systematic circuit analysis (t&r chap 3)
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MAE140 Linear Circuits40
Systematic Circuit Analysis (T&R Chap 3)
Node-voltage analysisUsing the voltages of the each node relative to a ground
node, write down a set of consistent linear equations forthese voltages
Solve this set of equations using, say, Cramer’s Rule
Mesh current analysisUsing the loop currents in the circuit, write down a set of
consistent linear equations for these variables. Solve.
This introduces us to procedures for systematicallydescribing circuit variables and solving for them
MAE140 Linear Circuits41
Nodal Analysis
Node voltagesPick one node as the ground nodeLabel all other nodes and assign voltages vA, vB, …, vN
and currents with each branch i1, …, iMRecognize that the voltage across a branch
is the difference between the end nodevoltagesThus v3=vB-vC with the directionas indicated
Write down the KCL relations at each nodeWrite down the branch i-v relations to express branch
currents in terms of node voltagesAccommodate current sourcesObtain a set of linear equations for the node voltages
v5
C
B
A
vC
vB
vA
v4
v3
v2
v1+
+
+
++
--
-
- -
MAE140 Linear Circuits42
Nodal Analysis – Example 3-1 p.72
Apply KCLNode A: i0-i1-i2=0Node B:i1-i3+i5=0Node C: i2-i4-i5=0
Write the element/branch eqnsi0=iS0 i3=G3vB
i1=G1(vA-vB) i4=G4vC
i2=G2(vA-vC) i5=iS2
Substitute to get node voltage equationsNode A: (G1+G2)vA-G1vB-G2vC=iS1
Node B: -G1vA+(G1+G3)vB=is2Node C: -G2vA+(G2+G4)vC=-iS2
Solve for vA, vB, vC then i0, i1, i2, i3, i4, i5
vCvB
vA
R4
R2R1
R3
iS2
iS1 i4i3
i2
i1i0
Reference node
i5
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Si
Si
Si
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Bv
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GGG
GGGG
MAE140 Linear Circuits43
Systematic Nodal Analysis
Writing node equations by inspectionNote that the matrix equation looks
just like Gv=i for matrix G and vector v and iG is symmetric (and non-negative definite)
Diagonal (i,i) elements: sum of all conductances connectedto node i
Off-diagonal (i,j) elements: -conductance between nodes iand j
Right-hand side: current sources entering node iThere is no equation for the ground node – the column sums
give the conductance to ground
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vA
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iS2
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i2
i1i0
Reference node
i5
MAE140 Linear Circuits44
Nodal Analysis Example 3-2 p.74
Node A:Conductances
G/2B+2GC=2.5GSource currents entering= iS
Node B:Conductances
G/2A+G/2C+2Gground=3G Source currents entering= 0
Node C:Conductances
2GA+G/2B+Gground=3.5G Source currents entering= 0
iS
vBvA vC
R/2
2R2R
R/2 R
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5.35.02
5.035.0
25.05.2 S
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v
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MAE140 Linear Circuits45
Nodal Analysis – some points to watch
1. The formulation given is based on KCL with thesum of currents leaving the node0=itotal=GAtoB(vA-vB) +GAtoC(vA-vC)+…+ GAtoGroundvA+ileavingA
This yields0=(GAtoB+…+GAtoGround)vA-GAtoBvB-GAtoCvC…-ienteringA
(GAtoB+…+GAtoGround)vA-GAtoBvB-GAtoCvC…=ienteringA
2. If in doubt about the sign of the current source,go back to this basic KCL formulation
3. This formulation works for independent currentsourceFor dependent current sources (introduced later) use your
wits
MAE140 Linear Circuits46
Nodal Analysis Exercise 3-2 p. 72
Solve this using standard linear equation solversCramer’s ruleGaussian eliminationMatlab
+
_
vA vB
500Ω1KΩ
2KΩ
iS1 iS2vO
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105.2105.0
105.0105.1
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S
B
A
i
i
v
v
MAE140 Linear Circuits47
Nodal Analysis with Voltage Sources
Current through voltage source is not computablefrom voltage across it. We need some tricks!They actually help us simplify things
Method 1 – source transformation
+_
Restof thecircuitvS
RS
vA
vB
Restof thecircuit
vA
vB
RSSR
Sv!
Then use standard nodal analysis – one less node!
