Communication Systems
2022
Q1
JEE Mains
Numerical
10 Mar 2026
A modulating signal $2 \sin \left(6.28 \times 10^{6}\right) t$ is added to the carrier signal $4 \sin \left(12.56 \times 10^{9}\right) t$ for amplitude modulation. The combined signal is passed through a non-linear square law device. The output is then passed through a band pass filter. The bandwidth of the output signal of band pass filter will be _________ $\mathrm{MHz}$.
Correct Answer: 2
Explanation:
$\mathrm{W}_{\mathrm{C}}=12.56 \times 10^{9}$
$ W_{m}=6.25 \times 10^{6} $
After amplitude modulation
Bandwidth frequency
$ =\frac{2 W_{m}}{2 \pi}=\frac{2 \times 6.28}{2 \pi} \times 10^{6}=2 \mathrm{MHz} $
$ W_{m}=6.25 \times 10^{6} $
After amplitude modulation
Bandwidth frequency
$ =\frac{2 W_{m}}{2 \pi}=\frac{2 \times 6.28}{2 \pi} \times 10^{6}=2 \mathrm{MHz} $
2022
Q2
JEE Mains
Numerical
10 Mar 2026
The required height of a TV tower which can cover the population of $6.03$ lakh is $h$. If the average population density is 100 per square $\mathrm{km}$ and the radius of earth is $6400 \mathrm{~km}$, then the value of $h$ will be _____________ $m$.
Correct Answer: 150
Explanation:
$r = \sqrt {{{(h + R)}^2} - {R^2}} \cong \sqrt {2hR} $
$A = {{6.03 \times {{10}^5}} \over {100}}$
$\pi {r^2} = 6.03 \times {10^3}$
$\pi 2Rh = 6.03 \times {10^3}$
$h = {{6.03 \times {{10}^3}} \over {2 \times \pi \times R}} = 0.015 \times 10 \times {10^3}$ m
$=150$ m
2022
Q3
JEE Mains
Numerical
10 Mar 2026
The height of a transmitting antenna at the top of a tower is 25 m and that of receiving antenna is, 49 m. The maximum distance between them, for satisfactory communication in LOS (Line-Of-Sight) is K$\sqrt5$ $\times$ 102 m. The value of K is ___________.
(Assume radius of Earth is 64 $\times$ 10+5 m) [Calculate upto nearest integer value]
Correct Answer: 192
Explanation:
$d = \sqrt {2{h_t}{R_e}} + \sqrt {2 \times {h_R}{R_e}} $
$ = \sqrt {2 \times 25 \times 64 \times {{10}^5}} + \sqrt {2 \times 49 \times 64 \times {{10}^5}} $
$ = 8000\sqrt 5 + 11200\sqrt 5 $ m
$ = 19200\sqrt 5 $ m
$ = 192\sqrt 5 + {10^2}$ m
2022
Q4
JEE Mains
Numerical
10 Mar 2026
An antenna is placed in a dielectric medium of dielectric constant 6.25. If the maximum size of that antenna is 5.0 mm, it can radiate a signal of minimum frequency of __________ GHz.
(Given $\mu$r = 1 for dielectric medium)
Correct Answer: 6
Explanation:
We know that v = f$\lambda$
Putting the values,
${{3 \times {{10}^8}} \over {\sqrt {6.25} }} = f \times 20 \times {10^{ - 3}}$
$ \Rightarrow f = 6 \times {10^9}$ Hz
2021
Q5
JEE Mains
Numerical
10 Mar 2026
A carrier wave with amplitude of 250 V is amplitude modulated by a sinusoidal base band signal of amplitude 150V. The ratio of minimum amplitude to maximum amplitude for the amplitude modulated wave is 50 : x, then value of x is ____________.
Correct Answer: 200
Explanation:
Given, the amplitude of the carrier wave, Ac = 250 V
The amplitude of the message wave, Am = 150 V
We know that,
The maximum amplitude, Amax = Ac + Am
Substituting the values in the above equation, we get
Amax = 250 + 150 = 400 V
We know that,
The minimum amplitude, Amin = Ac $-$ Am
Substituting the values in the above equation, we get
Amin = 250 $-$ 150 = 100 V
Thus, the ratio of the minimum amplitude to the maximum amplitude of the modulated wave is
${{{A_{\min }}} \over {{A_{\max }}}} = {{100} \over {400}} = {1 \over 4}$
Comparing with, 1 : x
The value of the x = 4.
