Systematic Circuit Analysis (T&R Chap...

45 MAE140 Linear Circuits 45

Systematic Circuit Analysis (T&R Chap 3)

Node-voltage analysis Using the voltages of the each node relative to a ground

node, write down a set of consistent linear equations for these voltages

Solve this set of equations using, say, Cramer’s Rule

Mesh current analysis Using the loop currents in the circuit, write down a set of

consistent linear equations for these variables. Solve.

This introduces us to procedures for systematically describing circuit variables and solving for them


Nodal Analysis

Node voltages Pick one node as the ground node Label all other nodes and assign voltages vA, vB, …, vN

and currents with each branch i1, …, iM

Recognize that the voltage across a branch is the difference between the end node voltages Thus v3=vB-vC with the direction as indicated

Write down the KCL relations at each node Write down the branch i-v relations to express branch

currents in terms of node voltages Accommodate current sources Obtain a set of linear equations for the node voltages

v5

C

B

A

vC

vB

vA

v4

v3

v2

v1 +

+

+

++

-

-

-

- -


Nodal Analysis – Ex 3-1 (T&R, 5th ed, p.72)

Apply KCL

Write the element/branch eqns

Substitute to get node voltage equations

Solve for vA, vB, vC then i0, i1, i2, i3, i4, i5

vC vB

vA

R4

R2 R1

R3

iS2

iS1 i4 i3

i2

i1 i0

Reference node

i5


Nodal Analysis – Ex 3-1 (T&R, 5th ed, p.72)

Apply KCL Node A: i0-i1-i2=0 Node B:i1-i3+i5=0 Node C: i2-i4-i5=0

Write the element/branch eqns i0=iS1 i3=G3vB

i1=G1(vA-vB) i4=G4vC

i2=G2(vA-vC) i5=iS2

Substitute to get node voltage equations Node A: (G1+G2)vA-G1vB-G2vC=iS1 Node B: -G1vA+(G1+G3)vB=is2

Node C: -G2vA+(G2+G4)vC=-iS2

Solve for vA, vB, vC then i0, i1, i2, i3, i4, i5

vC vB

vA

R4

R2 R1

R3

iS2

iS1 i4 i3

i2

i1 i0

Reference node

i5

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−

=

+−

+−

−−+

221

4202

03112121

SiSiSi

CvBvAv

GGGGGG

GGGG

MAE140 Linear Circuits


Systematic Nodal Analysis

Writing node equations by inspection Note that the matrix equation looks

just like Gv=i for matrix G and vector v and i G is symmetric (and non-negative definite) Diagonal (i,i) elements: sum of all conductances connected

to node i Off-diagonal (i,j) elements: -conductance between nodes i

and j Right-hand side: current sources entering node i There is no equation for the ground node – the column sums

give the conductance to ground

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−

=

+−

+−

−−+

221

4202

03112121

SiSiSi

CvBvAv

GGGGGG

GGGG

vC vB

vA

R4

R2 R1

R3

iS2

iS1 i4 i3

i2

i1 i0

Reference node

i5


Nodal Analysis Ex. 3-2 (T&R, 5th ed, p.74)

Node A: Conductances

Source currents entering

Node B: Conductances


Node C: Conductances


iS

vB vA vC

R/2

2R 2R

R/2 R


Nodal Analysis Ex. 3-2 (T&R, 5th ed, p.74)

Node A: Conductances

G/2B+2GC=2.5G Source 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

vB vA vC

R/2

2R 2R

R/2 R

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−−

−−

−−

00

5.35.025.035.025.05.2 S

C

B

A i

vvv

GGGGGGGGG



Nodal Analysis – some points to watch

1.  The formulation given is based on KCL with the sum of currents leaving the node 0=itotal=GAtoB(vA-vB) +GAtoC(vA-vC)+…+ GAtoGroundvA+ileavingA

This yields 0=(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 current sources For dependent current sources (introduced later) use your

wits


Nodal Analysis Ex. 3-2 (T&R, 5th ed, p. 72)

+

_

vA vB

500Ω 1KΩ

2KΩ

iS1 iS2 vO



Solve this using standard linear equation solvers Cramer’s rule Gaussian elimination Matlab

+

_

vA vB

500Ω 1KΩ

2KΩ

iS1 iS2 vO

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××−

×−×−−

−−

2

133

33

105.2105.0105.0105.1

S

S

B

Aii

vv



Nodal Analysis with Voltage Sources

Current through voltage source is not computable from voltage across it. We need some tricks! They actually help us simplify things

Method 1 – source transformation

+ _

Rest of the circuit vS

RS

vA

vB

Rest of the circuit

vA

vB

RS SRSv≡

Then use standard nodal analysis – one less node!


