IMO Practice Test: Communication Systems | Class 12 Physics

Question 1 of 14

An antenna radiates power proportional to $(L/\lambda)^2$. If the carrier frequency is doubled with antenna length fixed, the radiated power becomes:

half

unchanged

twice

four times

Explanation: $P\propto(Lf/c)^2$. Doubling $f$ multiplies power by $2^2=4$.

Question 2 of 14

A tower of height $h$ covers a circular area $A$. To double the covered area, the height must be:

halved

doubled

quadrupled

kept the same

Explanation: $A=\pi d^2=2\pi Rh\propto h$, so doubling $h$ doubles the area.

Question 3 of 14

A 100% modulated AM wave has $A_{\min}$ equal to:

$A_c$

$2A_c$

zero

$A_c/2$

Explanation: For $\mu=1$, $A_{\min}=A_c-A_m=A_c-A_c=0$; the envelope just touches zero.

Question 4 of 14

Two antennas of equal height $h$ give a line-of-sight range $d_M$. A single antenna of the same height $h$ gives range $d$. The ratio $d_M/d$ is:

$1$

$\sqrt{2}$

$2$

$4$

Explanation: $d_M=2\sqrt{2Rh}=2d$, so the ratio is $2$.

Question 5 of 14

An AM transmitter sends a carrier of $1.2\,\text{MHz}$ with message frequencies up to $8\,\text{kHz}$. The frequency range occupied is:

$1.192$ to $1.208\,\text{MHz}$

$1.2$ to $1.208\,\text{MHz}$

$8\,\text{kHz}$ wide only

$1.2\,\text{MHz}$ only

Explanation: Side bands span $f_c\pm f_m=1.2\,\text{MHz}\pm8\,\text{kHz}$, i.e. $1.192$ to $1.208\,\text{MHz}$ (bandwidth $16\,\text{kHz}$).

Question 6 of 14

The total power of an AM wave with carrier power $P_c$ and index $\mu$ is $P_t=P_c(1+\mu^2/2)$. For $\mu=1$ the fraction of total power in the side bands is:

$\tfrac{1}{2}$

$\tfrac{1}{3}$

$\tfrac{2}{3}$

$\tfrac{1}{4}$

Explanation: $P_t=P_c(1+0.5)=1.5P_c$; side-band power $=0.5P_c$, fraction $=0.5/1.5=1/3$.

Question 7 of 14

A wave of $40\,\text{MHz}$ is sent vertically into a layer of critical frequency $10\,\text{MHz}$. It will:

be reflected to Earth

pass through and escape into space

be fully absorbed

split into two side bands

Explanation: Since $40\,\text{MHz}>f_c=10\,\text{MHz}$, the wave penetrates the layer and is not reflected.

Question 8 of 14

Doubling the maximum electron density of an ionospheric layer changes its critical frequency by a factor of:

$2$

$4$

$\sqrt{2}$

$1/2$

Explanation: $f_c=9\sqrt{N_{\max}}\propto\sqrt{N}$, so doubling $N$ multiplies $f_c$ by $\sqrt{2}$.

Question 9 of 14

An audio band $20\,\text{Hz}$ to $20\,\text{kHz}$ amplitude-modulates a $1\,\text{MHz}$ carrier. The total bandwidth of the AM signal is:

$20\,\text{kHz}$

$40\,\text{kHz}$

$20\,\text{Hz}$

$1\,\text{MHz}$

Explanation: Bandwidth $=2f_{m,\max}=2\times20\,\text{kHz}=40\,\text{kHz}$.

Question 10 of 14

A signal loses $30\,\text{dB}$ in a fibre then is amplified by $50\,\text{dB}$. The net power ratio is:

$10$

$100$

$1000$

$10000$

Explanation: Net gain $=50-30=20\,\text{dB}$, so power ratio $=10^{2}=100$.

Question 11 of 14

An antenna is to be efficient at $\lambda/4$ length for a carrier of wavelength $3\,\text{m}$. Its length should be about:

$0.75\,\text{m}$

$3\,\text{m}$

$12\,\text{m}$

$1.5\,\text{m}$

Explanation: $L\approx\lambda/4=3/4=0.75\,\text{m}$.

Question 12 of 14

An AM wave $c_m(t)=(A_c+A_m\sin\omega_m t)\sin\omega_c t$ is over-modulated when:

$A_m=0$

$A_m

$A_m=A_c$

$A_m>A_c$

Explanation: Over-modulation ($\mu>1$) occurs when $A_m>A_c$; the envelope is clipped and the message distorts.

Question 13 of 14

A receiving antenna must add a horizon distance of $10\,\text{km}$ via $\sqrt{2Rh_R}=10\,\text{km}$. Its height $h_R$ is about ($R=6.4\times10^{6}\,\text{m}$):

$3.9\,\text{m}$

$7.8\,\text{m}$

$78\,\text{m}$

$390\,\text{m}$

Explanation: $h_R=d_R^2/(2R)=(10^{4})^2/(1.28\times10^{7})=10^{8}/1.28\times10^{7} \approx 7.8\,\text{m}$.

Question 14 of 14

Compared with AM of the same message, FM generally occupies a bandwidth that is:

smaller

the same

larger

zero

Explanation: FM spreads energy over many side bands, so it needs a larger bandwidth than AM, which is why FM uses the VHF band.

IMO Practice Test — Communication Systems