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TWO PORT TWO PORT NETWORKSNETWORKS

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SUB - TOPICSSUB - TOPICSZ – PARAMETERY – PARAMETERT (ABCD) – PARAMETERTERMINATED TWO PORT NETWORKS

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OBJECTIVESOBJECTIVES• TO UNDERSTAND ABOUT TWO – PORT

NETWORKS AND ITS FUNTIONS.• TO UNDERSTAND THE DIFFERENT

BETWEEN Z – PARAMETER, Y – PARAMETER, T – PARAMETER AND TERMINATED TWO PORT NETWORKS.

• TO INVERTIGATE AND ANALYSIS THE BEHAVIOUR OF TWO – PORT NETWORKS.

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TWO – PORT NETWORKSTWO – PORT NETWORKS• A pair of terminals through which a

current may enter or leave a network is known as a port.

• Two terminal devices or elements (such as resistors, capacitors, and inductors) results in one – port network.

• Most of the circuits we have dealt with so far are two – terminal or one – port circuits.

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• A two – port network is an electrical network with two separate ports for input and output.

• It has two terminal pairs acting as access points. The current entering one terminal of a pair leaves the other terminal in the pair.

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Linear network+V-

I

I

One – port network

Linear network+V1

-

I1

I1 I2

I2

+V2

-

Two – port network

7

• Two (2) reason why to study two port – network:Such networks are useful in

communication, control system, power systems and electronics.

Knowing the parameters of a two – port network enables us to treat it as a “black box” when embedded within a larger network.

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• From the network, we can observe that there are 4 variables that is I1, I2, V1and V2, which two are independent.

• The various term that relate these voltages and currents are called parameters.

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Z – PARAMETER Z – PARAMETER • Z – parameter also called as impedance

parameter and the units is ohm (Ω)• Impedance parameters is commonly used

in the synthesis of filters and also useful in the design and analysis of impedance matching networks and power distribution networks.

• The two – port network may be voltage – driven or current – driven.

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• Two – port network driven by voltage source.

• Two – port network driven by current sources.

Linear network

I1 I2

+−

+−V1 V2

I1 I2

+V1

-

Linear network+V2

-

11

• The “black box” is replace with Z-parameter is as shown below.

• The terminal voltage can be related to the terminal current as:

+V1

-

I1I2

+V2

-

Z11

Z21

Z12

Z22

2221212

2121111

IzIzV

IzIzV

+=+= (1)

(2)

12

• In matrix form as:

• The Z-parameter that we want to determine are z11, z12, z21, z22.

• The value of the parameters can be evaluated by setting:1. I1= 0 (input port open – circuited)

2. I2= 0 (output port open – circuited)

=

2

1

2221

1211

2

1

I

I

zz

zz

V

V

13

• Thus,

01

221

01

111

2

2

=

=

=

=

I

I

I

Vz

I

Vz

02

222

02

112

1

1

=

=

=

=

I

I

I

Vz

I

Vz

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• Where;z11 = open – circuit input impedance.

z12 = open – circuit transfer impedance from port 1 to port 2.

z21 = open – circuit transfer impedance from port 2 to port 1.

z22 = open – circuit output impedance.

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Example 1Example 1Find the Z – parameter of the circuit below.

40Ω

240Ω120Ω

+

V1

_

+

V2

_

I1I2

16

Solution Solution

i) I2 = 0(open circuit port 2). Redraw the circuit.

40Ω

240Ω

120Ω

+

V1

_

+

V2

_

I1Ia

Ib

17

Ω==∴

→

=

=

84

(2) (1) sub

)2......(400

280

)1.......(120

1

111

1

1

I

VZ

II

IV

b

b

Ω==∴

→

=

=

72

(3) (4) sub

)4.......(400

120

.......(3)240

1

221

1

2

I

VZ

II

IV

a

a

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ii) I1 = 0 (open circuit port 1). Redraw the circuit.

40Ω

240Ω120Ω

+

V1

_

+

V2

_

Iy I2

Ix

19

Ω==∴

→

=

=

96Z

(2)(1) sub

)2.......(400

160

)1.......(240

2

222

2

2

I

V

II

IV

x

x

Ω==∴

→

=

=

72

(3)(4) sub

)4.......(400

240

)3.......(120

2

112

2

1

I

VZ

II

IV

y

y

[ ]

=

9672

7284Z

In matrix form:

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Example 2Example 2Find the Z – parameter of the circuit below

+

_

+

V1

_

+

V2

_-j20Ω

10Ωj4Ω2Ω

10I2

I2I1

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Solution Solution i) I2 = 0 (open circuit port 2). Redraw the circuit.

