ABCD Parameters MCQ Quiz - Objective Question with Answer for ABCD Parameters - Download Free PDF

Explanation:

Two-Port Symmetric Bilateral Network

Definition: A two-port symmetric bilateral network is a type of electrical network that has two pairs of terminals, referred to as "ports." The network is termed symmetric when certain parameters are equal, and it is bilateral when the network behaves identically when the input and output ports are interchanged. These networks are often analyzed using transmission parameters (A, B, C, D), where the relationships between input and output voltages and currents are described as:

Equations:

    \( V_1 = A \cdot V_2 + B \cdot I_2 \)

    \( I_1 = C \cdot V_2 + D \cdot I_2 \)

For symmetric networks, the parameters satisfy the following conditions:

• \( A = D \) (Symmetry condition)
• \( A \cdot D - B \cdot C = 1 \) (Reciprocity condition)

Given:

• \( A = 3 \, \Omega \)
• \( B = 1 \, \Omega \)

We need to determine the value of the parameter \( C \).

Step 1: Use the Reciprocity Condition

From the reciprocity condition, we know:

    \( A \cdot D - B \cdot C = 1 \)

Since the network is symmetric, \( A = D \). Substituting \( A = 3 \, \Omega \) and \( B = 1 \, \Omega \), we get:

    \( 3 \cdot 3 - 1 \cdot C = 1 \)

    \( 9 - C = 1 \)

Step 2: Solve for \( C \)

Rearranging the equation:

    \( C = 9 - 1 \)

    \( C = 8 \, \text{s} \)

Thus, the value of the parameter \( C \) is \( 8 \, \text{s} \).

Step 3: Evaluate the Correct Option

From the options provided, the correct answer corresponds to:

• Option 1: \( 8 \, \text{s} \)

Important Information:

To analyze why the other options are incorrect, let us revisit the reciprocity condition:

    \( A \cdot D - B \cdot C = 1 \)

For a symmetric network:

• \( A = D = 3 \, \Omega \)
• \( B = 1 \, \Omega \)

Substituting these values into the equation, we calculated \( C = 8 \, \text{s} \). Any other value for \( C \) would violate the reciprocity condition. Let’s examine the incorrect options:

Option 2: \( C = 6 \, \text{s} \)

If \( C = 6 \, \text{s} \), substituting into the reciprocity condition:

    \( 3 \cdot 3 - 1 \cdot 6 = 9 - 6 = 3 \)

Here, the result is \( 3 \), which does not satisfy the reciprocity condition (\( A \cdot D - B \cdot C = 1 \)). Thus, this option is incorrect.

Option 3: \( C = 4 \, \text{s} \)

If \( C = 4 \, \text{s} \), substituting into the reciprocity condition:

    \( 3 \cdot 3 - 1 \cdot 4 = 9 - 4 = 5 \)

Here, the result is \( 5 \), which does not satisfy the reciprocity condition. Thus, this option is incorrect.

Option 4: \( C = 16 \, \text{s} \)

If \( C = 16 \, \text{s} \), substituting into the reciprocity condition:

    \( 3 \cdot 3 - 1 \cdot 16 = 9 - 16 = -7 \)

Here, the result is \( -7 \), which does not satisfy the reciprocity condition. Thus, this option is incorrect.

Option 5: \( C = 1 \, \text{s} \)

If \( C = 1 \, \text{s} \), substituting into the reciprocity condition:

    \( 3 \cdot 3 - 1 \cdot 1 = 9 - 1 = 8 \)

While the arithmetic is correct, the parameter \( C \) is defined as a reciprocal value in seconds (\( \text{s} \)), and the units here do not align with the requirements of the problem. Thus, this option is also incorrect.

Conclusion:

The correct value of \( C \) in the given two-port symmetric bilateral network is \( 8 \, \text{s} \), satisfying the reciprocity condition. This corresponds to Option 1. The other options fail to meet the necessary conditions or have incorrect units, as demonstrated in the analysis above.

