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Stacked dipole antennas are often found at the top of cellphone masts, particularly for large cells and operating frequencies below \(1 \mathrm{GHz}\). These antennas have an efficiency that is close to \(90 \%\). Consider an antenna that has \(40 \mathrm{~W}\) of input power, an antenna gain of \(10 \mathrm{dBi},\) and transmits a signal at \(900 \mathrm{MHz}\) (a) What is the EIRP in watts? (b) If the power density drops as \(1 / d^{3},\) where \(d\) is the distance from the transmit tower, what is the power density at \(1 \mathrm{~km}\) if the power density is \(100 \mathrm{~mW} / \mathrm{m}^{2}\) at \(10 \mathrm{~m} ?\)

Short Answer

Expert verified
EIRP is 360 W. Power density at 1 km is \( 10^{-7} \text{ W/m}^2 \).

Step by step solution

01

Calculate the Effective Radiated Power (ERP)

To find the Effective Radiated Power (ERP), use the formula: \[ \text{ERP} = \text{Input Power} \times \text{Efficiency} \] Given input power (\text{P}_{\text{in}}) is 40 W and efficiency is 90% (or 0.9), so: \[ \text{ERP} = 40 \times 0.9 = 36 \text{ W} \]
02

Convert Antenna Gain to Linear Scale

The antenna gain is given in dBi, which needs to be converted to a linear scale. Use the formula: \[ G = 10^{\frac{\text{Gain (dBi)}}{10}} \] Given gain is 10 dBi, so: \[ G = 10^{\frac{10}{10}} = 10 \]
03

Calculate the Effective Isotropic Radiated Power (EIRP)

To find the EIRP, use the formula: \[ \text{EIRP} = \text{ERP} \times G \] Substitute the values of ERP (36 W) and G (10) into the formula: \[ \text{EIRP} = 36 \times 10 = 360 \text{ W} \]
04

Find the Power Density at 10 m

The power density at 10 meters is given as 100 mW/m² (which is 0.1 W/m²).
05

Calculate the Power Density at 1 km

The power density drops as \( \frac{1}{d^{3}} \) where \( d \) is the distance. Use the ratio of distances to find the power density at 1 km. \[ \frac{S_{1}}{S_{2}} = \frac{d_{2}^3}{d_{1}^{3}} \] Given: \[ d_{1} = 10 \text{ m}, \ d_{2} = 1000 \text{ m}, \ S_{1} = 0.1 \text{ W/m}^{2} \] Substitute into the formula: \[ \frac{0.1}{S_{2}} = \frac{1000^3}{10^3} \] Solve for \( S_{2} \): \[ \frac{0.1}{S_{2}} = \frac{10^9}{10^3} \] \[ \frac{0.1}{S_{2}} = 10^6 \] \[ S_{2} = \frac{0.1}{10^6} = 10^{-7} \text{ W/m}^{2} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Effective Isotropic Radiated Power (EIRP)
When dealing with antennas, the concept of Effective Isotropic Radiated Power (EIRP) is crucial. EIRP measures the power radiated by an antenna as if it were an isotropic radiator, which means it spreads energy equally in all directions. To calculate EIRP, we multiply the Effective Radiated Power (ERP) by the antenna's gain in linear terms.
For instance, in our example, we have an input power of 40 W with an efficiency of 90%. So, the ERP is calculated as follows:
\[ \text{ERP} = 40 \times 0.9 = 36 \text{ W} \]
Next, convert the antenna gain from dBi to a linear scale using the formula:
\[ G = 10^{\frac{\text{Gain (dBi)}}{10}} \]
Given the gain is 10 dBi, we get:
\[ G = 10^{\frac{10}{10}} = 10 \]
Therefore, our EIRP is:
\[ \text{EIRP} = \text{ERP} \times G = 36 \times 10 = 360 \text{ W} \]
Understanding EIRP helps us determine how much power an antenna effectively radiates towards a specific direction compared to an ideal isotropic radiator.
Antenna Gain
Antenna gain is a key concept in radio communications that describes how well an antenna focuses energy in a particular direction. It is usually measured in dBi, where 'i' stands for isotropic, referring to an ideal antenna that distributes power equally in all directions. Higher dBi values indicate that the antenna is more effective at focusing energy.
The conversion from dBi to a linear scale is essential for accurate calculations. The formula we use for this conversion is:
\[ G = 10^{\frac{\text{Gain (dBi)}}{10}} \]
For example, if an antenna has a gain of 10 dBi, it means:
\[ G = 10^{\frac{10}{10}} = 10 \]
This antenna would concentrate the energy 10 times more effectively in a specific direction compared to an isotropic radiator. We utilize this gain factor to enhance signal strength and cover larger distances effectively. Remember, a higher gain translates to better performance of the antenna in a particular direction, but it does not increase the total power radiated by the antenna.
Power Density Drop
As radio waves travel from a transmitting antenna, the power density decreases with distance. This reduction is known as power density drop. It follows a specific drop-off pattern based on the distance. Here, we consider the power density drop proportional to \( \frac{1}{d^{3}} \).
In our situation, the power density at 10 meters is given as 100 mW/m², or 0.1 W/m². To find the power density at 1 km (1000 meters), we use the following ratio formula:
\[ \frac{S_{1}}{S_{2}} = \frac{d_{2}^3}{d_{1}^{3}} \]
Given:
\[ d_{1} = 10 \text{ m}, \ d_{2} = 1000 \text{ m}, \ S_{1} = 0.1 \text{ W/m}^{2} \]
We substitute these values into our equation:
\[ \frac{0.1}{S_{2}} = \frac{1000^3}{10^3} = 10^6 \]
Solving for \( S_{2} \) we get:
\[ S_{2} = \frac{0.1}{10^6} = 10^{-7} \text{ W/m}^{2} \]
This calculation shows that as the distance increases, the power density decreases rapidly. Understanding power density drop is crucial for designing effective communication systems, ensuring optimal coverage, and minimizing interference.

