Communication System
Chapter 15
COMMUNICATION SYSTEMS
MCQ I
15.1 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:
(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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Answer: (b)15.2 A 100m long antenna is mounted on a 500m tall building. The complex can become a transmission tower for waves with $\lambda$
(a) $\sim 400 m$.
(b) $\sim 25 m$.
(c) $\sim 150 m$.
(d) $\sim 2400 m$.
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Answer: (a)15.3 A $1 KW$ signal is transmitted using a communication channel which provides attenuation at the rate of $-2 dB$ per $km$. If the communication channel has a total length of $5 km$, the power of the signal received is [gain in $dB=10 \log (\frac{P_0}{P_i})$ ]
(a) $900 W$.
(b) $100 W$.
(c) $990 W$.
(d) $1010 W$.
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Answer: (b)15.4 A speech signal of $3 kHz$ is used to modulate a carrier signal of frequency $1 MHz$, using amplitude modulation. The frequencies of the side bands will be
(a) $1.003 MHz$ and $0.997 MHz$.
(b) $3001 kHz$ and $2997 kHz$.
(c) $1003 kHz$ and $1000 kHz$.
(d) $1 MHz$ and $0.997 MHz$.
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Answer: (a)15.5 A message signal of frequency $\omega_m$ is superposed on a carrier wave of frequency $\omega_c$ to get an amplitude modulated wave (AM). The frequency of the AM wave will be
(a) $\omega_m$.
(b) $\omega_c$.
(c) $\frac{\omega_c+\omega_m}{2}$.
(d) $\frac{\omega_c-\omega_m}{2}$.
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Answer: (b)15.6 $I-V$ characteristics of four devices are shown in Fig. 15.1
(i)
(ii)
(iii)
(iv)
Fig. 15.1
Identify devices that can be used for modulation:
(a) ‘i’ and ‘iii’.
(b) only ‘iii’.
(c) ‘ii’ and some regions of ‘iv’.
(d) All the devices can be used.
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Answer: (c)15.7 A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to
(a) poor selection of modulation index (selected $0<m<1$ )
(b) poor bandwidth selection of amplifiers.
(c) poor selection of carrier frequency
(d) loss of energy in transmission.
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Answer: (b)15.8 A basic communication system consists of
(A) transmitter.
(B) information source.
(C) user of information.
(D) channel.
(E) receiver.
Choose the correct sequence in which these are arranged in a basic communication system:
(a) ABCDE.
(b) BADEC.
(c) BDACE.
(d) BEADC.
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Answer: (b)15.9 Identify the mathematical expression for amplitude modulated wave:
(a) $A_c \sin [{\omega_c+k_1 v_m(t)} t+\phi]$.
(b) $A_c \sin {\omega_c t+\phi+k_2 v_m(t)}$.
(c) ${A_c+k_2 v_m(t)} \sin (\omega_c t+\phi)$.
(d) $A_c v_m(t) \sin (\omega_c t+\phi)$.
MCQ II
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Answer: (c)15.10 An audio signal of $15 kHz$ frequency cannot be transmitted over long distances without modulation because
(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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Answer: (a), (b), (d)15.11 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?
(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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Answer: (b), (d)15.12 A TV trasmission tower has a height of $240 m$. Signals broadcast from this tower will be received by LOS communication at a distance of (assume the radius of earth to be $6.4 \times 10^{6} m$ )
(a) $100 km$.
(b) $24 km$.
(c) $55 km$.
(d) $50 km$.
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Answer: (b), (c), (d)15.13 The frequency response curve (Fig. 15.2) for the filter circuit used for production of AM wave should be
(i)
(ii)
(iii)
(iv)
Fig. 15.2
(a) (i) followed by (ii).
(b) (ii) followed by (i).
(c) (iii).
(d) (iv).
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Answer: (a), (b), (c)15.14 In amplitude modulation, the modulation index $m$, is kept less than or equal to 1 because
(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.
VSA
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Answer: (b), (d)15.15 Which of the following would produce analog signals and which would produce digital signals?
(i) A vibrating tuning fork.
(ii) Musical sound due to a vibrating sitar string.
(iii) Light pulse.
(iv) Output of NAND gate.