The amplitude of the message wave, Am = 150 V
We know that,
The maximum amplitude, Amax = Ac + Am
Substituting the values in the above equation, we get
Amax = 250 + 150 = 400 V
We know that,
The minimum amplitude, Amin = Ac $-$ Am
Substituting the values in the above equation, we get
Amin = 250 $-$ 150 = 100 V
Thus, the ratio of the minimum amplitude to the maximum amplitude of the modulated wave is
${{{A_{\min }}} \over {{A_{\max }}}} = {{100} \over {400}} = {1 \over 4}$
Comparing with, 1 : x
The value of the x = 4.
2021
Q6
JEE Mains
Numerical
10 Mar 2026
A bandwidth of 6 MHz is available for A.M. transmission. If the maximum audio signal frequency used for modulating the carrier wave is not to exceed 6 kHz. The number of stations that can be broadcasted within this band simultaneously without interfering with each other will be __________.
Correct Answer: 500
Explanation:
Signal bandwidth = 2 fm = 12 kHz
$\therefore$ N = ${{6MHZ} \over {12kHZ}} = {{6 \times {{10}^6}} \over {12 \times {{10}^3}}} = 500$
$\therefore$ N = ${{6MHZ} \over {12kHZ}} = {{6 \times {{10}^6}} \over {12 \times {{10}^3}}} = 500$
2021
Q7
JEE Mains
Numerical
10 Mar 2026
If the sum of the heights of transmitting and receiving antennas in the line of sight of communication is fixed at 160 m, then the maximum range of LOS communication is ___________ km. (Take radius of Earth = 6400 km)
Correct Answer: 64
Explanation:
hT = hR = 160 ........ (i)
$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
$d = \sqrt {2R} \left[ {\sqrt {{h_T}} + \sqrt {{h_R}} } \right]$
$d = \sqrt {2R} \left[ {\sqrt x + \sqrt {160 - x} } \right]$
${{d(d)} \over {dx}} = 0$
${1 \over {2\sqrt x }} + {{1( - 1)} \over {2\sqrt {160 - x} }} = 0$
${1 \over {\sqrt x }} = {1 \over {\sqrt {160 - x} }}$
x = 80 m
${d_{\max }} = \sqrt {2 \times 6400} \left[ {\sqrt {{{80} \over {1000}}} + \sqrt {{{20} \over {1000}}} } \right]$
$ = {{80\sqrt 2 \times 2\sqrt {80} } \over {10\sqrt {10} }}$
$ = 8 \times 2 \times \sqrt 2 \times 2\sqrt 2 = 64$ km
$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
$d = \sqrt {2R} \left[ {\sqrt {{h_T}} + \sqrt {{h_R}} } \right]$
$d = \sqrt {2R} \left[ {\sqrt x + \sqrt {160 - x} } \right]$
${{d(d)} \over {dx}} = 0$
${1 \over {2\sqrt x }} + {{1( - 1)} \over {2\sqrt {160 - x} }} = 0$
${1 \over {\sqrt x }} = {1 \over {\sqrt {160 - x} }}$
x = 80 m
${d_{\max }} = \sqrt {2 \times 6400} \left[ {\sqrt {{{80} \over {1000}}} + \sqrt {{{20} \over {1000}}} } \right]$
$ = {{80\sqrt 2 \times 2\sqrt {80} } \over {10\sqrt {10} }}$
$ = 8 \times 2 \times \sqrt 2 \times 2\sqrt 2 = 64$ km
2021
Q8
JEE Mains
Numerical
10 Mar 2026
A transmitting antenna has a height of 320 m and that of receiving antenna is 2000 m. The maximum distance between them for satisfactory communication in line of sight mode is 'd'. The value of 'd' is ................. km.
Correct Answer: 224
Explanation:
${d_m} = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
${d_m} = \left( {\sqrt {2 \times 6400 \times {{10}^3} \times 320} + \sqrt {2 \times 6400 \times {{10}^3} \times 2000} } \right)$m
dm = 224 km
${d_m} = \left( {\sqrt {2 \times 6400 \times {{10}^3} \times 320} + \sqrt {2 \times 6400 \times {{10}^3} \times 2000} } \right)$m
dm = 224 km
2021
Q9
JEE Mains
Numerical
10 Mar 2026
An amplitude modulated wave is represented by
Cm(t) = 10(1 + 0.2 cos 12560t) sin(111 $\times$ 104t) volts. The modulating frequency in kHz will be ................. .
