Unit 5 | unit 5 two port network

Unit - 5

Two Port Networks

5.1 Two Port Networks: Impedance parameter

Open circuit impedance parameters

The Z parameters of a two-port for the positive directions of voltages and currents may be defined by expressing the port voltages V1 and V2 in terms of the currents I1 and I2. Here V1 and V2 are dependent variables, and I1, I2 are independent variables. The voltage at port 1-1’ is the response produced by the two currents I1 and I2. Thus,

Z-parameter:

v1 = Z11I1 + Z12I2

v2 = Z11I1 + Z22I2

Z11 = I2=0

Z12 = I1=0

Z21 = I2=0

Z22 = I1=0

I1 and I2 are excitations at port 1 & 2 respectively.

V1 and V2 are the responses at port 1 and 2 respectively.

Equivalent circuit:

Symmetry:

I2=0 =I1 = 0

Z11 = Z22

Reciprocal two port N/W:-

I2=0= I1=0

Z12 = Z21

I1& I2 should be independent

Que 1.

Find Z-parameter

Solution: I1 = -I2

Current dependent so Z-parameter doesn’t exist

Que 2.

Find z-parameter

Solution: V1 =R (I1 + I2)

V2 = R (I1 + I2)

Z11 = Z12 = Z21 = Z22 = R

Que 3.

Solution: V1 = I1Za + I1Zc + I2Zc

= (Za + Zc)I1 + ZcI2

V2 = I2Zb + I2Zc + I1Zc

= (Zb + Zc)I1 + ZcI1

Z11 = (Za + Zc)

Z12 = Zc = Z21

Z22 = (Zb + Zc)

Que 4.

Solution: V1 = Za(I1 - I)

(I - I1)Za+ IZc+ Zb(I + I2) = 0

I(Za + Zb + Zc) – I1Za + I2Zb = 0

I =

V1 = ZaI1 - Za

= I1 + I2

V2 = Zb(I2 + I)

= ZbI2 + Zb

= I2 + I2

Z11 =

Z12 = Z21 =

Z22 =

Can be solved by Y-A conversion

Que 1.

Solution:

Z11 = I2=0

V1 - (Za + Zb) = 0

= Z11 =

Z21 = I2=0

V2 - Zb +Za = 0

Z12 =

Z22 =

Que 2.

Find Z21?

Solution: Z21 = I2=0

I1/2 =

= I1

V2 = I1/2

= × I1

= I1

Z21 = I2 = 0 = I1 Ω

Que 3.

Solution: + + 2V1 = 0

2Vx – 2V1 + Vx – V2 + 2V1 = 0

3Vx = V2

Vx =

I1 = V1 +

= 3V1 – 2Vx

= 3V1 – 2

V1 =

+ 2V2 = I2

3V2 - = I2

V2 = I2

V1 =

Z11 =

Z12 =

Z21 = 0

Z22 =

5.2 Admittance Parameters

Short circuit Admittance parameters

Y-parameter:-

I1 = Y11V1 + Y12V2

I2 = Y21V1 + Y22V2

Y11 = V2=0

Y12 = V1=0

Y21 = V2=0

Y22 = V1=0

V1& V2 should be independent

Equivalent circuit: -

Symmetrical two port N/W: -

I2 = 0 = I1=0

Y12 = Y22

Reciprocal two port N/W: -

I1=0 =I2 = 0

Y12 = Y21

Que 1.

Solution: V1 – I1R – V2 = 0

V1 – V2 = I1R

I1 = V1 - V2

V2 = I2R + V1

I2 = - V1 + V2

Y11 =

Y12 = Y21 =

Y22 =

Que 2.

Solution: Y-parameter does not exist as V1 = V2

Que 3.

Solution: I1 = V1Ya + (V1 – V2)Yc

I1 = (Ya + Yc)V1 - YcV2

I2 = V2Yb + (V2 – V1)Yc

I1 = (Yb + Yc)V2 - YcV1

Y11 = Yb + Yc

Y12 = Y21 = - Yc

Y22 = Yb + Yc

Que 4.

Solution:

Que 5:

Find Y-parameter?

Solution:

-I1 + – 2V2 + + 2V1 = 0

I1= V1 + V1 - 3V2 + 2V1

I1= 4V1 - 3V2V

V2 + 2V2- 2V1 = 2(I2 + 2V1)

- 2V1 + 3V2 = 2(I2 + 2V1)

3V2 - 2V1 – 4V1 = 2I2

I2 = -3V1 + V2

Y11 = 4

Y12 = -3

Y21 = -3

Y22 =

5.3 Transmission Parameters

Transmission parameter [ABCD]

V1 = AV2 - BI2

I1 = CV2 - DI2

A = I2=0

B = V2=0

C = -I2=0

D = V2=0

Symmetrical two port N/W :-

I2 = 0 = I1=0

A = D

Reciprocal two port N/W :-

I1=0 =I2 = 0

AD – BC = 1

Que 1.

Solution:

-3V1 – I1 + + = 0

+ = I1

V1 – V2 = I1

V1 = V2 + I1

V1= V2- I1----------------(1)

I2 = 3V1 + V2 + V2 – V1

I2 = 2V1 + 2V2

2V1 = I2 - 2V2

2V1 = - 2V2 + I2

V1 = -V2 + I2 ----------------(2)

A = -1

B =

From (1) & (2)

-V2 + I2 = I1 - V2

V2 - V2 + I2 = I1

I1 = V2 + V2 - I2

I1 = V2 - I2

C =

D =

Que 2.

Solution: V1 = RI1 + V2 ----------------(1)

I2R = V2 – V1

I2 = V2 - V1

I2R = V2 – V1

V1 = I2R - V2 --------------------(2)

A = 1

B = R

From (2) in (1)

V2 - I2R = V2 + RI1

I2 = -I1

C = 0

D = 1

Que 3.

