Thinking Process

Mode of communication depend on the frequencies of a wave.

Answer

(b) Mode of communication frequency range

Ground wave propagation $-530 kHz$ to $1710 kHz$

Sky wave propagation $-1710 kHz$ to $40 MHz$

Space wave propagation $-54 MHz$ to $4.2 GHz$

Answer

(a) Given, length of the building $(l)$ is given by

$$ l=500 m $$

we know that, wavelength of the wave which can be transmitted by

$$ \lambda \sim 4 l=4 \times 100=400 m $$

Answer

(b) Given, power of signal transmitted is given $P_{i}=1 kW=1000 W$

Rate of attenuation of signal $=-2 dB / km$

Length of total path $=5 km$

Thus,

$$ \text { gain in } d B=5 \times(-2)=-10 dB $$

Also,

$$ \text { gain in } d B=10 \log \frac{P_0}{P_{i}} $$

Here $P_0$ is the power of the received signal.

Putting the given values in Eq. (i),

$$ -10=10 \log \frac{P_0}{P_{i}}=-10 \log \frac{P_{i}}{P_0} \newline \Rightarrow \log \frac{P_{i}}{P_0}=1 \Rightarrow \log \frac{P_{i}}{P_0}=\log 10 \newline \Rightarrow \frac{P_{i}}{P_0}=10 \Rightarrow 1000 W=10 P_0 \newline \Rightarrow P_0=100 W $$

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.

Answer

(a) Given, frequency of carrier signal is $\omega_{c}=1 MHz$

and $\quad$ frequency of speech signal $=3 kHz$

$$ \begin{aligned} & =3 \times 10^{-3} MHz \\ & =0.003 MHz \end{aligned} $$

Now, we know that,

Frequencies of side bands $=(\omega_{c} \pm \omega_{m})$

$$ =(1 \pm 0.003) $$

$$ =1.003 MHz \text { and } 0.997 MHz $$

Thinking Process

In amplitude modulation, the frequency of modulated wave is equal to the frequency of carrier wave.

Answer

(b) Here, according to the question, frequency of carrier wave is $\omega_{c}$.

Thus the amplitude modulated wave also has frequency $\omega_{c}$.

Thinking Process

A square law device is something where either current or voltage depends on the square of the other.

Answer

(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.

Thinking Process

The frequency of male voice less than that of a female voice.

Answer

(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.

Answer

(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

Thinking Process

An arbitrary change in phase angle of the modulating signal is given by $\phi$.

Answer

(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 $=2 \pi V_{m}=\phi V_{m}$

Also consider a sinusoidal carrier wave represented by $C(t)=A_{c} \sin \omega_{c} t$

Thus, modulated wave is given by

Here,

$$ \begin{aligned} C_{m}(t) & =(A_{c}+A_{m} \sin \omega_{m} t) \sin \omega_{c} t \\ & =A_{c}[1+\frac{A_{m}}{A_{c}} \sin \omega_{m} t) \sin \omega_{c} t \end{aligned} $$

$\Rightarrow$

$$ \frac{A_{m}}{A_{c}}=M $$

$\Rightarrow \quad C_{m}(t)=(A_{c}+A_{c} \times \mu \sin \omega_{m} t) \sin \omega_{c} t$

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 $\sin \omega_{c} t arrow \sin (\omega_{c} t+\phi)$

Thus,

$$ C_{m}(t)=(A_{c}+K V_{m})(\sin \omega_{c}+\phi) $$

Thinking Process

Transmission of a signal depends on three factors. These are size of antenna, medium of transmission and power of transmitted wave.

Answer

$(a, b, d)$

Given, frequency of the wave to be transmitted is

$$ \begin{aligned} v_{m} & =15 kHz=15 \times 10^{3} Hz \\ \text { Wavelength } \lambda_{m} & =\frac{c}{v_{m}}=\frac{3 \times 10^{8}}{15 \times 10^{3}}=\frac{1}{5} \times 10^{5} m \end{aligned} $$

Size of the antenna required, $\quad l=\frac{\lambda}{4}=\frac{1}{4} \times \frac{1}{5} \times 10^{5}$

$$ =5 \times 10^{3} m=5 km $$

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.