Cm(t) = 10(1 + 0.2 cos 12560t) sin(111 $\times$ 104t) volts. The modulating frequency in kHz will be ................. .
Correct Answer: 2
Explanation:
Wm = 12560 = 2$\pi$fm
${f_m} = {{12560} \over {2\pi }}$
= 2000 Hz
${f_m} = {{12560} \over {2\pi }}$
= 2000 Hz
2021
Q10
JEE Mains
Numerical
10 Mar 2026
The maximum amplitude for an amplitude modulated wave is found to be 12V while the minimum amplitude is found to be 3V. The modulation index is 0.6x where x is ____________.
Correct Answer: 1
Explanation:
Amax = Ac + Am = 12
Amin = Ac $-$ Am = 3
$\Rightarrow$ Ac = ${{15} \over 2}$ & Am = ${9 \over 2}$
modulation index = ${{{A_m}} \over {{A_c}}} = {{9/2} \over {15/2}} = 0.6$
$\Rightarrow$ x = 1
Amin = Ac $-$ Am = 3
$\Rightarrow$ Ac = ${{15} \over 2}$ & Am = ${9 \over 2}$
modulation index = ${{{A_m}} \over {{A_c}}} = {{9/2} \over {15/2}} = 0.6$
$\Rightarrow$ x = 1
2021
Q11
JEE Mains
Numerical
10 Mar 2026
The amplitude of upper and lower side bands of A.M. wave where a carrier signal with frequency 11.21 MHz, peak voltage 15 V is amplitude modulated by a 7.7 kHz sine wave of 5V amplitude are ${a \over {10}}V$ and ${b \over {10}}V$ respectively. Then the value of ${a \over b}$ is ____________.
Correct Answer: 1
Explanation:
${a \over {10}} = {b \over {10}} = {{\mu {A_C}} \over 2}$
$ \Rightarrow {a \over b} = 1$
2021
Q12
JEE Mains
Numerical
10 Mar 2026
A message signal of frequency 20 kHz and peak voltage of 20 volt is used to modulate a carrier wave of frequency 1 MHz and peak voltage of 20 volt. The modulation index will be :
Correct Answer: 1
Explanation:
Modulation index
$\mu = {{{A_m}} \over {{A_c}}} = {{20} \over {20}} = 1$
$\mu = {{{A_m}} \over {{A_c}}} = {{20} \over {20}} = 1$
2021
Q13
JEE Mains
Numerical
10 Mar 2026
A carrier wave Vc(t) = 160 sin(2$\pi$ $\times$ 106t) volts is made to vary between Vmax = 200 V and Vmin = 120 V by a message
signal Vm(t) = Am sin(2$\pi$ $\times$ 103t) volts. The peak voltage Am of the modulating signal is ___________.
signal Vm(t) = Am sin(2$\pi$ $\times$ 103t) volts. The peak voltage Am of the modulating signal is ___________.
Correct Answer: 40
Explanation:
Given, VC(t) = 160 sin(2$\pi$ $\times$ 106t) V
Vmax = 200 V, Vmin = 120 V and Vm(t) = Am sin(2$\pi$ $\times$ 103t) V.
Since, maximum amplitude,
Amax = Am + Ac
$\Rightarrow$ Vmax = Vm + Vc $\Rightarrow$ 200 = Vm + 160
$\Rightarrow$ Vm = 200 $-$ 160 $\Rightarrow$ Vm = 40
$\therefore$ Peak voltage, Am = 40
Vmax = 200 V, Vmin = 120 V and Vm(t) = Am sin(2$\pi$ $\times$ 103t) V.
Since, maximum amplitude,
Amax = Am + Ac
$\Rightarrow$ Vmax = Vm + Vc $\Rightarrow$ 200 = Vm + 160
$\Rightarrow$ Vm = 200 $-$ 160 $\Rightarrow$ Vm = 40
$\therefore$ Peak voltage, Am = 40
2021
Q14
JEE Mains
Numerical
10 Mar 2026
A TV transmission tower antenna is at a height of 20 m. Suppose that the receiving antenna is at.
