The 900MHz system here is read centrally. Your system is NOT wifi.
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Originally posted by Jim Frye View PostGas company meter reader was on the street yesterday, so I stopped him and asked what they use. It's a 900 MHz, frequencyhopping, spreadspectrum RF technology and he said the water utility was using the same tech,, but from a different vendor. I have no idea what that is, but it seems to work fine.
Guess it works OK if you can mount the unit outside with line of sight to the street, should get at least 200 feet of range that way. One of the issues in this area was that meter locations are often on the side or back of the house, or in the basement for water meters. Some of the meters could never be read due to location, and they still had to manually read them. The newer Wifi/CDA meters solved that limitation since they operate at higher frequency (better propagation through buildings) and have much greater range. Basically they will work anywhere a wificapable cell phone will work, which is nearly everywhere now. The readers are essentially just phone modules with different inputs, and it's the wide spread adoption of cell phones that made the cost viable.

Electrical Engineer by day, Woodworker by night
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Yeah, 900 systems nearly always have more range. I’ve installed both to miles, but 900 was able to do that long ago.

Slik Geek,
Friis shows that loss decreases with frequency and increases with wavelength, the proportional inverse of frequency.
SImple thought experiment to demonstrate intuitively that higher frequencies propagate better, then some math. Why is the range of propagation of a DC (0 frequency) signal zero? Why does a 60 Hz signal propagate poorly? Consider an AM radio, receives signals in the 1400 kHz frequency range with limited distance range (excluding bouncing off the ionosphere, considering only free space propagation). FM radio frequency band is an order of magnitude higher and propagates much better. Why does a microwave oven have to be shielded? Because microwaves propagate too well. If their propagation range was less than an inch, no shielding would be needed.
The classic Friis equation is Pr/Pt/=ArAt/(d^2*lambda*2). If we assume constant power and antenna aperture, that reduces to Pr/Pt is proportional to 1/(d^2* lambda^2). Recall c=f*lambda, so wavelength lambda is proportionally inverse to frequency, and higher frequency has shorter wavelength. Even more simply, Pr=Kf, where K combines the constants, showing that received power is proportional to frequency, and higher frequency results in higher received power.
At 900 MHz, wavelength is 0.33 m, 5.4 GHz is 0.06 m. All other factors the same, Friis shows that received power is about 15.5 dBm higher at 5.4 GHz than 2.4 GHz. To put it more simply, higher frequencies propagate better in free space.
Friis addresses free space transmission, but real world environments are cluttered with buildings, people, cars, etc. all of which attenuate signal. The shorter wavelength of higher frequency is a benefit here as well since it passes through a smaller aperture and is less attenuated by these objects. It's very environment dependent, but to put some practical numbers on it, we typically see at least a 6 dBm improvement going from 900 MHz to 2.4 GHz and range increase of at least four times in a typical suburban or urban environment.
There are other factors such as the effective aperture but generally we try to use the highest frequency possible since higher frequencies propagate better. Other factors can further obscure the optimal choice, for example max allowable unlicensed transmit power differs with frequency band. Antenna aperture and practical antenna size limitations are also considerations. However, the primary reason 2.7 GHz and 5.4 GHz systems have greater range than 900 MHz systems is mostly the benefit of the higher frequency.Last edited by woodturner; 10102018, 06:34 AM.


Originally posted by gsmittle View Post
Is that the technique Hedy Lamar invented?
The more widely used technique is Direct Sequence Spread Spectrum (DS SS). It's harder to explain in a simple way, but in DS SS the spectrum rather than the frequency is pseudorandomly changed periodically.

Electrical Engineer by day, Woodworker by night
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Given the recent discussion of IoT and wifi meter readers, I found some unrestricted information I can pass along to those who want to learn more.
A point of clarification that seems to be confusing some: IoT is "Internet of Things", a reference to connecting devices to the Internet. The Internet (technically TCP/IP) is a messaging protocol, a description of what information messages contain, how the messages are routed and name, and how those messages are separated into smaller units for communication.
Internet messages can travel over any communication medium  dial up, wireless, bluetooth, RF, wired Ethernet, etc. For example, wifi, CDA, and wired Ethernet are commonly used for Internet communications. One could use a local 900 MHz DSSS signal to a repeater or hub, for example, though as previously explained the short range of 900 MHz would likely make that impractical. There are existing commercially available systems that use 5.4 GHz local communications to a hub, then wifi or CDA to connect to the Internet.
Here are some links that explain more about some of the IoT meter readers currently in use as well as the smart meters and meter reading technology:
https://www.iotone.com/usecase/autom...readingamr/u9
https://www.huawei.com/minisite/iot/en/smartami.html
https://www.sierrawireless.com/appli...martmetering/
https://www.telit.com/blog/smartnb...rsenergyuse/
https://www.telenorconnexion.com/iot...uickoverview/

