Online Test: Communication Systems | Class 12 Physics

Question 1 of 20

The three essential blocks of any communication system are:

modulator, mixer, filter

transmitter, channel, receiver

source, noise, antenna

amplifier, repeater, detector

Explanation: Every communication system reduces to a transmitter, a channel and a receiver linking source to user.

Question 2 of 20

A microphone is an example of a:

modulator

transducer

rectifier

repeater

Explanation: A microphone converts sound energy into an electrical signal, so it is a transducer.

Question 3 of 20

An amplifier with output power $100$ times its input has a gain of:

$10\,\text{dB}$

$20\,\text{dB}$

$40\,\text{dB}$

$100\,\text{dB}$

Explanation: Gain $=10\log_{10}(100)=10\times2=20\,\text{dB}$.

Question 4 of 20

The approximate bandwidth needed for telephone-quality speech is:

$2.8\,\text{kHz}$

$20\,\text{kHz}$

$4.2\,\text{MHz}$

$6\,\text{MHz}$

Explanation: Speech for telephony uses about $300$ to $3100\,\text{Hz}$, a bandwidth of roughly $2.8\,\text{kHz}$.

Question 5 of 20

Which medium gives the largest bandwidth and least attenuation?

twisted-pair wire

coaxial cable

optical fibre

free space at LF

Explanation: Optical fibre works near $10^{14}\,\text{Hz}$, giving very large bandwidth with very low loss.

Question 6 of 20

Ground (surface) wave propagation is used for frequencies up to about:

$2\,\text{MHz}$

$30\,\text{MHz}$

$100\,\text{MHz}$

$1\,\text{GHz}$

Explanation: Surface waves follow the Earth up to roughly $2\,\text{MHz}$ (medium-wave AM).

Question 7 of 20

Long-distance short-wave radio reaches far places by:

ground wave

reflection from the ionosphere

line-of-sight space wave

diffraction at the poles

Explanation: Waves of $2$ to $30\,\text{MHz}$ are reflected back to Earth by the ionosphere (sky wave).

Question 8 of 20

The range of a single transmitting antenna of height $h$ is:

$\sqrt{Rh}$

$\sqrt{2Rh}$

$2\sqrt{Rh}$

$\sqrt{Rh/2}$

Explanation: The geometric horizon distance from height $h$ is $d=\sqrt{2Rh}$.

Question 9 of 20

A $90\,\text{m}$ tall antenna ($R=6.4\times10^{6}\,\text{m}$) has a range of about:

$24\,\text{km}$

$34\,\text{km}$

$48\,\text{km}$

$60\,\text{km}$

Explanation: $d=\sqrt{2\times6.4\times10^{6}\times90}=\sqrt{1.152\times10^{9}} \approx 3.39\times10^{4}\,\text{m} \approx 34\,\text{km}$.

Question 10 of 20

The maximum line-of-sight distance between two raised antennas is:

$\sqrt{2R(h_T+h_R)}$

$\sqrt{2Rh_T}+\sqrt{2Rh_R}$

$\sqrt{2Rh_T}-\sqrt{2Rh_R}$

$2\sqrt{Rh_Th_R}$

Explanation: Each antenna contributes its own horizon, so $d_M=\sqrt{2Rh_T}+\sqrt{2Rh_R}$.

Question 11 of 20

The critical frequency of a layer with $N_{\max}=9\times10^{10}\,\text{per m}^3$ is:

$0.9\,\text{MHz}$

$2.7\,\text{MHz}$

$9\,\text{MHz}$

$27\,\text{MHz}$

Explanation: $f_c=9\sqrt{9\times10^{10}}=9\times3\times10^{5}=2.7\,\text{MHz}$.

Question 12 of 20

Modulation is required mainly because, without it, the antenna would need to be:

very short

impractically long

made of fibre

buried underground

Explanation: Low-frequency audio has a huge wavelength, so $L\approx\lambda/4$ would be kilometres long; a high-frequency carrier shrinks the antenna.

Question 13 of 20

If $A_c=20\,\text{V}$ and $A_m=15\,\text{V}$, the modulation index is:

$0.25$

$0.5$

$0.75$

$1.33$

Explanation: $\mu=A_m/A_c=15/20=0.75$.

Question 14 of 20

An AM wave has $A_{\max}=10\,\text{V}$, $A_{\min}=2\,\text{V}$. Its modulation index is:

$0.33$

$0.5$

$0.67$

$0.8$

Explanation: $\mu=(A_{\max}-A_{\min})/(A_{\max}+A_{\min})=(10-2)/(10+2)=8/12 \approx 0.67$.

Question 15 of 20

A $1000\,\text{kHz}$ carrier modulated by a $5\,\text{kHz}$ tone has side bands at:

$5$ and $1000\,\text{kHz}$

$995$ and $1005\,\text{kHz}$

$500$ and $2000\,\text{kHz}$

$1000$ and $5000\,\text{kHz}$

Explanation: Side bands at $f_c\pm f_m=1000\pm5$, i.e. $995\,\text{kHz}$ and $1005\,\text{kHz}$.

Question 16 of 20

The bandwidth of an AM wave for a message of highest frequency $f_m$ is:

$f_m$

$2f_m$

$f_c$

$2f_c$

Explanation: Bandwidth is the spread between the side bands, $(f_c+f_m)-(f_c-f_m)=2f_m$.

Question 17 of 20

An envelope detector used to demodulate AM essentially consists of a:

square-law device only

diode followed by an RC filter

transformer and capacitor

band-pass amplifier only

Explanation: A diode rectifies the AM wave and the RC network smooths the carrier, leaving the message envelope.

Question 18 of 20

FM is preferred over AM for high-fidelity broadcast because it is:

narrower in bandwidth

less affected by noise

cheaper to transmit

lower in frequency

Explanation: Information is carried in frequency, so amplitude noise has little effect, giving FM better fidelity (at the cost of more bandwidth).

Question 19 of 20

A composite TV signal occupies a bandwidth of about:

$20\,\text{kHz}$

$200\,\text{kHz}$

$6\,\text{MHz}$

$60\,\text{MHz}$

Explanation: Video ($\sim4.2\,\text{MHz}$) plus audio gives a composite TV bandwidth of about $6\,\text{MHz}$.

Question 20 of 20

Each side band of an AM wave (index $\mu$, carrier amplitude $A_c$) has amplitude:

$\mu A_c$

$\dfrac{\mu A_c}{2}$

$2\mu A_c$

$\dfrac{A_c}{\mu}$

Explanation: Expanding the AM wave shows each side band has amplitude $\mu A_c/2$.

Online Test — Communication Systems