Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege
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Transcript of Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx1
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 43
RC & RL 1st Order
Ckts
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx2
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
C&L Summary
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx3
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Introduction Transient Circuits
In Circuits Which Contain Inductors & Capacitors, Currents & Voltages CanNOT Change Instantaneously
Even The Application, Or Removal, Of Constant Sources Creates Transient (Time-Dependent) Behavior
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx4
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1st & 2nd Order Circuits
FIRST ORDER CIRCUITS• Circuits That Contain
ONE Energy Storing Element– Either a Capacitor or an Inductor
SECOND ORDER CIRCUITS• Circuits With TWO Energy Storing
Elements in ANY Combination
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx5
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Circuits with L’s and/or C’s Conventional DC Analysis Using
Mathematical Models Requires The Determination of (a Set of) Equations That Represent the Circuit Response
Example; In Node Or Loop Analysis Of Resistive Circuits One Represents The Circuit By A Set Of Algebraic Equations
The Ckt The DC Math Model
Analysis
iv
G
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx6
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Ckt w/ L’s & C’s cont.
When The Circuit Includes Inductors Or Capacitors The Models Become Linear Ordinary Differential Equations (ODEs)
Thus Need ODE Tools In Order To Analyze Circuits With Energy Storing Elements• Recall ODEs from
ENGR25• See Math-4 for
More Info on ODEs
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx7
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
First Order Circuit Analysis
A Method Based On Thévenin Will Be Developed To Derive Mathematical Models For Any Arbitrary Linear Circuit With One Energy Storing Element
This General Approach Can Be Simplified In Some Special Cases When The Form Of The Solution Can Be Known BeforeHand• Straight-Forward ParaMetric Solution
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx8
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Basic Concept
Inductors & Capacitors Can Store Energy
Under Certain Conditions This Energy Can Be Released
RATE of Energy Storage/Release Depends on the parameters Of The Circuit Connected To The Terminals Of The Energy Storing Element
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx9
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example: Flash Circuit
The Battery, VS,Charges the CapTo Prepare for aFlash
Moving the Switch to the Right “Triggers” The Flash• i.e., The Cap
Releases its Stored Energy to the Lamp
Say“Cheese”
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx10
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Flash Ckt Transient Response The Voltage Across the Flash-Ckt Storage
Cap as a Function of TIME
Note That the Discharge Time (the Flash) is Much Less Than the Charge-Time
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx11
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
General Form of the Response Including the initial
conditions the model equation for the capacitor-voltage or the inductor-current will be shown to be of the form
xp(t) ANY Solution to the General ODE• Called the “Particular”
Solution
xc(t) The Solution to the General Eqn with f(t) =0• Called the “Complementary
Solution” or the “Natural” (unforced) Response
• i.e., xc is the Soln to the “Homogenous” Eqn
This is the General Eqn Now By Linear
Differential Eqn Theorem (SuperPosition) Let
tftxdt
tdx
0 txdt
tdx
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx12
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1st Order Response Eqns Given xp and xc
the Total Solution to the ODE
txtxtx cp Consider the Case
Where the Forcing Function is a Constant• f(t) = A
Now Solve the ODE in Two Parts
For the Particular Soln, Notice that a CONSTANT Fits the Eqn:
0
txdt
tdx
Atxdt
tdx
cc
pp
0
so and
1
dt
tdx
Ktx
p
p
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx13
