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Question

A Television transmitting antenna is 84 m tall. How much service area can it cover if the receiving antenna is at the ground level?

(Given \( \pi = \frac{22}{7} \))

The correct answer is

3379 km2

Understanding Television Antenna Service Area

This question asks us to determine the service area covered by a television transmitting antenna of a specific height, assuming the receiving antenna is at ground level. The service area refers to the geographical region where the signal from the antenna can be effectively received.

Key Concepts and Formulas

The maximum distance to the horizon or the maximum line-of-sight distance from a transmitting antenna to a receiving antenna at ground level is given by the formula:

\( d = \sqrt{2Rh_t} \)

Where:

  • \(d\) is the maximum distance to the horizon (or service range).
  • \(R\) is the radius of the Earth.
  • \(h_t\) is the height of the transmitting antenna.

The service area covered by the antenna is the area of a circle with radius \(d\) on the surface of the Earth (assuming a flat Earth for simplicity over this range, which is a common approximation in these problems). The formula for the area of a circle is:

\( A = \pi d^2 \)

Where:

  • \(A\) is the service area.
  • \(d\) is the maximum service range calculated above.
  • \( \pi \) is the mathematical constant Pi.

Step-by-Step Calculation of Service Area

We are given the height of the transmitting antenna \(h_t = 84 \text{ m}\) and \( \pi = \frac{22}{7} \). We need to use the radius of the Earth, which is approximately \(R = 6400 \text{ km}\). Before calculating, we must ensure all units are consistent. Let's convert the antenna height from meters to kilometers.

\( h_t = 84 \text{ m} = \frac{84}{1000} \text{ km} = 0.084 \text{ km} \)

Now, we can calculate the maximum service range \(d\):

\( d = \sqrt{2Rh_t} \)

Substitute the values:

\( d = \sqrt{2 \times 6400 \text{ km} \times 0.084 \text{ km}} \)

\( d = \sqrt{12800 \times 0.084} \text{ km} \)

\( d = \sqrt{1075.2} \text{ km} \)

Next, we calculate the service area \(A\) using the formula \(A = \pi d^2\). Note that \(d^2\) is simply the value inside the square root we just calculated.

\( A = \pi d^2 \)

\( A = \frac{22}{7} \times 1075.2 \text{ km}^2 \)

\( A = \frac{23654.4}{7} \text{ km}^2 \)

Performing the division:

\( A \approx 3379.2 \text{ km}^2 \)

Conclusion

The calculated service area is approximately \(3379.2 \text{ km}^2\). Comparing this value with the given options, the closest value is \(3379 \text{ km}^2\).

Therefore, a television transmitting antenna 84 m tall can cover a service area of approximately \(3379 \text{ km}^2\) if the receiving antenna is at ground level.

Revision Table: Antenna Service Area Calculation

Parameter Value Units
Antenna Height (\(h_t\)) 84 m
Antenna Height (\(h_t\)) 0.084 km
Earth Radius (\(R\)) 6400 km
Value for \(\pi\) \( \frac{22}{7} \) -
Maximum Distance Squared (\(d^2 = 2Rh_t\)) 1075.2 \( \text{km}^2 \)
Service Area (\(A = \pi d^2\)) \( \frac{22}{7} \times 1075.2 \approx 3379.2 \) \( \text{km}^2 \)

Additional Information: Factors Affecting Service Area

While the formula \( A = \pi (2Rh_t) \) provides a theoretical maximum service area based on the line-of-sight distance determined by the curvature of the Earth, actual television service area can be affected by many other factors:

  • Terrain: Hills, mountains, and even large buildings can block or reflect radio waves, reducing the actual coverage area, creating 'shadow' zones.
  • Atmospheric Conditions: Temperature inversions or other atmospheric effects can sometimes extend the range (ducting) or reduce it.
  • Obstacles: Trees, buildings, and other structures can attenuate (weaken) the signal.
  • Signal Frequency: Higher frequencies (like UHF for modern TV) are more susceptible to blocking by obstacles than lower frequencies (like VHF).
  • Transmitter Power: The power of the transmitting antenna influences how strong the signal is within the theoretical coverage area, affecting reception quality, especially towards the edges.
  • Receiving Antenna Height and Quality: A higher receiving antenna or a more sensitive/directional antenna can often pick up signals better, potentially extending the usable range for a specific location.

The calculation performed above provides a useful estimate based purely on antenna height and Earth's curvature, representing the maximum possible line-of-sight coverage in an ideal environment.

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Important Questions from Communication Systems

  1. The wavelength of radiation emitted when He+ makes a transition from the state n = 3 to the state n = 2 will be:

    (Take Rydberg constant R = 1.097 × 10⁷ m⁻¹)

  2. Match List - I with List - II 

    List-IList-II
    (A) Range(I) Range of frequencies over which communication system works
    (B) Band width(II) The largest distance between transmitter and receiver
    (C) Attenuation(III) Loss of strength of a signal during propagation
    (D) Transducer(IV) A device that receives an input in electrical form or provides an output in electrical form

    Choose the correct answer from the options given below:

  3. A carrier wave of peak voltage 14 V is used to transmit a message. What should be the peak voltage of the modulating signal in order to have a modulation index of 70%?

  4. Match List - I with List - II

    List-IList-II
    (A) Range(I) Range of frequencies over which communication system works
    (B) Band width(II) The largest distance between transmitter and receiver
    (C) Attenuation(III) Loss of strength of a signal during propagation
    (D) Transducer(IV) A device that receives an input in electrical form or provides an output in electrical form

    Choose the correct answer from the options given below:

  5. A carrier wave of peak voltage 14 V is used to transmit a message. What should be the peak voltage of the modulating signal in order to have a modulation index of 70%?

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