teori litar

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Two Port NetworksTwo Port Networks5.1 Is used to describe the performance of a circuit in terms of

the voltage and current at its input and output ports.5.2 Impedance parameters z(terminal voltages can be related

to the terminal current)5.3 Admittance parameters y (terminal currents can be

expressed in terms of the terminal voltages)5.4 Hybrid parameters h 5.5 Transmission parameters T (expressing the variables at the

input port in terms of the variables at the output port)

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Two Port NetworksTwo Port NetworksIt is a pair of terminals through which a current may enter or leave a network is known as a portTwo terminal devices (R,L &C) result in one-port NetworksMost of the circuits we have dealt with so far are two terminal or one port circuits represented in (a)

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Two Port NetworksTwo Port NetworksFour terminal or two port (one input and one output) circuits involving op amps, transistors, and transformer as shown in (b).Use of this building block is subject to 3 restrictions.1. No energy stored within the circuit2. No independent sources3. Current in = Current out4. No connection between port

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Two Port NetworksTwo Port NetworksOnly the terminal variables (I1, V1, I2 and V2) are of interest.The various terms that relate these voltages and currents are called parametersOur goal is to derive six sets of these parametersWe will show the relationship between these parameters and how two port networks can be connected in series, parallel or cascade

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One port or two terminal circuit

Two port or four terminal circuit

• It is an electrical network with two separate ports for input and output.

• No independent sources.

5.1 Introduction (2)5.1 Introduction (2)

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Impedance parametersImpedance parametersImpedance and admittance parameters are commonly used in the synthesis of filters Also useful in the design and analysis of impedance matching networks and power distribution networks

(a) Driven by voltage sources, (b) driven by current sources

Assume no independent source in the network

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5.2 Impedance parameters (1)5.2 Impedance parameters (1)

2221212

2121111

IzIzVIzIzV

2

1

2

1

2221

1211

2

1

II

zII

zzzz

VV

where the z terms are called the impedance parameters, or simply z parameters, and have units of ohms.

The terminal voltages can be related to the terminal currents as

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5.2 Impedance parameters5.2 Impedance parameters (2)(2)

z11 = impedance seen looking into port 1 when port 2 is openz21 = Transfer impedance. It is the ratio of the port 2 voltage to the port 1 current when port 2 is open

z12 = Transfer impedance. It is the ratio of the port 1 voltage to the port 2 current when port 1 is openz22 = impedance seen looking into port 2 when port 1 is open

0I1

221

0I1

111

22IVzand

IVz

0I2

222

0I2

112

11IVzand

IVz

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5.2 Impedance parameters5.2 Impedance parameters (2)(2)Therefore the impedance parameters may be either calculated or measured by first opening port 2 and determining the ratios V1/I1 and V2/I1, and then opening port 1 and determining the ratios V1/I2 and V2/I2

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5.2 Impedance parameters (2)5.2 Impedance parameters (2)

0I1

221

0I1

111

22IVz

andIVz

0I2

222

0I2

112

11IVzand

IVz

•When z11 = z22, the two-port network is said to be symmetrical. •When the two-port network is linear and has no dependent sources, the transfer impedances are equal (z12 = z21), and the two-port is said to be reciprocal.

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VV1 VV2

II2II1

Example 1

Determine the Z-parameters of the following circuit.

5.2 Impedance parameters 5.2 Impedance parameters

0I 1

221

0I 1

111

22

IV zand

IVz

0I 2

222

0I 2

112

11

IVzand

IVz

AnswerAnswer::

70404060

z

2221

1211

zzzz

z

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5.3 Admittance parameters (1)5.3 Admittance parameters (1)

Assume no independent source in the network

2221212

2121111

VyVyIVyVyI

2

1

2

1

2221

1211

2

1

VV

yVV

y yy y

II

where the y terms are called the admittance parameters, or simply y parameters, and they have units of Siemens.

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0V2

222

0V2

112

11

VIyand

VIy

5.3 Admittance parameters (2)5.3 Admittance parameters (2)

y11 = Short-circuit output admittancey21 = Short-circuit transfer admittance from port 2 to port 1

y12 = Short-circuit transfer admittance from port 1 to port 2y22 = Short-circuit input admittance

0V1

221

0V1

111

22

VIyand

VIy

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VV1 VV2

II1 II2

Example 2

Determine the y-parameters of the following circuit.