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Don't want to hijack this thread, but I felt compelled to respond to woodturner's comments, as I was puzzled by his analysis and thought it important to clarify.
The basic Friis equation is: Pr / Pt = Ar x At x ( wavelength / 4 x pi x distance ) ^ 2 Wavelength is in the numerator, indicating that received power increases with increasing wavelength
 Since wavelength is inversely proportional to frequency, showing that received power decreases with increasing frequency.
 Since loss is the reciprocal of gain, the Friis equation shows us that, stated in terms of loss, loss increases with increasing frequency. In other words, increase the frequency, and if all other aspects are kept constant, range can be expected to decrease (because the signal experiences greater losses).
The Friis equation shows that, all things otherwise held constant, received power is 7.1 dB higher at 2.4 GHz than at 5.4 GHz. Received power is 15.6 dB higher at 900 MHz than at 5.4 GHz.
For example an actual set of calculations result in the following received power signal levels:
5.4 GHz: 100.1 dBm
2.4 GHz: 93.0 dBm
900 MHz: 84.5 dBm
Note that this is a negative value, such that 84 dBm is a stronger signal than 93 dBm. 84.5  (100.1) = 15.6 and 93.0  (100.1) = 7.1
I agree that the shorter wavelength of a higher frequency signal passes through a smaller aperture. When it comes to foliage, most references show significantly increased attenuation with increased frequency.
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Originally posted by Slik Geek View Post
The basic Friis equation is: Pr / Pt = Ar x At x ( wavelength / 4 x pi x distance ) ^ 2
That may be the source of confusion, you are using a different variant of the Friis equation
The equation Friis published is:
Pr / Pt = Ar x At / ( wavelength^2* distance^2 )
(Friis, H.T. (May 1946). "A Note on a Simple Transmission Formula". IRE Proc.: 254–256.)
Wavelength is in the denominator  that's why, all other things held constant, received power INCREASES with frequency. As previously shown ( and easily verified experimentally) "All other factors the same, Friis shows that received power is about 15.5 dBm higher at 5.4 GHz than 2.4 GHz"
(I believe you may be confusing the Friis distance equation with the directivity equation
Edit: Slik Geek is using the gain relationship variant of Friis: Pr / Pt = Gr x Gt x ( wavelength / 4 x pi x distance ) ^ 2
Pr / Pt = Dr x Dt x ( wavelength / 4 x pi x distance ) ^ 2)(
Higher frequency signals are more directive, as the equation shows, but does not inform regarding distance.
(If you have a reference for your variant of the equation I'm interested in reviewing it.)
Edit: See Slik Geek's response for detailed explanation of the differencesLast edited by woodturner; 10202018, 06:32 AM. Reason: added reference to Slik Geek's posted and update to reflect differences, original text is in the smaller font
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Originally posted by woodturner View Post
The equation Friis published is:
Pr / Pt = Ar x At / ( wavelength^2* distance^2 )
(Friis, H.T. (May 1946). "A Note on a Simple Transmission Formula". IRE Proc.: 254–256.)
If you have a reference for your variant of the equation, assuming it was not simply a typo, I'm interested in reviewing it.
The version of the Friis equation usually cited in text books and various articles looks like this:
where P is power in Watts, R is the farfield lineofsight distance between the antennas, G is the antenna gain, lamda is the wavelength, and the subscripts r and t denote the receiver and transmitter, respectively.
This is the only version of the Friis equation with which I was familiar. I assumed that your version was simply using Ar and At as antenna gains, meaning we were discussing the same equation. But we were not.
You were citing the original equation form of the Friis equation:
where Ar and At describe the antenna effective area.
I'm still coming to terms with this revelation... but I see why the modified version is usually employed. The effective area of a thin antenna is proportional to the square of the wavelength. If one doubles the frequency of a signal, and adjust the antenna dimensions accordingly to keep the antenna identical in its characteristics (for example, keep it a halfwave dipole), the effective area of the antenna is decreased by a factor of 1/4 (onehalf squared)  but it's "isotropic gain" remains constant. This means that the signal strength at the higher frequency is actually reduced  not because of increased propagation loss, which one might erroneously conclude from the modified Friis equation (which is what I did), but because of the reduced size (effective area) of the antenna.
Thank you for the education.
Now, back to your statement, "that's why, all other things held constant, received power INCREASES with frequency". This is true as stated, in the context of the original Friis equation, the second equation shown above. However, in most wireless systems that I have seen, the implementation doesn't happen this way.
The key part of your statement is "all other things held constant", In reality, the antenna size usually isn't held constant, in order to adjust for the change in frequency. Because of the reduction in size of the antenna, in response to the increase in frequency from one band to another, the effective received power decreases with an increase in frequency. Stated differently, "keeping antenna gain and separation distance constant, received power DECREASES with frequency", as indicated by the first equation above. This is usually what transpires when radio systems move upward in frequency.
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Originally posted by Slik Geek View Post
Now I understand where you are coming from. That was mind blowing. Putting it in terms from my perspective...
The version of the Friis equation usually cited in text books and various articles looks like this:
where P is power in Watts, R is the farfield lineofsight distance between the antennas, G is the antenna gain, lamda is the wavelength, and the subscripts r and t denote the receiver and transmitter, respectively.
This is the only version of the Friis equation with which I was familiar.
The variants of the Friis equation were developed for different applications. From the communications perspective I prefer the original  usually I want to calculate power budgets when designing a new system, or quickly getting a rough idea whether an increase in frequency will increase distance enough to solve an application problem. An antenna designer likely would prefer the form you cited.
One benefit of higher frequency is also one can get better range from a smaller antenna, as you noted. Antenna size was a limiting factor in cell and cordless phone development  remember the early 27 MHz cordless phones where you had to pull out a collapsible antenna to get decent range? As frequency increased, a smaller antenna provided enough gain that we could use a stub antenna, and now with cell phones we can put the antenna inside the case.
Thanks for helping us work this out :)

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Originally posted by SerhijI contacted my provider and they tells me that it may reduce my router life, is this true?
"May" is the operative word. We really don't know either way what will happen to your specific router. Powering electronics off/on does cause a certain type of wear. Leaving them on causes a different type of wear. Which will win? Either way you're unlikely to kill the router within the 34 years of useful life before it becomes outdated and you want to replace it anyway.
Yesterday I found an interesting thing; my new access points have a lot of reporting and analytics. They showed that I'm being hit by doppler RADAR and it's causing 5GHz wifi interference. Time to do some more channel tuning. Which everyone should do, to find unused channels in the area. It's funny to see that five of my neighbors are all on one channel, and the speed must just suck for them.
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