3. methods of the circuit analysis · !topology( equations((r equazioni) (kcl:n-1 equations)...

1

B

J

Department of Electrical, Electronic, and Information Engineering (DEI) - University of Bologna

3. Methods of the Circuit Analysis

- Topology Equations (r equazioni)

(KCL: n-1 equations)

(KTL: r-n+1 equations)

- Element Equations (r equations)The number of these equations is equal to the number of the branches as they are the equation modeling each element of the circuit, and hence any branch.

The circuit analysis problem is described by 2r equations in 2r unknowns. The equations are the topology equations and the element equations. The unknowns are the branch tensions and the branch currents.

Circuit with n nodes and r branches

åå

=

=

m r

n r

0v0i


General method

i1 + i3 + i4 - i6 = 0i2 - i3 - i4 + i5 = 0i6 - i5 = 0- v1 + v2 + v3 = 0- v1 + v2 + v4 = 0v1 - v2 + v5 + v6 = 0

v1 = V0v2 = R2 i2v3 = R3 i3v4 = R4i4 + V1v5 = R5i5v6 = R6i6

Top.Eq.s

Elem.Eq.s

Circuit: r = 6, n = 4The circuit is described by 12 linear non-homogeneous equations and 12 unknowns. Therefore it has a unique solution.

H = G=E = F_ _ _

ExampleMethods of the Circuit Analysis

••

•

•B = C_

H = G=E = F_ _ _

DA2

1

5

6

43

Graph


A B C D

H G E F

i1

i2

i6i4

i3

i5

V0

V1R2

R6

R5

R3+-

• •• •

•

B = C_

-+

-

H = G=E = F_ _ _

R4

Ø General Method of the Circuit Analysis:(r branches, 2r unknowns, 2r equations)

Ø Method of the Tension Substitution:(r branches, r unknowns, r equations)

k0,kkk

m k

n k

ViRv

0v0i

+=

=

=

åå

These equations are for a generic element with a resistors and a tension source

åå

=+

=

m k0,kk

n k

0ViR0i

(n-1) eq.s

(r-n+1) e q.s

r eq.s

(n-1) eq.s

(r-n+1) eq.s

Methods of the Circuit Analysis


Superposition Principle:As a consequence of the linearity of the equations which describe the circuit the solution the equations of the Tension Substitution Method is given by the branch currents expressed by a linear combination of the independent sources of the circuit.

ir = Gr1V01 + Gr2V02 + … + GrlV0l + αr,l+1I01 + αr,l+2I02 +…+ αr,gI0g

We must stress that this is only valid in the linear case. In order to be in this case, the element equations must be linear.

V0k and I0k are the input of the circuit, ir is an output. Usually the source voltages and source currents are the inputs of the circuit. The branch voltages and branch currents are the outputs.

The superposition principle states that a branch current is the algebraic sum of the currents through the branch due to each independent source acting alone (the same statement holds for a branch voltage also).


As two inputs contribute to the operation of this circuit , their two contributions are considered separately to calculate the outputs (branch current and voltages). Firstly the circuit with the source V0 and I0 is switched off (I0 = 0 corresponds to an open circuit branch) is analyzed. Afterwards the circuit where V0 is switched (V0 = 0 corresponds to a closed circuit branch) and the only active source is I0.

Calculation of i3: i3 = i3’ + i3”

- To calculate i3’ when I0 = 0:( )

2

BG332BG

eq

010110BG

eq

01

432

4321eq01eq

R'v

'i 'i R 'v

RV

R - V 'i R V ' v RV

- 'i

RRR

RRRR R dove 0 V 'iR

=Þ=

=+=Þ=

úû

ùêë

é+++

+==+

02

toteq

2

BG30

toteq

"4

toteqBG

Beq

Aeq

Beq

Aeqtot

eq43Beq

21

21Aeq

IRR

R"v

"i I R i R "v

RR

RRR ;RR R ;

RRRR

R

==Þ=-=

+=+=

+=

•A B C D

H G E F

i1

i2

i6i4i3

i5

V0I0

R1

R4

R3

R2+-

B

i3’

V0

R1

R4

R3

R2

i1’

G

+-

B

G

i4”i3”I0

R1

R4

R3

R2

Example

- To calculate i3’’ when V0 = 0:

• •

•

• •

•

•

••

•

q The method is based on the node voltages, uk (k= 1,2,..,n-1), that are the potential differences between each non-reference node and the reference node (ground). Hence each node voltage is the voltage of that node with respect to the reference node (to be chosen).Ø KCL is applied to each non-reference node k, k = 1,2,…, n-1 (figure above):

i1+i2+….+ih = 0 (1)Ø KTL is applied to relate the node voltages to the branch voltages (figure below):

vr = uk-uh (2)

Ø The currents are expressed by the element equations

(3)

Ø By substituting eq. 3 into eq. 1 a set of n-1 equations in n-1 unknowns (uk, for k=1,2,…, n-1) is obtained.


