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
[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
\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] | |
\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] | |
\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] | |
\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] |
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}
(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
h_{11}=0, h_{12}=0, h_{21}= very high and h_{22}=0 | |
h_{11}= very high, h_{12}=0, h_{21}= very high and h_{22}=0 | |
h_{11}=0, h_{12}= very high, h_{21}= very high and h_{22}=0 | |
h_{11}=0, h_{12}=0, h_{21}= very high and h_{22}= very high |
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}.
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).
[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).
20 | |
80 | |
120 | |
180 |
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
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).
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).
3.2 | |
6.8 | |
4 | |
8 |
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
The admittance parameters, in siemens, are
y_{11}=2,\:y_{12}=-4,\:y_{21}=-4,\:y_{22}=2 | |
y_{11}=1,\:y_{12}=-2,\:y_{21}=-1,\:y_{22}=3 | |
y_{11}=2,\:y_{12}=-4,\:y_{21}=-1,\:y_{22}=2 | |
y_{11}=2,\:y_{12}=-4,\:y_{21}=-4,\:y_{22}=3 |
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.