Thinking Process

Here, in this question, options are giving the value of side band frequencies and band width of amplitude modulation. So, first of all find this quantities.

Answer

$(b, d)$

Given,

$$ \begin{aligned} \omega_{m} & =3 kHz \\ \omega_{c} & =1.5 MHz=1500 kHz \end{aligned} $$

Now, side band frequencies

$$ \begin{aligned} \omega_{c} \pm \omega_{m} & =(1500 \pm 3) \\ & =1503 kHz \text { and } 1497 kHz \end{aligned} $$

Also, bandwidth $=2 \omega_{m}=2 \times 3=6 kHz$

Thinking Process

$$ \text { Range } d_{T}=\sqrt{2 R h_{T}} $$

Answer

$(b, c, d)$

Given, $\quad$ height of tower $h=240 m$

For LOS (line of sight) communication.

The maximum distance on earth from the transmitter upto which a signal can be received is given by

$$ d=\sqrt{2 R h} $$

Here $R$ is the radius of the earth i.e., $R=6.4 \times 10^{6} m$

Putting all these values in Eq. (i), we get

$$ \begin{aligned} d & =\sqrt{2 R h}=\sqrt{2 \times 6.4 \times 10^{6} \times 240} \\ & =55.4 \times 10^{3} m=55.4 km \end{aligned} $$

Thus, the range of $55.4 km$ covers the distance $24 km, 55 km$ and $50 km$.

Answer

$(a, b, c)$

Here, for the production of amplitude modulated wave, bandwidth is given by =frequency of upper side band - frequency of lower side band

$$ =\omega_{USB}-\omega_{LSB}=(\omega_{c}+\omega_{m})-(\omega_{c}-\omega_{m}) $$

Answer

$(b, d)$

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

$$ \begin{aligned} m_{f} & =\frac{\Delta v_{\max }}{v_{m}(\max )} \\ & =\frac{\text { frequency deviation }}{\text { maximum frequency value of modulating wave }} \end{aligned} $$

If $m_{t}>1$, then $\Delta v_{\max }>v_{m}$. It means, there will be overlapping of both side bands of modulated wave resulting into loss of information.

Answer

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, (a) and

(b) would produce analog signal and

(c) and

(d) would produce digital signals.

Answer

A signal to be transmitted through sky waves must have a frequency range of $1710 kHz$ to $40 MHz$.

But, here the frequency of TV signals are $60 MHz$ which is beyond the required range.

So, sky waves will not be suitable for transmission of TV signals of $60 MHz$ frequency.

Answer

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.

Answer

Let $A_{c}$ and $A_{m}$ be the amplitudes of carrier wave and modulating wave respectively. So,

$$ \text { Maximum amplitude } arrow A_{\max }=A_{c}+A_{m}=15 V $$

$$ \text { Minimum amplitude } arrow A_{\min }=A_{c}-A_{m}=3 V $$

Adding Eqs. (i) and (ii), we get

$$ \begin{aligned} 2 A_{c} & =18 \\ A_{c} & =9 V \end{aligned} $$

$$ \text { and } \quad \begin{aligned} A_{m} & =15-9=6 V \\ \text { Modulating index of wave } \mu & =\frac{A_{m}}{A_{c}}=\frac{6}{9}=\frac{2}{3} \end{aligned} $$

Thinking Process

$$ \text { For tuned amplifier } f=\frac{1}{2 \pi \sqrt{L C}} $$

Answer

Given, the frequency of carrier wave is $1 MHz$.

Formula for the frequency of tuned amplifier,

$$ \begin{aligned} \frac{1}{2 \pi \sqrt{L C}} & =1 MHz \\ \sqrt{L C} & =\frac{1}{2 \pi \times 10^{6}} \\ L C & =\frac{1}{(2 \pi \times 10^{6})^{2}}=2.54 \times 10^{-14} S \end{aligned} $$

Thus, the product of $L C$ is $2.54 \times 10^{-14} s$.

Answer

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.