MAE140 Linear Circuits48
Nodal Analysis with Voltage Sources 2
Method 2 – grounding one node
Restof thecircuit
vA
vB
+_ vS
This removes the vB variable – simpler analysisBut can be done once per circuit
MAE140 Linear Circuits49
Nodal Analysis with Voltage Sources 3
Method 3Create a supernode
Act as if A and B were one nodeKCL still works for this node
Sum of currents enteringsupernode box is 0
Write KCL at all N-3 other nodesN-2 nodes less Ground node
using vA and vB as usualWrite one supernode KCLAdd the constraint vA-vB=vS
These three methods allow us to deal with all cases
Restof thecircuit
vA
vB
+_ vS
supernode
MAE140 Linear Circuits50
Nodal Analysis Example 3-4 p. 76
This is method 1 – transform the voltage sourcesApplicable since voltage sources appear in series with Rs
Now use nodal analysis with one node, A
+_ +__+
v0vS1 vS2R3
R2R1
1
1
R
Sv
2
2
R
SvR1 R2 R3
_
+
v0
vA
!
( )
321
2211
2211321
GGG
vGvGv
vGvGvGGG
SSA
SSA
++
+=
+=++
MAE140 Linear Circuits51
Nodal Analysis Example 3-5 p. 77
Rewrite in terms of vS, vB, vCThis is method 2
Solve
vS
vBvA vC
R/2
2R2R
R/2 R+_
Riniin
What is the circuit input resistance viewed through vS?
05.35.02
05.035.0
=+!!
=!+!
=
CvBGvAGv
CGvBGvAGv
SvAv
SGvCGvBGvSGvCGvBGv
25.35.0
5.05.03=+−
=−
RR
inR
R
Sv
R
CvSv
R
BvSv
ini
Sv
CvSv
Bv
872.075.11
25.10
25.10
75.11
2/2
25.10
25.6,
25.10
75.2
==
=!
+!
=
==
MAE140 Linear Circuits52
Nodal Analysis Example 3-6 p. 78
Method 3 – supernodesKCL for supernode:Or, using element equations
Now use
Other constituent relation
vBvA vC
vS1
R3R2
_
_
+
+
vS2 R4R1
reference
i4
i3i2
i1+
-vO
supernode
04321 =+++ iiii
04)(3)(21 =+!+!+ CvGBvCvGBvAvGAvG
2SvBv =
2)32()43()31( SvGGCvGGAvGG +=+++
1SvCvAv =!
Two equations in two unknowns
MAE140 Linear Circuits53
Summary of Nodal Analysis
1. Simplify the cct by combining elements in series or parallel
2. Select as reference node the one with most voltage sourcesconnected
3. Label node voltages and supernode voltages – do not labelthe nodes directly connected to the reference
4. Use KCL to write node equations. Express element currentsin terms of node voltages and ICSs
5. Write expressions relating node voltages and IVSs
6. Substitute from Step 5 into equations from Step 4Write the equations in standard form
7. Solve using Cramer, gaussian elimination or matlab
MAE140 Linear Circuits54
Solving sets of linear equations
Cramer’s Rule Thomas & Rosa Appendix B pp. A-2 to A-11
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100150250200
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MAE140 Linear Circuits55
Solving sets of linear equations (contd)
Notes:This Cramer is not as much fun as Cosmo Kramer in SeinfeldI do not know of any tricks for symmetric matrices
24114250340
210
34)3(
86
34)5(
86
2105
863
2105
345
2
=++!=
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107
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633
1075
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48.050
2422 ==
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68.150
8433 ==
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!=x
MAE140 Linear Circuits56
Solving Linear Equations – gaussian elimination
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Augmentmatrix
Rowoperationsonly
row2+row1row3+row1×3/5 row3×5
row3+row2×21/5
row3÷10 (row2+row3×5) ÷5 (row1+row2 ×2 + row3×3) ÷5
MAE140 Linear Circuits57
Solving Linear Equations - matlabA=[5 –2 –3; -5 7 –2; -3 –3 8]
A = 5 –2 –3 -5 7 –2 -3 –3 8
B=[4;-10;6]
B = 4 -10 6
inv(A)
ans = 1.0000 0.5000 0.5000 0.9200 0.6200 0.5000 0.7200 0.4200 0.5000
inv(A)*B
ans = 2.0000 0.4800 1.6800
A\B
ans = 2.0000 0.4800 1.6800
MAE140 Linear Circuits58
Mesh Current AnalysisDual of Nodal Voltage Analysis with KCL
Mesh Current Analysis with KVLMesh = loop enclosing no elements