This method is utilized in the AC regime where Ir , Uk and Zr replaces ir ,vr , and Rr .

r

hk

r

rr R

uuRvi -

==

Nodal Analysis (n-1 equations, n-1 unknowns)

k1h

32

uh

+-vr

h k

uk0

•

•

• •

•

•

• •


Ø The n-1 node voltages are determined by the solution of a linear non-homo-geneous system of n-1 equations. As the node voltages are known, the branch currents are obtained from eq. 3 and the branch voltages are derived from eq. 2.

Nodal Analysis (n-1 equations, n-1 unknowns)

Step to determine the node voltages1. Define the reference node and assign the n-1 node voltages which are the voltage of the non-reference nodes with respect to the reference one.

2. Apply KCL to each non-reference node.3. Apply KTL to relate the node voltages to the branch voltages.

4. Express the branch currents in terms of the node voltages through the element equations and substitute them in the cur-rents equations given by KCL in step 2.

5. Solve the resulting simultaneous equations to obtain the node voltages.

k1h

32

uh

+-vr

h k

uk0

•

• •

• •

•

•

•

v2 = u1v3 = u2v4 = u2 - u1i1 = - I1i5 = I2

v2 = R2 i2

v3 = R3 i3

v4 = R4 i4

i1 = - I1i5 = I2

i1 = - I1

i2 = u1 /R2

i3 = u2 /R3

i4 = (u2 - u1) /R4i5 = I2

ïïî

ïïí

ì

=--

+

=+-

-+-

0IRuu

Ru

0IRuu

RuI

24

12

3

2

24

12

2

11

îíì

=-+=+-+

0 i i i 0 i i i i

543

5421

( ) ( )( )î

íì

=++--=-+

24323413

214222142

IRRuRRuRIIRRuRuRR

i1

i4

i3i2

i5

I1

R4

R3R2

0

u1 u2

I2

Example

2. KCL application 3. KTL application

4. Use of element eq.s

ir as function of uk

ir expressed by uk in KCL eq.s

5. Solution of KCL eq.s in ukunknown (2 q.s and 2 unknowns)

1. Reference node definition

•

••


i1 i6i4i3

i5

V0I0

R1

R4

R3

R2+-

0

u1 u2

v1 = u1v3 = u1v4 = u1v5 = u2 - u1v6 = - u2

v1 = V0 + R1 i1

v3 = R2 i3

i4 = - I0v5 = R3i5v6 = R4i6

i1 = (u1 - V0)/ R1

i3 = u1 /R2

i4 = - I0

i5 = (u2-u1)/ R3

i6 = - u2/R4

îíì

==++

0 i - i0 i - i i i

65

5431

( )( )ïî

ïíì

=-+-

+=-++

0uRRuR

VRRIRRRuRRuRRRRRR

23414

032023212211213231

1. Reference node definition

Example

2. KCL application

3. KTL application 4. Use of elem. eq.s ir as function of uk

ir expr. by uk in KCL eq.s5. Solution of KCL eq.s in ukunknown (2 q.s and 2 unknowns)

•

•

••


ïïî

ïïí

ì

=+-

=-

--++

0Ru

Ruu

0RuuI

Ru

RVu

4

2

3

12

3

120

2

1

1

01 -

In the circuit there are branches that contain only voltage sources (independent or controlled sources). For these branches the currents cannot be expressed in terms of the voltages. A branch with only a voltage source can be incorporated into a closed surface. Thereafter KCL is applied to this surface. This branch is said supernode. As it results from KTL, the difference of the node voltages at the terminals of the source branch, is given by the source voltage:

uk-uh = V0

ïî

ïíì

==+

=++

032

5432

541

Vu-u0 i-i -i i

0 i i i

• •

i4

i2

R4

R2

0

u2 u3

i3

R3

-+

V0u1i5

R5

i1

R1

By expressing the 5 currents by means of the 3 node voltages, the system of equations is given by three equations with three unknowns which are the node voltages.

