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EASTERN MEDITERRANEAN UNIVERSITY EE 529 Circuit and Systems Analysis Lecture 6 Mustafa Kemal Uyguroğlu

EE 529 Circuit and Systems Analysis Lecture 6

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Mathematical Models of Electrical Components A. Nullator a b

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Page 1: EE 529 Circuit and Systems Analysis Lecture 6

EASTERN MEDITERRANEAN UNIVERSITY

EE 529 Circuit and Systems

AnalysisLecture 6

Mustafa Kemal Uyguroğlu

Page 2: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

A

B

O

( ) 0( ) 0v ti t

A. Nullator

a

b

( )( )v ti t

Page 3: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

A

B

0 0 ( )0 0 ( )

v ti t

0

B. Noratora

b

( )( )v ti t

Nullator and Norator are conceptual elements. They are used to represent some electrical elements in different ways.

Page 4: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

E C

B

1 2

1 2

1

2

1 1

1 1

kv t kv tES R CS

kv t kv tF ES CS

i t I e I e

i t I e I e

C. Three Terminal and two-port Circuit Elements

C.1 TRANSISTORe

b

c

1 2

Ebers-Moll Equations

Page 5: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

1 : n

C

i1 i2A B+

v 1

-

+

v2

-

C. Three Terminal and two-port Circuit Elements

C.2 IDEAL TRANSFORMER

a

c

b

1 2 1 1

2 2

10

1 0

v ini v

n

1 1

2 2

00

i vnv in

Page 6: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

1 : ni1 i2

A

B

C

D

+

v1

-

+

v2

-

C. Three Terminal and two-port Circuit Elements

C.2 IDEAL TRANSFORMER

1 1

2 2

10

1 0

v ini v

n

1 1

2 2

00

i vnv in

c

d

2

a

b

1

Page 7: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

C

i1 i2A B+

v 1

-

+

v2

-

k

C. Three Terminal and two-port Circuit Elements

C.3 IDEAL GYRATOR

a

c

b

1 2

1 1

2 2

00

v ikv ik

1 1

2 2

10

1 0

i vki v

k

Page 8: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

i1 i2A

k

B

C

D

+

v1

-

+

v2

-

C. Three Terminal and two-port Circuit Elements

C.3 IDEAL GYRATOR

1 1

2 2

00

v ikv ik

1 1

2 2

10

1 0

i vki v

k

c

d

2

a

b

1

Page 9: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

1 : ni1 i2

A

B

C

D

+

v1

-

+

v2

-

1 1

2 2

00

i vnv in

Representation of Ideal Transformer with Dependent Sourcesc

d

2

a

b

1

1 1 1

2 2 2

1 1 1

2 2 2

1 1 1 1

2 2 2 2

0 0 00 0 0

or

0 0 00 0 0

i v vnv i in

i i iv v v

i v i vnv i v in

Page 10: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

1 1 1

2 2 2

1 1 1

2 2 2

1 1 1 1

2 2 2 2

0 0 00 0 0

or

0 0 00 0 0

i v vnv i in

i i iv v vi v i vnv i v in

Representation of Ideal Transformer with Dependent Sources

A

B

C

D

i2

+

v'2

_

i'1

+

v1

_

nv1

A

B

C

D

i2

+

v''2

_

+

v1

_

i''1

ni2

i1

+

v1

-

Page 11: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

i1 i2A

B

C

D

+

v1

-

+

v2

-

nv1-ni2

1 1

2 2

00

i vnv in

Representation of Ideal Transformer with Dependent Sources

1 : ni1 i2

A

B

C

D

+

v1

-

+

v2

-

Page 12: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

i1 i2A

B

C

D

+

v1

-

+

v2

-

-ki1ki2

Representation of Ideal Gyrator with Dependent Sources

i1 i2A

k

B

C

D

+

v1

-

+

v2

-

1 1

2 2

00

v ikv ik

Page 13: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

i1 i2A

B

C

D

+

v1

-

+

v2

--1/k v2 1/k v1

Representation of Ideal Gyrator with Dependent Sources

i1 i2A

k

B

C

D

+

v1

-

+

v2

-

1 1

2 2

10

1 0

i vki v

k

Page 14: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical ComponentsOperational Amplifier

A1A2

A3

A0

1 1

2 2

3 3

0 0 00 0 0

0

i vi vv A A i

a1

1 2

a2 a3

a0

3

Page 15: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical ComponentsOperational Amplifier

A1A2

A3

A0

1 1

2 2

3 3

0 0 00 0 0

0

i vi vv A A i

a1

1 2

a2 a3

a0

3

A: open loop gain, very big!

2 1 very small!v t v t

Page 16: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical ComponentsOperational Amplifier

A1 A3

A0

1 1

3 3

0 00

i vv iA

a1

1

a3

a0

3

A: open loop gain, very big!

