# 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.