xc (t )  K1e  K 2 e
s1t
s2t
 The TOTAL Natural Response is thus a Decaying
Exponential plus a Constant
xTOT t   xc (t )  x p (t )
 K1e
  0 0  2 1  t


 K 2e
  0 0  2 1  t


A 
as s1, 2   0   202  02 and    0
Engineering-43: Engineering Circuit Analysis
19
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
2
0
Case 2: <1 → UNDERdamped
 Since <1 The Characteristic Eqn Yields
COMPLEX Roots as Complex Conjugates
• So with 𝑗 = −1
s1  0  j0 1   2    jn
s2  0  j0 1   2    jn
 Where
• n  Damped natural
Oscillation Frequency
• α  Damping Coefficient
xc t   e
t
 A1 cos nt  A2 sin nt 
Engineering-43: Engineering Circuit Analysis
20
 Then The
UnderDamped
UnForced (Natural)
Response Equation
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
UnderDamped Eqn Development
 Start w/ Soln to
Homogeneous Eqn
xc (t )  K1e s1t  K 2 e s2t
 Since K1 & K2 are
Arbitrary Constants,
Replace with NEW
Arbitrary Constants
A1  K1  K 2
 From Appendix-A;
The Euler Identity
e  jnt  cos nt  j sin nt
 Then
xc (t )  K1e (   jn )t  K 2 e  (  jn )t
A2  jK1  K 2 
 Sub A1 & A2 to Obtain
xc (t )  e t  A1 cos nt  A2 sin nt 

 K1e t e jdnt  K 2 e t e  jnt
 e t K1 cos nt  jK1 sin nt  K 2 cos nt  jK 2 sin nt 
 e t K1  K 2 cos nt  jK1  K 2 sin nt 
Engineering-43: Engineering Circuit Analysis
21
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
UnderDamped IC’s
 Find Under Damped
Constants A1 & A2
 Given “Zero Order” IC
x(0)  X 0
 Now dx/dt at any t
n A2  A1  cos nt 
dx
t 
e 





A


A
sin

t
dt
n 1
2
n 

 A  A1  cos n 0 
 With xp = D (const) then dx (0)  e  0  n 2





A


A
sin

0
dt
at t=0 for total solution
n 1
2
n 

x ( 0)  X 0 
dx
(0)  X 1  n A2  A1
e  0  A1 cos n 0  A2 sin n 0  D dt
 A1  X 0  D
 For 1st-Order IC
dx dt t 0  X 1
Engineering-43: Engineering Circuit Analysis
22
 Arrive at Two Eqns in
Two Unknowns
• But MUST have a
Number for X1
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Case 3: =1 →CRITICALLY damped
 The Damped Natural Frequencies, s1 and s2, are
REAL and EQUAL
 The Natural Response Described by Relation
xc (t )  B1e
 0t
 B2te
 Find Constants from
Initial Conditions and
TOTAL response
 EXERCISE
 0t
 B1  B2t e
 The Natural Response
is a Decaying
Exponential against The
the Equation of a LINE
• VERIFY that the Above
IS a solution to the
Homogenous Equation
Engineering-43: Engineering Circuit Analysis
23
t
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Example: Case Analyses
 Determine The General Form Of The Solution
(1)
d 2x
dx
(
t
)

4
(t )  4 x(t )  0
2
dt
dt
 Characteristic Eqn
s 2  4s  4  0
 02  4  0  2 S1
 2  20  4    2 &   1
 Factor The Char. Eqn
s 2  4s  4  0  ( s  2) 2  0
 Real, Equal Roots →
Critically Damped (C3)
x(t )  ( B1  B2t )e
st
x(t )  ( B1  B2t )e 2t
Engineering-43: Engineering Circuit Analysis
24
d 2x
dx
(2) 4 2 (t )  8 (t )  16 x(t )  0
dt
dt
 Recast To Std Form
d 2x
dx
(t )  2 (t )  4 x(t )  0
2
dt
dt
 Then The Undamped
Frequency and
Damping Ratio
02  4  0  2 S 1
2  2    1
20  2    0.5
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Example: Case Analyses cont.
 For Char. Eqn
Complete the Square


s 2  2s  4  s 2  2s  1  3
0  4  2
20  2    1 2
2  2  0  1 S 1
 ( s  1) 2  3  0
n  0 1   2  2 1  0.25  3S 1
 ( s  1) 2  3
n  02   2  4  1  3S 1
 s  1  j 3
 The Roots are Complex
and Unequal → an
Underdamped (Case 2)
System
 Then the Solution
x(t )  e t  A1 cos nt  A2 sin nt 

x(t )  e t A1 cos 3t  A2 sin 3t
• Find the Damped
Parameters
Engineering-43: Engineering Circuit Analysis
25
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx