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Answer: (i) analog
(ii) analog
(iii) digital
(iv) digital
15.16 Would sky waves be suitable for transmission of TV signals of $60 MHz$ frequency?
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Answer: No, signals of frequency greater than $30 MHz$ will not be reflected by the ionosphere, but will penetrate through the ionosphere.15.17 Two waves A and B of frequencies $2 MHz$ and $3 MHz$, respectively are beamed in the same direction for communication via sky wave. Which one of these is likely to travel longer distance in the ionosphere before suffering total internal reflection?
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Answer: The refractive index increases with increase in frequency which implies that for higher frequency waves, angle of refraction is less, i.e. bending is less. Hence, the condition of total internal relection is atained after travelling larger distance (by $3 MHz$ wave).15.18 The maximum amplitude of an A.M. wave is found to be $15 V$ while its minimum amplitude is found to be $3 V$. What is the modulation index?
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Answer: $A_c+A_m=15, A_c-A_m=3$
$\therefore 2 A_c=18,2 A_m=12$
$\therefore m=\frac{A_m}{A_c}=\frac{2}{3}$
15.19 Compute the $L C$ product of a tuned amplifier circuit required to generate a carrier wave of $1 MHz$ for amplitude modulation.
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Answer: $ \frac{1}{2 \pi \sqrt{L C}}=1 MHz$
$\sqrt{LC}=\frac{1}{2 \pi \times 10^{6}}$
15.20 Why is a AM signal likely to be more noisy than a FM signal upon transmission through a channel?
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Answer: In AM, the carrier waves instantaneous voltage is varied by modulating waves voltage. On transmission, noise signals can also be added and receiver assumes noise a part of the modulating signal.
However in FM, the carriers frequency is changed as per modulating waves instantaneous voltage. This can only be done at the mixing/ modulating stage and not while signal is transmitting in channel. Hence, noise doesn’t effect FM signal.
15.21 Figure 15.3 shows a communication system. What is the output power when input signal is of $1.01 mW$ ? (gain in $dB=10 \log _{10}$ $(P_o / P_i)$.
Fig. 15.3
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Answer: Loss suffered in transmission path
$=-2 dB km^{-1} \times 5 km=-10 dB$
Total amplifier gain $=10 dB+20 dB$
$ =30 dB $
Overall gain of signal $=30 dB-10 dB$
$=20 dB$
$ \begin{aligned} 10 \log (\frac{P_o}{P_i}) \quad=12 \text{ or } P_o & =P_i \times 10^{2} \\ & =1.01 mW \times 100=101 mW . \end{aligned} $
15.22 A TV transmission tower antenna is at a height of $20 m$. How much service area can it cover if the receiving antenna is (i) at ground level, (ii) at a height of $25 m$ ? Calculate the percentage increase in area covered in case (ii) relative to case (i).
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Answer: (i) Range $=\sqrt{2 \times 6.4 \times 10^{6} \times 20}=16 km$
Area covered $=803.84 km^{2}$
(ii) Range $=\sqrt{2 \times 6.4 \times 10^{6} \times 20}+\sqrt{2 \times 6.4 \times 10^{6} \times 25}$
$ =(16+17.9) km=33.9 km $
Area covered $=3608.52 km^{2}$
$\therefore$ Percentage increase in area
$ \begin{aligned} & =\frac{(3608.52-803.84)}{803.84} \times 100 \\ & =348.9 % \end{aligned} $
15.23 If the whole earth is to be connected by LOS communication using space waves (no restriction of antenna size or tower height), what is the minimum number of antennas required? Calculate the tower height of these antennas in terms of earths radius?
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Answer: $ d_m^{2}=2(R+h_T)^{2}$
$8 R h_T=2(R+h_T)^{2}$
$(\because d m=2 \sqrt{2 R h_T})$
$4 R h_T=R^{2}+h_T^{2}+2 R h_T$
$(R-h_T)^{2}=0$
$R=h_T$
Since space wave frequency is used, $\lambda«h_T$, hence only tower height is taken to consideration.
In three diamensions, 6 antenna towers of $h_T=R$ would do.
15.24 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 inonosphere. Estimate the maximum electron densities of the $F_1$ and $F_2$ layers on that day.
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Answer: For $F_1$ layer
$5 \times 10^{6}=9(N _{\max })^{1 / 2}$ or $N _{\max }=(\frac{5}{9} \times 10^{6})^{2}=3.086 \times 10^{11} m^{-3}$
For $F_2$ layer
$8 \times 10^{6}=9(N _{\max })^{1 / 2}$ or
$N _{\max }=(\frac{8}{9} \times 10^{6})=7.9 \times 10^{11} m^{-3}=7.9 \times 10^{11} m^{-3}$.
