# Two Port Networks

 Question 1
The S-parameters of a two port network is given as

$[S]=\left[\begin{array}{ll} S_{11} & S_{12} \\ S_{21} & S_{22} \end{array}\right]$

with reference to $Z_{0}$. Two lossless transmission line sections of electrical lengths $\theta_{1}=\beta l_{1}$ and $\theta_{2}=\beta l 2$ are added to the input and output ports for measurement purposes, respectively. The S-parameters $\left[S^{\prime}\right]$ of the resultant two port network is

 A $\left[\begin{array}{cc}S_{11} e^{-j 2 \theta_{1}} & S_{12} e^{-j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{-j 2 \theta_{2}}\end{array}\right]$ B $\left[\begin{array}{cc}S_{11} e^{j 2 \theta_{1}} & S_{12} e^{-j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{j 2 \theta_{2}}\end{array}\right]$ C $\left[\begin{array}{cc}S_{1} e^{j 2 \theta_{1}} & S_{12} e^{e j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{j 2 \theta_{2}}\end{array}\right]$ D $\left[\begin{array}{cc}S_{11} e^{-j 2 \theta_{1}} & S_{12} e^{e j\left(\theta_{1}+\theta_{2}\right)} \\ S_{21} e^{j\left(\theta_{1}+\theta_{2}\right)} & S_{22} e^{-j 2 \theta_{2}}\end{array}\right]$
GATE EC 2023   Network Theory
Question 1 Explanation:
Let us evaluate $S_{11}$ and $S_{21}$ first at $V_{2}^{+}=0$

(A) $S11:$
\begin{aligned} V_{1}&=V_{1}^{+}+V_{1}^{-}=\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}+\left(V_{1}^{-}\right) e^{+j \beta l_{1}} \\ \therefore \quad V_{1}^{+}&=\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}} \\ V_{1}^{-}&=\left(V_{1}^{-}\right) e^{+j \beta l_{1}} \\ S_{11}&=\frac{V_{1}^{-}}{V_{1}^{+}}=\frac{\left(V_{1}^{-}\right) e^{+j \beta l_{1}}}{\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}}=S_{11}^{\prime} e^{+j 2 \beta l_{1}} \\ \Rightarrow \quad S_{11}^{\prime}&=S_{11} e^{-j 2 \beta l_{1}}=S_{11} e^{-j 2 \theta_{1}} \\ \therefore \quad S_{11}^{\prime}&=S_{11} e^{-j 2 \theta_{1}} \end{aligned}

(B) $S_{21}$ :
$V_{2}=V_{2}^{+} e^{+j \beta l_{2}}+V_{2}^{-} e^{-j \beta l_{2}}=\left(V_{2}^{+}\right)^{\prime}+\left(V_{2}^{-}\right)^{\prime}$
Here, $V_{2}^{+}=0$
Hence,
$V_{2}=\left(V_{2}^{-}\right)^{\prime}=V_{2}^{-} e^{-j \beta l_{2}}$
$\Rightarrow \quad V_{2}^{-}=\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}}$

From previous discussion in $S_{11}$,
\begin{aligned} V_{1}^{+}&=\left(V_{1}^{+}\right) e^{-j l_{1}} \\ \therefore \quad S_{21}&=\frac{V_{2}^{-}}{V_{1}^{+}}=\frac{\left(V_{2}^{-}\right)^{\prime} e^{+j \beta l_{2}}}{\left(V_{1}^{+}\right)^{\prime} e^{-j \beta l_{1}}}=S_{21}^{\prime} e^{+j\left(\beta l_{2}+\beta l_{1}\right)} \\ \Rightarrow S_{21}^{\prime}&=S_{21} e^{-j\left(\beta l_{2}+\beta l_{1}\right)} \\ \Rightarrow \quad S_{21}^{\prime}&=S_{21} e^{-j\left(\theta_{1}+\theta_{2}\right)} \end{aligned}
 Question 2
The h-parameters of a two port network are shown below. The condition for the maximum small signal voltage gain $\frac{V_{\text {out }}}{V_{s}}$ is