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Most popular questions from this chapter

Describe Rayleigh fading in approximately 4 lines and without using a diagram.

On a resonant antenna a large current is established by creating a standing wave. The current peaking that thus results establishes a strong electric field (and hence magnetic field) that radiates away from the antenna. A typical dipole loses \(15 \%\) of the power input to it as resistive \(\left(I^{2} R\right)\) losses and has an antenna gain of \(10 \mathrm{dBi}\) measured at \(50 \mathrm{~m}\). Consider a base station dipole antenna that has \(100 \mathrm{~W}\) input to it. Also consider that the transmitted power density falls off with distance \(d\) as \(1 / d^{3}\). Hint, calculate the power density at \(50 \mathrm{~m}\). [Parallels Example 4.2\(]\) (a) What is the input power in \(\mathrm{dBm} ?\) (b) What is the power transmitted in \(\mathrm{dBm} ?\) (c) What is the power density at \(1 \mathrm{~km} ?\) Express your answer as \(\mathrm{W} / \mathrm{m}^{2}\). (d) What is the power captured by a receive antenna (at \(1 \mathrm{~km}\) ) that has an effective antenna aperture of \(6 \mathrm{~cm}^{2} ?\) Express your answer in first \(\mathrm{dBm}\) and then watts. (e) If the background noise level captured by the antenna is \(1 \mathrm{pW}\), what is the SNR in decibels? Ignore interference that comes from other transmitters.

Consider a point-to-point communication system. Parabolic antennas are mounted high on a mast so that ground effects do not exist, thus power falls off as \(1 / d^{2}\). The gain of the transmit antenna is \(20 \mathrm{dBi}\) and the gain of the receive antenna is \(15 \mathrm{dBi}\). The distance between the antennas is \(10 \mathrm{~km}\). The effective area of the receive antenna is \(3 \mathrm{~cm}^{2}\). If the power input to the transmit antenna is \(600 \mathrm{~mW},\) what is the power delivered at the output of the receive antenna?

An antenna operating at \(28 \mathrm{GHz}\) has an antenna gain of \(50 \mathrm{dBi}\). What is the effective aperture area of the antenna?

A transmitter and receiver operating at \(2 \mathrm{GHz}\) are at the same level, but the direct path between them is blocked by a building and the signal must diffract over the building for a communication link to be established. This is a classic knife-edge diffraction situation. The transmit and receive antennas are each separated from the building by \(4 \mathrm{~km}\) and the building is \(20 \mathrm{~m}\) higher than the antennas (which are at the same height). Consider that the building is very thin. It has been found that the path loss can be determined by considering loss due to free-space propagation and loss due to diffraction over the knife edge. (a) What is the additional attenuation (in decibels) due to diffraction? (b) If the operating frequency is \(100 \mathrm{MHz}\), what is the attenuation (in decibels) due to diffraction? (c) If the operating frequency is \(10 \mathrm{GHz}\), what is the attenuation (in decibels) due to diffraction?

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