Solution:

To find the Z parameters (Z₁₁, Z₁₂, Z₂₁, Z₂₂) for the given network, we need to understand the two-port network analysis. The Z-parameters or impedance parameters are defined by the following set of equations:

V₁ = Z₁₁I₁ + Z₁₂I₂

V₂ = Z₂₁I₁ + Z₂₂I₂

Where:

• V₁ is the input voltage
• V₂ is the output voltage
• I₁ is the input current
• I₂ is the output current

To determine the Z-parameters, we need to perform the following steps:

Step 1: Calculate Z₁₁

Z₁₁ is found by setting I₂ = 0 (open-circuit output). Under this condition, the input impedance is:

Z₁₁ = V₁ / I₁ (with I₂ = 0)

Step 2: Calculate Z₁₂

Z₁₂ is found by setting I₂ = 0 (open-circuit output). Under this condition, the reverse transfer impedance is:

Z₁₂ = V₁ / I₂ (with I₁ = 0)

Step 3: Calculate Z₂₁

Z₂₁ is found by setting I₁ = 0 (open-circuit input). Under this condition, the forward transfer impedance is:

Z₂₁ = V₂ / I₁ (with I₂ = 0)

Step 4: Calculate Z₂₂

Z₂₂ is found by setting I₁ = 0 (open-circuit input). Under this condition, the output impedance is:

Z₂₂ = V₂ / I₂ (with I₁ = 0)

Let's now solve these for the given options:

Correct Option Analysis:

The correct option is:

Option 3: 30Ω, 20Ω, 20Ω, 30Ω

We will validate this by calculating each of the Z parameters for the given network:

Z₁₁:

Given that I₂ = 0, V₁ = Z₁₁I₁. If Z₁₁ = 30Ω, then:

V₁ = 30Ω × I₁

Z₁₂:

Given that I₂ = 0 and assuming Z₁₂ = 20Ω, then:

V₁ = 20Ω × I₂

Z₂₁:

Given that I₁ = 0 and assuming Z₂₁ = 20Ω, then:

V₂ = 20Ω × I₁

Z₂₂:

Given that I₁ = 0, V₂ = Z₂₂I₂. If Z₂₂ = 30Ω, then:

V₂ = 30Ω × I₂

Thus, the Z-parameters are correctly given by option 3: Z₁₁ = 30Ω, Z₁₂ = 20Ω, Z₂₁ = 20Ω, Z₂₂ = 30Ω.

Let's analyze the other options to understand why they are incorrect:

Option 1: 40Ω, 20Ω, 20Ω, 40Ω

While Z₁₂ and Z₂₁ are the same as in the correct answer, Z₁₁ and Z₂₂ are different. For the given network, these values do not match the correct Z-parameter values.

Option 2: 40Ω, 30Ω, 30Ω, 40Ω

All the Z-parameter values here differ from the correct option. These values do not satisfy the given network's conditions.

Option 4: 30Ω, 30Ω, 30Ω, 30Ω

In this case, Z₁₂ and Z₂₁ are incorrectly given as 30Ω instead of 20Ω. This does not align with the correct Z-parameter values.

Conclusion:

By understanding the principles of Z-parameters and carefully analyzing the network, we have determined that the correct Z-parameters for the given network are Z₁₁ = 30Ω, Z₁₂ = 20Ω, Z₂₁ = 20Ω, and Z₂₂ = 30Ω, which corresponds to option 3.

Concept:

ABCD Parameter:

We know that the ABCD parameter of the transmission line

\(\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_2}}\\ { - {I_2}} \end{array}} \right]\ \)

A = D

AD – BC = 1

\(\left[{\begin{array}{*{20}{c}} A&B\\ C&D \end{array}}\right]\)= \(\left[\begin{array}{ll} 1 & 8 \\ 6 & 3 \end{array}\right]\)

So, B = 8

Conditions of reciprocity and symmetry in terms of different two-port parameters are:

Concept:

The T-parameters of the cascaded network are obtained by multiplying the T-parameters of the individual networks.

