Communication Systems

Step 1:
The power radiated by an antenna is proportional to the square of the length and inversely proportional to the square of the wavelength. Thus, the ratio of the effective powers radiated is: Substitute the given ratios of lengths and wavelengths : Thus, the ratio of the effective powers radiated is .

Step 1: Formula for Modulation Index:
The modulation index ( or ) is given by: where is maximum amplitude and is minimum amplitude.
Step 2: Substitution and Calculation:
Given V and V.
Step 3: Final Answer:
The modulation index is 1/2 (or 0.5).

Step 1: Identify the frequencies and the modulation type.
The problem involves a carrier wave ( ) and a message signal ( ), and asks for the sideband frequencies. This is an Amplitude Modulation (AM) scenario. Carrier frequency: . Message frequency: .

Step 2: Convert all frequencies to a single unit (kHz).
Since : . .

Step 3: State the formulas for the sideband frequencies.
In Amplitude Modulation (AM), two sideband frequencies are generated: an Upper Sideband Frequency ( ) and a Lower Sideband Frequency ( ).

Step 4: Calculate the sideband frequencies.
Upper Sideband Frequency: Lower Sideband Frequency: The frequencies of the sidebands are and .

The reflection of radio waves from the upper atmosphere allows for long-distance communication. This process is known as skywave propagation.
This reflection occurs in a region of the atmosphere called the ionosphere.
The ionosphere is not a distinct layer like the others but is a region characterized by a high concentration of ions and free electrons, created by the ionization of atmospheric gases by solar radiation.
The ionosphere overlaps with the upper parts of the Mesosphere and extends throughout the Thermosphere.
The different layers of the ionosphere (D, E, F1, and F2 layers) have varying properties. The D layer absorbs HF waves during the day and disappears at night. The E and F layers are responsible for reflecting the waves.
At night, the D and E layers become much weaker, and the F1 and F2 layers combine to form a single F layer at a higher altitude. This F layer, located in the Thermosphere, is very effective at reflecting high-frequency (HF) radio waves, enabling long-distance communication.
Since the primary reflective layers at night are located within the Thermosphere, the Thermosphere is the correct answer.

Step 1: Understanding the Concept:
This problem deals with the line-of-sight propagation of radio waves. The maximum distance a signal can travel from a transmitting antenna to the horizon (its range) is limited by the curvature of the Earth.
Step 2: Key Formula or Approach:
The formula for the maximum line-of-sight distance, or range ( ), from a transmitting antenna of height is given by:
where is the radius of the Earth. If there is also a receiving antenna of height , the total range is . In this case, the receiving antenna is on the ground, so , and the formula simplifies to just .
Step 3: Detailed Explanation:
First, ensure all units are consistent. It's best to convert everything to meters.
- Height of transmitting antenna, .
- Radius of the Earth, .
Now, substitute these values into the range formula:
To find , we can note that and the number ends in 4, so the root must end in 2 or 8. Let's try 112:
So, .
Finally, convert the range from meters to kilometers:
Step 4: Final Answer:
The range of the transmitting antenna is 112 km. Therefore, option (B) is correct.

1. The service area for a transmitting antenna is the area within the line-of-sight distance, approximated for a curved Earth.
2. The maximum distance to the horizon is , where is antenna height and is Earth's radius.
3. Given m, km = m.
4. Calculate m.
5. The service area is a circle: m = m .
6. Therefore, the correct option is (4) m .

• Optical fibres use light as the carrier signal.
• Frequency of visible light: , nm.
• Compute : m/s, m → Hz.
• Hence optical communication range = 1 THz – 1000 THz.