Communication Systems

To determine the minimum height of the receiving antenna required to receive the signal in line of sight at a distance of 4 km from the transmitting antenna, we need to consider the curvature of the Earth. Let's denote the height of the receiving antenna as $h$ and the distance between the antennas as $d = 4 \mathrm{~km}$.

We can use the Pythagorean theorem to relate the Earth's radius $R$, the height of the receiving antenna $h$, and the distance between the antennas $d$:

$(R + h)^2 = R^2 + d^2$

Now, we can plug in the given values for $R$ and $d$:

$(6400 + h)^2 = 6400^2 + 4^2$

Expanding the equation and subtracting $6400^2$ from both sides gives:

$2 \times 6400 \times h + h^2 = 16$

Since $h$ is much smaller than the Earth's radius, we can ignore the $h^2$ term:

$2 \times 6400 \times h \approx 16$

Now, we can solve for $h$:

$h \approx \frac{16}{2 \times 6400} = \frac{1}{800} \mathrm{~km}$

Converting to meters and multiplying by $10^2$ to match the desired format:

$x = h \times 10^3 \times 10^2 = \frac{1}{800} \times 10^5 = 125$

So, the value of $x$ is 125

The bandwidth of an amplitude-modulated wave is given by $2B$, where $B$ is the bandwidth of the modulating signal.

In this case, the modulating signal is a message signal of frequency $3~\mathrm{kHz}$, so the bandwidth of the amplitude modulated wave is $2(3~\mathrm{kHz})=6~\mathrm{kHz}$.

Therefore, the correct answer is $6~\mathrm{kHz}$.

The range (or distance to the horizon) for line-of-sight communication is given by the formula:

$d = \sqrt{2hR}$,

where:

• $d$ is the range,
• $h$ is the height of the antenna, and
• $R$ is the radius of the Earth.

Given that $h = 98 \, \text{m} = 98 \times 10^{-3} \, \text{km}$ and $R = 6400 \, \text{km}$, we can substitute these values into the formula to find $d$:

$d = \sqrt{2 \times 98 \times 10^{-3} \times 6400} \approx 35.42 \, \text{km}$.

The surface area $A$ covered by the transmitting antenna is approximately a circle with radius equal to the range $d$. The area of a circle is given by the formula $A = \pi r^2$, where $r$ is the radius. Therefore, we can substitute $d$ into this formula to find $A$:

$A = \pi d^2 = \pi (35.42)^2 \approx 3940 \, \text{km}^{2}$.

Therefore, the surface area covered by the transmitting antenna is approximately $3940 \, \text{km}^{2}$.

In the case of Amplitude Modulation, the modulation index (µ) can also be represented as:

$ \mu = \frac{A_{\max} - A_{\min}}{A_{\max} + A_{\min}} $

where:

• $A_{\max}$ is the maximum amplitude of the modulated wave,
• $A_{\min}$ is the minimum amplitude of the modulated wave.

Given that the modulation index is 60% or 0.6 and the minimum amplitude $A_{\min}$ is 3 V, we can rearrange this formula to solve for the maximum amplitude $A_{\max}$:

$ 0.6 = \frac{A_{\max} - 3}{A_{\max} + 3} $

Solving this equation gives:

$ A_{\max} = 12 \, \text{V} $

So, the correct answer is 12 V.

The range of transmission (R) of a TV tower (or any radio tower) depends on the height of the tower (h). This relationship is given by the formula:

$ R = \sqrt{2hR_E} $

where $R_E$ is the radius of the Earth. This formula comes from the geometry of the situation, considering the curvature of the Earth.

Now, if the height of the tower is increased, this will increase the range of transmission. Specifically, because the height appears under the square root in the formula, the range of transmission is proportional to the square root of the height. So, if the height is multiplied by some factor, the range will be multiplied by the square root of that factor.

In this case, the height of the tower is increased by 21%, which means the new height is $h_2 = 1.21h_1$. If we substitute this into the formula for the range, we get:

$ R_2 = \sqrt{2h_2R_E} = \sqrt{2(1.21h_1)R_E} = \sqrt{1.21} \sqrt{2h_1R_E} = 1.1R_1 $

This shows that the new range is 1.1 times the original range, or in other words, the range has increased by 10%.

For transmitting a signal, antenna should have a size comparable to the wavelength of the signal i.e., at least ${1 \over 4}$ in dimension, so that the antenna properly senses the time variation of the signal.

The amplitude of the carrier wave is varied in accordance with the message signal is termed as amplitude modulation.

So, statement I is correct but statement II is incorrect.

Theoretical attenuation $\to$ (IV)

Transducer $\to$ (III)

Demodulation $\to$ (II)

Repeater $\to$ (I)

Amplitude of each side band = $\frac{A_{\mathrm{message}}}{2}$

$\mathrm{A_{carrier}+A_{message}=120}$ ............ (1)

$\mathrm{A_{carrier}-A_{message}=80}$ .............. (2)

From (1) and (2)

$\mathrm{A_{message}=20~V}$

$\therefore$ Amplitude of each side band = 10 V

The modulation index (m) of an amplitude-modulated wave is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier wave. It is a measure of the degree of modulation of the carrier wave and is given by the following equation:

$A_{m}=$ Amplitude of modulating wave $=1$ volt

$A_{c}=$ Amplitude of carrier wave $=2$ volt

Modulation index $=\frac{A_{m}}{A_{c}}=\frac{1}{2}=\frac{1}{2}$

In an Amplitude Modulated (AM) signal, the amplitude of the carrier signal is varied in accordance with the information being sent. The extent of this variation is given by the modulation index, often denoted by $\mu$.

