GATE EC 2012

We know that
\begin{array}{l} r_{\pi}=(\beta+1) r_{e} \\ r_{\pi}=(\beta+1) \frac{V_{T}}{I_{e}} \\ r_{\pi}=(\beta+1) \frac{V_{T}}{(\beta+1) I_{b}} \\ r_{\pi}=\frac{V_{T}}{I_{b}} \end{array}
Where I_{b} is d.c. current through base so
I_{b}=1 \mathrm{mA}
V_{T}=25 \mathrm{mV} at room temperature
So, r_{\pi}=\frac{25 \times 10^{-3}}{1 \times 10^{-3}}=25 \Omega

We know that E\left[X^{2}(t)\right] represent the total power in random signal. So
\begin{aligned} P=E\left[X^{2}(t)\right]&=2 \times \frac{1}{2 \pi} \int_{9 \times 10^{3}}^{11 \times 10^{3}} S_{x}(\omega) d \omega \\ E\left[X^{2}(t)\right]&=\frac{1}{\pi}\left[400+\frac{1}{2} \times 6 \times 2 \times 10^{3}\right] \\ E\left[X^{2}(t)\right]&=\frac{6400}{\pi} \end{aligned}
At \omega=0, there is no any frequency component present, hence dc value of the process is zero.

\begin{aligned} \mathrm{BW} &=\frac{R_{b}}{2}(1+\alpha) \\ 3500 &=\frac{R_{b}}{2}(1+0.75) \\ R_{b} &=4000 \mathrm{bits} / \mathrm{sec} \end{aligned}
In base band transmission \rightarrow
Symbol rate = Bit rate =4000 symbols/sec

Incident wave propagation \rightarrow-3 \hat{a}_{x}+4 \hat{a}_{y}
Reflected wave propagation \rightarrow 3 \hat{a}_{x}+4 \hat{a}_{y}
E_{i} incident direction \rightarrow 8 \hat{a}_{x}+6 \hat{a}_{y}+5 \hat{a}_{z}
\left.\begin{array}{l}E_{i} \text { tangential } \rightarrow 6 \hat{a}_{y}+5 \hat{a}_{z} \\ E_{r} \text { tangential } \rightarrow-6 \hat{a}_{y}-5 \hat{a}_{z}\end{array}\right\} E_{\text {tang }}=0
\left.\begin{array}{rl}E_{i} \text { normal } & =8 \hat{a}_{x} \\ E_{r} \text { normal } & =8 \hat{a}_{x}\end{array}\right\} E_{\text {normal }}=E_{\max }
E_{r} reflected direction =8 \hat{a}_{x}-6 \hat{a}_{y}-5 \hat{a}_{z}

Out of phase by 90^{\circ}and equal amplitude the wave is circularly polarized.
Check the trace of E with the time which gives left circularly polarized.
\begin{aligned} a_{y}&=\sin \omega t \\ a_{z}&=\cos \omega t \end{aligned}

Given flipflop is S-R flipflop with A= Sand B = R.
In S-R flipflop race around condition does not occur.

Output will be 1 if A>B.
If B=00 then there will be three combinations for which output will be 1 i.e. when A=01,10 , or 11.
If B=01 there will be two conditions i.e. A=10 and 11.
If B=10 there will be one condition i.e. A=11.
So total 6 combinations are there for which output will be 1 .

Applying KVL in given circuit we have
10-1000 i-v=0 \qquad\ldots(1)
since 10 V is greater than 0.7 V so current will flow through diode and is
i=\frac{v-0.7}{500}\qquad\ldots(2)
from (1) and (2) we have,
\begin{aligned} 10-2(v-0.7)-v &=0 \\ 10+1.4-3 v &=0 \\ 3 v &=11.4 \\ v &=\frac{11.4}{3}=3.8 v\\ \text{so}\quad i &=\frac{v-0.7}{500}=\frac{3.8-0.7}{500} \\ &=6.2 \mathrm{mA} \end{aligned}

Since there is no resistance so time constant will be zero. That means as soon as the switch will be closed voltages at C_{1} and C_{2} will become equal and capacitor allows sudden change of voltage only if impulse of current will pass through it.

Average power is same as RMS power.
\begin{array}{l} P=I_{\mathrm{rms}}^{2} R=\left(\frac{5}{\sqrt{2}}\right)^{2} \times 4 \\ =\frac{25}{2} \times 4=50 \mathrm{W} \end{array}
Note: Power is consumed only by resistance i.e. by real part of impedance.