Concept:

Application:

For the given network, Y₁₂ ≠ Y₂₁. So the given network is non-reciprocal.

Y₁₁ = Y₂₂, so the given network is symmetric.

z₁₁ is obtained by open-circuiting the output voltage, i.e.

\({z_{11}} = {\left. {\frac{{{V_1}}}{{{I_1}}}} \right|_{{I_2} = 0}}\)

On open circuiting the output terminal, we observe that the 20 Ω and 40 Ω resistances are connected in series.

Also, 10 Ω and 30 Ω resistances are connected in series.

The circuit is redrawn as:

\(\frac{{{V_1}}}{{{R_1}}} = {z_{11}}\) = Parallel combination of the two resistancs, i.e.

z₁₁ = (10 + 30)||(20 + 40)

z₁₁ =  40 || 60

\({z_{11}} = \frac{{40\; \times \;60}}{{100}}\;{\rm{\Omega }}\)

z₁₁ = 24 Ω

Concept:

• States can be considered as variables that carry sufficient information about the history of the system.\
• A purely resistive network has no states at all as it has no storage property.
• Only a network with energy storage elements or reactive elements will have states as they store the information about the past states.

​Number of the states required = Order of the network

Also, the Order of the network = Number of energy storage elements

Since the given circuit has two energy storage elements (L and C), the minimum number of required states will be 2.

h-parameter equation:

V₁ = h₁₁ I₁ + h₁₂ V₂

I₂ = h₂₁ I₁ + h₂₂ V₂

When Port-2 is short circuited:

\({h_{11}} = {\left. {\frac{{{V_1}}}{{{I_1}}}} \right|_{{V_2} = 0}} = {R_1}||{R_2}\)

⇒ R₁||R₂ = 4 Ω 

Concept:

Thevenin voltage is the open-circuit voltage at the terminal, i.e. V_th is obtained by open circuiting the terminal and calculating the voltage across it.

Application:

To evaluate the Thevenin’s voltage, the given port network is redrawn as:

The z-parameter equation for a two-port network is:

V₁ = z₁₁ I₁ + z₁₂ I₂         ---(1)

V₂ = V_th = z₂₁ I₁ + z₂₂ I₂        ---(2)

Substituting I₂ = 0 in equation (1), we get:

V₁ = z₁₁ I₁   ---(3)

From the given network, we can write:

V_s – Z_s I₁ = V₁

Using Equation (3), this can be written as:

V_s - Z_s I₁ = z₁₁ R₁

V_s = R₁ (z₁₁ + z_s)

\({I_1} = \frac{{{V_s}}}{{{z_{11}} + {Z_s}}}\)

Substituting the above in equation (2) we get:

\({V_{th}} = {z_{21}}\left( {\frac{{{V_s}}}{{{z_{11}} + {Z_s}}}} \right) + 0\)

\({V_{th}} = \frac{{{V_s}\;{z_{21}}}}{{{z_{11}} + {Z_s}}}\)

Concept:

The inductive impedance for a frequency ω is given by:

jX_L = jωL

The capacitive impedance for a frequency ω is given by:

\(j{X_L} = \frac{1}{{j\omega L}}\)

For ω = 3 rad/s, the circuit is redrawn as:

Method I:

Applying KVL at the input side, we get:

V₁ – (6 + 6j) I₁ + 4j (I₁ + I₂) = 0

V₁ = (6 + 6j – 4j) I₁ – 4j I₂

V₁ = (6 + 2j) I₁ – 4j I₂    ---(1)

Applying KVL at two output side, we get:

V₂ – (-4j)(I₁ + I₂) = 0

⇒ V₂ = -4j I₁ – 4j I₂     ---(2)

Comparing equations (1) and (2) with standard z-parameter equations, we get:

z₁₁ = 6 + 2j

z₁₂ = -4j

z₂₁ = -4j

z₂₂ = -4j

Alternate Method:

Since the two port is linear and has no dependent source, it is a reciprocal network, i.e. z₁₂ = z₂₁

Also, for a reciprocal network, the T-equivalent circuit is drawn as:

Comparing this with the given two-port T - network, we can write:

z₁₂ = z₂₁ = -4j Ω

z₁₁ – z₁₂ = 6 + 6j

with z₁₂ = -4j

z₁₁ + 4j = 6 + 6j

z₁₁ = 6 + 2j

Also, z₂₂ – z₁₂ = 0

z₂₂ = z₁₂ = -4j Ω  

Concept:

The voltage across the inductor is given by:

\(V_L=L\frac{di}{dt}\)

