ARRL FILES REGULATION-BY-BANDWIDTH PETITION

This is what I sent to the man in charge of enforcement of PART97 for amateurs, Mr. Riley Hollingsworth:

riley.hollingsworth@fcc.gov

 

Some definitions that have been used in the latest discussion.

Crest Factor - The crest factor of any current or
voltage waveform can be defined as
the ratio of peak to rms. To express in dB use 20 log (crest factor).

Peak Power - The output power averaged over that cycle of an electromagnetic wave having the maximum peak value that can occur during transmission.

Average Power - The average power consumed by a two-terminal electrical device is a function of the root mean square values of the sinusoidal voltage across the terminals and the sinusoidal current passing through the device.

Peak Envelope Power - The average power supplied to the antenna transmission line by a transmitter during one radio frequency cycle at the crest of the modulation envelope taken under normal operating conditions.

Peak to Average Ratio (Power) The ratio of peak power to average power. To express in dB use 10 log (peak/average).

73,

Mark N5RFX

 

AB0WR:"One thing everyone needs to be careful about is that although the term "average power" is bandied about a lot *I* always consider it to be "rms" average power unless otherwise specified. Your Mileage May Vary (YMMV)."

Tim,

Yes there is a bit of ambiguity when using the term average power. When I use the term I am speaking of power derived by using the RMS voltage multiplied by the RMS current.

AB0WR:"I'm surprised that your program doesn't analyze a straight sine wave right but shows the right figure for a two-tone test."

I believe that it does analyze a "straight" sine wave correctly. If you agree that it shows the "right figure" for a two tone test, I don't understand why you think it does not analyze a sine wave correctly. A formula for calculating the average power when the peak power is known is 1/n, where n is the number of tones. For a 2 tone test n=2 and 1/2=.5. When the peak power of a two tone test is 1, the average power is .5. The peak to average ratio is 3dB. When a single tone is used, the peak to average ratio is 1, or 0dB. This is from Single Sideband Principles and Circuits pages 2 and 3 written by Pappenfus, Bruene and Schoenike. The program is working well.

73,

Mark N5RFX

 

I subtracted 1.414 from 3 because you said: "BTW to convert from peak to average ratio to crest factor, add 3 dB."

You can't calculate peak power to average power in a sine wave the way you have.

First you calculated "average power" as .707^2. Then you just defined peak average power as the same thing because the "envelope" is "flat". I'm not sure exacly what a "flat envelope" is defined as being.

You have defined the crest factor as 0db just by your definition.

Crest factor, as a power ratio for a sine wave, is defined as the (peak power)/average power. The peak power would be (1^2/R) and the average power would be (.707^2/R).

When you divide these you get (1/.4998) which is approximately 2. To calculate the crest factor as a power ratio you then take 10log(2) = 3db. The crest factor itself is (1/.707) = 1.414. You can convert this directly to a power ratio by squaring it (approx = 2) and doing 10log(2) = 3db.

If Cool Edit is not calculating it in this manner it is not doing something quite right.

When it is actually calculating a two-tone test does it show the peak average power as 0db and an average power as -3db by any chance?


tim ab0wr

 

See my comments. All I am interested in is the real world performance of digital on digital, analog voice on analog voice, digital on analog voice and analog voice on digital. Bandwidth, occupancy rates, necessary separation, interference potentials, close proximity (contest) performance. My comments above were my predictive theory based on testing I did for digital voice a few years ago.

 

Mark,

I think you will find your definition stems from the test results. If you have two tones of equal amplitude applied to an amplifier, the total output voltage when both are in phase will be 2A. The average power at that point will PEP=(2A*.707)^2/Rload. The average power of each tone at that point will be Ptone=(A*.707)^2/Rload. The total tone power will be 2*Ptone. So if we divide the two PEP/(2*Ptone) we get 2. 10log(2) = 3db.

If you have three equal amplitude tones, when all three are in phase you will have a peak voltage of 3A. The average power at that point will be PEP=(3A*.707)^2/Rload. The power in each tone will be (A*.707)^2/Rload. The total tone power will be 3Ptone. So PEP/(3*Ptone)=3 or 4.8db.

This is not truly a definition of crest factor. Crest factor is peak power versus average power, not Peak Envelope Power (an average power itself) versus Average Power.

If you want the crest factor for a two-tone test you would take the peak power as (2A)^2/Rload. The average power would still be 2*[(.707*A^2)/Rload]. This would give you a crest factor of 4 (2 times .707^2 = 1). This is a power crest factor of 6db not 3db.

