Question 1 |
The transfer function \frac{V_{2}\left ( s \right )}{V_{1}\left ( s \right )} of the circuit shown below is
\frac{0.5s+1}{s+1}
| |
\frac{3s+6}{s+2}
| |
\frac{s+2}{s+1}
| |
\frac{s+1}{s+2}
|
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
The value of the load resistance R_{L} is
\frac{R}{4} | |
\frac{R}{2} | |
R | |
2R |
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
The component values are
L = 5 H,R = 0.5 \Omega,C = 0.1 F | |
L = 0.1H,R = 0.5 \Omega,C = 5 F | |
L = 5 H,R = 2 \Omega,C = 0.1 F | |
L = 0.1H,R = 2 \Omega,C = 5 F |
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
4 | |
1 | |
1/2 | |
1/4 |
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}
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 low-pass filter | |
a high-pass filter | |
a band-pass filter | |
a band-reject filter |
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
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,
|R_{neg}|\leq Re Z_{1}(j\omega ),\forall \omega | |
|R_{neg}|\leq |Z_{1}(j\omega )|,\forall \omega | |
|R_{neg}|\leq Im Z_{1}(j\omega ),\forall \omega | |
|R_{neg}|\leq \angle Z_{1}(j\omega ),\forall \omega |
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}
\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
RL network only | |
RC network only | |
LC network only | |
RC as well as RL networks |
Question 7 Explanation:
RC impedance function has
(1) 1^{st} critical frequency due to pole
(2) Last critical frequency due to pole
(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 zero and a pole | |
a zero and a zero | |
a pole and a pole | |
a pole and a zero |
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
\frac{3(s+3)}{s^{2}+2s+3}
| |
\frac{2(s+3)}{s^{2}+2s+2} | |
\frac{3(s-3)}{s^{2}-2s-2} | |
\frac{2(s-3)}{s^{2}-2s-3} |
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.