15.25 On radiating (sending out) an AM modulated signal, the total radiated power is due to energy carried by $\omega_c, \omega_c-\omega_m & \omega_c+\omega_m$. Suggest ways to minimise cost of radiation without compromising on information.
LA
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Answer: Of $\omega_c-\omega_m, \omega_c$ and $\omega_m+\omega_m$, only $\omega_c+\omega_m$ or $\omega_c-\omega_m$ contains information.
Hence cost can be reduced by transmitting $\omega_c+\omega_m, \omega_c-\omega_m$, both $\omega_c+\omega_m$ $& \omega_c-\omega_m$
15.26 (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 per cent after a distance of $(\frac{\ln 4}{\alpha})$
(ii) Attenuation of a signal can be expressed in decibel (dB) according to the relation $dB=10 \log _{10}(\frac{I}{I_o})$ What is the attenuation in $dB / km$ for an optical fibre in which the intensity falls by 50 per cent over a distance of $50 km$ ?
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Answer: (i) $\frac{I}{I_o}=\frac{1}{4}$, so $\ln (\frac{1}{4})=-\alpha x$
or $\ln 4=$ ax or $x=(\frac{\ln 4}{\alpha})$
(ii) $lolog _{10} \frac{I}{I_o}=-\alpha x$ where $\alpha$ is the attunation in $dB / km$.
Here $\frac{I}{I_o}=\frac{1}{2}$
or $10 \log (\frac{1}{2})=-50 \alpha$ or $\log 2=5 \alpha$
or $\alpha=\frac{\log 2}{5}=\frac{0.3010}{5}=0.0602 dB / km$
15.27 A $50 MHz$ sky wave takes $4.04 ms$ to reach a receiver via re-transmission from a satellite $600 km$ above earth’s surface. Assuming re-transmission time by satellite negligible, find the distance between source and receiver. If communication between the two was to be done by Line of Sight (LOS) method, what should size and placement of receiving and transmitting antenna be?
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Answer: $ \frac{2 x}{\text{ time }}=$ velocity
$2 x=3 \times 10^{8} m / s \times 4.04 \times 10^{-3} s$
$x=\frac{12.12 \times 10^{5}}{2} m=6.06 \times 10^{5} m=606 km$
$d^{2}=x^{2}-h_s^{2}=(606)^{2}-(600)^{2}=7236 ; d=85.06 km$
Distance between source and receiver $=2 d \cong 170 km$
$d_m=2 \sqrt{2 R h_T}, 2 d=d_m, \quad 4 d^{2}=8 R h_T$
$\frac{d^{2}}{2 R}=h_T=\frac{7236}{2 \times 6400} \approx 0.565 km=565 m$.
15.28 An amplitude modulated wave is as shown in Fig. 15.4. Calculate (i) the percentage modulation, (ii) peak carrier voltage and, (iii) peak value of information voltage.
Fig. 15.4
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Answer: From the figure
$V _{\text{max }}=\frac{100}{2}=50 V, V _{\text{min }}=\frac{20}{2}=10 V$.
(i) Percentage modulation
$ \mu(%)=\frac{V _{\max }-V _{\min }}{V _{\max }+V _{\min }} \times 100=(\frac{50-10}{50+10}) \times 100=\frac{40}{60} \times 100=66.67 % $
(ii) Peak carrier voltage $=V_c=\frac{V _{\max }+V _{\min }}{2}=\frac{50+10}{2}=30 V$
(iii) Peak information voltage $=V_m=\mu V_c=\frac{2}{3} \times 30=20 V$.
Exemplar Problems-Physics
15.29 (i) Draw the plot of amplitude versus ’ $\omega$ ’ for an amplitude modulated wave whose carrier wave $(\omega_c)$ is carrying two modulating signals, $\omega_1$ and $\omega_2(\omega_2>\omega_1)$. [Hint: Follow derivation from Eq 15.6 of NCERT Textbook of XII]
(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 sol to the above problem. In the process can one understand another advantage of modulation in terms of bandwidth?