The same thing applies for a sine wave. You can't just define "peak power" as "average power". If that is what is being done in the "Single Sideband Principles and Circuits" book, then I disagree with their terminology.

A sine wave is E=Asin(wt). Emax=A. So maximum power is A^2/Rload. RMS average voltage is .707*A. So average power is A^2/(2*Rload). The power crest factor is .5 or 3db. The voltage crest factor is still 1.414.

Please note that in Single-Sideband Systems and Circuits they define a true crest factor for speech. They use instantaneous power versus average power as being 14.5db. That's actually a power crest factor of about 25 to 1 or a voltage crest factor of about 5 to 1.

I was probably off in my earlier comparisons of voice versus data total power.

The actual Peak power output of a 100watt PEP transmitter should be about 140 watts. The actual average power output from a typical voice would be 25 times less than this or about 5.6 watts. I think I said 4watts before. Not a significant difference.

I guess I'm curious now. What is the instantaneous peak power of an mt63 signal compared to its average power? 10db?

I'm still going to work on how to do this on-the-air. I think that is the only way to get a test that will actually tell just how much interference is caused to SSB by an mt63 signal.

tim ab0wr

 

Tim,

Ok so we got that behind us. I reported that MT63 had a peak (PEP) to average ratio of 9 dB, simply add 3 dB to that number and you have a 12 dB crest factor by my measurement.

73,

Mark N5RFX

 

AB0WR:"Please note that in Single-Sideband Systems and Circuits they define a true crest factor for speech. They use instantaneous power versus average power as being 14.5db. That's actually a power crest factor of about 25 to 1 or a voltage crest factor of about 5 to 1.

I was probably off in my earlier comparisons of voice versus data total power.

The actual Peak power output of a 100watt PEP transmitter should be about 140 watts. The actual average power output from a typical voice would be 25 times less than this or about 5.6 watts. I think I said 4watts before. Not a significant difference. "

Tim,

I can't agree with your math here. A 100W PEP transmitter has a peak instantaneous power of nearly 200 Watts. Your math earlier proved this. PEP is the peak average power. If my peak average power is 100W, then the RMS voltage required to do this (in a 50 Ohm system) is nearly 71 volts. The the peak voltage would be around 100 volts. The peak instantaneous power would be 200 watts. A signal with a crest factor of 14.5 dB would produce an average power of around 7 watts. Good luck measuring that 100 volt peak or that 200 watt peak with an OFDM signal on any legacy test equipment. You will need some pretty fine stuff to measure the instantaneous peak. Look on the WEB for peak to average power discussions for OFDM signals and you will see what I mean. That is why traditionally in radio work the peak to average power ratio has always been peak average power to average power ratio. Crest factor was always derived from that kind of measurement in the past, until the advent of sophisticated DSP technology. Cool Edit Pro wasn't cheap either.

73,

Mark N5RFX

 

AG4YO:"My comments above were my predictive theory based on testing I did for digital voice a few years ago."

Was part of your charter analyzing whether digital voice and analog voice can co-exist?

73,

Mark N5RFX

 

You're right about the peak power. That's what I get for doing this off the top of my head. 1.414^2 is 2. So peak power would be 200watts

The average power of a signal with a 25:1 crest factor would actually be 8watts.

I don't think you need fancy equipment to measure the instantaneous peak. Since the instantaneous peak will occur when all tones are in phase, the maximum instantaneous peak will just be the sum of the voltages of all the tones. In fact, I could take an mt63 signal, monitor it on my old Heathkit monitor scope and just mark the highest points on the face with a grease pencil. I could then generate a CW signal at that same value, measure the RF voltage and know directly what the maximum instantaneous peak power is.

From there, knowing the *typical* crest factor for mt63 I can calculate an average power level. I emphasize *typical* because with this complex of a waveform,, the actual average power level is going to be at least somewhat dependent on what you are sending.

I would agree that it is going to be hard to actually measure peak and average power at the same time on the waveform. Actually determining the crest factor is going to require some fancy equipment.

However, *I* don't need to be able to do that to do tests if I use the typical crest factor that has already been determined. In fact, I don't even need to be able to actually measure average power. All I need to be able to measure is peak instantaneous power for either the mt63 signal or for the voice signal in order to make a pretty good determination of what the average power is for each.