Nodal AnalysisSupernode

•

•


When supernode contains the reference node the voltage of the non-reference node inside the supernode is given by the source voltage:

uk = V0ïî

ïí

ì

==+-=++

02

432

541

Vu0 i i i0 i i i

• •

i4R4

0

u2 u3

i3

R3V0

u1i5

R5

i1

R1 +-

i2R2

Nodal AnalysisSupernode

In this case also by expressing the 5 currents by means of the 3 node voltages, the system of equations is given by 3 equations with three unknowns which are the node voltages.

• •

•

•


ïïî

ïïí

ì

=-=-

=+--=++

543

21

5432

521

i9uu02uu

0 ii i i0 10-i i i

•

i5

0

u2 u3u1

3Ω

i1 i3

4Ω

i2

i4

1Ω

6 Ω-+-+

u4

10A

3 v520V

i1 = u1/2i2 = (u2-u3)/6i3 = u3/4i4 = u4/1i5 = (u1- u4)/3

ïïïï

î

ïïïï

í

ì

-=-

=-

=-

+---

=-

+-

+

3uu9uu

20uu

03uu

1u

4u

6uu

103uu

6uu

2u

4143

21

414332

41321

ïïî

ïïí

ì

=--=-

=--+=--+

0u2u3u20uu

0u8u5,2u2u60u2uu5u

431

21

4321

4321

2Ω

Example

•••

•


Tellegen’s Theoremq The Tellegen’s theorem states that in an insulated circuit (not connect to other circuits or networks) the algebraic sum of the power calculated for each branch is equal to zero.

q Alternatively it can be stated that the total power delivered by the sources is equal to the power absorbed by the loads.

0 iv 1k

kk =å=

r

Tellegen’s theorem is a consequence of the energy conservation principle. It fulfills the topology equations (KCL and KTL).



•

• • •AB

C

D

1

2 3

4 5 6

The following quantities are given.i1 = 1; i2 = 2; i3 = 3v4 = 4; v5 = 5; v6 = 6

From KCL: i4 = - i1 - i2 = - 3i5 = i2 - i3 = - 1i6 = i3 + i1 = 4

From KTL: v1 = v4 - v6 = - 2v2 = v4 - v5 = - 1v3 = v5 - v6 = - 1

The quantities given by the problem and those obtained by KCL and KTL verify Tellegen’s theorem.

i1v1 + i2v2 + i3v3 + i4v4 + i5v5 + i6v6 = 0

Example

•

• • •


In Out

=Out F In

Methods of the Circuit AnalysisTransfer FunctionIn a circuit we will distinguish between input and output. The inputs are the independent current and voltage sources, also said excitations. The output are the branch currents and the tensions (branch voltages, node voltages or any potential difference between two nodes).

In a linear, time independent circuit for an input-output pair a transfer function (or network function) is defined. The transfer function is the ratio between an output and an input when the other excitations, except the one considered, are switched off.

Linear, time-independent

network


The transfer function can be defined in the time domain [voltages and currents: v(t) and i(t)], in the frequency domain [voltages and currents: �̇� and �̇� ] or in the Laplace transform-domain.

Methods of the Circuit AnalysisTransfer FunctionIn a linear, time independent circuit for an input-output pair the transfer function does not depend on the values assumed by the input and the output. Therefore the transfer function does not varies when those quantities are varying. This property is a consequence of the homogeneity property of the transfer functions of a linear circuit. For this property if the input is multiplied for a constant, the output results to be multiplied for the same constant:

Example - linear resistor: v = Ri → kv = R ki (k real constant)

In a linear, time independent circuit the additivity property is also fulfilled. For this property the output corresponding to the sum of two inputs is equal to the sum of the outputs corresponding to the separately applied inputs:

Example - linear resistor: v1 = Ri1 ; v2 = Ri2

v = Ri = R(i1 + i2) = Ri1 + Ri2 = v1 + v2


Following from the homogeneity and the additivity properties, for the superposition principle in a linear time independent circuit any voltage vr and any current is can be expressed as a linear combination of the p independent tension sources and the qindependent current sources :

vr = αr1 V01 + αr2 V02 + …. + αrp V0p + Rr1 I01 + Rr2 I02 + …. + Rrq I0q

is = Gs1 V01 + Gs2 V02 + …. + Gsp V0p + βs1 I01 + βs2 I02 + …. + βsq I0q

The coefficients αri, Rrj, Gsi, βsj are the transfer functions of the r voltages and the stensions when coupled two by two to the p tension sources and the q current sources. The transfer functions αri and βsj are dimensionless. The Rrj have the dimension of a resistance or an impedance (Ω). The Gsi have the dimension of a conductance or an admittance (S = 1/Ω).