UnderDamped Parallel RLC Exmpl
 Find Damping Ratio
and Undamped Natural
Frequency given
• R =1 Ω
• L=2H
• C=2F
 The Homogeneous Eqn
from KCL (1-node Pair)
d 2 v 1 dv v
C 2 
 0
dt
R dt L
d 2 v 1 dv v
2 2 
 0
dt
1 dt 2
Engineering-43: Engineering Circuit Analysis
26
 Or, In Std From
d 2 v 1 dv v

 0
2
dt
2 dt 4
 Recognize Parameters
1 1
1
1
0 
 ; 20    
4 2
2
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Parallel RLC Example cont
 Then: Damping Factor,
Damped Frequency
1
  0 
4
1
1
3
n  0 1    1  
2
4
4
2
 Then The Response Equation

3
3 
vc (t )  e  A1 cos
t  A2 sin
t 
4
4 


t
4
 If: v(0)=10 V, and dv(0)/dt = 0 V/S, Then Find:
A1  10
A2  10
 Plot on Next Slide
Engineering-43: Engineering Circuit Analysis
27
3

3
1
3 

vc (t )  10e  cos
t
sin
t 
4
4 
3


t
4
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Underdamped Parallel RLC Circuit Response
10.0
Envelope (top)
v(t) (V)
Envelope (bot)
7.5
5.0
vO (V)
2.5
0.0
0
2
4
6
8
10
12
14
16
18
-2.5
-5.0

3
1
3 
vc (t )  10V e  cos
t
sin
t 
4
4
3


-7.5

t
4
-10.0
file = Engr44_Lec_06-1_Last_example_Fall03..xls
Engineering-43: Engineering Circuit Analysis
28
Time (s)
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
20
Determine Constants Using ICs
 Standardized form of
the ODE Including the
FORCING FCN “A”
d 2x
dx
2
(
t
)

2

(
t
)


x(t )  A
0
2
dt
dt
 Case-1 → OverDamped
x(t ) 
A

2
0
x (0  ) 
 K1e s1t  K 2 e s2t
x(t ) 
A

A

2
0
 K1  K 2
Engineering-43: Engineering Circuit Analysis
2
0
 e t  A1 cos nt  A2 sin nt 
x (0  ) 
A

2
0
 A1
dx
(0 )  A1  n A2
dt
 Case-3 → Crit. Damping
x(t ) 
dx
(0)  s1 K1  s2 K 2
dt
29
 Case-2 → UnderDamped
A
02
x (0  ) 
 B1  B2t e t
A

2
0
 B1
dx
(0 )  B1  B2
dt
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
KEY to 2nd Order → [dx/dt]t=0+
 Most Confusion in 2nd
Order Ckts comes in
the from of the
First-Derivative IC
dx dt t 0  X 1
 If x = iL, Then Find vL
diL
L
dt
 vL 0  
t 0
or X 1  vL 0   L
Engineering-43: Engineering Circuit Analysis
30
 If x = vC, Then Find iC
dvC
C
dt
 iC 0  
t 0
or X 1  iC 0   C
 MUST Find at t=0+
vL or iC
 Note that THESE
Quantities CAN Change
Instantaneously
• iC (but NOT vC)
• vL (but NOT iL)
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
[dx/dt]t=0+ → Find iC(0+) & vL(0+)
 If this is needed
dv
dt t 0
di
dt t 0
 Then Find a CAP
and determine the
Current through it
 

dv
i0

dt t 0
C
Engineering-43: Engineering Circuit Analysis
31
 If this is needed
 Then Find an IND
and determine the
Voltage through it
 

di
v0

dt t 0
L
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
WhiteBoard → Find vO(t)


24V
Engineering-43: Engineering Circuit Analysis
32
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
33
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
34
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
35
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
36
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
37
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Engineering-43: Engineering Circuit Analysis
38
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Numerical Example
 For The Given 2nd
Order Ckt Find for t>0
• io(t), vo(t)

KVL
24V
 From Ckt Diagram
Recognize by Ohm’s
Law
v0 (t )  18i0 (t )  12(V )
 KVL at t>0
 1 t

di
4
i
(
x
)
dx

v
(
0
)

2
(t )  18i (t )  12  0

C

1
/
36
dt
0


 Taking d(KVL)/dt → ODE
2
2
d i
di
(
t
)

18
(t )  36i (t )  0
2
dt
dt
Engineering-43: Engineering Circuit Analysis
39
 The Char Eqn & Roots
Ch. Eq. : s 2  9s  18  0
0  s  3s  6
REAL roots : s  3,  6
 The Solution Model
io (t )  K1e 3t  K 2 e 6t ; t  0
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx

Numerical Example cont
 Steady State for t<0


24V  12V
i
(
0
)

 0 .5 A
vC (0)  0
L
6  18
24V
The Analysis at t = 0+
io (0)  iL (0)  0.5( A)
dio
diL
v L (0 )  L
(0  )  L
(0  )
dt
dt
 KVL at t=0+ (vc(0+) = 0)
 4  vL (0)  18iL (0)  12  0
vL 0   4 12  180.5 A  17V
iL (0)

v C ( 0)