If I am fully utilizing the 100w PEP transmitter, my average power output for a typical voice will be about 8 watts. An mt63 signal that is 6db below this will have an average power output of about 2 watts. With a peak to average crest factor of 12db, I would see peak powers in the range of 30watts. So I would expect to see peak RF voltages (assuming a 50ohm system) of about 100v. For the mt63 signal I would expect to see RF voltages of about 40v.

I've got an old Tempo 2020 transmitter I can hook up to a dummy load and transmit voice. If I can get a PC with a sound card to hook up to my 751a I can try sending an mt63 signal into a dummy load. Using the above voltage levels should give me a good basis to see just how much interference is generated by the mt63 signal.

The big question here is how many mt63 users turn their transmitters down to get a peak instantaneous power of 30watts and an average power output of 2 watts. If they don't do this they will interfere with qso's occuring in their own geographical area and will also interfere with qso's in other geographical areas.

Do you have a feel for how many mt63 users run typically run their transmitters at this low of a level?

tim ab0wr

 

Tim,

Like I said good luck seeing those instantaneous peaks on your old Heathkit monitor scope. Believe me I have been down that road. I think you are better off measuring average power, it is much easier. I agree you will have no problem seeing those peaks with the voice signal, but MT63, no way.

As far as the MT63 signal and average power most operators set their power just below where the ALC kicks in. Set your RF power out wide open, and adjust your audio till you get some ALC action, then back it off so there is no ALC action. That will work fine. That is the way most operators of sound card modes operate their rigs. Run it with max ALC to see what happens. I think you will very surprised.

You have 3 bandwidth choices for MT63, 500 Hz, 1K and 2K. On 20M 1K is typically used with a long interleave, on 40 M 500 Hz is used with a short interleave. The base band tests I have been running have been using the 1K mode with long interleave.

I did run another baseband test with MT63. I generated two signals at the same power level and overlapped them by 750 Hz. That means that there was only 250 Hz of each signal that was unique. I have perfect copy on both signals. OFDM signals are pretty amazing.

73,

Mark N5RFX

 

Yes Mark it was. The answer is that today in Cellular, analog voice and digital voice (data signal) are kept segregated. They very much interfere with each other. Although the analog voice is FM, the frequency reuse plan causes havoc to digital and/or analog voice when they are not segregated.

Cellular on 850mhz is actually cleaner than HF is, so the interference would be greater in Amateur Radio. As an example, given a perfectly clear frequency and the ability of digital to decode weak signals, if an analog voice QSO came up out of range of both digital QSO parties on the same frequency, it would still inject bit error rates to digital. Likewise the digital QSO could appear slightly out of range of an analog voice QSO and appear as a raised noise floor on SSB perhaps interfereing with the QSO.

A characteristic inherent to FM makes reuse easier with analog voice. In the case of a reused frequency (also being used by digital voice), the FM receiver capture effect grabs the stronger analog voice signal signal. The digital voice coders see the analog signal as interference. So my point is that analog voice on FM handles reuse better. SSB would not do as well. Digital does not like either.

For digital to get a fair shot, it needs the least amount of interfering signal as possible. So all the whoopla about digital giving us more range was hooey. It only did so in non-interference situations. Practically, digital voice range was 10% less than analog voice given the same C/I.

 

Charlie,

Thanks or the explanation, it helps me to understand the tests that you wanted in the previous post. I am curious, was your digital experience, CDMA and TDMA?

The nice thing about the ARS is that the protocols are constantly changing and protocols that don't meet the challenge are dropped pretty quickly. That is the nice thing about being able to experiment, which we are able to do very liberally today. I am not as familiar with the digital voice protocols used in the ARS. We are living in the mixed environment today, what problems have occurred in this mixed environment in the ARS? I am not aware of any complaints to the FCC, nor have I read any comments or warnings from the FCC about the mixed environment that we have today. I am interested in hearing what the complaints are that are generating the concern over mixing analog and digital, when in fact we are mixing analog and digital today in the phone/image subbands right now. I keep hearing about region 1 calling for the separation of digital and analog signals, but I have not been able to find any papers discussing why the IARU feels this way. Since they did not change their bandplan for 2006 to prohibit mixing of analog and digital signals, there may be more to the story.

Thanks again for sharing your experience Charley.

73,

Mark N5RFX

 

I tested CDMA, TDMA, and GSM. Each has it's own reasons why Analog should be segregated.

 

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