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Author Topic: Mathematical correlation between Continuous and PEP  (Read 4194 times)
G3RZP
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Posts: 1313




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« Reply #15 on: February 16, 2019, 05:18:54 AM »

There is a mathematical relationship in the case of COFDM - Coded Orthogonal Frequency Division Multiplex - where there are a number of digitally modulated carriers. If there are signals which lead to all the carriers being in phase at the same time, the PEP capability of the transmitter has to be able to handle this, and by suitably coding the input signals so as to prevent this situation occurring, the required PEP capability can be significantly reduced. I heard of one claim (by a now retired professor) of it 'winning' as much as 15dB, allowing the average power to be increased by that much.
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W1BR
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« Reply #16 on: February 16, 2019, 07:17:14 AM »

I'd expect the PEP reading made on CW or SSB  to be a tad higher than the reading for a continuous CW carrier for most tube ham amplifiers.

The high voltage usually sags under key down.  Peak readings on SSB may be influenced by higher average DC plate voltages.

Pete
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G3RZP
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« Reply #17 on: February 16, 2019, 10:30:11 AM »

Quote
I'd expect the PEP reading made on CW or SSB  to be a tad higher than the reading for a continuous CW carrier for most tube ham amplifiers.

I agree, but another item that can cause (or maybe just add to) that is ALC overshoot.

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AC2RY
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« Reply #18 on: February 16, 2019, 02:49:18 PM »

Quote
I'd expect the PEP reading made on CW or SSB  to be a tad higher than the reading for a continuous CW carrier for most tube ham amplifiers.

I agree, but another item that can cause (or maybe just add to) that is ALC overshoot.



In random type of signal like speech you can measure so called dynamic range - difference between average and peak amplitudes. For normal speech it will be 10 - 15 dB. Amplifier should be able to handle these peaks without much distortion (clipping). Using speech processors (compressors) you can reduce that difference to 5 dB or less. Obviously reception will depend mostly on average amplitude. That is why processing on the transmission side helps.

For CW signal dynamic range is about 6 dB (2 times) in amplitude.

In all cases amplifier should be able to handle peak volume without clipping and have thermal design appropriate for handling average power. For tube amplifiers running in SE class B there will be little difference between average and peak dissipation because and lower power level efficiency drops significantly.

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KD9IQO
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« Reply #19 on: February 17, 2019, 07:05:41 AM »

To All:

PEP is a misnomer.  PEP defines the process instead of defining the end result. 
(1) The process begins by measuring the peak amplitude of the energy envelope.
(2) Then you divide that value by the square-root of 2.  This gives you the root-mean-square (average) value (for a sine-wave only!).
(3) The PROCESS demands that you continue to call this value Peak when you just converted it to average.

This is the crux of the confusion, gentlemen.
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W9IQ
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« Reply #20 on: February 17, 2019, 07:42:29 AM »

The general difficulty, from both a definition and measurement perspective, is that the modulating signal is not a constant amplitude nor a constant frequency when a voice is modulating an SSB transmitter. This means that there is no reliable mathematical relationship between average and peak power in this case. Certainly nothing to do with the square root of 2.

A peak reading meter would ideally measure and then hold on the display the peak power that it has observed. This would mean the needle or display would hold the peak value until a new peak arrived or the display was reset. The result would be a steady or increasing value but never a decreasing one until reset. What we see instead is peak meters that gyrate up and down with the modulation - resulting in a quasi peak reading at best.

- Glenn W9IQ
« Last Edit: February 17, 2019, 07:45:16 AM by W9IQ » Logged

- Glenn W9IQ

I never make a mistake. I thought I did once but I was wrong.
KD9IQO
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Posts: 11




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« Reply #21 on: February 17, 2019, 08:34:38 AM »

Choose the correct answer:

Your house voltage is:             (A)   120 Vac (rms)
                                           (b)   170 Vac (peak)
                                           (c)   340 Vac (peak-to-Peak)
                                           (d)   all of the above

hint:  the answer is "d"
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KD9IQO
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« Reply #22 on: February 17, 2019, 08:52:12 AM »

Quote
The general difficulty, from both a definition and measurement perspective, is that the modulating signal is not a constant amplitude nor a constant frequency when a voice is modulating an SSB transmitter. This means that there is no reliable mathematical relationship between average and peak power in this case. Certainly nothing to do with the square root of 2.

A peak reading meter would ideally measure and then hold on the display the peak power that it has observed. This would mean the needle or display would hold the peak value until a new peak arrived or the display was reset. The result would be a steady or increasing value but never a decreasing one until reset. What we see instead is peak meters that gyrate up and down with the modulation - resulting in a quasi peak reading at best.

- Glenn W9IQ



The test is not performed with voice modulation.  The test is performed with tone.  The square-root-of 2 is the conversion factor between peak and rms.  This is not open for debate. Please review the basics of rf measurement on page 4-7 of the General class study manual.
« Last Edit: February 17, 2019, 08:54:15 AM by KD9IQO » Logged
KD9IQO
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« Reply #23 on: February 17, 2019, 09:07:25 AM »

PEP calculator from ARRL General Class Study Manual:

http://u.cubeupload.com/KD9IQO/PEP2.jpg
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W9IQ
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Posts: 3523




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« Reply #24 on: February 17, 2019, 09:49:19 AM »

PEP calculator from ARRL General Class Study Manual:

http://u.cubeupload.com/KD9IQO/PEP2.jpg

The topic has many more nuances than that graphic depicts. You are correct from a license question perspective but not from a true application perspective.

