RF interference related question

Could some RF guru please clarify the following points ? Keeping in mind the ideas of co-channel interference, cross-channel interference and FCC's recent white space broadcast spectrum release,

  1. Most of the unlicensed band devices are specified to have output poer like 40mW, so would it not be the case that a 100 W TV transmitter could easily overwhelm one of the devices, and not the other way around ?
  2. In the same way, what about local cell phone base stations overwhelming these unlicensed band devices ?
  3. Are there any free simulators (OpNet maybe ? ) that could help one to simulate this type of heterogeneous network ? Thanks in advance for your help.
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That depends on how close the TV transmitter is both spatially and in frequency, and how close the intended transmitter is (mostly spatially, as we'd assume the frequency was dead on).

If, after the TV signal was attenuated both by distance and by the device's various filters, it was still stronger than the device's intended receive signal, then yes, that intended signal would get stomped on.

(And most commercial TV is way more than 100W).

Same answer.

This isn't hard stuff. If you have enough knowledge to operate such a simulation tool, then you have enough knowledge to do this -- at least roughly -- with pencil and paper. If you can't do it with pencil and paper, then when your simulation tool gives you absurd answers, you'll never know it.

Reply to
Tim Wescott

here is a crash course;

the first wavelength worth of distance gives you about 22dB of loss.. from there, every DOUBLING of the distance gives you an additional 6 dB loss.


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The OP also wasn't at all clear if he was concerned about powerful transmitters on _adjoining_ frequencies, or if he was concerned about transmitters _on_ the frequency of interest.

It does make a 'slight' difference.

Tim Wescott
Wescott Design Services
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Tim Wescott

The idea here is that if you and your next door neighbor are, say, 20 miles from that 100W TV transmitter, the signal is weak enough by the time it hits your neighbor's antenna that your little 40mW transmitter only 20 *feet* away can overwhlem it.

But if your and your neighbor live next door to the TV station then, yes, the TV transmitter will squash your whitespace device's ability to operate on that frequency.

Same answer... although note that TV transmitters are far, far more powerful than cell phone basestation transmitters.

Mmm... yes, but for a first-order approximation just use a spreadsheet or similar using 1/r^2 to determine the losses. (Look up the Friis Equation if you want to get a bit fancier and take antenna gains into account.)


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Joel Koltner







on if


If you would like to calculate the loss for large distance (but within horizon distance), you can better use the two-ray propagation formula. That formula takes earth reflection into account that modifies the loss to 1/r^4. Please read the limitations on its use before using it.

Other option is to get some (obsolete) ITU propagation charts that are/ were used for frequency coordination.


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And white space means white, meaning you are s'posed to transmit when there is a primary user signal (TV). Because you'd only have secondary user status on the pecking order. Else the Federales might come marching in.

From what I can see in the spectrum out here they have really crammed them in there, not much white space left. DTC is brick after brick after brick.

Yet not as good as NTSC anymore :-(

Regards, Joerg

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/Hello, / /If you would like to calculate the loss for large distance (but within /horizon distance), you can better use the two-ray propagation formula. /That formula takes earth reflection into account that modifies the /loss to 1/r^4. Please read the limitations on its use before using /it. / /other option is to get some (obsolete) ITU propagation charts that are/ /were used for frequency coordination.

Free space path loss

96.4 + 20 log (Freq GHz) + 20 log (Dist Miles)
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Hello tm,

We live on a large reflector (called earth). Because of this, Free space path loss cannot be applied in many cases in VHF and UHF propagation problems.

Using Friis formula will result in too optimistic results when mother earth is well within the first Fresnel zone.

With kind regards,


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OK, but isn't 1/R^4 too pessimistic? I've been told that 1/R^3 is a good compromise. :-)

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Joel Koltner

"Daku is so Dumb"

** It's usually the other way around.

TV transmitters ( plus FM broadcast, cell towers and the like) and their antenna systems are ENGINEERED to spread the energy evenly over the coverage area. By siting antennas high up and using narrow, horizontal beams aimed at the horizon, signal levels are kept within sensible limits at all usable locations on the ground.

OTOH - a portable 40mW transmitter might be located less than 10 metres from you and the signal strength high enough to swamp the front end of a receiver so blotting out the other, wanted signal.

.... Phil

Reply to
Phil Allison

TV whitespace devices are required to have either a database lookup capability, or spectrum-sensing technology such that they will operate only in areas where there are no TV incumbents in the band. Therefore, you would not (normally) have a 100 watt TV station and a whitespace device in the same market.

As for whether a whitespace device could interfere with a TV broadcast (in a geographic area permitted by the database, and/or allowed according to the spectrum-sensing requirement), this requires further engineering study on a case-by-case basis. For example, 1/r^2 or Friis, as others have mentioned. However, strictly from a regulatory perspective, use of the whitespace device would be permissible in such circumstance, pursuant to FCC rules, and specifically the recent FCC Order from OET in ET Docket 04-186.

After a quick review of the licenses, I think the only semi-legitimate concern here is for Cellular South (the largest privately-held wireless carrier in the US with substantial lower A-block 700 MHz spectrum holdings). Since their spectrum is adjacent to TV channel

51, they are rightly concerned about the proliferation of TV whitespace devices adjacent to their licenses, both in terms of frequency and geography.

To your second question, cellular phones operate in the 824-894 MHz band (which spectrum also includes some of Nextel's SMR holdings and certain other land-mobile users under FCC Part-90 rules). This spectrum is significantly removed from the TV whitespaces and therefore unlikely to cause interference. That said, if a TV whitespace device is located in VERY close proximity to a cell site, and has poor or no front end filtering, it could be overwhelmed by the cell site emissions (i.e., out-of-band power.) But of course, even a cheap whitespace device probably develops at least 30dB of attenuation (protection) through antenna aperature, conversion and spectral displacement.

Keep in mind this situation would be true of any transmitter (FM, other TV, railroad, taxicab, paging, even PCS bands, etc...) In other words, not unique to cellular.

Finally, keep in mind that technically, "cellular" refers only to the FCC Part-22 824-894 MHz band. There are also PCS (1900 MHz), SMR (Sprint/Nextel - 850 MHz Band), and AWS Bands (with various spectrum authorized). Laypeople often refer to all of these bands as "cell" phones, but they are in fact governed by different FCC Rules. That's why I think you could restate your question #2 above, because really, you are asking if the lower 700 MHz stuff could interfere with whitespace devices (and vice-versa), as opposed to the 824-894 MHz stuff.

Hope this helps.


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Hello Joel,

When you look to the full two-ray formula, you have a (close) range where the asymptote is 1/r^2 (that is the free space part). With increasing distance the asymptote will be 1/r^4 (that is the two-ray part). So in between you have (small) range where there is a 1/r^3 ratio.

You are right, when you start from the beginning with a 1/r^4 ratio, you get highly unreliable results. This is the reason to mention that one has to read the limitations on its use.

When it comes to real interference assessment for frequency planning, it becomes more complicated as you may have to consider abnormal conditions also (for example ducting far over the horizon).

Best regards,


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