Let the relevant bandwidth $(B)$ of a digital communication system be 1 MHz and $kT=-174$ dBm/Hz, where $k$ is Boltzmann's constant and $T$ is the equivalent noise temperature of the receiver. The power $(S)$ of signal received through an additive Gaussian channel is $-80$ dBm.
Which of the following options is/are TRUE about Shannon capacity $(C)$ of the channel?

Question

An NMOS transistor operating in the linear region has $I_{DS} = 5 \, \mu{A}$ at $V_{DS} = 0.1 \, {V}$. Keeping $V_{GS}$ constant, the $V_{DS}$ is increased to $1.5 \, {V}$. Given that: \[ \mu_n C_{ox} \frac{W}{L} = 50 \, \mu{A}/{V}^2, \] the transconductance at the new operating point (in $\mu{A}/{V}$, rounded off to two decimal places) is $\_\_\_\_$.

Answer 1

Step 1: Get the noise floor in dB, and stay in dB as long as possible.
Noise power in the 1 MHz band is $N=kT+10\log_{10}B=-174+10\log_{10}(10^6)=-174+60=-114$ dBm.

Step 2: Get the SNR in dB.
\[ SNR_{dB}=S-N=-80-(-114)=34\text{ dB} \]

Step 3: Use the high-SNR shortcut for capacity.
When $SNR\gg1$, $\log_2(1+SNR)\approx\log_2(SNR)$, and $\log_2(SNR)=\dfrac{SNR_{dB}}{10}\log_2(10)$, since $SNR_{dB}=10\log_{10}SNR$ gives $\log_{10}SNR=SNR_{dB}/10$ and $\log_2 x=\log_{10}x\times3.32$. So
\[ \log_2(1+SNR)\approx\frac{34}{10}\times3.32=3.4\times3.32\approx11.3 \]

Step 4: Multiply by the bandwidth.
\[ C\approx B\times11.3=11.3\times10^6\text{ bits/s} \]
This is a ratio of $C/B\approx11.3$.

Step 5: Test each claim against this ratio.
$C=B$ needs a ratio of $1$, $C=2B$ needs $2$, and $C<B$ needs a ratio below $1$. None of these match $11.3$. Only $C>3B$ is consistent with a ratio of about $11.3$.

Step 6: Conclude.
\[ \boxed{C\approx11.3B>3B} \]

Show Hint

The Correct Option is C

Solution and Explanation

Questions Asked in GATE EC exam