Voltage Gain: αrp = vr

V0p V0i = 0 per i≠pI0j = 0 ∀ j

Transf. Imped.: Rrq = vr

I0q V0i = 0 ∀ iI0j = 0 per j ≠ q

Transf. Admitt.: Gsp = is

V0p V0i = 0 per i≠pI0j = 0 ∀ j

Current Gain: βsq = isI0q V0i = 0 ∀ i

I0j = 0 per j ≠ q

Transfer Function


Equivalent CircuitsA circuit N+N1 is constituted by two parts, the circuit N and the circuit N1connected through a port only. The circuit N is linear, time-independent, with stationary sources. The circuit N is connected to the circuit N1 through the port AB. For a given value the voltage v of the port AB the current i flowing from A to B through N is determined by the circuit N itself. The relation between v and i, that is the element equation of the circuit N, can be determined.

The one port circuit N, characterized by i and v, can be considered a two terminal element. This can be done even if the circuit N is constituted by many circuit elements. The relation between i and v only depends on N. The determination of this relation is the main problem of the equivalent circuit determination.


Circuit N1

•

A

B

v

iCircuit NLinear, time-independentnetwork with stationary sources

•


Equivalent Circuits

The problem of the equivalent circuit determination is to find an elementary circuit, the operation of which simulates the operation of the circuit N. For this elementary circuit, that is the equivalent circuit, at a given voltage v the same current i flows through it as it would flows through the circuit N when the voltage v between the terminals A and B would be the same.


•

A

B

v

iCircuit NLinear, time-independentnetwork with stationary sources

•

This problem is solved by the Thévenin’s theorem and by the Norton theorem.

The equivalent Thévenin circuit is constituted by the series of an independent voltage source with a resistor.

The equivalent Norton circuit is constituted by the parallel of an independent current source with a resistor.


Consider an independent current source connected to the port AB of circuit N. Due to the linearity of N the voltage v is given by the linear combination of the pvoltage sources V0i, the q current sources I0j, which are inside N, and the current source i. The linear combination coefficients are the transfer functions:

j 0 Ii 0 Veq0 ieq

q

1j0jj

p

1i0iieqeqeq

q

1j0jj

p

1i0iieq

oj0i

iv R v V

I R V V V i R v

I R V i R v

"="==

==

==

==

+=+=Þ

++=

åå

åå

;or:

where a

a

Equivalent CircuitsThévenin’s Theorem

•A

B

v i

Circuit N

•


Therefore the following relation is obtained:

This relation is the element equation that describes the series between an independent voltage source and a resistor. It defines the current controlled Thévenin equivalent circuit.

Thévenin’s theorem:q A linear, time-independent circuit N with a single port in evidence is considered. The circuit is equivalent to an independent voltage source in series with a resistor. The voltage of the source is the open circuit voltage between A and B. The resistor is the equivalent resistor seen from the port AB when all independent sources of N are switched off.

Thévenin’s Theorem

eqeq V i R v +=

•

• B

A

v

i

+-

Thévenin’s equivalentcircuit

•

•

A

B

v

Circuit N


j 0 Ii 0 Veq0 veq

q

1j0jj

p

1i0iieqeqeq

q

1j0jj

p

1i0iieq

oj0i

vi G i I

I V G I I vG i

I V G vG i

"="==

==

==

==

+=+=Þ

++=

åå

åå

;or

where b

b

Norton’s TheoremConsider an independent voltage source connected to the port AB of circuit N. Due to the linearity of N the current i is given by the linear combination of the pvoltage sources V0i, the q current sources I0j, which are inside N, the by the voltage source v. The linear combination coefficients are the transfer functions:

•A

B

v

i

+-

Circuit N

Equivalent Circuits

•


eqeq I vG i +=

Norton’s TheoremTherefore the following relation is obtained:

This relation is the element equation that describes the parallel between an independent current source and a resistor. It defines the current controlled Norton equivalent circuit.

Norton’s theorem.q A linear, time-independent circuit N with a single port in evidence is considered. The circuit N is equivalent to a circuit element constituted by the parallel between an independent current source and a resistor. The current of the source is the current flowing through N when the port AB is short circuited. The resistor is the equivalent resistor seen from the port AB when all independent sources of N are switched off.