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1st Order Response Eqns cont Sub Into the General
(Particular) Eqn xp and dxp/dt
AK
AK
1
1
or
0
Next, Divide the Homogeneous (RHS=0) Eqn by ∙xc(t) to yield
Next Separate the Variables & Integrate
1
tx
dttdx
c
c
tdtdxtx c
c
11
Recognize LHS as a Natural Log; so
const where
ln
c
ct
txc
Next Take “e” to The Power of the LHS & RHS
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx14
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1st Order Response Eqns cont Then
tc
tcctc
eKtx
eeetx
2
Note that Units of TIME CONSTANT, , are Sec
Thus the Solution for a Constant Forcing Fcn
For This Solution Examine Extreme Cases• t =0• t → ∞
The Latter Case (K1) is Called the Steady-State Response• All Time-Dependent
Behavior has dissipated
/
21t
cp
eKKtx
txtxtx
121
210
KeKKtx
KKx
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx15
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Effect of the Time ConstantTangent reaches x-axis in one
time constant
Decreases 63.2% after One Time Constant time
Drops to 1.8% after 4 Time Constants 0183.04 e
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx16
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Large vs Small Time Constants Larger Time Constants Result in Longer
Decay Times• The Circuit has a Sluggish Response
Quick to Steady-State
Slow to Steady-State
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx17
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1st Order Ckt Solution Plan
1. Use some form of KVL/KCL, and to develop an ODE
2. Isolate the “0th” order term which reveals the • Time Constant• Forcing Function
3. “EyeBall” the ODE to “Guesstimate the Particular Solution, or
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx18
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1st Order Ckt Solution Plan
4. CHECK that the particular Solution Satisfies the ODE
5. Separate the Variables in the Homogeneous Equation and Integrate to obtain the Complementary Solution or
6. A the particular and complementary solutions to obtain the total Solution;
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx19
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
1st Order Ckt Solution Plan
The Solution Should include a Negative Exponential with an UnKnown PreFactor
7. Use the initial condition in the total solution to determine the Prefactor which completes the total solution
8. Check the total Solution for extreme cases:
• t = 0+ • t → ∞
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx20
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example 1st Order Ckt Soln
For the Ckt Below Find for
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx21
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx22
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx23
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx24
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx25
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx26
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx27
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx28
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx29
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Time Constant Example Charging a Cap
Use KCL at node-a
Now let• vC(t = 0 sec) = 0 V
• vS(t)= VS (a const)
ReArrange the KCL Eqn For the Homogenous Case where Vs = 0
Thus the Time Constant
vS
R S a
b
C
+
vc
_
dt
dvC C
S
SC
R
vv
S
S
S
CC
S
SCC
R
v
R
v
dt
dvC
R
vv
dt
dvC
0
SC
C
CRv
dtdv 1
SCR
t
cc
c eKtxtx
dttdx 2
1
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx30
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Time Constant Example cont Charging a Cap
The Solution Can be shown to be
“Fully” Charged Criteria• vC >0.99VS OR
vS
R S a
b
C
+
vc
_
t
SSC eVVtv
)(
6.401.0ln
01.0
t
ore t
CRS tSC eVtv 1)(Or
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx31
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Differential Eqn Approach Conditions for Using This Technique
• Circuit Contains ONE Energy Storing Device• The Circuit Has Only CONSTANT,
INDEPENDENT Sources• The Differential Equation For The Variable Of
Interest is SIMPLE To Obtain – Normally by Using Basic Analysis Tools; e.g., KCL, KVL,
Thevenin, Norton, etc.
