Question 1 |
The current i_{b} through the base of a silicon npn transistor is 1 + 0.1 cos(10000 \pi t) mA. At 300 K, the r_{\pi} in the small signal model of the transistor is
250 \Omega | |
27.5 \Omega | |
25 \Omega | |
22.5 \Omega |
Question 1 Explanation:
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
\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
Question 2 |
The power spectral density of a real process X(t) for positive frequencies is shown below. The
values of E[X^{2}(t)] and |E[X(t)]|, respectively, are
6000/\pi,0 | |
6400/\pi,0 | |
6400/\pi,20(\pi /\sqrt{2}) | |
6000/\pi,20(\pi /\sqrt{2}) |
Question 2 Explanation:
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.
Question 3 |
In a baseband communications link, frequencies upto 3500 Hz are used for signaling. Using a
raised cosine pulse with 75% excess bandwidth and for no inter-symbol interference, the maximum possible signaling rate in symbols per second is
1750 | |
2625 | |
4000 | |
5250 |
Question 3 Explanation:
\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
In base band transmission \rightarrow
Symbol rate = Bit rate =4000 symbols/sec
Question 4 |
A plane wave propagating in air with \vec{E}=(8\hat{a}_{x}+6\hat{a}_{y}+5\hat{a}_{z})e^{j(\omega t+3x-4y)} V/m is incident on a perfectly conducting slab positioned at 0 \leq x. The E field of the reflected wave is
(-8\hat{a}_{x}-6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t+3x+4y)} V/m | |
(-8\hat{a}_{x}+6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t+3x+4y)} V/m | |
(-8\hat{a}_{x}-6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t-3x-4y)} V/m | |
(-8\hat{a}_{x}+6\hat{a}_{y}-5\hat{a}_{z})e^{j(\omega t-3x-4y)} V/m |
Question 4 Explanation:
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}
Question 5 |
The electric field of a uniform plane electromagnetic wave in free space, along the positive
x direction, is given by \vec{E}=10(\hat{a}_{y}+j\hat{a}_{z})e^{-j25x}. The frequency and polarization of the wave, respectively, are
1.2 GHz and left circular | |
4 Hz and left circular | |
1.2 GHz and right circular | |
4 Hz and right circular |
Question 5 Explanation:
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}
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}
There are 5 questions to complete.