5.3 Admittance parameters (3)5.3 Admittance parameters (3)

0V1

221

0V1

111

22

VIyand

VIy

S 0.6250.5

0.50.75y

AnswerAnswer::

0V2

222

0V2

112

11

VIyand

VIy

S yyyy

y2221

1211

15

VV1 VV2

II2II1=i

Example 3

Determine the y-parameters of the following circuit.

5.3 Admittance parameters (4)

2221212

2121111

VyVyIVyVyI

S 0.250.250.050.15

y

AnswerAnswer::

212

211

0.25V0.25VI0.05V0.15VI

)I2(I)I4(2iV)I2(I8IV

2122

2111

Apply KVL

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5.4 Hybrid parameters (1)5.4 Hybrid parameters (1)

The h-parameter equivalent network of a two port network

2221212

2121111

VhIhIVhIhV

2

1

2

1

2221

1211

2

1

VI

hVI

hhhh

IV

where the h terms are called the hybrid parameters, or simply h parameters, and each parameter has different units, refer above.

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5.4 Hybrid parameters (2)5.4 Hybrid parameters (2)

0V1

221

0V1

111

2

2

IIh

IVh

0I2

222

0I2

112

1

1

VIh

VVh

h11= short-circuit input impedance ()

h21 = short-circuit forward current gain

h12 = open-circuit reverse voltage-gain

h22 = open-circuit output admittance (S)

Assume no independent source in the network

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VV1 VV2

II1 II2

Example 4

Determine the h-parameters of the following circuit.

5.4 Hybrid parameters (3)5.4 Hybrid parameters (3)

0V1

221

0V1

111

22

IIhand

IVh

S4Ω

h91

32

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AnswerAnswer::

0I2

222

0I2

112

11

VIhand

VVh

ShhhΩh

h2221

1211

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5.5 Hybrid parameters (1)5.5 Hybrid parameters (1)

Assume no independent source

in the network

2221212

2121111

IgVgVIgVgI

2

1

2

1

2221

1211

2

1

IV

gIV

gggg

VI

where the g terms are called the inverse hybrid parameters, and each parameter has different units.

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5.5 Transmission parameters (1)5.5 Transmission parameters (1)

Assume no independent source

in the network

221

221

DICVIBIAVV

2

2

2

2

1

1

IV

TI

V

DCBA

IV

where the T terms are called the transmission parameters, or simply T or ABCD parameters, and each parameter has different units.

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5.5 Transmission parameters (2)5.5 Transmission parameters (2)

0I2

1

0I2

1

2

2

VIC

VVA

0V2

1

0V2

1

2

2

IID

IVB

A=open-circuit voltage ratio

C= open-circuit transfer admittance

(S)

B= negative short-circuit transfer impedance ()

D=negative short-circuit current ratio

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VV11 VV22

Example 5

Determine the T-parameters of the following circuit.

5.5 Transmission parameters (3)5.5 Transmission parameters (3)

1.1760.059S15.294Ω1.765

TAnswerAnswer::

221

221

DICVIBIAVV

221

221

I1720V

171I

I17260V

1730V

)I20(I3IV)I20(I10IV

2112

2111

Apply KVL

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VV11 VV22

II11 II22

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

Example 6

The ABCD parameters of the two-port network below are

T =

5.5 Transmission parameters (4)5.5 Transmission parameters (4)

20.1S

20Ω4

AnswerAnswer: V: VTH = 10V V; R = 10V V; RL = 8 = 8; Pm = 3.125W.; Pm = 3.125W.

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5.5 Transmission parameters (1)5.5 Transmission parameters (1)

112

112

dIcVIbIaVV

1

1

1

1

2

2

IV

tI

V

dcba

IV

where the t terms are called the inverse transmission parameters, and each parameter has different units.

t parameter may be defined by expressing the variables at the output port in terms of the variables at the input port.

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5.5 Relationship between 5.5 Relationship between parameters (1)parameters (1)

• Since the six sets of parameters relate the same input and output terminal variables of the same two port network, they should be interrelated.

