Network Functions - PracticePaper

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Question 1

A series

R L C

circuit has a quality factor

Q

of 1000 at a center frequency of

10^{6} \mathrm{rad} / \mathrm{s}

. The possible values of

R, L

and

C

are

A	$R=1 \Omega, L=1 \mu \mathrm{H}$ and $C=1 \mu \mathrm{F}$
B	$R=0.1 \Omega, L=1 \mu \mathrm{H}$ and $C=1 \mu \mathrm{F}$
C	$R=0.01 \Omega, L=1 \mu \mathrm{H}$ and $C=1 \mu \mathrm{F}$
D	$R=0.001 \Omega, L=1 \mu \mathrm{H}$ and $C=1 \mu \mathrm{F}$

GATE EC 2023 Network Theory

Question 1 Explanation:

Given:

Q=1000

and

\omega_{0}=10^{6} \mathrm{rad} / \mathrm{sec}

We know, for series

R L C

circuit,

\begin{aligned} Q & =\frac{\omega_{0} L}{R} \\ \omega_{0} & =\sqrt{\frac{1}{L C}} \\ Q & =\frac{1}{\sqrt{L C}} \times \frac{1}{R}=\frac{1}{R} \sqrt{\frac{L}{C}} \end{aligned}

So,

L=1 \mu \mathrm{H}, C=1 \mu \mathrm{F}

and

R=0.001

Question 2

The transfer function

\frac{V_{2}\left ( s \right )}{V_{1}\left ( s \right )}

of the circuit shown below is

A	$\frac{0.5s+1}{s+1}$
B	$\frac{3s+6}{s+2}$
C	$\frac{s+2}{s+1}$
D	$\frac{s+1}{s+2}$

GATE EC 2013 Network Theory

Question 2 Explanation:

\begin{aligned} \frac{V_{2}(s)}{V_{1}(s)}&=\frac{10 \times 10^{3}+\frac{1}{100 \times 10^{-6} s}}{10 \times 10^{3}+\frac{1}{100 \times 10^{-6} s}+\frac{1}{100 \times 10^{-6} s}} \\ \frac{V_{2}(s)}{V_{1}(s)}&=\frac{s \times 10^{4}+10^{4}}{s \times 10^{4}+10^{4}+10^{4}}=\frac{10^{4}(1+s)}{10^{4}(s+2)} \\ \frac{v_{2}(s)}{V_{1}(s)}&=\frac{s+1}{s+2} \end{aligned}

Question 3

If the transfer function of the following network is

\frac{V_o(s)}{V_i(s)}=\frac{1}{2+sCR}

The value of the load resistance

R_{L}

A	$\frac{R}{4}$
B	$\frac{R}{2}$
C	R
D	2R

GATE EC 2009 Network Theory

Question 3 Explanation:

\begin{aligned} Z &=\frac{R_{L}}{1+S R_{L} C} \\ H(s) &=\frac{Z}{Z+R}=\frac{R_{L}}{\left(R_{L}+R\right)+S R R_{L} C} \\ \text{if}\quad R &=R_{L} \\ H(s) &=\frac{1}{2+s R C} \end{aligned}

Question 4

The driving point impedance of the following network is given by

Z(s)=\frac{0.2s}{s^{2}+0.1s+2}

The component values are

A	L = 5 H,R = 0.5 $\Omega$ ,C = 0.1 F
B	L = 0.1H,R = 0.5 $\Omega$ ,C = 5 F
C	L = 5 H,R = 2 $\Omega$ ,C = 0.1 F
D	L = 0.1H,R = 2 $\Omega$ ,C = 5 F

GATE EC 2008 Network Theory

Question 4 Explanation:

\begin{aligned} Z(s) &=\frac{0.2 s}{s^{2}+0.1 s+2} \\ r(s) &=\frac{s^{2}+0.1 s+2}{0.2 s} \\ &=\frac{s}{0.2}+\frac{1}{2}+\frac{2}{0.2 s} \\ &=5 s+0.5+\frac{10}{s}\\ \text{Comparing with}\\ n(s) &=C s+\frac{1}{R}+\frac{1}{L s} \\ C &=5 F, R=\frac{1}{0.5}=2 \Omega \\ L &=\frac{1}{10}=0.1 \mathrm{H} \end{aligned}

Question 5

Two series resonant filters are as shown in the figure. Let the 3-dB bandwidth of Filter 1 be

B_1

and that of Filter 2 be

B_2

. The value

\frac{B_{1}}{B_{2}}

A	4
B	1
C	$1/2$
D	$1/4$

GATE EC 2007 Network Theory

Question 5 Explanation:

Bandwidth of series RLC circuit is R/L
Bandwidth of filter 1;

B_{1}=\frac{R}{L_{1}}

Bandwidth of filter 2;

B_{2}=\frac{R}{L_{2}}

=\frac{R}{L_{1} / 4}=\frac{4 R}{L_{1}}

So,

\frac{B_{1}}{B_{2}}=\frac{1}{4}

There are 5 questions to complete.