• The INITIAL CONDITION For The Differential Equation is Known, Or Can Be Obtained Using STEADY STATE Analysis Prior to Switching– Based On: Cap is OPEN, Ind is SHORT
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx32
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example Given the RC Ckt At
Right with• Initial Condition (IC):
– v(0−) = VS/2
Find v(t) for t>0
Looks Like a Single E-Storage Ckt w/ a Constant Forcing Fcn• Assume Solution of Form
Model t>0 using KCL at v(t) after switch is made
)0();(
0,)(
211
21
xKKxK
teKKtxt
0
tdt
dvC
R
Vtv S
Find Time Constant; Put Eqn into Std Form• Multiply ODE by R
sVtvtdt
dvRC
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx33
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example cont Compare Std-Form with
Model• Const Force Fcn Model
Note: the SS condition is Often Called the “Final” Condition (FC)
In This Case the FC
Now Use IC to Find K2
1Kxdt
dx
• In This Case
sVtvtdt
dvRC )()(
Next Check Steady-State (SS) Condition• In SS the Time Derivative
goes to ZERO
s
s
V K
Vtvt
1 :confirms
)(
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx34
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Example cont.2 At t = 0+
• The Model Solution
The Total/General Soln
Recall The IC: v(0−) = VS/2 = v(0+) for a Cap• Then
12
021
)0(
)0(
KvK
eKKv
2
2
)0(
2
2
12
S
SS
VK
VVK
KvK
RC
t
S eVtv 5.01)(
Check sS VeVv 5.05.01)0( 0
sS VeVv 5.01)(
or0,21
teKKtvt
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx35
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inductor Exmpl Find i(t) Given
• i(0−) = 0
Recognize Single E-Storage Ckt w/ a Constant Forcing Fcn• Assume Solution of Form
In This Case x→i
)0();(
0,)(
211
21
xKKxK
teKKtxt
To Find the ODE Use KVL for Single-Loop ckt
0)0()0( ii
)(ti
0;)( 21
teKKtit
Rv
Lv)(ti
KVL
)()( tdt
diLtRivvV LRS
Now Consider IC• By Physics, The Current
Thru an Inductor Can NOT Change instantaneously
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx36
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inductor cont Casting ODE in
Standard form
Recognize Time Const
Next Using IC (t = 0+)
)(ti
RL
Thus the ODE Solution
R
Vtit
dt
di
R
L S )()(
Also Note FC
RVtit s )(
RVK
RVK
KiK
S
S
2
2
12
0
)0(
tL
Rs
t
eR
VeKKti 1)( 21
RVK s1
Thus
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx37
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Solution Process Summary
1. ReWrite ODE in Standard Form• Yields The Time-Constant,
2. Analyze The Steady-State Behavior• Finds The Final Condition Constant, K1
3. Use the Initial Condition• Gives The Exponential PreFactor, K2
4. Check: Is The Solution Consistent With the Extreme Cases
• t = 0+• t →
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx38
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Solutions for f(t) ≠ Constant1. Use KVL or KCL to Write the ODE
2. Perform Math Operations to Obtain a CoEfficient of “1” for the “Zeroth” order Term. This yields an Eqn of the form
3. Find a Particular Solution, xp(t) to the FULL ODE above• This depends on f(t), and may require
“Educated Guessing”
) (yields tfxdt
dx
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx39
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Solutions for f(t) ≠ Constant
4. Find the total solution by adding the COMPLEMENTARY Solution, xc(t) to previously determined xp(t). xc(t) takes the form:• The Total Solution
at this Point →
5. Use the IC at t=0+ to find K; e.g.; IC = 7
• Where M is just a NUMBER
tc Ketx
txKetx pt
MKxKe p 707 0
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx40
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Capacitor Example
At t=0+ Apply KCL Assume a Solution of the Form for vc
C
1R
2R
)0();(
0,)(
211
21
CC
t
C
vKKvK
teKKtv
0)()(
0)(
21
21
cC
CC
vtdt
dvCRR
RR
vt
dt
dvC
For Ckt Below Find vo(t) for t>0 (note f(t) = const; 12V)
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx41
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Cap Exmp cont Step1: By Inspection of
