exemplar-phy-12-communication-system

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Ques. Three waves A, B and C of frequencies 1600 kHz, 5 MHz and 60 MHz, respectively are to be transmitted from one place to another. Which of the following is the most appropriate mode of communication?
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(a) A is transmitted via space wave while B and C are transmitted via sky wave
(b) A is transmitted via ground wave, B via sky wave and C via space wave
(c) B and C are transmitted via ground wave while A is transmitted via sky wave
(d) B is transmitted via ground wave while A and C are transmitted via space wave
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Thinking Process
The amplitude modulated signal consists of the carrier wave of frequency $\omega_c$ with two additional sinusoidal waves, one of frequency $(\omega_c-\omega_m)$ and other of frequency $(\omega_c+\omega_m)$. These two waves are called side bands and their frequencies are called side band frequency.
Ans.
(a) Given, frequency of carrier signal is $\omega_c=1$ MHz and frequency of speech signal = 3 kHz
<Success>Sol. to be continued …</Success>
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Thinking Process
A square law device is something where either current or voltage depends on the square of the other.
Ans.
(c) The device which follows square law is used for modulation purpose. Characteristics shown by (i) and (iii) corresponds to linear devices.
Characteristics shown by (ii) corresponds to square law device. Some part of (i) also follow square law.
Hence, (ii) and (iv) can be used for modulation.
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Thinking Process
The frequency of male voice less than that of a female voice.
Ans.
(b) Here, in this question, the frequency of modulated signal received becomes more, which is possible with the poor bandwidth selection of amplifiers.
This happens because bandwidth in amplitude modulation is equal to twice the frequency of modulating signal.
But, the frequency of male voice is less than that of a female.
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Ans.
(b) A communication system is the set-up used in the transmission and reception of information from one place to another.
The whole system consist of several elements in a sequence. It can be represented as the diagram given below
<img src="https://cdn.mathpix.com/cropped/2024_01_24_b3449e08884b7d728783g-257.jpg?height=393&width=1121&top_left_y=399&top_left_x=323" height="180px">
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Thinking Process
An arbitrary change in phase angle of the modulating signal is given by $\phi$.
Ans.
(c) Consider a sinusoidal modulating signal represented by
$m(t)=A_m \sin \omega_m t$
where, $A_m=$ Amplitude of modulating signal $\omega_m=$ Angular frequency $=2 \pi V_m=\phi V_m$
<Success>Sol. to be continued …</Success>
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$\Rightarrow C_m(t)=(A_c+A_c \times \mu \sin \omega_m t) \sin \omega_c t$
Now, we know that $A_c \times \mu=K$ [wave constant] and $\sin \omega_m t=V_m$ [wave velocity]
Thus, Eq. (iii) becomes
$C_m(t)=(A_c+K \times V_m) \sin \omega_c t
$
Now, consider a change in phase angle by $\phi$ then $\sin \omega_c t \to \sin (\omega_c t+\phi)$
Thus,
$C_m(t)=(A_c+K V_m)(\sin \omega_c+\phi)$
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Ques. An audio signal of 15 kHz frequency cannot be transmitted over long distances without modulation, because
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(a) the size of the required antenna would be at least 5 km which is not convenient
(b) the audio signal can not be transmitted through sky waves
(c) the size of the required antenna would be at least 20 km, which is not convenient
(d) effective power transmitted would be very low, if the size of the antenna is less than 5 km
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Thinking Process
Transmission of a signal depends on three factors. These are size of antenna, medium of transmission and power of transmitted wave.
Ans. (a, b, d)
Given, frequency of the wave to be transmitted is
$ v_m =15 kHz=15 \times 10^{3}$ Hz
Wavelength $\lambda_m =\frac{c}{v_m}=\frac{3 \times 10^{8}}{15 \times 10^{3}}=\frac{1}{5} \times 10^{5} m $
<Success>Sol. to be continued …</Success>
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Size of the antenna required, $ l=\frac{\lambda}{4}=\frac{1}{4} \times \frac{1}{5} \times 10^{5}$
$=5 \times 10^{3} m$ = 5 km
The audio signals are of low frequency waves. Thus, they cannot be transmitted through sky waves as they are absorbed by atmosphere.
If the size of the antenna is less than 5 km, the effective power transmission would be very low because of deviation from resonance wavelength of wave and antenna length.
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Ques. Audio sine waves of 3 kHz frequency are used to amplitude modulate a carrier signal of 1.5 MHz. Which of the following statements are true?
