CBSE Class 12 Physics Chapter 15: Communication System NCERT Solutions

For communication that extends beyond the horizon, the radio waves need to travel long distances. This is often achieved by reflecting signals off the ionosphere, a phenomenon known as sky wave propagation.

Let's analyze the options:

• 10 kHz: These very low frequencies require extremely large antennas for efficient radiation and are not typically used for long-distance sky wave communication.
• 10 MHz: Frequencies in this range are well-suited for reflection by the ionosphere. They can travel long distances by bouncing off the ionospheric layers, enabling beyond-the-horizon communication.
• 1 GHz and 1000 GHz: These very high frequencies tend to penetrate the ionosphere rather than being reflected by it. They are more suitable for line-of-sight communication (space waves).

Therefore, 10 MHz is the most suitable frequency for beyond-the-horizon communication using sky waves.

Ultra High Frequency (UHF) waves fall within the frequency range of 300 MHz to 3 GHz. Due to their very high frequencies:

• They are too high to be effectively reflected by the ionosphere (sky waves).
• They do not follow the curvature of the Earth effectively (ground waves or surface waves).

Consequently, UHF waves primarily travel in straight lines. This mode of propagation, where the waves travel directly from the transmitting antenna to the receiving antenna without significant reflection or diffraction, is known as space wave propagation or line-of-sight communication.

Let's analyze each statement regarding digital signals:

• (i) Do not provide a continuous set of values: This is true. Digital signals represent information using distinct, separate levels, typically two (like 0 and 1), rather than a continuous range.
• (ii) Represent values as discrete steps: This is also true. The distinct levels of a digital signal form discrete steps.
• (iii) Can utilize binary system: This is a fundamental characteristic. Digital signals are most commonly represented and processed using the binary system (base-2), with digits 0 and 1.
• (iv) Can utilize decimal as well as binary systems: This statement is false in the context of how digital signals are fundamentally processed and transmitted. While a digital signal's value might be *interpreted* or *converted* from/to decimal for human understanding, the signal itself is inherently discrete and most efficiently handled by the binary system. Using the decimal system directly for digital signal representation and processing is not standard and would be less efficient than binary.

Therefore, statements (i), (ii), and (iii) are true, while (iv) is not accurate for the core nature of digital signals.

For line-of-sight communication, it is not necessary for the transmitting and receiving antennas to be at the same height. Line-of-sight communication requires a clear, unobstructed path between the transmitter and receiver. The range is primarily limited by the curvature of the Earth and any physical obstructions. Increasing the height of either antenna generally increases the communication range.

Given:

• Height of the transmitting antenna, $h\; =\; 81\; \backslash text\{\; m\}$
• Radius of the Earth, $R\; =\; 6.4\; \backslash times\; 10^6\; \backslash text\{\; m\}$

The maximum distance ($d$) that a signal can travel from a transmitting antenna of height $h$ to a receiving antenna at ground level is given by the formula derived from the tangent to the Earth's curvature:

$$d\; \backslash approx\; \backslash sqrt\{2Rh\}$$

The service area ($A$) covered by the antenna is the area of a circle with radius $d$:

$$A\; =\; \backslash pi\; d^2\; =\; \backslash pi\; (2Rh)$$

Now, let's calculate the service area:

$A\; =\; \backslash pi\; \backslash times\; (2\; \backslash times\; 6.4\; \backslash times\; 10^6\; \backslash text\{\; m\}\; \backslash times\; 81\; \backslash text\{\; m\})$

$A\; =\; 3.14159\; \backslash times\; (1036.8\; \backslash times\; 10^6\; \backslash text\{\; m\}^2)$

$A\; \backslash approx\; 3257.2\; \backslash times\; 10^6\; \backslash text\{\; m\}^2$

To express this in square kilometers (km²), we use the conversion $1\; \backslash text\{\; km\}\; =\; 1000\; \backslash text\{\; m\}$, so $1\; \backslash text\{\; km\}^2\; =\; (1000\; \backslash text\{\; m\})^2\; =\; 10^6\; \backslash text\{\; m\}^2$:

$A\; \backslash approx\; 3257.2\; \backslash text\{\; km\}^2$

Therefore, the TV transmitting antenna can cover a service area of approximately 3257.2 square kilometers.