the ReGrouped KCL Eqn Recognize
Now Examine the Reln Between vo and vC
• a V-Divider
s
mFk
CRR
6.0
1.06
)( 21
)(3
1)(
42
2)( tvtvtv CCO
Step-2: Consider The Steady-State• In This Case After the
Switch Opens The Energy Stored in the Cap Will be Dissipated as HEAT by the Resistors
With xp = K1 = SSsoln
0)()( tvtv Co
0)(1 CvK
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx42
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Cap Exmp cont Now The IC If the Switch is Closed for a Long Time before t =0, a
STEADY-STATE Condition Exists for NEGATIVE Times
vo(0−) by V-Divider Recall Reln Between vo and vC for t ≥ 0
Vv
VVv
O
O
3
8)0(
129
212
423
2)0(
VVvv
tvtv
CC
OC
8129
6)0()0(
)(3)(
Recall: Cap is OPEN to DC
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx43
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Cap Exmp cont Step-3: Apply The IC
Now have All the Parameters needed To Write The Solution
)0,80)(
0,)(
6.0
21
tVetv
teKKtv
s
t
C
t
C
VvK
vK
vKK
C
C
C
8)0(
0)( and
)0(
2
1
21
Note: For f(t)=Const, puttingthe ODE in Std Form yields and K1 by Inspection:
0)()( 21 cC vt
dt
dvCRR
1K
Recall vo = (1/3)vC
0,V3
8)( 6.0
tetv s
t
o
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx44
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Thevenin/Norton Techniques Obtain The Thevenin Voltage Across The Capacitor,
Or The Norton Current Through The Inductor
With This approach can Analyze a SINGLE-LOOP, or SINGLE-NODE Ckt to Find• Time Constant using RTH
• Steady-State Final Condition using vTH (if vTH a constant)
Circuitw ith
resistancesand
sources
I nductororCapacitor
a
b
R epresentation of an arbitrarycircuit w ith one storage elem ent
ThéveninVT H
R T H
I nductororCapacitor
a
b
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx45
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Thevenin Models for ODE
KVL for Single Loop
VT H
R T H a
b
C
+
vc
_
Case 1.1Voltage across capacitor
ciRi
0 Rc iidt
dvCi C
c
TH
THCR R
vvi
0
TH
THCC
R
vv
dt
dvC
THCC
TH vvdt
dvCR
VT H
R T H a
b
L
iL
Case 1.2Current through inductor
Rv
Lv
THLR vvv LTHR iRv
dt
diLv L
L THLTHL viR
dt
diL
NTH
THL
L
TH
IR
vi
dt
di
R
L
KCL at node a
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx46
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Find ODE Soln By Thevenin
Break Out the Energy Storage Device (C or L) as the “Load” for a Driving Circuit
Analyze the Driving Ckt to Arrive at it’s Thévenin (or Norton) Equivalent
ReAttach The C or L Load Use KCL or KVL to arrive at ODE Put The ODE in Standard Form
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx47
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Find ODE Soln By Thevenin
Recognize the Solution Parameters For Capacitor
• = RTHC
• K1 (Final Condition) = vTH = vOC = xp
For Inductor• = L/RTH
• K1 (Final Condition) = vTH/RTH = iSC = iN = xp
In Both Cases Use the vC(0−) or iL(0−)Initial Condition to Find Parameter K2
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx48
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inductor Example
The Variable Of Interest Is The Inductor Current
The Thévenin model
Since this Ckt has a CONSTANT Forcing Function of 0V (the 24V source is switched OUT at t = 0), Then The Solution Is Of The Form
66
6
6
H3
V24
)(tiO
0t
0t;(t)iFind O
TH
THO
O
TH R
vi
dt
di
R
L
0;)( 21
teKKtit
O
Next Construct the Thévenin Equivalent for the Inductor “Driving Circuit”
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx49
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inductor Example cont The f(t)=const ODE in
Standard Form
s
H
R
L
R
v
TH
TH
TH
3.010
3
106666
0
The Solution Substituted into the ODE at t > 0
66
6
6V24
0t
Thevenin for t>0at inductor terminals
a
b
0;03.0 tidt
dis O
O
03.0
3.0 3.021
3.02
tt
eKKeK
0;)( 3.02
teKti
t
O
Combining the K2 terms shows shows K1 = 0, thus
From This Ckt Observe
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx50
