Methods of the Circuit Analysis - die.ing.unibo.it · 3 As two inputs contribute to the operation...

1

! Topology(Equations( (r equations)

(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

!!

=

=

m r

n r

0v0i

Methods of the Circuit Analysis

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

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 i2

v3 = R3 i3

v4 = R4i4 + V1

v5 = R5i5

v6 = R6i6

Top.Eq.s

Elem.Eq.s

Circuit:22r = 6, n = 4The$circuit$is$described$by$12$linear$non4homogeneous$equations$and$12$unknowns.$Therefore$it$has$a$unique$solution.$$$

H � G�E � F_ _ _

Example


••

•

•B � C_

H � G�E � F_ _ _

DA2

1

5

6

43

Graph

Department)of)Electrical,)Electronic,)and)Information)Engineering)(DEI))6 University)of)Bologna

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

2

! General'Method'of'the'Circuit'Analysis:(r'branches,*2r'unknowns,*2r'equations)

! Method'of'the'Tension'Substitution:(r'branches,*r'unknowns,*r'equations)

k0,kkk

m k

n k

ViRv

0v0i

+=

=

=

!!

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

!!

=+

=

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


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

Superposition+Principle:As#a#consequence#of#the#linearity#of#the#equations#which#describe#the#circuit#the#solution#of#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#any#branch#current#is#the#algebraic#sum#of#the#currents#through#the#branch#due#to#each#independent#source#acting#alone#(the#same#statement#holds#for#the#branch#voltages#also).#


3

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

=!=

=+=!=

"#

$%&

'+++

+==+

02

toteq

2

BG30

toteq

"4

toteqBG

Beq

Aeq

Beq

Aeqtot

eq43Beq

21

21Aeq

IR

R

R"v

"i I R i R "v

RR

RRR ;RR R ;

RRRR

R

==!="=

+=+=

+=

•A B C D

H G E F

i1

i2

i6i4i3

i5

V0 I0

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:

• •

•

• •

•

•

••

•

! The$method$is$based$on$the$node%voltages,$uk (k=$1,2,..,n61),$that$are$the$potential$differences$between$each$non6reference$node$and$the$reference$node$(ground).$Hence$each$node$voltage$is$the$voltage$of$that$node$with$respect$to$the$reference$node." KCL$$is$applied$to$each$non6reference$node$k,$k$=$1,2,…,$n61$(figure$above):

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

vr = uk-uh (2)

" The$currents$are$expressed$by$the$element$equations

(3)

" By$substituting$eq.$3$into$eq.$1$a$set$of$n61$equations$in$n61$unknowns$(uk,$for$k=1,2,…,$n61)$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

+6vr

h k

uk0

•

•

• •

•

•

• •

4


! 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/n71/node/voltages/which/are/the/voltage/of/the/non7reference/nodes/with/respect/to/the/reference/one.

2.///Apply/KCL/to/each/non7reference/node.3.///Apply/KTL/to/refer/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/cur7rents/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 = u2

v4 = u2 - u1

i1 = - 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

!!"

!!#

$

=%%

+

=+%

%+%

0IRuu

Ru

0IRuu

RuI

24

12

3

2

24

12

2

11

!"#

=$+=+$+

0 i i i 0 i i i i

543

5421

( ) ( )( )!

"#

=++$$=$+

24323413

214222142

IRRuRRuRIIRRuRuRR

i1i4

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

•

••


5

i1 i6i4i3

i5

V0 I0

R1

R4

R3

R2+"

0

u1 u2

v1 = u1v3 = u1v4 = u1

v5 = u2 - u1

v6 = - u2

v1 = V0 + R1 i1

v3 = R2 i3

i4 = - I0

v5 = R3i5v6 = R4i6

i1 = (u1 - V0)/ R1

i3 = u1 /R2

i4 = - I0

i5 = (u2-u1)/ R3

i6 = - u2/R4

!"#

==++

0 i - i0 i - i i i

65

5431

( )( )!"

!#$

=%+%

+=%++

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)

•

•

••


!!"

!!#

$

=+%

=%

%%++

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

!"

!#$

==+

=++

032

5432

541

Vu-u0 i-i -i i

0 i i i

• •

i4

i2

R4

R2

0

u2 u3

i3

R3

-+

V0u1 i5 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

•

•••

Department-of-Electrical,-Electronic,-and-Information-Engineering-(DEI)-8 University-of-Bologna

6

When%%supernode%contains%the%reference%node%the%voltage%of%the%non4reference%node%inside%the%supernode%is%given%by%the%source%voltage:

uk = V0!"

!#

$

==+%=++

02

432

541

Vu0 i i i0 i i i

• •

i4 R4

0

u2 u3

i3

R3V0

u1 i5 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.

• •

•

•


!!"

!!#

$

=%=%

=+%%=++

543

21

5432

521

i9uu02uu

0 ii i i0 10-i i i

•

i5

0

u2 u3u1

3Ω

i1 i34Ω

i2

i41Ω

6 Ω-+-+

u4

10!

3 v520V

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

!!!!

"

!!!!

#

$

%=%

=%

=%

+%%%

=%

+%

+

3uu9uu

20uu

03uu

1u

4u

6uu

103uu

6uu

2u

4143

21

414332

41321

!!"

