What factors need to be considered in the propagation of electromagnetic waves?
In "How are electromagnetic waves transmitted?" 》We introduced the three basic ways of space electromagnetic wave propagation: ground wave propagation, sky wave propagation and line-of-sight propagation. As the frequency of electromagnetic waves increases, line-of-sight propagation is currently the most important propagation method.
With line-of-sight transmission, does it mean that as long as you can see it, you can receive the signal? Is the attenuation of electromagnetic waves propagating in space large? What factors need to be considered in the propagation of electromagnetic waves?
The magical thing about electromagnetic waves is that they don’t need to pass through a medium to propagate. As long as the electric field and the magnetic field can be turned on alternately, we can make great strides forward and propagate at the speed of light.
As shown below.
This kind of energy that can be transmitted without relying on a carrier can be transmitted without loss in free space. Since there is no loss, can it be transmitted continuously? I think the answer should be yes, but no loss does not mean that the strength can remain unchanged. The propagation of electromagnetic waves is divergent. As the distance increases, the electromagnetic waves continue to spread and their intensity will gradually weaken.
Therefore, the most feared thing about the propagation of electromagnetic waves is distance. Therefore, even if high-energy electromagnetic waves like sunlight reach the surface of the earth, they can be divided into spring, summer, autumn and winter with different hot and cold conditions according to the distance.
No.1 free space transmission loss
First, we assume that there are two antennas. The transmitting power of the transmitting antenna is Pt, the antenna gain is Gt, the gain of the receiving antenna is Gr, and the distance from the transmitting antenna is R. This R is far enough that in the far field area of the two antennas . So what is the power Pr received by the receiving antenna?
We assume that the transmitting antenna is omnidirectional and lossless, and the power density at the receiving antenna at a distance R is:
Considering that the gain of the transmitting antenna is Gt, the power density at the receiving antenna is:
Assuming that the effective area of the receiving antenna is Ae, then the power received by the receiving antenna is
The effective area of the antenna is a physical quantity related to the operating wavelength and gain of the antenna:
In this way, the power received by the receiving antenna can be calculated.
this is famousFriis transmission equation Friis transmission equationthis formula was first derived by Harald T. Friss of Bell Labs in 1945, and was elaborated in the 1946 article "Note on a Simple Transmission Formula".
This formula is also known as the electromagnetic wave free space loss equation.
If the ratio of the receiver's input power Pr and the transmitter's output power Pt is defined as the propagation loss, this propagation loss can be directly derived from the above formula.
Converted to dB form is
If the gains of the transmitting antenna and the receiving antenna are not considered, the free space attenuation formula of electromagnetic waves can be further simplified to
This is also the famous Frith transmission formula.
If the speed of light C and distance R are in km, and f is in MHz, the above equation can be converted into something we are more familiar with.
Note that the above formula is the famous free space loss formula.
Give a chestnut
Assume that the transmit power of a 5G base station is 100W, the antenna gain is 20dB, and the operating frequency is 3.5GHz. How much power does the mobile phone receive at a distance of 2km?
The spatial transmission attenuation Lfr is:
Generally, mobile phone antennas are omnidirectional antennas. Usually the built-in antenna gain is 0dB-1dB. Assuming it is 0dB, then the power received by the mobile phone is:
That's about 0.0000011 watts, which is equal to 11 microwatts.
It is often said in wireless communications that "when the power increases by 6dB, the transmission distance doubles." Where does the 6dB here come from? You can calculate it based on the free space loss formula, mainly considering the middle term 20log (R_KM). Does it mean that when the distance is doubled, the attenuation increases by exactly 6dB?
No.2 Atmospheric Attenuation
However, the Earth's surface is different from free space. In addition to the intensity attenuation due to distance, there is also the influence of the atmosphere. The atmosphere and precipitation absorb and scatter electromagnetic waves, especially electromagnetic waves with frequencies above 1GHz.
In some specific frequency ranges, attenuation peaks occur due to molecular resonance phenomena. The attenuation rate curve caused by atmospheric gases is given in "ITU-R P.676-11", as shown in the figure below. At approximately 23 GHz, the first resonance point due to water vapor absorption appears, the second resonance point due to dry air (mainly oxygen) absorption appears at 62 GHz, and the third resonance point due to dry air absorption appears at 120 GHz. The resonance point and the two absorption frequencies of water vapor behind it are 180GHz and 350GHz.
Therefore, when communicating in the atmosphere, the above-mentioned frequency bands with severe attenuation should be used in a closed manner.
In addition, precipitation will also have a relatively large impact on electromagnetic waves above 10GHz.
Of course, the attenuation model in reality is much more complicated. The Frisian attenuation formula can be used to quickly evaluate the coverage distance of a wireless station. Of course, the impact of radiation from this station can also be quickly evaluated.
Review Editor: Huang Fei
#factors #considered #propagation #electromagnetic #waves
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