Ω=∴=

Ω+==∴

+=

0 Z

circuit)(short 0 V

j4)(2 I

V Z

j4) (2 I V

21

2

1

111

11

+

V1

_

j4Ω2ΩI1

+

V2

_

I2 = 0

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ii) I1 = 0 (open circuit port 1). Redraw the circuit.

Ω==∴

+=

+−

=

Ω==∴

=

j8)-(16 I

V Z

10

1

20

j V 2I

10

10I - V

j20

V I

10 I

V Z

10I V

2

222

22

2222

2

112

21

[ ]

+=

j8) - (1610

0j4)(2 Z

form;matrix In

+

_

+

V1

_

+

V2

_-j20Ω

10Ω

10I2

I2I1 = 0

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Y - PARAMETERY - PARAMETER• Y – parameter also called admittance

parameter and the units is siemens (S).• The “black box” that we want to replace

with the Y-parameter is shown below.

+V1

-

I1I2

+V2

-

Y11

Y21

Y12

Y22

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• The terminal current can be expressed in term of terminal voltage as:

• In matrix form:

2221212

2121111

VyVyI

VyVyI

+=+= (1)

(2)

=

2

1

2221

1211

2

1

V

V

yy

yy

I

I

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• The y-parameter that we want to determine are Y11, Y12, Y21, Y22. The values of the parameters can be evaluate by setting:i) V1 = 0 (input port short – circuited).

ii) V2 = 0 (output port short – circuited).

• Thus;

01

221

01

111

2

2

=

=

=

=

V

V

V

IY

V

IY

02

222

02

112

1

1

=

=

=

=

V

V

V

IY

V

IY

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Example 1Example 1

Find the Y – parameter of the circuit shownbelow.

5Ω

15Ω20Ω

+

V1

_

+

V2

_

I1I2

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Solution Solution

i) V2 = 0

5Ω

20Ω

+

V1

_

I1

I2

Ia

SV

IY

II

IV

a

a

4

1

(2)(1) sub

)2.......(25

5

)1.......(20

1

111

1

1

==∴

→

=

=

SV

IY

IV

5

1

5

1

221

21

−==∴

−=

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ii) V1 = 0

In matrix form;

5Ω

15Ω

+

V2

_

I1I2

Ix

SV

IY

II

IV

x

x

15

4

(4)(3) sub

)4.......(25

5

)3.......(15

2

222

2

2

==∴

→

=

=

SV

IY

IV

5

1

5

2

112

12

−==∴

−=

[ ] SY

−

−=

15

4

5

15

1

4

1

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Example 2 (circuit with Example 2 (circuit with dependent source)dependent source)

Find the Y – parameters of the circuitshown.

+

_

+

V1

_

+

V2

_-j20Ω

10Ωj4Ω2Ω

10I2

I2I1

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Solution Solution i) V2 = 0 (short – circuit port 2). Redraw the circuit.

+

_

+

V1

_

10Ωj4Ω2Ω

10I2

I2I1

S 0 V

I Y

S j0.2) - (0.1 j4 2

1

V

I Y

j4)I (2 V

0 I

1

221

1

111

11

==∴

=+

==∴

+==

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ii) V1 = 0 (short – circuit port 1). Redraw the circuit.

)2.......(j20-

1

10

1 V 2I

10

10I - V

j20-

V I

)........(1j42

10I- I

22

2222

21

+=

+=

+=

+

_

+

V2

_-j20Ω

10Ωj4Ω2Ω

10I2

I2I1

[ ] S j0.025 0.050

j0.075 1.0j0.2 0.1 Y

form;matrix In

S j0.075) (-0.1 V

I Y

(1) (2) sub

S j0.025) (0.05 V

I Y

2

112

2

222

++−+

=∴

+==

→

+==∴

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T (ABCD) PARAMETERT (ABCD) PARAMETER• T – parameter or ABCD – parameter is a

another set of parameters relates the variables at the input port to those at the output port.

• T – parameter also called transmission parameters because this parameter are useful in the analysis of transmission lines because they express sending – end variables (V1 and I1) in terms of the receiving – end variables (V2 and -I2).

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• The “black box” that we want to replace with T – parameter is as shown below.

• The equation is:

+V1

-

I1I2

+V2

-

A11

C21

B12

D22

)2.......(

)1.......(

221

221

DICVI

BIAVV

−=−=

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• In matrix form is:

• The T – parameter that we want determine are A, B, C and D where A and D are dimensionless, B is in ohm (Ω) and C is in siemens (S).