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inductor Example cont.2
Now Find K2
Assume Switch closed for a Long Time Before t = 0• Inductor is SHORT to DC
Analyzing Ckt with 3H Ind as Short Reveals
• The Reader Should Verify the Above
0;3
16
36
24
2
1
6
24
tA
VViO
0;33.5)( 3.0
teAti s
t
O
Then the Entire Solution
66
6
6
H3
V24
)(tiO
0t
0t;(t)iFind O
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx51
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Untangle to find iO(0−)
Untangle
Analyz
e
Be Faithful to Nodes
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx52
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
First Order Inductor Ciruit Transient Solution
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
5.5
-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8
Time (s)
Cu
rren
t (A
)
i(t) (A)
file = Engr44_Lec_06-1_Last_example_Fall03..xls
0;333.5)( 3.0
teAti s
t
O
s
A
s
A
s
A
dt
di
s
i
dt
di
s
i
dt
di
s
i
dt
dii
dt
dis
t
O
O
t
OOO
OOO
O
78.1733
1016
103
316
103
0
103
3.003.0
0
0
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx53
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard Work Let’s Work This
Problem
t= 0
24V 2
4 4
86 VC ( t)
(1 /4)H
+
-
Well, Maybe NEXT time…
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx54
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
WhiteBoard Work
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx55
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Bruce Mayer, PELicensed Electrical & Mechanical Engineer
Engineering 43
AppendixDE
Approach
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx56
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Differential Eqn Approach cont Math Property
• When all Independent Sources Are CONSTANT, then for ANY variable y(t); i.e., v(t) or i(t), in The Circuit The Solution takes the Form
t
eKKty
21
The Solution Strategy• Use The DIFFERENTIAL EQUATION And The
FINAL & INITIAL Conditions To Find The Parameters K1 and K2
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx57
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Differential Eqn Approach cont If the ODE for y is
Known to Take This Form
We Can Use This Structure to Find The Unknowns. If:
Then Sub Into ODE
Equating the TRANSIENT (exponential) and CONSTANT Terms Find
01
0
120
1 0
a
AK
a
aeKa
a t
0
01
)0( yy
Ayadt
dya
t
t
eK
dt
dy
teKKty
2
21 0,)(
AKaeKaa
AeKKaeK
a
t
tt
10201
2102
1
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx58
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Differential Eqn Approach cont Up to Now
Next Use the Initial Condition
So Finally
If we Write the ODE in Proper form We can Determine By Inspection and K1
012
0
aa
t
eKa
Aty
102
021
00
21
0
or
)0(
)0(
KyK
yKK
yeKKy
yy
01
00
0
aa
t
ea
Ay
a
Aty
00
101 a
Ay
dt
dy
a
aAya
dt
dya
1K
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx59
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inductor Example
The Model for t>0 → KVL on single-loop ckt
Rewrite In Std Form
L
)(1 ti
Lv
S9
1
Recognize Time Const
0)()(18
21
1 titdt
di
)0();(
0,)(
12111
211
iKKiK
teKKtit
0)(126 11 ti
dt
diL
For The Ckt Shown Find i1(t) for t>0
Assume Solution of the Form
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx60
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inductor Example cont Examine Std-Form Eqn
to Find K1
For Initial Conditions Need the Inductor Current for t<0
Again Consider DC (Steady-State) Condition for t<0
The SS Ckt Prior to Switching• Recall An Inductor is a
SHORT to DC
0
0)()(9
1
1
11
K
titdt
di
)0(1 i
SoA
Vi 1
12
12)0(1
[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx61
Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis
Inductor Example cont.1 Now Use Step-3 To
Find K2 from IC• Remember Current Thru
an Inductor Must be Time-Continuous
• Recall that K1 was zero
Construct From the Parameters The ODE Solution
2
2111
][1
)0()0(
KA
KKii
0],[)(
0],[)(9
1
91
1
tAeti
tAetit
t The Answer
0