Which of the following transformation between the z (impedance) and h (hybrid) parameters is correct?
z-h Parameter Transformation Explained
This section details the transformation between z-parameters (impedance parameters) and h-parameters (hybrid parameters) used in analyzing electrical circuits.
Understanding Network Parameter Definitions
To find the transformation, let's first recall the defining equations for both parameter sets:
z-Parameter Equations
The z-parameters relate the input and output voltages ($V_1, V_2$) to the input and output currents ($I_1, I_2$):
- \( V_1 = z_{11} I_1 + z_{12} I_2 \)
- \( V_2 = z_{21} I_1 + z_{22} I_2 \)
In matrix form:
| \( \begin{bmatrix} V_1 \\ V_2 \end{bmatrix} \) | = | \( \begin{bmatrix} z_{11} & z_{12} \\ z_{21} & z_{22} \end{bmatrix} \) | \( \begin{bmatrix} I_1 \\ I_2 \end{bmatrix} \) |
h-Parameter Equations
The h-parameters relate the input voltage ($V_1$) and output current ($I_2$) to the input current ($I_1$) and output voltage ($V_2$):
- \( V_1 = h_{11} I_1 + h_{12} V_2 \)
- \( I_2 = h_{21} I_1 + h_{22} V_2 \)
In matrix form:
| \( \begin{bmatrix} V_1 \\ I_2 \end{bmatrix} \) | = | \( \begin{bmatrix} h_{11} & h_{12} \\ h_{21} & h_{22} \end{bmatrix} \) | \( \begin{bmatrix} I_1 \\ V_2 \end{bmatrix} \) |
Deriving z-parameters from h-parameters
We aim to express the z-parameters ($z_{11}, z_{12}, z_{21}, z_{22}$) using the h-parameters ($h_{11}, h_{12}, h_{21}, h_{22}$). Let's start by manipulating the h-parameter equations.
From the second h-parameter equation, we can solve for \( V_2 \):
\( I_2 = h_{21} I_1 + h_{22} V_2 \)
Rearranging to isolate \( V_2 \):
\( h_{22} V_2 = I_2 - h_{21} I_1 \)
Assuming \( h_{22} \neq 0 \), we get:
\( V_2 = \frac{1}{h_{22}} I_2 - \frac{h_{21}}{h_{22}} I_1 \)
Now, substitute this expression for \( V_2 \) into the first h-parameter equation ($V_1 = h_{11} I_1 + h_{12} V_2$):
\( V_1 = h_{11} I_1 + h_{12} \left( \frac{1}{h_{22}} I_2 - \frac{h_{21}}{h_{22}} I_1 \right) \)
Distribute \( h_{12} \):
\( V_1 = h_{11} I_1 + \frac{h_{12}}{h_{22}} I_2 - \frac{h_{12} h_{21}}{h_{22}} I_1 \)
Group terms involving \( I_1 \) and \( I_2 \):
\( V_1 = \left( h_{11} - \frac{h_{12} h_{21}}{h_{22}} \right) I_1 + \left( \frac{h_{12}}{h_{22}} \right) I_2 \)
Combine the terms for \( I_1 \) into a single fraction:
\( V_1 = \left( \frac{h_{11} h_{22} - h_{12} h_{21}}{h_{22}} \right) I_1 + \left( \frac{h_{12}}{h_{22}} \right) I_2 \)
By comparing this equation with the z-parameter definition \( V_1 = z_{11} I_1 + z_{12} I_2 \), we can identify:
- \( z_{11} = \frac{h_{11} h_{22} - h_{12} h_{21}}{h_{22}} \)
- \( z_{12} = \frac{h_{12}}{h_{22}} \)
Now, let's compare the expression we derived for \( V_2 \) with the second z-parameter equation ($V_2 = z_{21} I_1 + z_{22} I_2$):
\( V_2 = \frac{1}{h_{22}} I_2 - \frac{h_{21}}{h_{22}} I_1 \)
Rearranging to match the standard form:
\( V_2 = \left( \frac{-h_{21}}{h_{22}} \right) I_1 + \left( \frac{1}{h_{22}} \right) I_2 \)
By comparison, we identify:
- \( z_{21} = \frac{-h_{21}}{h_{22}} \)
- \( z_{22} = \frac{1}{h_{22}} \)
Transformation Matrix Representation
Combining these results, the transformation of z-parameters in terms of h-parameters is given by the following matrix:
| \( \begin{bmatrix} z_{11} & z_{12} \\ z_{21} & z_{22} \end{bmatrix} \) | = | \( \begin{bmatrix} \rm \frac{h_{11}h_{22}-h_{12}h_{21}}{h_{22}}&\rm \frac{h_{12}}{h_{22}} \\\ \rm \frac{-h_{21}}{h_{22}}&\rm \frac{1}{h_{22}} \end{bmatrix} \) |
Identifying the Correct Transformation
Comparing this derived matrix with the options provided in the question, the correct transformation is:
| \( \begin{bmatrix} \rm \frac{h_{11}h_{22}-h_{12}h_{21}}{h_{22}}&\rm \frac{h_{12}}{h_{22}} \\\ \rm \frac{-h_{21}}{h_{22}}&\rm \frac{1}{h_{22}} \end{bmatrix} \) |
This corresponds to option 2.
Important Questions from Two Port Networks
A short-circuit admittance matrix of a two-port network is
\(\left[ {\begin{array}{} 0\\ {\frac{1}{2}} \end{array}\begin{array}{} { - \frac{1}{2}}\\ 0 \end{array}} \right]\)
The two-port network is
A two-port network has scattering parameters given \(\left[ s \right] = \left[ {\begin{array}{*{20}{c}} {{s_{11}}}&{{s_{12}}}\\ {{s_{211}}}&{{s_{22}}} \end{array}} \right]\). If the port 2 of the two-port is short-circuited, the s11 parameter for the resultant one-port network is
With 10 V dc connected at port A, the current drawn by 7 Ω connected at port B is
With 6 V dc connected at port A, 1 Ω connected at port B draws 7/3 A. If 8 V dc is connected to port A, the open circuit voltage at port B is
In a linear two – port network, when 10 V is applied to Port 1, a current of 4 A flows through Port 2 when it is short-circuited. When 5 V is applied to Port, a current of 1.25 A flows through a 1 Ω resistance connected across Port 2. When 3 V is applied to Port 1, then current (in Ampere) through a 2 Ω resistance connected across Port 2 is __________.