Solution: V1 = R(I1 + I2)

V2 = R(I1 + I2)

V1 = V2 + 0I2

A = 1

B = 0

V2 = RI1 + RI2

RI1= V2 - RI2

I1 = V2 – I2

C = , D = 1

5.4 Hybrid Parameters

H-parameter:

V1 = h11I1 + h12V2

I2 = h21I1 + h22V2

Equivalent circuit for H-parameter:

Fig: Equivalent circuit for H-parameter

h11 = V2=0

h12 = I1=0

h21 = V2=0

h22 = I1=0

Symmetrical two port N/W: -

I2 = 0 = I1=0

∆h = 0

Reciprocal two port N/W: -

I1=0 =I2 = 0

h12 = h21

Que. Find all h-parameter?

Solution: + = I1

V1 - = I1

V1 = + I1

- + 3I1 = I2

3I1 + V2 = I2

From (1)

I2 = 3I1 + V2 – [ I1 + V2]

I2 = I1 + V2

h11 =

h12 =

h21 =

h22 =

Key takeaway

Relationship of two-port variables

Fig: 2 Port network

Fig: 1 port network

Fig: 3 port network

z-parameter open circuit impedance

y-parameter short circuit admittance

h-parameter hybrid parameter

g-parameter inverse hybrid parameter

ABCD-parameter transmission parameter

A’B’C’D’ parameter inverse transmission

s-parameter scattering

Used for very high frequency application

Relationship of Two Part Variables

Condition for Reciprocity and Symmetry

Sr No.	Parameter	Reciprocity	Symmetry
1.	Z Parameter	Z12 = Z21	Z11= Z22
2.	Y Parameter	Y12 = Y21	Y12 = Y21
3.	h Parameter	h12 = h21	Δ = 1
4.	ABCD Parameter	AD-BC = 1	A = D
5.	Inverse h Parameter	g12 = g21	Δ = 1
6.	Inverse Transmission	A1D1 – B1 C1= 1	A1 = D1

5.5 Interconnection of two port networks: Series, Parallel, cascade, Series Parallel, Parallel series

Series connection

V1 = V1a +V1b

I1 = I1a = I1b

V2 = V2a +V2b

I2 = I2a = I2b

Parallel connection

V1 = V1a = V1b

Series parallel connection

I1 = I1a = I1b

V1 = V1a +V1b

If two 2-ports are connected in series parallel then overall h-parameter is sum of individual h-parameter

Parallel series connection

Cascade connection

V1 = V1a , V2a = V1b , V2 = V2b

I1 = I1a , I2a = -I1b , I2 = I2b

Transmission parameter for N/W (A)

Transmission parameter for N/W (B)

Overall transmission parameter

Que 1.

Find out overall transmission parameter?

Solution:

Que 2.

Find overall Y-parameter?

Solution:

Key takeaway

Relation between Two- Port Parameters:

Δ = X11 X22 – X12 X21 , ΔT = AD – BC

[z]

[y]

[h]

[T]

[z]

Z11 Z12

y22 – y12

Δy Δy

Δh – h12

h22 h22

A – ΔT

C C

Z21 Z22

y21 – y11

Δy Δy

-h21 h

h22 h22

1 D

C C

[y]

Z22 -Z12

Δz Δz

y11 y12

1 -h12

h11 h11

D – ΔT

B B

-Z21 Z11

Δz Δz

y21 y22

h21 Δn

h11 h11

-1 A

B B

[h]

Δz Z12

Z22 Z22

1 -y12

y11 y11

h11 h12

B ΔT

D D

-Z21 -1

Z22 Z22

y21 Δy

y11 y11

H21 h22

-1 C

D D

[T]

Z21 ΔZ

Z21 Z21

-y22 -1

y21 y21

-Δh -h11

h21 h21

A B

1 Z22

Z21 Z21

-Δy -y11

y21 y21

-h22 -1

h21 h21

C D

5.5 Lattice network

One of the common four-terminal two-port network is the lattice, or bridge network shown in Figure. Lattice networks are used in filter sections and are also used as attenuaters. Lattice structures are sometimes used in preference to ladder structures in some special applications. Za and Zd are called series arms, Zb and Zc are called the diagonal arms. It can be observed that, if Zd is zero, the lattice structure becomes a p-section. The lattice network is redrawn as a bridge network as shown in Figure.

Z11 = I2=0

When

If the network is symmetric, then and

When is the voltage across

Substituting the value of from Eq, we have

If the network is symmetric,

When the input port is open,

The network can be redrawn as shown in Fig.

Substituting the value of in Eq, we get

If the network is symmetric,

We have

If the network is symmetric,

From the above equations,

And

Q) Obtain the lattice equivalent of a symmetrical T network figure?

A) A two-port network can be realised as a symmetric lattice if it is reciprocal and symmetric. The z parameter of the network.

Z11= 3

Z12=Z21= 2

Z22=3

Since, Z11=Z22

Z12=Z21

The given network is symmetrical and reciprocal. The parameters of the lattice network are

Za=Z11-Z12=1

Zb=Z11+Z12=5

The lattice network is shown below

Q) Obtain the lattice equivalent of a symmetric p-network shown in Figure?

A) The Z parameters of the given network are

Z11=6 =Z22

Z12=Z21=4

Hence, the parameters of the lattice network are

Za=Z11-Z12=2

Zb=Z11+Z12= 10

5.6 Ladder Network

For this type of network series are represented as impedance and about cum represent admittance.

Now we can find the transfer function of above network KCL and KVL and then substituting the value in each equation the equation is reached which relative the output to input