For circuit:

\(T=\left[ \begin{matrix} 1 & {{Z}_{1}} \\ 0 & 1 \\ \end{matrix} \right]\;\)

For circuit

\(T=\left[ \begin{matrix} 1 & 0 \\ {{y}_{2}} & 1 \\ \end{matrix} \right]\;\)

ABCD parameters of the transformer circuit

\(T = \left[ {\begin{array}{*{20}{c}} {\frac{1}{a}}&0\\ 0&a \end{array}} \right]\)

Calculation:

\(T = \begin{bmatrix} 1 & 0\\ 1/5 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix} \begin{bmatrix} 1/2 & 0\\ 0 & 2\end{bmatrix} \begin{bmatrix} 1 & 0\\ 1/5 & 1\end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1\end{bmatrix}\)

\(T = \begin{bmatrix} 1 & 2\\ 1/5 & 7/5\end{bmatrix} \begin{bmatrix} 1/2 & 0\\ 0 & 2\end{bmatrix} \begin{bmatrix} 1 & 2\\ 1/5 & 7/5\end{bmatrix}\)

\(T = \begin{bmatrix} 1 & 2\\ 1/5 & 7/5\end{bmatrix} \begin{bmatrix} 1/2 & 1\\ 2/5 & 14/5\end{bmatrix}\)

\(T=\begin{bmatrix} 1.3 & 6.6\\ 0.66 & 4.12\end{bmatrix}\)

Comparing with T-parameter

\(T=\begin{bmatrix} A & B\\ C & D\end{bmatrix}\)

A = 1.3

B = 6.6 

C = 0.66

D = 4.12

ABCD or Transmission Parameter :

Dependent Variable  : V₁ and I₁

Independent Variable : V₂ and I₂

V₁ = A V₂ - BI₂

I₁ = CV₂ - DI₂

abcd or Inverse Transmission Parameter :

Dependent Variable  : V2 and I2

Independent Variable : V1 and I1

V2 = a V1 - b I1

I2 = cV1 - dI1

Concept:

Transmission or ABCD parameters

General two-port network and Transmission parameters two-port network is shown below.

Transmission 2 port network

The direction of I2 is opposite to the standard form so it is taken as negative.

Transmission parameters are defined as:

\(\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_2}}\\ { - {I_2}} \end{array}} \right]\)

V1 and I1 are dependent and V2 and I2 are independent.

Calculation:

From the ABCD parameters matrix, we get the following equations.

V1 = A V2 – B I2   ---(1)

I1 = C V2 - D I2   ---(2)

To calculate C, we open circuit the output terminal as shown:

Using Equation (1), we get:

I1 = C V2 

\(C=\frac{I_1}{V_2}\)

Since I₂ = 0, the current across Z_b will be I₁, i.e.

V₂ = I₁ × Z_b

\(\frac{I_1}{V_2}=C=\frac{1}{Z_b}\)

Conditions for symmetry and reciprocity:

The two-port networks of n : 1 ideal transformer is shown below.

V1 = nV2

\({I_2} = - n{I_1} \Rightarrow {I_1} = \frac{{ - 1}}{n}{I_2}\)

ABCD parameters of a two-port network can be represented as follows.

V1 = A V2 – B I2

I1 = C V2 – D I2

The equations of ideal transformer are

V1 = n V2 – 0 I2

\({I_1} = 0{V_2} - \frac{1}{n}{I_2}\)

By comparing with standard equations,

A = n, B = 0, c = 0, \(D = \frac{1}{n}\)

\(\left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} n&0\\ 0&{\frac{1}{n}} \end{array}} \right]\)

The given matrix describes the transformer in terms of ABCD parameters.

Concept:

We know that the ABCD parameter of transmission line 

\(\rm \left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_2}}\\ { - {I_2}} \end{array}} \right] \)

For symmetrical network:

A = D

For reciprocity network:

AD – BC

Analysis:

For symmetrical network A = D = 3

For bilateral AD – BC = 1

9 – C = 1

C = 8

Conditions of reciprocity and symmetry in terms of different two-port parameters are:

Concept:

ABCD Parameter:

We know that the ABCD parameter of the transmission line

\(\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_2}}\\ { - {I_2}} \end{array}} \right]\ \)

A = D

AD – BC = 1

\(\left[{\begin{array}{*{20}{c}} A&B\\ C&D \end{array}}\right]\)= \(\left[\begin{array}{ll} 1 & 8 \\ 6 & 3 \end{array}\right]\)

So, B = 8

Conditions of reciprocity and symmetry in terms of different two-port parameters are:

Concept:

There are four variables V₁, V₂, I₁, and I₂ in a two-port network as shown in the figure:

\(\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_2}}\\ { - {I_2}} \end{array}} \right]\)

Where,

\(A= {\left. {\frac{{{V_1}}}{{{V_2}}}} \right|_{{I_2} = 0}}\)

