Understanding Physics Topics is essential for solving complex problems in many fields, including engineering and medicine.
What is Broadband Communication?
Two factors are to be ascertained first to decide how to transmit a data signal from one place to another.
i) Medium: Atmosphere, coaxial cable, or optical fibre are used as medium. Sometimes combination of different types of media are also used.
ii) Range of frequency: What part of the electromagnetic spectrum would be used as carrier wave, is also to be ascertained. The corresponding range of the frequency is called transmission band.
The arrangement for transmission of data signal from one point to another with a combination of an appropriate medium and a carrier wave of appropriate frequency is called a communication channel. As for example, the carrier waves, of frequency in the range of approximate 300 kHz to 10 MHz, can propagate as sky wave through the atmosphere from one place to another. Hence it is an effective communication channel. The efficiency of a communication channel is determined by the total number of discrete data signals that can be transmitted through it, without distortion.
As for example, a carrier wave of frequency 1 MHz carries an audio signal. Naturally, the frequency of the audio signal lies between 20 Hz to 20000 Hz. If the carrier wave is amplitude modulated by the audio signal, then the bandwidth of the amplitude modulated wave would be = 2 × 20000 = 40000 Hz (see section 1.5.1). This band lies symmetrically on either side of the main frequency and its location is between (106-20000) Hz or 980000 Hz to (106 + 20000) or 1020000 Hz. Now if an audio signal is transmitted from another transmitting station through a carrier wave of frequency 1.05 MHz, the location of the band would be between the frequency 1030000 Hz to 1070000 Hz.
Clearly, there is a distinct difference between the former band and the present band and they will not mix up together, so the receiver can clearly distinguish the two audio signals. It is clear from this example that, since the bandwidth of the audio signal is 40 kHz hence, if the difference of frequencies between two carrier waves be more than 40 kHz, the audio signals can be received distinctly by the receiving machine.
Now we can consider the propagation of sky wave through atmosphere. Here the propagation of effective frequency is from 100 kHz to 10 MHz, i.e., the width of the transmission band is 107 – 105 ≈ 107 Hz = 10 MHz. Now, for clear and distinct transmission, the difference of frequency between the carrier waves, should be at Least 0.04 MHz. For example, if this difference is taken as 0.05 MHz, then through the 10 MHz width of transmission band, (10 ÷ 0.05) or, 200 carrier waves can pass together, which means 200 audio signals can be transmitted distinctly through the sky wave band.
In reality, since frequencies higher than 4000 Hz are almost absent in audio signal, the bandwidth of amplitude modulated wave ¡s very narrow. Hence, it is sufficient if the difference of frequencies between the carrier waves be 0.01 MHz (or 10 kHz). In that case, more audio signals can be transmitted by sky wave communication channel through atmosphere.
In the context of the modern day progress of telecommunication and internet, the use of some super-efficient communication channel has become essential. Microwave communication, use of artificial satellite, application of optical fibre as medium—all these steps develop a channel communication system of infinite range, through which an extremely large number of data signals can pass. This Is called broadband communication system.
Numerical Examples
Example 1.
If the height of a television tower is 300 m, how far would the TV transmission be possible? (Given, the radius of earth = 6400 km)
Solution:
Given the height of tower, h = 300 m
Radius of earth, R = 6400 km = 6.4 × 106 m
∴ The required distance,
d = \(\sqrt{2 R h}\)
= \(\sqrt{2 \times 6.4 \times 10^6 \times 300}\) ≈ 6.2 × 104m = 62 km
Example 2.
At what height is the transmitting antenna to be placed to make a TV transmission up to a distance of 32 km? The radius of earth = 6.4 × 106 m.
Solution:
Given radio horizon,
d = 32 km = 32 × 103 m
Let the transmitting antenna be placed at height h.
We know, d = \(\sqrt{2 R h}\)
or, h = \(\frac{d^2}{2 R}\) = \(\frac{\left(32 \times 10^3\right)^2}{2 \times 6.4 \times 10^6}\) = 80 m
Example 3.
Equation of a wave: V = 10 sin (106t + 0.4 sin 1000t). What is its Index of modulation?
Solution:
The given wave is a frequency modulated wave.
General equation of the wave, V = V0sin(Ωt + βsinωt)
Comparing this with the given equation we get,
modulation index, β = 0.4
Example 4.
Electron number density in a layer of ionosphere 4 × 105 cm-3. For an electromagnetic wave of frequency 40 MHz, what would be the refractive Index of that layer?
Solution:
Frequency, f = 40 MHz = 40 × 106 Hz
Electron number density,
n = 4 × 105 cm-3 = 4 × 1o5 × 106 m-3 = 4 × 1011 m-3
∴ Refractive index,
μ = (1 – \(\frac{81 n}{f^2}\))1/2 = [1 – \(\frac{81 \times 4 \times 10^{11}}{\left(40 \times 10^6\right)^2}\)]2
= (1 – 0.02025)1/2 = \(\sqrt{0.97975}\) = 0.99
Example 5.
Modulation index of an amplitude modulated wave is 50% and power dissipated in transmission 18 kW. What is the rate of energy dissipation for each side band?
Solution:
Modulation index, β = 50% = \(\frac{1}{2}\)
We know, PAM = PC(1 + \(\frac{\beta^2}{2}\)) or, 18 = PC[1 + \(\frac{\left(\frac{1}{2}\right)^2}{2}\)] = \(\frac{9}{8}\)PC
i.e., Power of carrier wave,
PC = 18 × \(\frac{8}{9}\) = 16 kW
Increase in power, due to modulation = 18 – 16 = 2 kW.
This power is distributed equally in the two sidebands.
∴ Rate of energy dissipation for each sideband
= \(\frac{2}{2}\) = 1 kW