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(a) The side band frequencies are 1506 kHz and 1494 kHz
(b) The bandwidth required for amplitude modulation is 6 kHz
(c) The bandwidth required for amplitude modulation is 3 MHz
(d) The side band frequencies are 1503 kHz and 1497 kHz
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Ques. In amplitude modulation, the modulation index $m$, is kept less than or equal to 1 because
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(a) $m>1$, will result in interference between carrier frequency and message frequency, resulting into distortion
(b) $m>1$, will result in overlapping of both side bands resulting into loss of information
(c) $m>1$, will result in change in phase between carrier signal and message signal
(d) $m>1$, indicates amplitude of message signal greater than amplitude of carrier signal resulting into distortion
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Ans. (b, d)
The modulation index $(m)$ of amplitude modulated wave is
$m=\frac{\text{ amplitude of message signal }(A_m)}{\text{ amplitude of carrier signal }(A_c)}$
If $m>1$, then $A_m>A_c$.
In this situation, there will be distortion of the resulting signal of amplitude modulated wave. Maximum modulation frequency $(m_f)$ of $A_m$ wave is
<Success>Sol. to be continued …</Success>
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Ans.
Analog and digital signals are used to transmit information, usually through electric signals. In both these technologies, the information such as any audio or video is transformed into electric signals.
The difference between analog and digital technologies is that in analog technology, information is translated into electric pulses of varying amplitude. In digital technology, translation of information is into binary formal (zero or one) where each bit is representative of two distinct amplitudes.
Thus, (i) and
(ii) would produce analog signal and
(iii) and
(iv) would produce digital signals.
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Ans.
As the frequency of wave $B$ is more than wave $A$, it means the refractive index of wave $B$ is more than refractive index of wave $A$ (as refractive index increases with frequency increases).
For higher frequency wave (i.e., higher refractive index) the angle of refraction is less i.e., bending is less. So, wave $B$ travel longer distance in the ionosphere before suffering total internal reflection.
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Ans.
In case of AM, the instantaneous voltage of carrier waves is varied by the modulating wave voltage. So, during the transmission, nosie signals can also be added and receiver assumes noise a part of the modulating signal.
In case of FM, the frequency of carrier waves is changed as the change in the instantaneous voltage of modulating waves. This can be done by mixing and not while the signal is transmitting in channel. So, noise does not affect FM signal.
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Ans.
The distance travelled by the signal is 5 km Loss suffered in path of transmission = 2 dB/km
So, total loss suffered in 5 km = $-2 \times 5=-10$ dB
Total amplifier gain $=10 dB+20 dB=30 dB$
Overall gain in signal $=30-10=20 dB$
According to the question, gain in $dB=10 \log _{10} \frac{P_0}{P_j}$
<Success>Sol. to be continued …</Success>
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Area covered A $=\pi(\text{ range })^{2} $
$ =3.14 \times 16 \times 16=803.84 km^{2} $
(ii) At a height of H = 25 m from ground level
$ \text{ Range } =\sqrt{2 h R}+\sqrt{2 H R} $
$ =\sqrt{2 \times 20 \times 6.4 \times 10^{6}}+\sqrt{2 \times 25 \times 6.4 \times 10^{6}} $
$ =16 \times 10^{3}+17.9 \times 10^{3} $
<Success>Sol. to be continued …</Success>
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$\Rightarrow 8 h_t R =2(R+h_t)^{2} $
$\Rightarrow 4 h_t R =R^{2}+2 R h_t+h_t^{2} $
$\Rightarrow R^{2}-2 h_t R+h_t^{2} =0 $
$\Rightarrow (R-h_t)^{2} =0 $
$\Rightarrow R =h_t$
Since, space wave frequency is used so $\lambda«h_t$, hence only tower height is to be taken into consideration. In three dimensions of earth, 6 antenna towers of each of height $h_t=R$ would be used to cover the entire surface of earth with communication programme.
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Ques. The maximum frequency for reflection of sky waves from a certain layer of the ionosphere is found to be $f _{\max }=9(N _{\max })^{1 / 2}$, where $N _{\max }$ is the maximum electron density at that layer of the ionosphere.
On a certain day it is observed that signals of frequencies higher than 5 MHz are not received by reflection from the $F_1$ layer of the ionosphere while signals of frequencies higher than 8 MHz are not received by reflection from the $F_2$ layer of the ionosphere. Estimate the maximum electron densities of the $F_1$ and $F_2$ layers on that day.