L = Inductance of the inductor

Also, the current across a capacitor is expressed as:

\(i=C\frac{dV}{dt}\)

Calculation:

Applying KVL from the inductor voltage to the capacitor voltage V_c(t), we can write:

\(\frac{1}{5}\frac{{d{i_L}}}{{dt}} = \frac{1}{2}\left[ {i\left( t \right) - {i_L}\left( t \right)} \right] + {V_C}\left( t \right)\)

The above equation can be expressed as:

\({{\bar i}_L} = - \frac{5}{2}{i_L} + 5\;{V_C} + \frac{{5i}}{2}\)

Similarly, the current across the capacitor will be:

\(\frac{{d{V_C}}}{{dt}} = i\left( t \right) - {i_L}\left( t \right) - 2{V_C}\left( t \right)\)

This can be written as:

\({{\bar V}_C} = - {i_L} - 2{V_C} + i\)

The state matrix is, therefore:

\(\\ \left[ {\begin{array}{*{20}{c}} {{{\bar i}_L}}\\ {{{\bar V}_C}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \frac{5}{2}}&5\\ { - 1}&{ - 2} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{i_L}}\\ {{V_C}} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {\frac{5}{2}}&1 \end{array}} \right]i \)

Concept:

ABCD parameters are defined as:

V1 = A V2 – B I2

I1 = C V2 – D I2

Application:

With A = 1, B = -j, C = 1 and D = 1 - j, the ABCD equations can be written as:

V1 = V2 - (- j) I2   ---(1)

I1 = V2 – (1 - j) I2   ---(2)

The given two-port network is redrawn as:

Z₁ can be obtained by short-circuiting the output, i.e.

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

Applying KVL at the output, we get

V₁ + I₂Z₁ = 0

V₁ = -I₂Z₁

\(B = {Z_1} = \frac{{ {V_1}}}{{{-I_2}}}\)

Comparing this with the given ABCD matrix of Equation (1), we get:

B = Z₁ = - j

∴ Z₁ = - j Ω 

Concept:

The z-parameter equations for a two-port network is given by:

V₁ = z₁₁ I₁ + z₁₂ I₂

V₂ = z₂₁ I₂ + z₂₂ I₂

Where the voltages and current can be in phasors as well.

Application:

V₁ = 100 ∠0

And V₂ = -10 I₂

Where bold letter V/I represents the phasor voltage and current respectively.

Substituting V₁ and V₂ in the standard 2-port z-parameter equations, we get:

100 = 40 I₁ + j20 I₂      ---(1)

-10 I₂ = j30 I₁ + 50 I₂

-60 I₂ = j30 I₁

\({I_2} = \frac{{ - j}}{2}{I_1}\)

Substituting I₂ in equation (1), we get:

\(\Rightarrow 100 = 40\;{I_1} + j20\;\left( {\frac{{ - j}}{2}{I_1}} \right)\)

⇒ 100 = 40 I₁  + 10I₁

\({I_1} = \frac{{100\angle 0^\circ }}{{50}} = 2\angle 0^\circ \;A\)

\({I_2} = 2\angle 0^\circ \times \frac{{ - j}}{2}\)

\(= \frac{{2\angle 0^\circ }}{2} \times \angle - 90^\circ \)

I₂ = 1∠-90° A

Concept:

The transmission parameter relates the variables at the input port to those at the output port.

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

V1 = AV2 - BI2

I1 = CV2 - DI2

Also, given ABCD parameters, the z and y-parameter matrix is evaluated as:

\(\left[ z \right] = \left[ {\begin{array}{*{20}{c}} {{z_{11}}}&{{z_{12}}}\\ {{z_{21}}}&{{z_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{A}{C}}&{\frac{{AD - BC}}{C}}\\ {\frac{1}{C}}&{\frac{D}{C}\;} \end{array}} \right]\)  

\(\left[ y \right] = \left[ {\begin{array}{*{20}{c}} {{y_{11}}}&{{y_{12}}}\\ {{y_{21}}}&{{y_{22}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{D}{B}}&{ - \frac{{AD - BC}}{B}}\\ {\frac{{ - 1}}{B}}&{\frac{A}{B}} \end{array}} \right]\)

Application:

For a transformer:

\(\frac{{{V_1}}}{{{V_2}}} = n\)

V₁ = n V₂

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

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

We observe that it is easier to represent this in terms of ABCD parameters as:

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

Since \({z_{11}} = \frac{A}{C}\) and \({y_{11}} = \frac{D}{B}\), with B = C = 0, neither z nor y parameters are defined for this network.