Impedance parameters

InputPort

Output Port

+

_ _

+V1 V2

I1 I2

2221

1211

zz

zz

ZPARAMETERS

IMPEDANCE

Copyright © 2012 Ahmad Nauman. All Rights Reserved

LinearNetwork

V1 V2

I1 I2

INTRODUCTIONA two-port network may be voltage-driven as

or current-driven as

LinearNetwork

I1 I2

+

_V1_

+V2


From previous Figures

Only two of the four variables (V1, V2, I1 and I2) are independent/known.The other two can be found using equation (a)

2121111 Iz Iz V

2221212 Iz Iz V (a)

And in matrix form as

2

1

2221

1211

2

1

I

I

zz

zz

V

V V


2

1

2

1

I

I z

V

V

or

Where z terms are called impedance parameters or simply z-parameters

zz

zz z

2221

1211


LinearNetwork

V1

I1 I2=0

_

+V2

The values of the parameters can be evaluated by setting I2 = 0(output port open-circuited).

0I1

111

2I

V z

0I1

221

2I

V z


LinearNetwork

V1

I1=0

The values of the parameters can be evaluated by setting I1 = 0 (output port open-circuited).

0I2

112

1I

V z

0I2

222

1I

V z

V2

I2

_

+


impedance.input circuit -Openz11

impedance.output circuit -Openz22

12z

21z

Open Circuit Transfer impedance from port 1 to port 2.

Open Circuit Transfer impedance from port 2 to port 1.


• Since the z-parameters are obtained by open-circuiting the input or output port, they are also called the open-circuit impedance parameters.

• Since they are obtained as a ratio of voltage and current.

• And the parameters are obtained by open-circuiting port 2 ( I2 = 0) or port1 ( I1 = 0).

• Units of z-parameter are in ohms’ (Ω)


Impedance parameters are commonly used;• In the synthesis of filter.• They are also useful in the design and analysis

of impedance-matching networks.• And power distribution networks.

USAGE:


T-NETWORK

+

_ _

+V1 V2

I1 I21z

2z

3z

2

1

2221

1211

2

1

I

I

zz

zz

V

V

General form is


T-NETWORK

+

_ _

+

V1V2

1z

2z

3zI1 I2

Find the z-parameters.


+

_

V1I1

1z

3z

Applying KVL in loop 1

)I (I z Iz V 213111

I2

Iz )Iz (z 23131

But we also know that

2121111 Iz Iz V


_

+

V2

2z

3zI1 I2


)I (I z Iz V 213222

)Iz (z )I(z 23213

But we also know that

2221212 Iz Iz V

2

1

323

331

2

1

I

I

zzz

zzz

V

V

+

_ _

+V1 V2

I1 I21z

2z

3z

And the z-parameters are

zzz

zzz z

323

331


D-PROBLEM 17.8(A) -HAYT

+

_ _

+

V1V2

20 50

25

Figure 17.19 (a)

Find the z-parameters.


+

_ _

+

V1V2

20 50

25

Solution:

As we know that z-parameters in T Network are

zzz

zzz z

323

331

255025

252520 z

So from above figure


+

_ _

+

V1V2

20 50

25

Solution:

7525

2545 z

AnsweR


SERIES CONNECTION

Network A+

_ _

+

V1 V2

I1=I1A

Network B

I1=I1B

I1

+

+

_

_

I2=I2A

I2=I2B

+

+

_

_

I2

V1A

V1B

V2A

V2B

If each two-port has common reference node for its input and output, and if the references are connected together as indicated in Figure.


SERIES CONNECTIONI1 flows through the input ports of the two networks in series. A similar statement holds for I2. Thus, ports remain ports after the interconnection. It follows that I = IA = IB.

And

So

BBAABA Iz Iz V V V

I z )Iz (z BA

WhereBA z z z


_

+

V21V

I2

g11s ZI V V

THÉVENIN EQUIVALENTgZ

outZ

From this circuit

(a)


As we know that

From eq. (a)

g1212111 I Iz Iz ZVs

2121g11 Iz I)(z Z Vs

)(z

Iz

)(z I

g11

212

g111 ZZ

V s

We also know that

(b)

(c)2221212 I z Iz V

outz

2121111 Iz I z V

inz


222g11

22112

g11

21 Iz)(z

Izz

)(z

z V

ZZ

Vs2

)(z

zz - z

g11

211222 Z

Z out

Thus output impedance in terms of z-parameters

2g11

211222

g11

21 I )(z

zzz

)(z

z V

ZV

Z s2


_

+

V2

g11

211222out zz

zzzZ

sV

I2

g

th zz

zz

11

21

outZ

If the generator impedance is zero, the simpler expression

0zg

11

211222out z

zzzZ

11

z

11

21122211

z

zzzz


D-PROBLEM 17.9 - HAYT

+

_ _

+

V1 V2

20 50

25

Figure 17.19 (c)

2V5.0

I1I2



+

_

V1

20

25

2V5.0

I1I2

2211

22111

0.5V 25I 45I V

0.5V )I (I 25 20I V

(a)

Solution:


_

+

V2

50

25

2V5.0

I1I2


212

2122

22122

150I I50V

75I 25I V5.0V

0.5V )I (I 25 50I V

(b)

Solution:


Substituting equation (b) into (a), we get;

211

211

21211

I100 70I V

I)75(25 25)I(45 V

)I1500.5(50I 25I 45I V

From equation (c) and (b), we get;

(c)

2

1

2

1

I

I

15050

10070

V

V

As we know that

2

1

2221

1211

2

1

I

I

zz

zz

V

V

Solution:


We know z-parameters are

zz

zz z

2221

1211

So [z] will be

15050

10070 z

AnsweR

Solution:


Impedance parameters

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