(i) ground level
(ii) a height of 5 m.
The increase in antenna range in case (ii) relative to case (i) is n%.
The value of n, to the nearest integer, is ___________.
(i) ground level
(ii) a height of 5 m.
The increase in antenna range in case (ii) relative to case (i) is n%.
The value of n, to the nearest integer, is ___________.
Correct Answer: 50
Explanation:
For calculation of Range from tower.
$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
$ \therefore $ hT = height of tower
& hR = height of receiver
For 1st case : hT = 20m, hR = 0
$ \because $ ${d_1} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} $ = 16 km
For 2nd case : hT = 20m, hR = 5m
$ \because $ ${d_2} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} + \sqrt {2 \times 6400 \times {{10}^3} \times 5} $
= 16 + 8 = 24 km
$ \therefore $ % change in range
$ = {{{d_2} - {d_1}} \over {{d_1}}} \times 100$
$ = {{24 - 16} \over {16}} \times 100$
$ = {8 \over {16}} \times 100$ = 50%
$d = \sqrt {2R{h_T}} + \sqrt {2R{h_R}} $
$ \therefore $ hT = height of tower
& hR = height of receiver
For 1st case : hT = 20m, hR = 0
$ \because $ ${d_1} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} $ = 16 km
For 2nd case : hT = 20m, hR = 5m
$ \because $ ${d_2} = \sqrt {2 \times 6400 \times {{10}^3} \times 20} + \sqrt {2 \times 6400 \times {{10}^3} \times 5} $
= 16 + 8 = 24 km
$ \therefore $ % change in range
$ = {{{d_2} - {d_1}} \over {{d_1}}} \times 100$
$ = {{24 - 16} \over {16}} \times 100$
$ = {8 \over {16}} \times 100$ = 50%
2021
Q15
JEE Mains
Numerical
10 Mar 2026
For VHF signal broadcasting, ___________ km2 of maximum service area will be covered by an antenna tower of height 30 m, if the receiving antenna is placed at ground. Let radius of the earth be 6400 km. (Round off to the Nearest Integer) (Take $\pi$ as 3.14)
Correct Answer: 1206
Explanation:
$d = \sqrt {2hR} $
area = $\pi$d2
Area = $\pi$(2hR) = 3.14 $\times$ 2 $\times$ 30 $\times$ 6400 $\times$ 103 . m2
= 1205.76 km2
$ \approx $ 1206 km2
area = $\pi$d2
Area = $\pi$(2hR) = 3.14 $\times$ 2 $\times$ 30 $\times$ 6400 $\times$ 103 . m2
= 1205.76 km2
$ \approx $ 1206 km2
2021
Q16
JEE Mains
Numerical
10 Mar 2026
If the highest frequency modulating a carrier is 5 kHz, then the number of AM broadcast stations accommodated in a 90 kHz bandwidth are _________.
Correct Answer: 9
Explanation:
B. W. (Bandwidth) = 2 $\times$ maximum frequency at modulating signal
= 2 $\times$ 5kHz
= 10 kHz
$ \therefore $ No of stations accommodate
= ${{90} \over {10}}$ = 9
= 2 $\times$ 5kHz
= 10 kHz
$ \therefore $ No of stations accommodate
= ${{90} \over {10}}$ = 9
2021
Q17
JEE Mains
Numerical
10 Mar 2026
The maximum and minimum amplitude of an amplitude modulated wave is 16V and 8V respectively. The modulation index for this amplitude modulated wave is x $\times$ 10$-$2. The value of x is __________.
Correct Answer: 33
Explanation:
${A_m} = {{{A_{\max }} - {A_{\min }}} \over 2}$
${A_c} = {{{A_{\max }} + {A_{\min }}} \over 2}$
Modulation index (mi) = ${{{A_m}} \over {{A_c}}} = {{{{{A_{\max }} - {A_{\min }}} \over 2}} \over {{{{A_{\max }} + {A_{\min }}} \over 2}}} = {{{A_{\max }} - {A_{\min }}} \over {{A_{\max }} + {A_{\min }}}}$
mi = ${{16 - 8} \over {16 + 8}} = {8 \over {24}} = {1 \over 3} = 0.33$
mi = 33 $\times$ 10$-$2
$ \therefore $ x = 33
${A_c} = {{{A_{\max }} + {A_{\min }}} \over 2}$
Modulation index (mi) = ${{{A_m}} \over {{A_c}}} = {{{{{A_{\max }} - {A_{\min }}} \over 2}} \over {{{{A_{\max }} + {A_{\min }}} \over 2}}} = {{{A_{\max }} - {A_{\min }}} \over {{A_{\max }} + {A_{\min }}}}$
mi = ${{16 - 8} \over {16 + 8}} = {8 \over {24}} = {1 \over 3} = 0.33$
mi = 33 $\times$ 10$-$2
$ \therefore $ x = 33
2021
Q18
JEE Mains
Numerical
10 Mar 2026
A signal of 0.1 kW is transmitted in a cable. The attenuation of cable is $-$5 dB per km and cable length is 20 km. The power received at receiver is 10$-$x W. The value of x is ________.