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Ans.
In amplitude modulated signal, only side band frequencies contain information. Thus only $(\omega_c+\omega_m)$ and $(\omega_c-\omega_m)$ contain information.
Now, according to question, the total radiated power is due to energy carried by
$
\omega_c,(\omega_c-\omega_m) \text{ and }(\omega_c+\omega_m) \text{. }
$
Thus to minimise the cost of radiation without compromising on information $\omega_c$ can be left and transmitting. $(\omega_c+\omega_m),(\omega_c-\omega_m)$ or both $(\omega_c+\omega_m)$ and $(\omega_c-\omega_m)$.
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Ques. (i) The intensity of a light pulse travelling along a communication channel decreases exponentially with distance $x$ according to the relation $I=I_0 e^{-\alpha x}$, where $I_0$ is the intensity at $x=0$ and $\alpha$ is the attenuation constant.
Show that the intensity reduces by $75 %$ after a distance of $\frac{\ln 4}{\alpha}$.
(ii) Attenuation of a signal can be expressed in decibel (dB) according to the relation $dB=10 \log _{10} \frac{I}{I_0}$. What is the attenuation in dB/km for an optical fibre in which the intensity falls by $50 \%$ over a distance of 50 km ?
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$x =\frac{\ln 4}{\alpha} $
Therefore, at distance $x=\frac{\ln 4}{\alpha}$, the intensity is reduced to $75 \%$ of initial intensity.
(ii) Let $\alpha$ be the attenuation in dB/km. If $x$ is the distance travelled by signal, then
$10 \log_{10} \frac{I}{I_0}=-\alpha x $
where, $I_0$ is the intensity initially.
According to the question,
<Success>Sol. to be continued …</Success>
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Ques. (i) Draw the plot of amplitude versus $\omega$ for an amplitude modulated were whose carrier wave $(\omega_c)$ is carrying two modulating signals, $\omega_1$ and $\omega_2$ $(\omega_2>\omega_1)$.
(ii) Is the plot symmetrical about $\omega_c$ ? Comment especially about plot in region $\omega<\omega_c$.
(iii) Extrapolate and predict the problems one can expect if more waves are to be modulated.
(iv) Suggest solutions to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
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(ii) From figure, we note that frequency spectrum is not symmetrical about $\omega_c$. Crowding of spectrum is present for $\omega<\omega_t$.
(iii) If more waves are to be modulated then there will be more crowding in the modulating signal in the region $\omega<\omega_c$. That will result more chances of mixing of signals.
(iv) To accommodate more signals, we should increase bandwidth and frequency carrier waves $\omega_t$. This shows that large carrier frequency enables to carry more information (i.e., more $\omega_m$ ) and the same will in turn increase bandwidth.
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Demodulation by a diode is possible if the condition $\frac{1}{f_c} \ll R C<\frac{1}{f_m}$ is satisfied
Thus,
$\frac{1}{f_c} =\frac{1}{20 \times 10^{6}}=0.5 \times 10^{-7} $
$\frac{1}{f_m} =\frac{1}{1.5 \times 10^{3}} Hz=0.7 \times 10^{-3} s $
and
Now, gain through all the options of $R$ and $C$ one by one, we get
<Success>Sol. to be continued …</Success>
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(i) $R C=1 k \Omega \times 0.01 \mu F=10^{3} \Omega \times(0.01 \times 10^{-6} F)=10^{-5} S$
Here, condition $\frac{1}{f_c} \ll R C<\frac{1}{f_m}$ is satisfied.
Hence it can be demodulated.
(ii) $R C=10 K \Omega \times 0.01 \mu F=10^{4} \Omega \times 10^{-8} F=10^{-4} S$
Here condition $\frac{1}{f_c} \ll R C<\frac{1}{f_m}$ is satisfied.
Hence, it can be demodulated.
<Success>Sol. to be continued …</Success>
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(iii) $R C=10 k \Omega \times 1 \mu \mu F=10^{4} \Omega \times 10^{-12} F=10^{-8} S$
Here, condition $\frac{1}{f_c}>R C$, so this cannot be demodulated.
The emf induced across $P Q$ due to its motion or change in magnetic flux linked with the loop change due to change of enclosed area.
For nth minima to be formed on the screen path difference between the rays coming from $S_1$ and $S_2$ must be $(2 n-1) \frac{\lambda}{2}$.
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