Network Functions

Question 1
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 1 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 2
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} is
A
\frac{R}{4}
B
\frac{R}{2}
C
R
D
2R
GATE EC 2009   Network Theory
Question 2 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 3
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 3 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 4
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}} is
A
4
B
1
C
1/2
D
1/4
GATE EC 2007   Network Theory
Question 4 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}
Question 5
The RC circuit shown in the figure is
A
a low-pass filter
B
a high-pass filter
C
a band-pass filter
D
a band-reject filter
GATE EC 2007   Network Theory
Question 5 Explanation: 
At \omega \rightarrow \infty , Capacitor \rightarrow short circuited
Circuit looks like,


at\omega \rightarrow 0, Capacitor \rightarrow open circuited
Circuit looks like


So frequency response of the circuit will be,

So the circuit is Band pass filter
Question 6
A negative resistance R_{neg} is connected to a passive network N having driving point impedance Z_{1}(s) as shown below. For Z_{2}(s) to be positive real,
A
|R_{neg}|\leq Re Z_{1}(j\omega ),\forall \omega
B
|R_{neg}|\leq |Z_{1}(j\omega )|,\forall \omega
C
|R_{neg}|\leq Im Z_{1}(j\omega ),\forall \omega
D
|R_{neg}|\leq \angle Z_{1}(j\omega ),\forall \omega
GATE EC 2006   Network Theory
Question 6 Explanation: 
For Z_{2}(s) to be positive real,
\begin{aligned} \operatorname{Re}(Z(s)) & \geq\left|R_{\operatorname{nog}}\right| \\ \Rightarrow \quad\left|R_{\operatorname{neg}}\right| & \leq \operatorname{Re}\left\{Z_{1}(j \omega)\right\}, \text { for all } \omega \end{aligned}
Question 7
The first and the last critical frequencies (singularities) of a driving point impedance function of a passive network having two kinds of elements, are a pole and a zero respectively. The above property will be satisfied by
A
RL network only
B
RC network only
C
LC network only
D
RC as well as RL networks
GATE EC 2006   Network Theory
Question 7 Explanation: 
RC impedance function has
(1) 1^{st} critical frequency due to pole
(2) Last critical frequency due to pole
Question 8
The first and the last critical frequency of an RC-driving point impedance function must respectively be
A
a zero and a pole
B
a zero and a zero
C
a pole and a pole
D
a pole and a zero
GATE EC 2005   Network Theory
Question 8 Explanation: 
For stability Poles and zero interlace on real axis. Since its RC, first pole should come and zero at last.
Question 9
The driving point impedance Z(s) of a network has the pole-zero locations as shown in the figure. If Z(0) = 3, then Z(s) is
A
\frac{3(s+3)}{s^{2}+2s+3}
B
\frac{2(s+3)}{s^{2}+2s+2}
C
\frac{3(s-3)}{s^{2}-2s-2}
D
\frac{2(s-3)}{s^{2}-2s-3}
GATE EC 2003   Network Theory
Question 9 Explanation: 
\begin{array}{l} Z(s)=\frac{K(s-z)}{\left(s-p_{1}\right)\left(s-p_{2}\right)}=\frac{K(s+3)}{(s+1+j)(s+1-j)} \\ Z(s)=\frac{K(s+3)}{(s+1)^{2}-j 2}=\frac{K(s+3)}{(s+1)^{2}+1} \\ Z(0)|_{\omega=0}=0 \\ \Rightarrow \frac{3 K}{2}=3 \\ \Rightarrow \quad K=2 \\ \therefore \quad Z(s)=\frac{2(s+3)}{s^{2}+2 s+2} \end{array}
There are 9 questions to complete.
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