# Peak Envelope Power (PEP) = Transmitter Power?

I need to determine the Peak Envelope Power for the VSAT station. Given that the modulation is not AM, would it be correct to assume that the PEP will be the same as transmitter power?

• Please add this information to your question: What modulation is being used? What is “the VSAT station”, or rather, what is VSAT? – Kevin Reid AG6YO Sep 23 '15 at 14:39
• How accurate do you want the PEP to be? You could use a PEP reading meter or even make one yourself if the power is not too high (which requires more special design & construction methods). – K7PEH Sep 23 '15 at 15:14
• Thank you all for your comments. I am not doing it for the concrete station, but rather I want to establish a general rule on how to calculate PEP. Also, I have no access to the said station, I only have the technical details of it. – gsimonen Sep 24 '15 at 6:33

There are modulations that would not be called "AM" which still modulate amplitude. For example, BPSK reverses phase at a zero amplitude point. BPSK could also be considered a digital amplitude modulation where the amplitude varies between 1 and -1.

Thus, you must be careful to specify what you mean by "not AM". If you mean to exclude all modulations that modulate power at all, then peak and average power could be considered the same in practice. MSK is an example that would qualify.

This would only apply to the extent that the transmitted waveform has an RMS average equivalent to a sine wave. For example consider that a square wave has a higher RMS average, and thus a higher power, than a sine wave of equal amplitude. In practice the difference will be small or nonexistent for most modulations.

• Phil , I think you have a misconception that multiplying by ±1 would be considered AM. It isn't, because the amplitude is not changed. Only the phase changes by 180 degrees. I'll grant that in a finite bandwidth system, there would be some (generally negligible) envelope modulation, but in modulation theory, it's classified as phase modulation. Similarly, QPSK can be considered phase modulation in multiples of 90 degrees (if you don't like the quadrature carrier model). and still has a constant envelope (ignoring BW limiting). BTW, I've been designing PSK systems for nearly forty years. – Jim Feb 26 '17 at 6:32
• @Jim No real PSK system used in amateur radio I've seen multiplies by just ±1. Rather, it employs a pulse shaping filter, usually an RRC filter or something close to it. Your minor concession to a bandwidth limited system is nearly always the case for radio, with the exception of inexpensive, low power remotes which don't need to secure primary frequency allocations, and for which cost is the driving concern for design. Here's a reference: slideshare.net/sristykp/modulation-46756444 and another: photonteck.com/upfile/2014/05/27/20140527165836_983.pdf (see 3.7) – Phil Frost - W8II Feb 26 '17 at 14:06

For all intents and purposes they are the same, yes.

Here is a good quote from W8JI that explains it.

"Average power is same as equivalent work power or heating power of each cycle averaged over a significant time compared to the time when power level changes. With an unchanging (during the measurement period) power level, such as a steady unmodulated carrier applied to a constant resistance load, average power and peak envelope power are the same. If we close and hold-closed a manual telegraph key on a good stable CW transmitter, we will see the average power displayed on a power meter. It will not be the "RMS power". It is also the peak envelope power, because it is the maximum stable heating power level over some period of time that we hold the key."

This is from his website on AM, here: Amplitude Modulation

For VSAT (Very Small Aperture Terminal), I can't find anything that specifically identifies the modulation used other than to say that it uses TDMA (Time Division Multiple Access). But my experience with this type of equipment suggests that it is probably a constant envelope type of waveform while transmitting. (Such as QPSK). This is because the amplifiers used for these systems are typically non-linear and are designed to run at one power output level.

So PEP would be equivalent to Average Power for a steady transmission.

But your average power calculation might need to take into account the duty factor, since it is a TDMA signal. If that's important to you, you should say so in your question. If you do that, we can readdress your question.

• Are you sure QPSK is constant envelope? – Phil Frost - W8II Feb 25 '17 at 4:06
• Absolutely. It is two BPSK signals in quadrature (sine and cosine), The sum of the two will always be a constant magnitude. The individual channels do not possess this characteristic by themselves. – Jim Feb 25 '17 at 4:11
• Why would one channel increase in amplitude when the other decreases to keep the sum amplitude equal? If one channel is not performing any phase reversals (say that means all 0 bits) and the other does perform a phase transition, how is that a constant amplitude? – Phil Frost - W8II Feb 25 '17 at 17:12
• Because they are two equal amplitude BPSK channels in quadrature (sine and cosine). The RF voltage produced is the RSS (root sum square) from the two channels. The RSS os sine and cosine is (from a well known trigonometric identity) equal to a constant. Note that the phase shifts (sign) don't affect the result. – Jim Feb 26 '17 at 2:10
• It's not just the phase shifts: it's that the phase reversal requires (if you want to do it in less than infinite bandwidth) the amplitude to go through zero. Adding an orthogonal carrier changes nothing: $\cos(t) + A \sin(t)$ is not equal to a constant amplitude if A varies between 1 and -1. And let's not forget the other carrier can change phase also. And they can change phase at the same time. – Phil Frost - W8II Feb 26 '17 at 3:13