In the problem statement, we are given a peak-to-peak variation in the signal of 8V. This is essentially twice the amplitude of the modulating signal, because the peak-to-peak value represents the total variation from the minimum to the maximum. Thus, we find the amplitude of the modulating signal ($A_m$) as follows:

$A_m = \frac{8V}{2} = 4V$

The maximum amplitude of the AM wave is 9V. This includes the amplitude of the carrier signal and the amplitude of the modulating signal. We therefore find the amplitude of the carrier signal ($A_c$) as follows:

$A_c = 9V - A_m = 9V - 4V = 5V$

Finally, the modulation index $\mu$ is defined as the ratio of the amplitude of the modulating signal to the amplitude of the carrier signal:

$\mu = \frac{A_m}{A_c} = \frac{4V}{5V} = 0.8$

Thus, the modulation index of the AM signal is 0.8, meaning that the amplitude of the carrier signal is varied by 80% of its original amplitude to encode the information being sent. The correct answer among the provided options is therefore Option A: 0.8.

Bandwidth of FM wave $ = 2(\Delta f + {f_m})$

${{\Delta f} \over {{f_m}}} = 10$ (Given)

$\Delta f = {f_m}(10) = 20 \times 10 = 200$ kHz

$BW = 2(200 + 20)$ kHz

$ = 440$ kHz

For effective modulation,

$\mu$ $\le$ 1

$\mu = {{{A_m}} \over {{A_c}}}$

$ = {1 \over 5} = 0.2$

$\therefore$ ${r_m} = \sqrt {2Rh} $

$ \Rightarrow {{{r_1}} \over {{r_2}}} = \sqrt {{{{h_1}} \over {{h_2}}}} $

$ \Rightarrow {1 \over 3} = \sqrt {{{100} \over {{h_2}}}} $

$ \Rightarrow {h_2} = 900$ m

Percentage modulation

$\mu = {{{V_{\max }} - {V_{\min }}} \over {{V_{\max }} + {V_{\min }}}} \times 100$

$ = {{60 - 20} \over {60 + 20}} \times 100$

$ = 50\% $

${v_1} = 6 \times {10^6}$ Hz

$ \Rightarrow {\lambda _1} = {{3 \times {{10}^8}} \over {6 \times {{10}^6}}} = 50$ m

${v_2} = 10 \times {10^6}$ Hz

$ \Rightarrow {\lambda _2} = {{3 \times {{10}^8}} \over {10 \times {{10}^6}}} = 30$ m

$\Rightarrow$ Wavelength band with

$ = |{\lambda _1} - {\lambda _2}| = 20$ m

A_max = 6 V

A_min = 2 V

$\mu = {{{A_{\max }} - {A_{\min }}} \over {{A_{\max }} + {A_{\min }}}} = {{6 - 2} \over {6 + 2}} = 0.5$

$\mu = 50\% $

Range $R = \sqrt {2h{\mathop{\rm Re}\nolimits} } $

Let the height be h' to double the range so

$2R = \sqrt {2h'{\mathop{\rm Re}\nolimits} } $

On solving h' = 4h

h' = 500 m

So $\Delta$h = 375 m

$v = f\lambda $

$ \Rightarrow f = {v \over \lambda } = {{3 \times {{10}^8}} \over {1000 \times {{10}^{ - 9}}}}$ Hz $ = 3 \times {10^{14}}$ Hz

$\Rightarrow$ Channels $ = {{{2 \over {100}} \times 3 \times {{10}^{14}}} \over {8 \times {{10}^3}}} = 75 \times {10^7}$

Bandwidth = 2 $\times$ f_m

= 2 $\times$ 10⁴ Hz = 20 kHz

Television signal $\Rightarrow$ 6 MHz

Radio signal $\Rightarrow$ 2 MHz

High Quality music $\Rightarrow$ 20 kHz

Human speech $\Rightarrow$ 3 kHz

For longer wavelength, size of antenna would increase. Also, mixing of signals needs to be avoided.

Also, we can use modulation to send low frequency signal by superimposing them with high frequency signals.

Modulate signal $s(t) \equiv [1 + 20\sin (1000\pi t)]\sin ({10^7}\pi t)$

$ \equiv \sin ({10^7}\pi t) + 10\cos ({10^7}\pi t - {10^3}\pi t) + 10\cos ({10^7}\pi t + {10^3}\pi t)$

$\Rightarrow$ Required amplitude = 10

The correct match is :

Optical fibres supports frequency of electromagnetic waves in the range 10¹⁴ Hz to 10¹⁵ Hz.

${v _c} = 3.5 \times {10^9}$ Hz

$\therefore$ $\lambda = {c \over {{v _c}}} = {{3 \times {{10}^8}} \over {3.5 \times {{10}^9}}}$

$\therefore$ Size of antenna $ = {\lambda \over 4}$

$ = {{8.57 \times {{10}^{ - 2}}} \over 4}$

$ = 21.4$ mm