\(B=- {\left. {\frac{{{V_1}}}{{{I_2}}}} \right|_{{V_2} = 0}}\)

\(C= {\left. {\frac{{{I_1}}}{{{V_2}}}} \right|_{{I_2} = 0}}\)

\(D=\; - {\left. {\frac{{{I_1}}}{{{I_2}}}} \right|_{{V_2} = 0}}\)

• T parameters are called as the transmission parameters or ABCD parameters (Reason is true).
• The parameters, A and D do not have any units, since these are dimensionless.
• The units of parameters, B and C are ohm and mho respectively.
• These parameters are used for the analysis of an electrical network.
•  It is also used for determining the performance of input voltage, output voltage, and current of the transmission network. (Assertion is true).
• ABCD parameters are also known as transmission parameters because they provide a link between supply and receiving end voltages and currents considering the circuit elements to be linear. (Statement 2 is not the correct explanation to statement 1)

Given that a two port network N has transmission parameters \(\left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]\)

Then the equations will be

\({V_1} = A{V_2} + B{I_2}\)

\({I_1} = C{V_2} + D{I_2}\)

The input impedance of the network at port 1 is,

\(\frac{{{V_1}}}{{{I_1}}}\;at\;{I_2} = 0\)

Concept:

Application:

Given \({\rm{T}} = \left[ {\begin{array}{*{20}{c}} {\rm{A}}&{\rm{B}}\\ {\rm{C}}&{\rm{D}} \end{array}} \right]\)

The two-port network is symmetric if A = D and reciprocal if AD – BC = 1

Concept:

ABCD parameter equation is defined as:

V1 = AV2 – BI2

I1 = CV2 – DI2

Application:

\({\rm{B}} = {\left. {\frac{{{{\rm{V}}_1}}}{{ - {{\rm{I}}_2}}}} \right|_{{{\rm{V}}_2} = 0}}\)

From the above network, we can write:

\({{\rm{I}}_1} = \frac{{{{\rm{V}}_1}}}{{{{\rm{R}}_{{\rm{eq}}}}}} = \frac{{{{\rm{V}}_1}}}{{2 + 5||2}}\)

\(- {{\rm{I}}_2} = {{\rm{I}}_1} \times \frac{5}{7} \)

\(-I_2= \frac{{5{{\rm{V}}_1}}}{{24}}\)

\(-I_2= \frac{5}{7} \times \frac{{{{\rm{V}}_1}}}{{2 + \frac{{10}}{7}}} \)

\(\frac{{{{\rm{V}}_1}}}{{ - {{\rm{I}}_2}}} = \frac{{24}}{5} = 4.8 = {\rm{B}}\)

Concept:

The equivalent transmission parameter for the cascade network is the product of the individual transmission parameter.

\({\left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]_{Total}} = {\left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]_1}{\left[ {\begin{array}{*{20}{c}} A&B\\ C&D \end{array}} \right]_2}\)

Calculation:

We can write as follows.

\(\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{A_1}}&{{B_1}}\\ {{C_1}}&{{D_1}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{A_2}}&{{B_2}}\\ {{C_2}}&{{D_2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_3}}\\ {{I_3}} \end{array}} \right]\)

\(\left[ {\begin{array}{*{20}{c}} {{V_1}}\\ {{I_1}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\left( {{A_1}{A_2} + {B_1}{C_2}} \right)}&{\left( {{A_1}{B_2} + {B_1}{D_2}} \right)}\\ {\left( {{C_1}{A_2} + {D_1}{C_2}} \right)}&{\left( {{C_1}{B_2} + {D_1}{D_2}} \right)} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_3}}\\ {{I_3}} \end{array}} \right]\)

For Z_th, V₁ = 0

\(\Rightarrow {Z_{th}} = \frac{{{V_3}}}{{\left( { - {I_3}} \right)}}\)

V₁ = (A₁A₂ + B₁C₂)V₃ + (A₁B₂ + B₁D₂)I₃ = 0

\({Z_{th}} = \frac{{{V_3}}}{{\left( { - {I_3}} \right)}} = \frac{{{A_1}{B_2} + {B_1}{D_2}}}{{{A_1}{A_2} + {B_1}{C_2}}}\)

For V_th, I₃ = 0, V₃ = V_th

V₁ = (A₁A₂ + B₁C₂)V_th