Nodal Analysis with Voltage Sources 2

Method 2 – grounding one node

Rest of the circuit

vA

vB

+ _ vS

This removes the vB variable (plus we know vA=vS) – simpler analysis But can be done once per circuit


Nodal Analysis with Voltage Sources 3

Method 3 Create a supernode

Act as if A and B were one node KCL still works for this node

Sum of currents entering supernode box is 0

Write KCL at all N-3 other nodes (N-2 nodes less Ground node)

using vA and vB as usual +Write one supernode KCL +Add the constraint vA-vB=vS

These three methods allow us to deal with all cases

Rest of the circuit

vA

vB

+ _ vS

supernode



This is method 1 – transform the voltage sources Applicable since voltage sources appear in series with Resist

Now use nodal analysis with one node, A

+ _ + _ _ + v0 vS1 vS2 R3

R2 R1

11RSv

22

RSvR1 R2 R3

_

+

v0

vA

≡



This is method 1 – transform the voltage sources Applicable since voltage sources appear in series with Resist

Now use nodal analysis with one node, A

+ _ + _ _ + v0 vS1 vS2 R3

R2 R1

11RSv

22

RSvR1 R2 R3

_

+

v0

vA

≡

( )

321

2211

2211321

GGGvGvGv

vGvGvGGG

SSA

SSA

+++

=

+=++




vS

vB vA vC

R/2

2R 2R

R/2 R + _

Rin iin

What is the circuit input resistance viewed through vS?



Rewrite in terms of vS, vB, vC This is method 2

Solve

vS

vB vA vC

R/2

2R 2R

R/2 R + _

Rin iin

What is the circuit input resistance viewed through vS?

05.35.02

05.035.0

=+−−

=−+−

=

CvBGvAGvCGvBGvAGv

SvAv

S Gv C Gv B Gv S Gv C Gv B Gv

2 5 . 3 5 . 0 5 . 0 5 . 0 3

= + - = -

RRinR

RSv

RCvSv

RBvSv

ini

SvCv

SvBv

872.075.1125.10

25.1075.11

2/2

25.1025.6

,25.1075.2

==

=−

+−

=

==




Method 3 – supernodes

vB vA vC

vS1

R3 R2

_

_

+

+

vS2 R4 R1

reference

i4

i3 i2

i1 +

- vO

supernode



Method 3 – supernodes KCL for supernode: Or, using element equations Now use Other constituent relation

vB vA vC

vS1

R3 R2

_

_

+

+

vS2 R4 R1

reference

i4

i3 i2

i1 +

- vO

supernode

04321 =+++ iiii

04)(3)(21 =+−+−+ CvGBvCvGBvAvGAvG

2SvBv =

2)32()43()31( SvGGCvGGAvGG +=+++

1SvCvAv =−

Two equations in two unknowns MAE140 Linear Circuits


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 sources connected

3.  Label node voltages and supernode voltages – do not label

the reference node

4.  Use KCL to write node equations. Express element currents in terms of node voltages and ICSs

5.  Write expressions relating node voltages and IVSs

6.  Substitute from Step 5 into equations from Step 4

Write the equations in standard form

7.  Solve using Cramer, Gaussian elimination or matlab


Solving sets of linear equations

Cramer’s Rule Thomas & Rosa Appendix B pp. A-2 to A-11

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−−

−−

−−

6104

833275325

3

2

1

xxx

( ) ( ) ( )

5075125250

)3(7)2()2(3)3()3(8)2(5)2()3(875

2732

)3(8332

)5(8327

5833275325

=−−=

−×−−×−−−×−−×−+−×−−×=

−

−−−+

−

−−−−

−

−=

−−

−−

−−

=Δ

( ) ( ) ( )

100150250200

)3(7)2()2(6)3()3(8)2(10)2()3(874

2732

68332

)10(8327

48362710324

1

=+−=

−×−−×−+−×−−×−+−×−−×=

−

−−+

−

−−−−

−

−=

−

−−

−−

=Δ2

501001

1 ==Δ

Δ=x


Solving sets of linear equations (contd)

Notes: This Cramer is not as much fun as Cosmo Kramer in Seinfeld I do not know of any tricks for symmetric matrices

24114250340

21034

)3(8634

)5(86210

58632105345

2

=++−=

−−

−−+

−−−

−−=

−

−−−

−

=Δ

8424060

10742

)3(6342

)5(63107

56331075425

3

=+−=

−

−−+

−

−−−

−

−=

−−

−−

−

=Δ

48.050242

2 ==Δ

Δ=x

68.150843

3 ==Δ

Δ=x


Solving Linear Equations: Gaussian elimination

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−−

54264

531

5210

550325

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508450

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100010325

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508450

2450

100

100010001

Augment matrix Row

operations only

row2+row1 row3+row1×3/5 row3×5 row3+row2×21/5

row3÷10 (row2+row3×5) ÷5 (row1+row2 ×2 + row3×3) ÷5


Solving Linear Equations - matlab A=[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


Mesh Current Analysis Dual of Nodal Voltage Analysis with KCL

Mesh Current Analysis with KVL Mesh = loop enclosing no elements

Restricted to Planar Ccts – no crossovers (unless you are really clever)