1 0v

Page 17: EE 529 Circuit and Systems Analysis Lecture 6

Mathematical Models of Electrical Components

A1 A3

A0

O

A1 A3A2

A0A2

A1 A3

A0

O

A1 A3

A0

• Representation of OP-AMP with Nullator and Norator.

Page 18: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components The terminal equations of resistors are

The terminal equations of multi-terminal components are similar to two-terminal components but the coefficient matrices are full.

,or

,

where , , , and are diagonal matrices

t t t t

t t t t

b b b c c c

b b b c c c

b c b c

v R i v R i

i G v i G v

R R G G

Page 19: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

or

t tt t

t tt t

b bb bc b

c cb cc c

b bb bc b

c cb cc c

v R R iv R R i

i G G vi G G v

where

vb : branch voltages

vc : Chord voltages

ib : branch currents

ic : Chord currents

Page 20: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

(A) Branch Voltages Method

1 2

2

or

where : current sources

tt

t

tt

t

cb

s

b1 s

c

s

ii A A

i

iU A A i

i

i

2 ......................(1)t

tt

bb bc b1 s

cb cc c

G G vU A A i

G G v

By using the terminal equations of the multiterminal components, the above equation can be written as

Page 21: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components On the other, the chord voltages can be

written in terms of branch voltages by using the fundamental circuit equations.

If Eq.(2) is substituted into (1) and the known quantities are collected on the right hand side then the following equation is obtained:

or

.................(2)

tt

t

t tt t

sc 1 2

b

b s

c b1 2

vv B B

v

v v0 Uv vB B

Page 22: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

21

1 1

where

t t t

bb bc bc T1 b 1 1 s sT

cb cc cc

T

T1 1

G G GUU A v U A A v A i

G G GB

B A

A B

Page 23: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

Example : In the following figure, the circuit contains a 3-terminal component. The terminal equation of the 3-terminal component is:

Using the branch voltages method, obtain the circuit equations

1 11 12 1

2 21 22 2

i g g vi g g v

3-terminal

A B

C D

Ra

Rb

Is

Vs 1 2

a b

c

1 2

a b

c

IS

VS

(Ra)(Rb)

Page 24: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

Example : In the following figure, the circuit contains a 3-terminal component. The terminal equation of the 3-terminal component is:

Using the branch voltages method, obtain the circuit equations

1 11 12 1

2 21 22 2

.........(1)i g g vi g g v

3-terminal

A B

C D

Ra

Rb

Is

Vs 1 2

a b

c

1 2

a b

c

IS

VS

(Ra)(Rb)

Page 25: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

The fundamental cut-set equations for tree branches 1 and 2:

1 2

a b

c

IS

VS

(Ra)(Rb)

1

2

........(2)a s

s b

i IiI ii

The terminal equations of the resistors:

............(3)a a a

b b b

i G vi G v

Subst. of Eqs.(3) and (1) into (2) yields:

Page 26: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

va and vb can be expressed in terms of branch voltages using fundamental circuit equations.

Subst. of Eq.(5) into (4) gives:

11 12 1

21 22 2

1.........(4)

1a a

sb b

G vg g vI

G vg g v

1 1

2 2

01 0.....(5)

0 1a

b s s

v v vv v V Vv

11 12 1 1

21 22 2 2

0 0 10 1a

s sb b

Gg g v vV I

G Gg g v v

or

11 12 1

21 22 2

0 11

a s

b b s

g G g Vvg g G G Iv

Page 27: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

Example : In the following figure, the circuit contains a 2-port gyrator and a 3-terminal voltage controlled current source. The terminal equations of these components are:

Using the branch voltages method, obtain the circuit equations

1 1 3 3

2 2 4 4

0 0 0, ...........(1)

0 0i v i vi v i vk

B

Ra

Rb

i4(t)

A

Vs

CD

a

d

a

d

bb

c

1 2

3

4

Page 28: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

B

Ra

Rb

i4(t)

A

Vs

CD

B

Ra

Rb

A

Vs

CD

2 - port

3-terminal

1 2

3

4

Vs

(Ra)

(Rb)

Page 29: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

31

3 42

........(2)a b

b

i i iii i ii

1 2

3

4

Vs

(Ra)

(Rb)

The terminal equations of the resistors:

..........(3)a a a

b b b

i G vi G v

Subst. of Eqs.(3) and (1) into (2) yields:

Page 30: EE 529 Circuit and Systems Analysis Lecture 6

Analysis of Circuits Containing Multi-terminal Components

13

2

0 0.............(4)

00a b a

b b

G G vvv

G vv k

va , vb and v3 can be expressed in terms of branch voltages using

fundamental circuit equations.

11

1 22

3 1 2

1 0 01 1 .....(5)1 1

a

b s s

s s

v vv

v v V v Vv

v v V v V

•Subst. of Eq.(5) into (4) gives:

1 1 1

2 2 2

1 1

2 2

00 1 0 01 1

00 1 1

00

a bs

b s

a b b bs

b b b

G Gv v vV

G Vv v vk

G G G Gv vV

G k G k G kv v