[Gain in $dB = 10{\log _{10}}\left( {{{{P_o}} \over {{P_i}}}} \right)$]
[Gain in $dB = 10{\log _{10}}\left( {{{{P_o}} \over {{P_i}}}} \right)$]
Correct Answer: 8
Explanation:
Given, power of transmitted signal, Pi = 0.1 kW = 0.1 $\times$ 103W = 102 W
Rate of attenuation, R = $-$5 dB/km
Length of cable, l = 20 km
Power received at receiver, Px = 10$-$x W
Total loss, $\beta$ = R $\times$ l = $-$5 $\times$ 20 = $-$ 100 dB
$\because$ Gain ($\beta$) = $10{\log _{10}}{{{P_0}} \over {{P_i}}}$
$\therefore$ $\beta = - 100 = - 10{\log _{10}}{{{P_0}} \over {{P_i}}}$
$ \Rightarrow - 10 = {\log _{10}}{{{P_0}} \over {{P_i}}} \Rightarrow {10^{ - 10}} = {{{P_0}} \over {{P_i}}}$
$ \Rightarrow {P_0} = {10^{ - 10}}{P_i} = {10^{ - 10}} \times {10^2} = {10^{ - 8}} \Rightarrow {P_0} = {10^{ - 8}}V$
Hence, x = 8
Rate of attenuation, R = $-$5 dB/km
Length of cable, l = 20 km
Power received at receiver, Px = 10$-$x W
Total loss, $\beta$ = R $\times$ l = $-$5 $\times$ 20 = $-$ 100 dB
$\because$ Gain ($\beta$) = $10{\log _{10}}{{{P_0}} \over {{P_i}}}$
$\therefore$ $\beta = - 100 = - 10{\log _{10}}{{{P_0}} \over {{P_i}}}$
$ \Rightarrow - 10 = {\log _{10}}{{{P_0}} \over {{P_i}}} \Rightarrow {10^{ - 10}} = {{{P_0}} \over {{P_i}}}$
$ \Rightarrow {P_0} = {10^{ - 10}}{P_i} = {10^{ - 10}} \times {10^2} = {10^{ - 8}} \Rightarrow {P_0} = {10^{ - 8}}V$
Hence, x = 8
2021
Q19
JEE Mains
Numerical
10 Mar 2026
An audio signal vm = 20 sin 2$\pi$(1500t) amplitude modulates a carrier vc = 80 sin 2$\pi$(100,000t).
The value of percent modulation is _________.
The value of percent modulation is _________.
Correct Answer: 25
Explanation:
Given, audio signal,
Vm = 20 sin 2$\pi$(1500t) .... (i)
Carrier signal, Vc = 80 sin 2$\pi$(100000t) .... (ii)
We know that, modulation index,
${m_f} = {{{A_m}} \over {{A_c}}}$
From Eqs. (i) and (ii), we get
Am = 20, Ac = 80
Percentage of modulation index,
${m_f} = {{{A_m}} \over {{A_c}}} \times 100 = {{20} \over {80}} \times 100 = 25\% $
Vm = 20 sin 2$\pi$(1500t) .... (i)
Carrier signal, Vc = 80 sin 2$\pi$(100000t) .... (ii)
We know that, modulation index,
${m_f} = {{{A_m}} \over {{A_c}}}$
From Eqs. (i) and (ii), we get
Am = 20, Ac = 80
Percentage of modulation index,
${m_f} = {{{A_m}} \over {{A_c}}} \times 100 = {{20} \over {80}} \times 100 = 25\% $