Answer

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 $dB=10 \log _{10} \frac{P_0}{P_j}$

$$ \therefore 20 =10 \log_{10} \frac{P_0}{P_i} $$ $$ \text {or} \log _{10} \frac{P_0}{P_i}=2 $$

Here, $P_{i}=1.01 mW$ and $P_0$ is the output power.

$$ \begin{matrix} \therefore & \frac{P_0}{P_{i}}=10^{2}=100 \\ \Rightarrow & P_0=P_{i} \times 100=1.01 \times 100 \\ \text { or } & P_0=101 mW \\ \text { Thus, the output power is } 101 mW . \end{matrix} $$

Answer

Given, height of antenna $h=20 m$

Radius of earth $=6.4 \times 10^{6} m$

At the ground level,

(i) Range $=\sqrt{2 h R}=\sqrt{2 \times 20 \times 6.4 \times 10^{6}}$

$$ \begin{aligned} & =16000 m=16 km \\ \text { Area covered } A & =\pi(\text { range })^{2} \\ & =3.14 \times 16 \times 16=803.84 km^{2} \end{aligned} $$

(ii) At a height of $H=25 m$ from ground level

$$ \begin{aligned} \text { Range } & =\sqrt{2 h R}+\sqrt{2 H R} \\ & =\sqrt{2 \times 20 \times 6.4 \times 10^{6}}+\sqrt{2 \times 25 \times 6.4 \times 10^{6}} \\ & =16 \times 10^{3}+17.9 \times 10^{3} \\ & =33.9 \times 10^{3} m \\ & =33.9 km \end{aligned} $$

$$ \begin{aligned} \text { Area covered } & =\pi(\text { Range })^{2} \\ & =3.14 \times 33.9 \times 33.9 \\ & =3608.52 km^{2} \\ \text { Percentage increase in area } & =\frac{\text { Difference in area }}{\text { Initial area }} \times 100 \\ & =\frac{(3608.52-803.84)}{803.84} \times 100 \\ & =348.9 % \end{aligned} $$

Thus, the percentage increase in area covered is $348.9 %$

Answer

Consider the figure given below to solve this question

Suppose the height of transmitting antenna or receiving antenna in order to cover the entire surface of earth through communication is $h_{t}$ and radius of earth is $R$

Then, maximum distance

$$ \begin{matrix} d_m^{2} =(R+h_{t})^{2}+(R+h_{t})^{2} \newline =2(R+h_{t})^{2} \newline \therefore d_{m} =\sqrt{2 h_{t} R}+\sqrt{2 h_{t} R}=2 \sqrt{2 h_{t} R} \newline \Rightarrow 8 h_{t} R =2(R+h_{t})^{2} \newline \Rightarrow 4 h_{t} R =R^{2}+2 R h_{t}+h_t^{2} \newline \Rightarrow R^{2}-2 h_{t} R+h_t^{2} =0 \newline \Rightarrow (R-h_{t})^{2} =0 \newline \Rightarrow R =h_{t} \end{matrix} $$

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.

Answer

The maximum frequency for reflection of sky waves

$$ f_{\max }=9(N_{\max })^{1 / 2} $$

where, $N_{\text {max }}$ is a maximum electron density.

$$ \begin{aligned} & \text { For } F_1 \text { layer, } \\ & \begin{aligned} f_{\max } & =5 MHz \\ 5 \times 10^{6} & =9(N_{\max })^{1 / 2} \end{aligned} \end{aligned} $$

Maximum electron density

$$ \begin{aligned} & N_{\max }=\frac{5}{9} \times 10^{6}{ }^{2}=3.086 \times 10^{11} / m^{3} \\ & f_{\max }=8 MHz \\ & 8 \times 10^{6}=9(N_{\max })^{1 / 2} \end{aligned} $$

Maximum electron density

$$ N_{\max }=\frac{8 \times 10^{6}}{9}=7.9 \times 10^{11} / m^{3} $$

Answer

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 $(\omega_{c}-\omega_{m})$.

Answer

(a) Given, the intensity of a light pulse $I=I_0 e^{-\alpha x}$

where, $I_0$ is the intensity at $x=0$ and $\alpha$ is constant.