 Then Find The
Constants from ICs
dio
17 A
(0  )  
 3K1  6 K 2
dt
2 S
io (0)  0.5 A  K1  K 2
 Then di0/dt by vL = LdiL/dt  Solving for K and K
1
2
dio
11
14
(0)  17V / 2 H
K1   A; K 2  A
dt
6
6
Engineering-43: Engineering Circuit Analysis
40
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Numerical Example cont.2
 Return to the ODE
d 2i
di
(
t
)

9
(t )  18i(t )  0
2
dt
dt
 Yields Char. Eqn Roots


24V
s 2  9s  18  0
0  s  3s  6
s1  3
s2  6
 Write Soln for i0
io (t )  K1e
io (t )  
3t
6 t
 K 2e ; t  0
11 3t 14 6t
e  e ;t 0
6
6
Engineering-43: Engineering Circuit Analysis
41
 And Recall io & vo reln
vo (t )  18io (t )  12(V )
 So Finally
vo (t )  33e 3t  42e 6t  12V ; t  0
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
General Ckt Solution Strategy
 Apply KCL or KVL depending on Nature
of ckt (single: node-pair? loop?)
 Convert between VI using
• Ohm’s Law
v R  iR R
iR  v R R
• Cap Law
ic  C
dvc
dt
1 t
vc   ic  x dx vc t0 
C t0
• Ind Law
diL
vL  L
dt
1 t
iL   vL  x dx iL t0 
L t0
 Solve Resulting Ckt Analytical-Model
using Any & All MATH Methods
Engineering-43: Engineering Circuit Analysis
42
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
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
d 2v 1 dv v diS
C 2
 
dt R dt L dt
Differentiating
Engineering-43: Engineering Circuit Analysis
43
d 2i
di i dv
L 2 R   S
dt
dt C
dt
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
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
Engineering-43: Engineering Circuit Analysis
44
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
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
st
s t
x

K
e

K
e
 xp
 Solve for the 2
1
2
Unknowns to
x  e st mt  b   x p
Complete the x  e t  A1 cos nt  A2 sin nt   x p
Solution Process
1
Engineering-43: Engineering Circuit Analysis
45
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
All Done for Today
nd
2
Order
IC is
Critical!
Series 
diL
 L
Case 
dt
 vL 0  
t 0
Engineering-43: Engineering Circuit Analysis
46
Parallel 
dvC
 C
Case 
dt
 iC 0  
t 0
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Complete the Square -1
 Consider the General
2nd Order Polynomial
• a.k.a; the Quadratic Eqn
 Next, Divide by “a” to
give the second order
term the coefficient of 1
b
c
x  x
a
a
ax  bx  c  0
2
• Where a, b, c are
CONSTANTS
 Solve This Eqn for x by
Completing the Square
 First; isolate the Terms
involving x
ax  bx  c
2
Engineering-43: Engineering Circuit Analysis
47
2
 Now add to both Sides of
the eqn a “quadratic
supplement” of (b/2a)2
2
2
b
b a b a c
x  x
 
 
a
 2   2  a
2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Complete the Square -2
 Now the Left-Hand-Side  Use the Perfect Sq
(LHS) is a PERFECT
Expression
2
2
Square
b   b  c

2
2
x    
b
b a b a c
2
x  x
 
 
2a   2a  a

a
 2   2  a
2
2
b   b  c

x    
2a   2a  a

 Solve for x; but first let
2
b a c

   RHS  D
 2  a
b 2a  F
Engineering-43: Engineering Circuit Analysis
48
or
x  F 2  D
 Finally Find the Roots of
the Quadratic Eqn
x  F 
2
D
x  F   D or
x1 , x2   F  D
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Derive Quadratic Eqn -1
 Start with the
PERFECT SQUARE
Expression
2
2
b   b  c

x    
2a   2a  a

 Take the Square Root
of Both Sides
2
b
 b  c
x
   
2a
 2a  a
Engineering-43: Engineering Circuit Analysis
49
 Combine Terms inside
the Radical over a
Common Denom
b
b2 c
x


2
2a
4a
a
b
b 2 c 4a
x


2
2a
4a
a 4a
b
b 2  4ac
x

2a
4a 2
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Derive Quadratic Eqn -2
 Note that Denom is,
itself, a PERFECT SQ
b
b 2  4ac
x

2a
4a 2
b
b 2  4ac
x

2a
2a
 Next, Isolate x
b
b 2  4ac
x 
2a
2a
Engineering-43: Engineering Circuit Analysis
50
 Now Combine over
Common Denom
 b  b  4ac
x
2a
2
 But this the Renowned
QUADRATIC FORMULA
 Note That it was
DERIVED by
COMPLETING the
SQUARE
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
Complete the Square
Engineering-43: Engineering Circuit Analysis
51
Bruce Mayer, PE
[email protected] • ENGR-43_Lec-04b_2nd_Order_Ckts.pptx
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