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Answer: (a) $v(t)=A(A _{m_1} \sin \omega _{m_1} t+A _{m_2} \sin \omega _{m_2} t+A_c \sin \omega_c t)$
$ +B(A _{m_1} \sin \omega _{m_1} t+A _{m_2} \sin \omega _{m_2} t+A_c \sin \omega_c t)^{2} $
$=A(A _{m_1} \sin \omega _{m_1} t+A _{m_2} \sin \omega _{m_2} t+A_c \sin \omega_c t)$
$+B((A _{m_1} \sin \omega _{m_1} t+A _{m_2} t)^{2}+A_c^{2} \sin ^{2} \omega_c t.$
$+2 A_c(A _{m_1} \sin \omega _{m 1} t+A _{m_2} \sin \omega_c t)$
$=A(A_1 \sin \omega _{m_1} t+A _{m_2} \sin \omega _{m_2} t+A_c \sin \omega_c t)$
$+B[A _{m_1}{ }^{2} \sin ^{2} \omega _{m_1} t+A _{m_2}{ }^{2} \sin ^{2} \omega _{m_2} t+2 A _{m_1} A _{m_2} \sin \omega _{m_1} t \sin \omega _{m_2} t.$
$+A_c^{2} \sin ^{2} \omega_c t+2 A_c(A _{m_1} \sin \omega _{m_1} t \sin \omega_c t+A _{m_2} \sin \omega _{m_2}+\sin \omega_c t]$
$=A(A _{m_1} \sin \omega _{m_1} t+A _{m_2} \sin \omega _{m_2} t+A_c \sin \omega_c t)$
$ +B[A_{m 1} ^2 \sin^2 \omega_{m_1} t+A_{m_2}^{2} \sin ^{2} \omega _{m_2} t+A_c^{2} \sin ^{2} \omega_c t.$
$+\frac{\not 2 A _{m_1} A _{m_2}}{\not 2}[\cos (\omega _{m_2}-\omega _{m_1}) t-\cos (\omega _{m_1}+\omega _{m_2}) t]$
$+\frac{\not 2 A_c A _{m_2}}{\not 2}[\cos (\omega_c-\omega _{m_1}) t-\cos (\omega_c+\omega _{m_1}) t]$
$+\frac{\not{Z} A_c A _{m_1}}{\not 2}[\cos (\omega_c-\omega _{m_2}) t-\cos (\omega_c+\omega _{m_2}) t]$
$\therefore$ Frequencies present are
$ \omega _{m_1}, \omega _{m_2}, \omega_c $
$(\omega _{m_2}-\omega _{m_1}),(\omega _{m_1}+\omega _{m_2})$
$ \begin{aligned} & (\omega_c-\omega _{m_1}),(\omega_c+\omega _{m_1}) \\ & (\omega_c-\omega _{m_2}),(\omega_c+\omega _{m_2}) \end{aligned} $
(i) Plot of amplitude versus $\omega$ is shown in the Figure.
(ii) As can be seen frequency spectrum is not symmetrical about $\omega_c$. Crowding of spectrum is present for $\omega<\omega_c$.
(iii) Adding more modulating signals lead to more crowding in $\omega<\omega_c$ and more chances of mixing of signals.
(iv) Increase band-width and $\omega_c$ to accommodate more signals. This shows that large carrier frequency enables to carry more information (more $\omega_m$ ) and which will in turn increase bandwidth.
15.30 An audio signal is modulated by a carrier wave of $20 MHz$ such that the bandwidth required for modulation is $3 kHz$. Could this wave be demodulated by a diode detector which has the values of $R$ and $C$ as
(i) $R=1 k \Omega, C=0.01 \mu F$.
(ii) $R=10 k \Omega, C=0.01 \mu F$.
(iii) $R=10 k \Omega, C=0.1 \mu F$.
Show Answer
Answer: $ f_m=1.5 kHz, \frac{1}{f_m}=0.7 \times 10^{-3} s$
$f_c=20 MHz, \frac{1}{f_c}=0.5 \times 10^{-7} s$
(i) $R C=10^{3} \times 10^{-8}=10^{-5} s$
So, $\frac{1}{f_c}<R C<\frac{1}{f_m}$ is satisfied
So it can be demodulated.
(ii) $R C=10^{4} \times 10^{-8}=10^{-4} s$.
Here too $\frac{1}{f_c}«R C<\frac{1}{f_m}$.
So this too can be demodulated
(iii) $R C=104 \times 10^{-12}=10^{-8} s$.
Here $\frac{1}{f_c}>R C$, so this cannot be demodulated.