• The values can be evaluated by settingi) I2 = 0 (input port open – circuit)

ii) V2 = 0 (output port short circuit)

−

=

2

2

1

1

I

V

DC

BA

I

V

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• Thus;

• In term of the transmission parameter, a network is reciprocal if;

02

1

02

1

2

2

=

=

=

=

I

I

V

IC

V

VA

02

1

02

1

2

2

=

=

=

=

V

V

I

ID

I

VB

1 BC - AD =

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Example Example

Find the ABCD – parameter of the circuitshown below.

2Ω

10Ω

+

V2

_

I1 I2

+

V1

_

4Ω

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Solution Solution i) I2 = 0,

2Ω

10Ω

+

V2

_

I1

+

V1

_

2.1

5

6

102

2

1.0

10

2

1

222

1

211

2

1

12

==∴

=+

=

+=

==∴

=

V

VA

VVV

V

VIV

SV

IC

IV

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ii) V2 = 0,

( )

Ω=−=∴

+

−=

+=++=

=−=∴

−=

8.6

1010

1412

1012

102

4.1

14

10

2

1

221

211

2111

2

1

12

I

VB

IIV

IIV

IIIV

I

ID

II2Ω

10Ω

I1 I2

+

V1

_

4Ω

I1 + I2

[ ]

=

4.11.0

8.62.1T

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TERMINATED TWO – PORT TERMINATED TWO – PORT NETWORKSNETWORKS

• In typical application of two port network, the circuit is driven at port 1 and loaded at port 2.

• Figure below shows the typical terminated 2 port model.

+V1

-

I1 I2

+V2

-

+−

Zg

ZLVg

Two – port network

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• Zg represents the internal impedance of the source and Vg is the internal voltage of the source and ZL is the load impedance.

• There are a few characteristics of the terminated two-port network and some of them are;

gV

V

V

V

I

I

I

V

I

V

2g

1

2v

1

2i

2

2o

1

1i

A gain, voltageoverall v)

A gain, voltageiv)

A gain,current iii)

Zimpedance,output ii)

Zimpedance,input i)

=

=

=

=

=

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• The derivation of any one of the desired expression involves the algebraic manipulation of the two – port equation. The equation are:1) the two-port parameter equation either Z or Y or ABCD.

For example, Z-parameter,

)2.......(IZIZV

)1.......(IZIV

2221212

2121111

+=+= Z

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2) KVL at input,

3) KVL at the output,

• From these equations, all the characteristic can be obtained.

.......(3)ZIVV g1g1 −=

)4.......(ZIV L22 −=

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Example 1 Example 1 For the two-port shown below, obtain the suitable value of Rs such that maximum power is available at the input terminal. The Z-parameter of the two-port network is given as

With Rs = 5Ω,what would be the value of

=

44

26

2221

1211

ZZ

ZZ

s

2

V

V

+V1

-

I1 I2

+V2

-

+−

Rs

4ΩVs

Z

44

Solution Solution 1) Z-parameter equation becomes;

2) KVL at the output;

Subs. (3) into (2)

)2.......(44

)1.......(26

212

211

IIV

IIV

+=+=

)3.......(4 22 IV −=

)4.......(21

2

II −=

45

Subs. (4) into (1)

For the circuit to have maximum power,

)5.......(5 11 IV =

Ω==∴ 51

11 I

VZ

Ω== 51ZRs

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To find at max. power transfer, voltagedrop at Z1 is half of Vs

From equations (3), (4), (5) & (6)Overall voltage gain,

s

2

V

V

)6.......(21sVV =

5

12 ==s

g V

VA

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Example 2Example 2The ABCD parameter of two – port network shown below are.

The output port is connected to a variable load for a maximum power transfer. Find RL and the maximum power transferred.

Ω20.1S

204

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ABCD parameter equation becomesV1 = 4V2 – 20I2

I1 = 0.1V2 – 2I2

At the input port, V1 = -10I

(1)

(2)

(3)

Solution

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(3) Into (1)-10I1 = 4V2 – 20I2

I1 = -0.4V2 2I2

(2) = (4)0.1V2 – 2I2 = -0.4V2 + 2I2

0.5V2 = 4I2

From (5);ZTH = V2/I2 = 8Ω

(4)

(5)

(6)

50

But from Figure (b), we know that V1 = 50 – 10I1 and I2 =0

Sub. these into (1) and (2)50 – 10I1 = 4V2

I1 = 0.1V2

(8)

(7)

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Sub (8) into (7)V2 = 10

Thus, VTH = V2 = 10V

RL for maximum power transfer,RL = ZTH = 8Ω

The maximum power P = I2RL = (VTH/2RL)2 x RL = V2

TH/4RL = 3.125W