!!#

$

=%%=%

=%%+=%%+

0u2u3u20uu

0u8u5,2u2u60u2uu5u

431

21

4321

4321

2Ω

Example

•• ••

•


7

Tellegen’s Theorem! 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.

! 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 =!=

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

•

• • •


8

In Out

=Out$$F$ $In


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: V and I ] or in the Laplace transform-domain.

Transfer 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#sources,#except#the#one#considered,#are#switched#off.##

Linear,,time.independent

network

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

As#it#had#been#stated#by#the#superposition#principle,#in#a#linear#circuit#any#voltage#vrand#any#current#is can#be#expressed#as#a#linear#combination#of#the#p independent#tension#sources#and#the#q independent#current#sources#:

vr =# αr1 V01 +#αr2 V02 +#….#+#αrp V0p +#Rr1 I01 +#Rr2 I02 +#….#+#Rrq I0qis =#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


9

A"circuit"N+N1 is"constituted"by"the"circuit"N"and"the"circuit"N1 connected"through"a"port."The"circuit"N"is"linear,"time;independent"(time"independent"properties"of"itacircuit"elements),"with"stationary"sources."The"circuit"N"is"connected"to"the"circuit"N1 through"the"port"AB.""For"a"given"value"of"the"voltage"v of"port"AB"the"current"iflowing"from"A"to"B"through"N"is"determined"by"the"characteristics"of"the"circuit"N."The"v;i relation,"as"an"uniqu element"equation"of"circuit"N,"can"be"determined.

Hence"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"i-v relation"only"depends"on"N."The"determination"of"this"relation"is"the"equivalent*circuit*determination"problem.


Circuit'N1

•

A

B

v

iCircuit'NLinear,'time0independentnetwork'with'stationary'sources

•

Department*of*Electrical,*Electronic,*and*Information*Engineering*(DEI)*9 University*of*Bologna

Equivalent Circuits

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.

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

10

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)p voltage)sources)V0i (i=1,..,p),)the)q current)sources)I0j (j=1,2,…,q))of 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

!=!==

==

==

==

+=+="

++=

##

##

;or:

where $

$

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:

! A#linear,#time-independent#circuit#N#with#an#highlighted#port#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 +=

Req

Veq•

• B

A

v

i

+-

Thévenin’s equivalentcircuit

•

•

A

B

v

Circuit'N

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

11

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

!=!==

==

==

==

+=+="

++=

##

##

;or

where $

$

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 (i=1,..,p),)the)qcurrent)sources)I0j (j=1,2,…,q))of N,)and)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.! A#linear,#time-independent#circuit#N#with#a#highlighted#port#is#considered.#The#circuit#N#is#equivalent#to#the#parallel#between#an#independent#current#source#and#a#resistor.#The#current#of#the#source#is#that##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#off.#

Norton’s equivalent circuit

eq eqi G v I= +

GeqIeq

•

• B

A

v

i

Geq = 1/Req

•

•

A

B

i

v

Circuit'N

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

12

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 a circuit 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%Veq of%Thévenin’s circuit%is%known,%Ieq of%Norton’s%circuit%can%be%derived%by%subtracting%eq.%1%from%eq.%2%multiplied%by%Req.%Alternatively%from%Ieq of%Norton’s%circuit,%Veq of%Thévenin’s circuit%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;

!"==

!"==

Equivalent Circuits


13

Maximum Power Transfer

Req

Veq

•

•B

A

RL

i

+-

RL=Req

P (RL)

RL

PMax

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:

! Maximum$power$is$transferred$to$the$load$when$the$load$resistance$is$equal$to$the$Thévenin equivalent$resistance$.

2

eq2L L

eq L

Vp R i R

R R! "

= = # $# $+% &

•

•

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

( ) ( )( )

( )

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

=!=

="+

!=##$

%

&&'

(

+

"+=

=##$

%

&&'

(

+

+"+=

!=

Hence

Maximum Power Transfer

Req

Veq

•

•B

A

RL

i

+-

Req

P (RL)

RL

PMax

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

•


14

Department)of)Electrical)Engineering)– University)of)Bologna27

adjustable*load*resistance

��

branch*current ��

branch*tension ��

element equation ��

equivalent*circuit ��

general*method*of*the*circuit*analysis

��(��)

input ��

load*resistance ��

maximum*power*transfer*theorem

��

method*of*the tensionsubstitution

��

network*function ��

nodal*analysis ��

node*voltage ��

Norton’s*theorem ��

output ��

reference*node ��

resonance*frequency ��

supernode ��

superpositionprinciple

��

Tellegen’s theorem ��

Thévenin’sequivalent*circuit

��

Terminology English – Chinese

Department)of)Electrical)Engineering)– University)of)Bologna28

Thévenin’s theorem ��

topology2equations ��

transfer2admittance ��

transfer2function ��

transfer2impedance ��

unknown ��

voltage2gain ��

Terminology English – Chinese

15

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 di0Thévenin

topology0equations equazioni0topologiche

transfer0admittance ammettenza0di0trasferimento

transfer0function funzione0di0trasferimento

transfer0impedance impedemza di0trasferimento0

unknown incognita

voltage0gain guadagno0di0tensione


Methods of the Circuit Analysis - die.ing.unibo.it · 3 As two inputs contribute to the operation...

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Transcript of Methods of the Circuit Analysis - die.ing.unibo.it · 3 As two inputs contribute to the operation...