 A $h_{11}=0, h_{12}=0, h_{21}=$ very high and $h_{22}=0$ B $h_{11}=$ very high, $h_{12}=0, h_{21}=$ very high and $h_{22}=0$ C $h_{11}=0, h_{12}=$ very high, $h_{21}=$ very high and $h_{22}=0$ D $h_{11}=0, h_{12}=0, h_{21}=$ very high and $h_{22}=$ very high
GATE EC 2023   Network Theory
Question 2 Explanation:
Dependent current source should have $h_{21} I_{1}$ instead of $h_{21} V_{1}$ according to $h$-parameter.
$A_{V}=\frac{V_{\text {out }}}{V_{s}}=\frac{-h_{21} I_{1} \times\left(\frac{1}{h_{22}} \| R_{L}\right)}{h_{11} I_{1}+h_{12} V_{2}}$

To achieve maximum $\frac{V_{\text {out }}}{V_{s}}$
\begin{aligned} & h_{11}=0, \quad h_{12}=0 \\ & h_{21}=\text { Very high, } \quad h_{22}=0 \end{aligned}

Hence, answer should be (A) according to $h_{21} I_{1}$.

 Question 3
For the two port network shown below, the $[Y]$-parameters is given as

$[Y]=\frac{1}{100}\left[\begin{array}{cc} 2 & -1 \\ -1 & 4 / 3 \end{array}\right] S$

The value of load impedance $Z_{L}$, in $\Omega$, for maximum power transfer will be ____ (rounded off to the nearest integer).

 A 20 B 80 C 120 D 180
GATE EC 2023   Network Theory
Question 3 Explanation:
$[Y]=\left[\begin{array}{cc} \frac{2}{100} & -\frac{1}{100} \\ -\frac{1}{100} & \frac{4}{300} \end{array}\right]$
For the given $Y$-parameter the two-port network is

$Y_{11}=Y_{a}+Y_{b}=\frac{2}{100}$[
$Y_{12}=Y_{21}=-Y_{b}=-\frac{1}{100}$
$Y_{22}=Y_{b}+Y_{c}=\frac{4}{300}$

On solving,
$\quad Y_{b}=\frac{1}{100} S$
$Y_{a}=\frac{1}{100} S$
$Y_{c}=\frac{1}{300} S$

The network becomes,

Converting $\Delta$ - to $\lambda$

$Z_{Th}=60+[(20+10)||60]=60+\frac{30 \times 60}{30+60}=80 \Omega$

For maximum power transfer,
$Z_{L}=Z_{\text {Th }}=80 \Omega$
 Question 4
A linear 2-port network is shown in Fig. (a). An ideal DC voltage source of 10 V is connected across Port 1. A variable resistance R is connected across Port 2. As R is varied, the measured voltage and current at Port 2 is shown in Fig. (b) as a $V_2$ versus $-I_2$ plot. Note that for $V_2=5V,I_2=0mA$, and for $V_2=4V,I_2=-4mA$.
When the variable resistance R at Port 2 is replaced by the load shown in Fig. (c), the current $I_2$ is _______ mA (rounded off to one decimal place).

 A 3.2 B 6.8 C 4 D 8
GATE EC 2022   Network Theory
Question 4 Explanation:

CASE-1:
$I_2=0$
$V_{oc}=5V$

CASE-2: With help of linearity
$\frac{4-5}{4-0}=\frac{I_{sc}-5}{I_{sc}-0}\Rightarrow I_{sc}=20mA$
$R_{th}=\frac{V_{oc}}{I_{sc}}=\frac{5}{20}k=250\Omega =0.25k\Omega$

CASE-3

$I_2=\frac{10-5}{1.25}=4mA$
 Question 5
Consider the two-port network shown in the figure.

The admittance parameters, in siemens, are
 A $y_{11}=2,\:y_{12}=-4,\:y_{21}=-4,\:y_{22}=2$ B $y_{11}=1,\:y_{12}=-2,\:y_{21}=-1,\:y_{22}=3$ C $y_{11}=2,\:y_{12}=-4,\:y_{21}=-1,\:y_{22}=2$ D $y_{11}=2,\:y_{12}=-4,\:y_{21}=-4,\:y_{22}=3$
GATE EC 2021   Network Theory
Question 5 Explanation:

Write KCL at $V_{1}$
\begin{aligned} I_{1}&=\frac{V_{1}}{1}+\frac{V_{1}-V_{2}}{1}-3 V_{2} \\ I_{1}&=2 V_{1}-4 V_{2} \\ \text{Write KCL at }V_{2}\qquad I_{2}&=\frac{V_{2}}{1}+\frac{V_{2}-V_{1}}{1}\\ I_{2}&=-V_{1}+2 V_{2} \\ [y]&=\left[\begin{array}{cc} 2 & -4 \\ -1 & 2 \end{array}\right] \mho \end{aligned}

There are 5 questions to complete.