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Friis Equation Calculator

Friis Equation:

\[ P_r = P_t \times G_t \times G_r \times \left( \frac{\lambda}{4 \pi d} \right)^2 \]

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1. What is the Friis Equation?

The Friis transmission equation is used in telecommunications engineering to calculate the power received by one antenna from another antenna under ideal conditions. It provides a fundamental relationship between transmitted power, antenna gains, wavelength, and distance.

2. How Does the Calculator Work?

The calculator uses the Friis equation:

\[ P_r = P_t \times G_t \times G_r \times \left( \frac{\lambda}{4 \pi d} \right)^2 \]

Where:

Explanation: The equation shows how received power decreases with the square of distance and is proportional to the square of wavelength.

3. Importance of Friis Equation

Details: The Friis equation is fundamental to wireless communication system design, helping engineers predict signal strength, plan network coverage, and optimize antenna placement for various applications.

4. Using the Calculator

Tips: Enter all values in appropriate units. All input values must be positive numbers. The calculator provides results in watts (W).

5. Frequently Asked Questions (FAQ)

Q1: What are the assumptions of the Friis equation?
A: The equation assumes free space propagation, no obstructions, perfect antenna alignment, impedance matching, and polarization matching between antennas.

Q2: How does wavelength relate to frequency?
A: Wavelength (λ) = speed of light (c) / frequency (f). For radio waves, c ≈ 3×10⁸ m/s.

Q3: What are typical values for antenna gains?
A: Isotropic antennas have gain of 1 (0 dBi). Practical antennas range from 2-20 (3-13 dBi) for common applications, with highly directional antennas reaching much higher gains.

Q4: When is the Friis equation not applicable?
A: It's not suitable for near-field calculations, environments with multipath propagation, through obstacles, or when atmospheric absorption is significant.

Q5: How accurate is this equation in real-world scenarios?
A: While providing a theoretical maximum, real-world conditions typically result in lower received power due to various losses and environmental factors not accounted for in the ideal equation.

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