Restricted to Planar Ccts – no crossovers (unless you are reallyclever)
Key Idea: If element K is contained in both mesh i and mesh j thenits current is ik=ii-ij where we have taken the reference directionsas appropriate
Same old tricks you already know
++ +++ + +- --
-
-
-
-
R1 R2
R3vS1 vS2v4v3
v2v1
v0
iA iB
Mesh A: v0-v1-v3=0Mesh B: v3-v2-v4=0
(R1+R3)iA-R3iB=vS1-R3iA+(R2+R3)iB=-vS2
v1=R1iA v0=vS1v2=R2iB v4=vS2v3=R3(iA-iB)
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Sv
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RRR
RRR
MAE140 Linear Circuits59
Mesh Analysis by inspectionMatrix of Resistances R
Diagonal ii elements: sum of resistances around loopOff-diagonal ij elements: - resistance shared by loops i and j
Vector of currents iAs defined by you on your mesh diagram
Voltage source vector vSSum of voltage sources assisting the current in your mesh
If this is hard to fathom, go back to the basic KVL to sortthese directions out
SvRi =
-
+
+
-vS1
vS2R1
R3R2R4
iA iB
iC
+
-v0 !
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A
v
v
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MAE140 Linear Circuits60
Mesh Equations with Current Sources
Duals of tricks for nodal analysis with voltage sources1. Source transformation to equivalentT&R Example 3-8 p. 91
+-5V
2KΩ
3KΩ
1KΩ
5KΩ
4KΩ2mA
iO
+-
2KΩ
3KΩ
1KΩ
5KΩ
iA iB
iO
+-
4KΩ
8V5V
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20006000
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MAE140 Linear Circuits61
Mesh Analysis with ICSs – method 2
Current source belongs to a single mesh
+-5V 2KΩ
3KΩ
1KΩ
5KΩ
4KΩ 2mA
iO
iA iB iC
mA2
04000110002000
520006000
!=
=!+!
=!
C
CBA
BA
i
iii
ii
Same equations!Same solution
Same example
MAE140 Linear Circuits62
Mesh Analysis with ICSs – Method 3
Supermeshes – easier than supernodesCurrent source in more than one mesh and/or not in parallel
with a resistance1. Create a supermesh by eliminating the whole branch
involved2. Resolve the individual currents last
R4
iS1iS2
R3
R2R1
vO
+
-iA iC
iB
Supermesh
Excluded branch
0)(342)(1 =!+++! AiCiRCiRBiRAiBiR
1SiAi =
2SiCiBi =!
MAE140 Linear Circuits63
Summary of Mesh Analysis
1. Check if cct is planar or transformable to planar
2. Identify meshes, mesh currents & supermeshes
3. Simplify the cct where possible by combiningelements in series or parallel
4. Write KVL for each mesh
5. Include expressions for ICSs
6. Solve for the mesh currents
MAE140 Linear Circuits64
Linearity & Superposition
Linear cct – modeled by linear elements andindependent sourcesLinear functions
Homogeneity: f(Kx)=Kf(x)Additivity: f(x+y)=f(x)+f(y)
Superposition –follows from linearity/additivityLinear cct response to multiple sources is the sum of the
responses to each source1. “Turn off” all independent sources except one and
compute cct variables2. Repeat for each independent source in turn3. Total value of all cct variables is the sum of the values
from all the individual sources
MAE140 Linear Circuits65
Superposition
Turning off sourcesVoltage source
Turned off when v=0 for all ia short circuit
Current source
Turned off when i=0 for all van open circuit
We have already used this inThévenin and Norton equiv
+_vS
i+
v
-
i
v
i+
v
-
i
v
iS
i+
v
-
i
v
i
v
i+
v
-
MAE140 Linear Circuits66
Where are we now?
Finished resistive ccts with ICS and IVSTwo analysis techniques – nodal voltage and mesh current
Preference depends on simplicity of the case at handThe aim has been to develop generalizable techniques for
access to analytical tools like matlab
Where to now?Active ccts with resistive elements – transistors, op-amps
Life starts to get interesting – design introducedCapacitance and inductance – dynamic ccts
Frequency response – s-domain analysis
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