Key Idea: If element K is contained in both mesh i and mesh j then its current is ik=ii-ij where we have taken the reference directions as appropriate

Same old tricks you already know

+ + + + + + + - - -

-

-

-

-

R1 R2

R3 vS1 vS2 v4 v3 v2 v1

v0 iA iB

Mesh A: -v0+v1+v3=0 Mesh B: -v3+v2+v4=0

(R1+R3)iA-R3iB=vS1 -R3iA+(R2+R3)iB=-vS2

v1=R1iA v0=vS1 v2=R2iB v4=vS2 v3=R3(iA-iB)

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−=

+−

−+

21

323331

SvSv

BiAi

RRRRRR


Mesh Analysis by inspection Matrix of Resistances R

Diagonal ii elements: sum of resistances around loop Off-diagonal ij elements: - resistance shared by loops i and j

Vector of currents i As defined by you on your mesh diagram

Voltage source vector vS Sum of voltage sources assisting the current in your mesh

If this is hard to fathom, go back to the basic KVL to sort these directions out

S v Ri =

-

+

+

- vS1

vS2 R1

R3 R2

R4

iA iB

iC

+

- v0


Mesh Analysis by inspection Matrix of Resistances R

Diagonal ii elements: sum of resistances around loop Off-diagonal ij elements: - resistance shared by loops i and j

Vector of currents i As defined by you on your mesh diagram

Voltage source vector vS Sum of voltage sources assisting the current in your mesh

If this is hard to fathom, go back to the basic KVL to sort these directions out

S v Ri =

-

+

+

- vS1

vS2 R1

R3 R2

R4

iA iB

iC

+

- v0 !

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+−−

−+

−+

1

2

2

3232

343

2210

0

S

S

S

C

B

A

vvv

iii

RRRRRRRRRR



Mesh Equations with Current Sources

Duals of tricks for nodal analysis with voltage sources 1. Source transformation to equivalent T&R, 5th ed, Example 3-8 p. 91

+ - 5V

2KΩ

3KΩ

1KΩ

5KΩ

4KΩ 2mA

iO + -

2KΩ

3KΩ

1KΩ

5KΩ

iA iB

iO + -

4KΩ

8V 5V

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85

11000200020006000

B

Aii iA=0.6290 mA

iB=-0.6129 mA iO=iA-iB=1.2419 mA


Mesh Analysis with ICSs – method 2

Current source belongs to a single mesh

+ - 5V 2KΩ

3KΩ

1KΩ

5KΩ

4KΩ 2mA

iO

iA iB iC

mA204000110002000520006000

−=

=−+−

=−

C

CBA

BA

iiiiii

Same equations! Same solution

Same example


Mesh Analysis with ICSs – Method 3

Supermeshes – easier than supernodes Current source in more than one mesh and/or not in parallel

with a resistance 1.  Create a supermesh by eliminating the whole branch

involved 2.  Resolve the individual currents last

R4

iS1 iS2

R3

R2 R1

vO

+

- iA iC

iB

Supermesh

Excluded branch

0)(342)(1 =−+++− AiCiRCiRBiRAiBiR

1SiAi =

2SiCiBi =−


Exercise from old midterm

€

i2− i

3= 3

10i3+ 20i

3+ 20(i

2− i

1) = 0

Set up mesh analysis equations Do not use any source transformation!

Supermesh

€

20i1+ 20(i

1− i

2) = 20

Remaining mesh



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 combining elements in series or parallel

4.  Write KVL for each mesh

5.  Include expressions for ICSs

6.  Solve for the mesh currents


Linearity & Superposition

Linear cct – modeled by linear elements and independent sources Linear functions

Homogeneity: f(Kx)=Kf(x) Additivity: f(x+y)=f(x)+f(y)

Superposition –follows from linearity/additivity Linear cct response to multiple sources is the sum of the

responses to each source 1.  “Turn off” all independent sources except one and

compute cct variables 2.  Repeat for each independent source in turn 3.  Total value of all cct variables is the sum of the values

from all the individual sources


Superposition

Turning off sources Voltage source

Turned off when v=0 for all i a short circuit

Current source

Turned off when i=0 for all v an open circuit

We have already used this in Thévenin and Norton equiv

+ _ vS

i +

v

-

i

v

i +

v

-

i

v

iS

i +

v

-

i

v

i

v

i +

v

-


Superposition

Find using superposition

€

v0

€

v01

=R

2

R1+ R

2

vS

€

v02

=R

1R

2

R1+ R

2

iS

€

v0=

R2

R1+ R

2

vS+

R1R

2

R1+ R

2

iS



Where are we now?

Finished resistive ccts with ICS and IVS Two analysis techniques – nodal voltage and mesh current

Preference depends on simplicity of the case at hand The aim has been to develop general techniques for access

to tools like matlab Where to now?

Active ccts with resistive elements – transistors, op-amps Life starts to get interesting – getting closer to design

Capacitance and inductance – dynamic ccts

Frequency response – s-domain analysis Filters

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