Communication Systems

Digital signal uses binary system, that means $(0,1)$. This leads to discrete wave produce from digital signal.

They are in the form of rectangular waveform. Digital signal cannot provide a continuous set of value because it is not using Hexadecimal system, which gives continuous values.

Hence, the correct option is (c).

Given,

Amplitude modulated wave is

$ C_m(t)=10[1+0.6 \sin (1250 t)] \sin \left(10^8 t\right) $

Compare this wave equation with standard wave equation for amplitude modulation.

$ C_m(t)=A_c\left[1+\frac{A_m}{A_c} \sin \left(\omega_m t\right)\right] \sin \left(\omega_c t\right) $

where $A_c=$ Amplitude of carrier wave.

$A_m=$ Amplitude of modulating wave.

Modulation index is given by

$ M=A_m / A_c \Rightarrow M=0.6 $

Hence, option (d) is correct option.

Carrier frequency

$ \begin{aligned} f_c & =20 \mathrm{GHz} \\ & =20 \times 10^9 \mathrm{~Hz} \\ & =2 \times 10^{10} \mathrm{~Hz} \end{aligned} $

Utilised carrier frequency,

$ \begin{aligned} f_c^{\prime} & =20 \% \text { of } f_C \\ & =\frac{20}{100} \times 2 \times 10^{10} \\ & =4 \times 10^9 \mathrm{~Hz} \end{aligned} $

Bandwidth of one channel.

$ \begin{aligned} f_m & =5 \mathrm{kHz} \\ & =5 \times 10^3 \mathrm{~Hz} \end{aligned} $

No. of telephonic channel

$ \begin{aligned} n & =\frac{f_c^{\prime}}{f_m}=\frac{4 \times 10^9}{5 \times 10^3} \\ & =8 \times 10^5 \mathrm{~Hz} \end{aligned} $

The loss of strength of a signal white propagating through a medium is known as attenation.

Modulation is the process in which characteristics of carrier wave is changed according to modulating signal.

Demodulation is the process of separating the original message signal from the modulated signal.

Noise is unwanted signal or unpleasant signal.

Modulation : It is a process in which the message signal is mixed with carrier signal and then sent it. The need for modulation is, Original (Message) signal distorts at very less distance and cannot be sent over large distance.

So to overcome this issue signal is modulated in carrier signal and then sent over large distance.

The given problem involves calculating the sideband frequencies produced when a message signal modulates a carrier wave.

Carrier wave frequency: $ 6 \text{ MHz} = 6000 \text{ kHz} $

Message signal frequency: $ 10 \text{ kHz} $

Let $ C $ denote the carrier wave frequency and $ M $ represent the message signal frequency. The sideband frequencies are calculated as follows:

Upper Sideband Frequency:

$ C + M = 6000 \text{ kHz} + 10 \text{ kHz} = 6010 \text{ kHz} $

Lower Sideband Frequency:

$ C - M = 6000 \text{ kHz} - 10 \text{ kHz} = 5990 \text{ kHz} $

Thus, the final sideband frequencies are $ 6010 \text{ kHz} $ and $ 5990 \text{ kHz} $.

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

Given modulation index, $\mu=0.5$

Maximum amplitude, $A_{\text {max }}=10 \mathrm{~V}$

Minimum amplitude, $A_{\text {min }}=$ ?

We know that,

$ \begin{aligned} & \mu=\frac{A_{\max }-A_{\min }}{A_{\max }+A_{\min }} \Rightarrow 0.5=\frac{10-A_{\min }}{10+A_{\min }} \\ \Rightarrow & 5+0.5 A_{\min }=10-A_{\min } \\ \Rightarrow & 1.5 A_{\min }=5 \\ \Rightarrow & A_{\min }=\frac{5}{1.5}=\frac{1}{0.3}=\frac{10}{3}=3.33 \mathrm{~V} \end{aligned} $

Peak voltage of carrier wave,

$ A_c=10 \mathrm{~V} $

Modulation index,

$ \mu=80 \%=0.8 $

If $A_m$ be the peak voltage of modulating signal, then

$ \begin{aligned} \mu & =\frac{A_m}{A_c} \\ \Rightarrow \quad A_m & =\mu A_c \\ & =0.8 \times 10 \\ & =8 \mathrm{~V} \end{aligned} $

The power radiated from a linear antenna is directly proportional to square of frequency. Thus, power radiated from the linear antenna increases with increasing frequency.

The range of frequency bands used for standard AM broadcast is $(540-1600) \mathrm{kHz}$.

Given, frequency of message signal,

$ f_m=15 \mathrm{kHz}=15 \times 10^3 \mathrm{~Hz} $

Let $f_c$ be the carrier frequency, then upper side band frequency $=f_c+f_m$

$ \begin{array}{ll} \Rightarrow & 1515=f_c+f_m \\ \Rightarrow & f_c+f_m=1515\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

Similarly, lower side band frequency

$ f_c-f_m=1485 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

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

$ \begin{aligned} 2 f_c & =3000 \\ \Rightarrow \quad f_c & =1500 \mathrm{kHz} \\ & =1.5 \times 10^6 \mathrm{~Hz} \\ & =1.6 \mathrm{MHz} \end{aligned} $

Given, height of transmission antenna,

$ h=40 \mathrm{~m}, $

Radius of earth,

$ \begin{aligned} R & =6400 \mathrm{~km} \\ & =6.4 \times 10^6 \mathrm{~m} \end{aligned} $

The maximum distance up to which signal can be broadcasted,

$ \begin{aligned} d_{\max } & =\sqrt{2 R h} \\ \therefore \text { Service area, } A & =\pi d_{\max }^2 \\ = & \pi \times 2 R h \\ = & \pi \times 2 \times 6.4 \times 10^6 \times 40 \\ = & 512 \pi \times 10^6 \mathrm{~m}^2 \end{aligned} $