• If we know one set of parameters, we can derive all the other sets from the known set

• Given the z parameters, let us obtain the y parameters

or

2

1

2

1

2221

1211

2

1

II

zII

zzzz

VV

2

1

2

1

VV

zII 1-

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5.5 Relationship between 5.5 Relationship between parameters (1)parameters (1)

• Also from eq.

• Comparing with eq.

shows that

The adjoint of the [z] matrix is

and its determinant is zy -1

2

1

2

1

2221

1211

2

1

VV

yVV

y yy y

II

2

1

2

1

VV

zII 1-

1121

1222

zzzz

21122211 zzzzz

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5.5 Relationship between 5.5 Relationship between parameters (1)parameters (1)

• Substituting these into eq. we get

Equating terms yields

zy -1

z2221

1211 zzzz

y yy y

1121

1222

z

zy

2211

z

zy

1212

z

zy

2121

z

zy

1122

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• In the typical application of a two port model, the circuit is driven at port 1 and loaded at port 2.• This Fig. shows a typically terminated two port model.

• Zg= internal impedance of the source• Vg= internal voltage of the source• ZL= load impedance• Analysis of this circuit involves expressing the terminal currents and voltages as functions of two

port parameters, Vg, Zg, and ZL

Zg

ZLVg

I1 I2

+

-

+

-V1 V2

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• Six characteristics of the terminated two port circuit define its terminal behavior

• The input impedance Zin=V1/I1• The output current I2

• The Thevenin voltage and impedance with respect to port 2• The current gain I2 /I1• The voltage gain V2/V1

• The voltage gain V2/Vg

Zg

ZLVg

I1 I2

+

-

+

-V1 V2

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• To shows how these six characteristics are derived, we develop the expressions using z parameters to model the two port portion of the circuit

• The derivation of any one of the desired expressions involves the algebraic manipulation of the two port eqs. along with the two constraint eqs. imposed by the terminations

Zg

ZLVg

I1 I2

+

-

+

-V1 V2

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• The impedance seen looking into port 1, that is Zin=V1/I1• Replace V2 with –I2ZL and solve the resulting expression for I2

• Then substitute this eq. into and solve for Zin

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

22

1212 zZ

IzIL

21111 IzIzV 12

Lin Zz

zzzZ

22

211211

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• To find I2, first solve for I1 after replacing

V1 with the RHS of , the result is

Now replace eq. into and solve

the resulting eq. for I2

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

21111 IzIzV 12

g

g1 Zz

IzVI

11

212gg ZIVV 11

g

g1 Zz

IzVI

11

212

22

1212 zZ

IzIL

21122211

21

zzZzZzVz

ILg

g2

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• The Thevenin voltage with respect to port 2 equals V2 when I2=0• With I2=0. eqs. and

combine to yield

But and

Therefore

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

21111 IzIzV 12 2221212 IzIzV

11

12112102 z

VzIzVI2

gg ZIVV 11 g

g1 Zz

VI

11

gg

ThI2 VzZ

zVV11

2102

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• The Thevenin, or output, impedance is the ratio V2/I2 when Vg is replaced by a short circuit. When Vg is zero eq.

• reduces to

• Substituting eq. into eq.

• gives

• Use eq. to replace

I1 in

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

gg ZIVV 11 gZIV 11

gZIV 11 21111 IzIzV 12

g1 Zz

IzI

11

212

g1 Zz

IzI

11

212

2221212 IzIzV

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• With the result that

• The current gain I2/I1 comes directly from eq.

For the voltage gain V2/V1, replace I2 ineq. with its value from

eq. ; thus

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

gTh

V2

2

ZzzzzZ

IV

g

11

211222

0

22

1212 zZ

IzIL

22

21

1

2

zZz

II

L

2221212 IzIzV

LZIV 22

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• Next solve eq. for I1 as a function of V1

• and V2

• or

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

L

2221212 Z

V-zIzV

21111 IzIzV 12

L

2111 Z

-VzVIz 121

L1 Zz

VzzVI

11

212

11

1

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• Now replace I1 in eq.