The general tendency to use sine wave conversion factors for an amateur radio transmitter or linear is not well founded. Most of our amplifier stages are class AB through class C depending on the mode. These stages do not deliver a sine wave as the carrier so using the sine wave relationships is only an approximation. There can be significant errors in this assumption. The good news is that the errors make the calculated P.E.P overly conservative.

As an example of the potential error, consider using an oscilloscope and measuring the peak voltage of a CW transmitter at 100 volts into a 50+j0 ohm load. If the carrier is a true sine wave, we would calculate P.E.P as:

     P.E.P. = (100 * 0.707)2/50 = 100 watts P.E.P.

However, a class C amplifier has only an ~50% conduction angle. Assuming the original wave that is amplified by this class C amplifier is a pure sine wave, then we would calculate P.E.P. as:

     P.E.P. = (100 * 0.51)2/50 = 52 watts P.E.P.

Why is this the case? For one half of the original sine wave, the amplifier output remains at 0 watts (and thus 0 volts). When it reaches 1/2π radians, it begins to follow the exciting sine wave times the gain of the amplifier. When the sine wave reaches 1.5π radians, there again is no output from the amplifier. The definition of P.E.P. is the average power supplied to the antenna transmission line by the transmitter during one RF cycle at the crest of the modulation envelope, under normal operating conditions. The normal 0.707 multiplier assumes that the full sine wave is present but in this case it is not. So the RMS conversion factor must necessarily be reduced since the full sine wave is not present.

Often the output circuit of the amplifier will tend to smooth out some of the output waveform so the conversion factor will not be as extreme but the general principal applies.

The same can be applied for a typical class AB amplifier with a greater conduction angle - only in this case, the RMS factor will be much closer to 0.707.

The formula referenced for the license exams is basically assuming a no distortion, class A amplifier which is hardly ever used in amateur applications due to its inefficiency.

- Glenn W9IQ
« Last Edit: February 17, 2019, 10:02:29 AM by W9IQ » Logged

- Glenn W9IQ

I never make a mistake. I thought I did once but I was wrong.
K6BRN
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Posts: leet




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« Reply #25 on: February 17, 2019, 10:16:27 AM »

Glen (W9IQ):

Quote
The general difficulty, from both a definition and measurement perspective, is that the modulating signal is not a constant amplitude nor a constant frequency when a voice is modulating an SSB transmitter. This means that there is no reliable mathematical relationship between average and peak power in this case. Certainly nothing to do with the square root of 2.

Yes.  Very succinct and relevant summary.

Thank you.

Brian - K6BRN
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KD9IQO
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Posts: 11




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« Reply #26 on: February 17, 2019, 10:25:20 AM »

Amplifier conduction angle only establishes efficiency and INPUT power.  If you exceed 387 peak Volts, into a 50Ω load then you have violated the 1500 Watt limit no matter how the amplifier is modulated.
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K6BRN
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« Reply #27 on: February 17, 2019, 10:52:00 AM »

Really?  What if you have a lossy coax?  (Sigh!)  Perfection is SO hard to achieve!
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KD9IQO
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« Reply #28 on: February 17, 2019, 10:55:07 AM »

You must be able to first recognize perfection.
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W9IQ
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Posts: 3523




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« Reply #29 on: February 17, 2019, 11:04:39 AM »

Amplifier conduction angle only establishes efficiency and INPUT power.

Unfortunately that is not a complete perspective and it can lead you to errant conclusions. The efficiency and input power are closely related but you have overlooked the cause of these which is the non-linearity of the amplifier.

From https://www.electronics-tutorials.ws/amplifier/amplifier-classes.html:

Quote
Amplifier Classes represent the amount of the output signal which varies within the amplifier circuit over one cycle of operation when excited by a sinusoidal input signal. The classification of amplifiers range from entirely linear operation (for use in high-fidelity signal amplification) with very low efficiency, to entirely non-linear (where a faithful signal reproduction is not so important) operation but with a much higher efficiency, while others are a compromise between the two.

From the 2019 ARRL Handbook:

Quote
Amplifiers are grouped by their operating class that describes the way in which the input signal is amplified.... The analog class designators specify over how much of the input cycle the active device is conducting current.

From https://wikipedia.org/wiki/Power_amplifier_classes:

Quote
The classes are related to the time period that the active amplifier device is passing current, expressed as a fraction of the period of a signal waveform applied to the input.

The point is that the efficiency and input power are largely a result of what percent of the input signal the amplifier amplifies.

- Glenn W9IQ

« Last Edit: February 17, 2019, 11:11:27 AM by W9IQ » Logged

- Glenn W9IQ

I never make a mistake. I thought I did once but I was wrong.
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