Norton’s equivalentcircuit

eq eqi G v I= +

•

• B

A

v

i

Geq = 1/Req

•

•

A

B

i

v

Circuit N

When dependent sources are present in the circuit, in order to calculate Req of the Thévenin or the Norton circuit (figure above, circuit where port AB is consi-dered), the independent sources are switched off and the dependent sources are left on. Req is expressed by the relation:

Equivalent CircuitsDependent Sources in Thévenin’s and Norton’s Theorems

where V0 is a voltage source connected to the port considered for the evaluation of the equivalent circuit. It is used to calculate i0 to determine Req(figure below).

0

0eq i

V R =

The method described is used to determine Req (for acircuit where dependent sources are present (above figure where the port AB is considered) . Req for the equivalent circuit is Req = V0 /i0 .

• •

I 4Ω

2vx

+-

•

•

2Ω 2Ω

6Ωvx

A

B

• •i3

4Ω

2vx

+-

•

•

i02Ω 2Ω

6Ωvx V0+-

A

B

When in a circuit no dependent sources are present and the sources are all independent Req can be calculated from the serie and parallel relations.

Summary

(2) I vG i :circuit equivalent sNorton'

(1) V i R v:circuit equivalent sThévenin'

eqeq

eqeq

+=

+=

When eq. 1 is divided by Req, eq. 2 is obtained. When eq. 2 is divided by Geqeq. 1 is obtained. If Thévenin’s circuit is known, Norton’s circuit specifications can be derived. Alternatively when Norton’s circuit is known, Thévenin’s circuit specifications can be derived.

0G ( GI

V G1 R

0R ( RV

I R1 G

eqeq

eqeq

eqeq

eqeq

eqeq

eqeq

)ifnin n to Thévefrom Norto;

)ifton nin to Norfrom Théve;

¹-==

¹-==

Equivalent Circuits


Maximum Power Transfer

•

•B

A

RL

i

+-

RL=Req

P (RL)

RL

P Max

The Thévenin equivalent circuit can be used in finding the maximum power which a linear circuit can deliver to a load.

The entire circuit is replaced by the Théveninequivalent except for the load which is an adjustable load resistor RL. The power delivered to the load is

For a given circuit, Veq and Req are fixed. By varying RL, the power delivered to the load varies. The power is small or large for small or large values of RL. The maximum power transfer theorem states that:

q Maximum power is transferred to the load when the load resistance equal the Théveninequivalent resistance .

2

eq2L L

eq L

Vp R i R

R Ræ ö

= = ç ÷ç ÷+è ø


( ) ( )( )

( )

eq

2eq

MaxeqL

LLeq

3Leq

LLeq2eq

4Leq

LeqL2

Leq2eq

L

L

R 4V

p R R

0 R2RR

0 RR

R2RRV

RR

RRR2RRV

dRdp

0 dRdp

=Þ=

=-+

Þ=úúû

ù

êêë

é

+

-+=

=úúû

ù

êêë

é

+

+-+=

Þ=

Hence

Maximum Power Transfer

•

•B

A

RL

i

+-

Req

P (RL)

RL

P Max

To prove the maximum power transfer theorem the following has to be done:


29

adjustable load resistance

resistenza di carico variabile

branch current corrente di ramo

branch tension tensione di ramo

element equation equazione descrittiva dell’elemento circuitale

equivalent circuit circuito equivalente

general method of the circuit analysis

metodo generale di analisi circuitale

input ingresso

load resistance resistenza di carico

maximum power transfer theorem

teorema di massimo trasferimento di potenza

method of the tensionsubstitution

metodo di sostituzione delle tensioni

network function funzione di rete

nodal analysis analisi nodale

node voltage potenziale di nodo

Norton’s theorem teorema di Norton

output uscita

reference node nodo di riferimento

resonance frequency frequenza di risonanza

supernode supernodo

superpositionprinciple

principio di sovrapposizione degli effetti

Tellegen’s theorem teorema di Tellegen

Thévenin’sequivalent circuit

circuito equivalente di Thévenin

Terminology


30

Terminology

Thévenin’s theorem teorema di Thévenin

topology equations equazioni topologiche

transfer admittance ammettenza di trasferimento

transfer function funzione di trasferimento

transfer impedance impedemza di trasferimento

unknown incognita

voltage gain guadagno di tensione


3. methods of the circuit analysis · !topology( equations((r equazioni) (kcl:n-1 equations)...

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