According to the question, $I=25 %$ of $I_0=\frac{25}{100} \cdot I_0=\frac{I_0}{4}$

Using the formula mentioned in the question,

or

$$ \begin{aligned} I & =I_0 e^{-\alpha x} \\ \frac{I_0}{4} & =I_0 e^{-\alpha x} \\ \frac{1}{4} & =e^{-\alpha x} \end{aligned} $$

Taking log on both sides, we get

$$ \begin{aligned} \ln 1-\ln 4 & =-\alpha x \ln e \quad(\because \ln e=1) \\ -\ln 4 & =-\alpha x \\ x & =\frac{\ln 4}{\alpha} \end{aligned} $$

Therefore, at distance $x=\frac{\ln 4}{\alpha}$, the intensity is reduced to $75 %$ of initial intensity.

(b) 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, $I=50 %$ of $I_0=\frac{I_0}{2}$ and $x=50 km$

Putting the value of $x$ in Eq. (i), we get

$$ \begin{aligned} 10 \log _{10} \frac{I_0}{2 I_0} & =-\alpha \times 50 \\ 10[\log 1-\log 2] & =-50 \alpha \\ \frac{10 \times 0.3010}{50} & =\alpha \end{aligned} $$

$\therefore$ The attenuation for an optical fibre

$$ \alpha=0.0602 dB / km $$

Answer

Let the receiver is at point $A$ and source is at $B$.

Velocity of waves $=3 \times 10^{8} m / s$

Time to reach a receiver $=4.04 ms=4.04 \times 10^{-3} s$

Let the height of satellite is

$h_{s}=600 km$

Radius of earth $=6400 km$

Size of transmitting antenna $=h_{T}$

We know that $\frac{\text { Distance travelled by wave }}{\text { Time }}=$ Velocity of waves

$$ \begin{aligned} \frac{2 x}{4.04 \times 10^{-3}} & =3 \times 10^{8} \\ x & =\frac{3 \times 10^{8} \times 4.04 \times 10^{-3}}{2} \\ & =6.06 \times 10^{5}=606 km \end{aligned} $$

or

Using Phythagoras theorem,

or

$$ d^{2}=x^{2}-h_s^{2}=(606)^{2}-(600)^{2}=7236 $$

So, the distance between source and receiver $=2 d$

$$ =2 \times 85.06=170 km $$

The maximum distance covered on ground from the transmitter by emitted EM waves

$$ \text { or } $$

$$ d=\sqrt{2 R h_{T}} $$

$$ \frac{d^{2}}{2 R}=h_{T} $$

or

$$ \text { size of antenna } h_{T}=\frac{7236}{2 \times 6400} $$

$$ =0.565 km=565 m $$

Answer

From the diagram,

$$ \begin{aligned} & \text { Maximum voltage } V_{\max }=\frac{100}{2}=50 V \\ & \text { Minimum voltage } V_{\min }=\frac{20}{2}=10 V \end{aligned} $$

(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, $\quad V_{c}=\frac{V_{\max }+V_{\min }}{2}=\frac{50+10}{2}=30 V$

(iii) Peak value of information voltage,

$$ V_{m}=\mu V_{c}=\frac{66.67}{100} \times 30=20 V $$

Answer

(i) The plot of amplitude versus $\omega$ can be shown in the figure below

(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.

Answer

Given, carrier wave frequency $f_{c}=20 MHz$

$$ =20 \times 10^{6} Hz $$

Bandwidth required for modulation is

$$ \begin{aligned} & 2 f_{m}=3 kHz=3 \times 10^{3} Hz \\ \Rightarrow & f_{m}=\frac{3 \times 10^{3}}{2}=1.5 \times 10^{3} Hz \end{aligned} $$

Demodulation by a diode is possible if the condition $\frac{1}{f_{c}} \ll R C<\frac{1}{f_{m}}$ is satisfied

Thus,

$$ \begin{aligned} \frac{1}{f_{c}} & =\frac{1}{20 \times 10^{6}}=0.5 \times 10^{-7} \\ \frac{1}{f_{m}} & =\frac{1}{1.5 \times 10^{3}} Hz=0.7 \times 10^{-3} s \end{aligned} $$

$$ \text { and } $$

Now, gain through all the options of $R$ and $C$ one by one, we get

(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.

(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 $(2 n-1) \frac{\lambda}{2}$.