• with eq. and solve the resulting

• expression for V2/V1

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

zZzZz

zzzzZzZz

VV

L

L21

L

L

1

2

11

2112221111

21

L

2221212 Z

V-zIzV

L1 Zz

VzzVI

11

212

11

1

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• To derive the voltage ratio V2/Vg, combine eqs.

• and to find I1 as a function • of V2 and Vg

• Now use eq.

• and eq. in conjunction with

21111 IzIzV 12

g

g

gL1 Zz

VZzZ

VzI

1111

212

)(

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

gg ZIVV 11 LZIV 22

g

g

gL1 Zz

VZzZ

VzI

1111

212

)(

LZIV 22

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5.5 Analysis of the terminated two 5.5 Analysis of the terminated two port circuitport circuit

• Eq. To derive an expression involving

• only V2 and Vg; that is

• Which can manipulate to get the desired voltage ratio:

L

gg

2221212

21111

ZIV

ZIVVIzIzVIzIzV

22

11

12

21122211

21

)( zzZzZzZz

VV

Lg

L

g

2

2221212 IzIzV

222

11

21

11

212212 )()(

VZz

ZzVz

ZzZVzzV

Lg

g

gL

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5.6 Interconnected two port circuit5.6 Interconnected two port circuit• Two port circuits may be interconnected five ways:In

cascade, in series, in parallel, in series parallel and parallel series

1 2

1

2

1

2

1

2

1

2

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5.6 Interconnected two port circuit5.6 Interconnected two port circuit• Consider the series connection of two port networks shown below

• For network 1 For network 2

2a22a1a21a2a

2aa1a11a1a

IzIzVIzIzV

12

I1

1

2

+-

+-

+-

+-+

-

+

-

V1

V1a

I1a I2I2a

V1b

V2a

V2b

V2I1b I2b

2b22b1b21b2b

2bb1b11b1b

IzIzVIzIzV

12

ba1 III 11 ba2 III 22

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5.6 Interconnected two port circuit5.6 Interconnected two port circuit• and that

• Thus, the z parameters for all the overall network are

• Or showing that the z • parameters for the overall network are the sum of the z parameters for the individual networks

222ba121b21a2ba2

212ba111b11a1ba1

)Izz)Iz(zVVV)Izz)Iz(zVVV

222

121

((

baba

baba

zzzzzzzz

zzzz

22222121

12121111

2221

1211

ba zzz

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5.6 Interconnected two port circuit5.6 Interconnected two port circuit• Consider the parallel connection of two port networks shown below

• For network 1 For network 2

2a22a1a21a2a

2aa1a11a1a

VyVyIVyVyI

12

2b22b1b21b2b

2bb1b11b1b

VyVyIVyVyI

12

ba1 III 11 ba2 III 22

1

2

+-

+-

+-

+-

+

-

+

-

I1 I1a I2a

V1a V2a

V2bV1b

I1b I2bV1 V2

I2

ba1 VVV 11 ba2 VVV 22

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5.6 Interconnected two port circuit5.6 Interconnected two port circuit• and that

• Thus, the y parameters for all the overall network are

• Or showing that the y • parameters for the overall network are the sum of the y parameters for the individual networks

222ba121b21a2

212ba111b11a1

)Vyy)Vy(yI)Vyy)Vy(yI

22

12

((

baba

baba

yyyyyyyy

yyyy

22222121

12121111

2221

1211

ba yyy

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5.6 Interconnected two port circuit5.6 Interconnected two port circuit• Consider the cascaded connection of two port networks shown below

• For network 1 For network 2

1a

1a

1

1

IV

IV I1

1 2+-V1 V1a

I1a I2I2a

V1bV2a V2b V2

I1b I2b

+-+-

+-

+-

+-

1b

1b

2a

2a

IV

I-V

2

2

2b

2b

I-V

I-V

2a

2a

aa

aa

1a

1a

IV

DCBA

IV

2b

2b

bb

bb

1b

1b

IV

DCBA

IV

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5.6 Interconnected two port circuit5.6 Interconnected two port circuit• and that

• Thus, the transmission parameters for all the overall network are the product of the transmission parameters for the individual transmission parameters

• or

2

2

bb

bb

aa

aa

1

1

IV

DCBA

DCBA

IV

DCBA

DCBA

DCBA

bb

bb

aa

aa

ba TTT