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Author Topic: Efffective PSK31 Power Calculation  (Read 3507 times)

Posts: 243

« on: February 15, 2001, 07:56:46 AM »

Is the following method of comparing effective power of PSK31 to SSB valid??

Divide 2500hz (SSB bandwidth) by 31hz (PSK bandwidth)for a result of about 80. Then multiply your PSK31 power by 80 to arrive at the equivilent effective power if you were operating SSB.

Thus if you are running 20 watts of PSK31, 1600 watts would be needed to arrive at the same signal level if you were using SSB HuhHuh?

Dick Boley  N3HKN

Posts: 550


« Reply #1 on: February 15, 2001, 11:52:51 AM »

Would not that assume that the total bandwidth of the RX be 31 Hz (IF and audio)? I don't know... What about if you were to compare CW (morse) to PSK31 power levels...? For sure increasing the selectivity of the signal does increase the sensitivity.

Posts: 893


« Reply #2 on: February 15, 2001, 02:47:13 PM »

It depends on what you mean by "effective power." What you're doing can be best described in terms of what is referred to as power spectral density (PSD) in the communications industry. It's a measure of how much power is in each Hz of spectrum, and is the mathematical representation of what you see on a spectrum analyzer. Divide the transmitted power by the occupied bandwidth, and you come up with an average density, assuming the power is equally distributed over a fixed bandwidth. BPSK actually has a sin (x) / x spectrum ("sine x over x"--it's hard to do equations in text mode) over the long term, so it really isn't flat.

When you do this with PSK and SSB signals, you can see why PSK is more power efficient--the power is concentrated in a smaller bandwidth. Of course you are limited in what you can transmit--you can't push a voice signal through that narrow a bandwidth, at least not practically. (Communications theory says that with enough power, you can transmit almost any signal through any bandwidth, but the power levels are usually not practical. This theorem is called the Shannon Channel Capacity theorem; it tells you what is theoretically possible, but doesn't tell you how. So the signal processing you'd have to do is undefined.)

This is also the reason that CW can get through on lower powers than SSB. The power is concentrated in a smaller bandwidth. The actual bandwidth is related to the speed of the CW--faster CW (actually the shorter dits) requires more bandwidth.

This is similar, but unrelated, to "effective power" as it is used in "effective radiated power", which is the composite of transmit power and antenna gain. ERP is a geometric concentration of power, while PSD refers to the spectral concentration.

It really isn't fair to do the calculation the way you did, to estimate how much power would be needed in an SSB signal to achieve the same PSD as you achieve with PSK. The signal-to-noise ratio needed for intelligible communications is different for the modes, so you're only comparing one of the necessary factors.

Posts: 243

« Reply #3 on: February 16, 2001, 08:22:25 AM »

Not being in the field I forgot about the power density. That is what I was driving at. The narrow signals indeed offer the ability to communicate with less power over distances similar to wider bandwidth modes. ATV is a good example of the other extreme.

Thanks for the formula. It does show the non-linear aspect of narrowing the bandwidth vs power.

I have seen some info that says that PSK31 takes less bandwidth than 20wpm CW. As you said the short dits need more space.

Finally, I believe that it is critical for Hams to understand the power density issue so they are not driven to buy continuos duty 1.5kw amps to be "competitive". If people use their credit cards, instead of the info you provided, the mode will become just another free-for-all using the new formula that expresses an  -   EGO to POWER RATIO

Dick  N3HKN  20 watts or less!

Posts: 3

« Reply #4 on: April 22, 2002, 01:47:14 PM »

The relationship between bandwidth and power, for fixed S/N, is fairly straightforward and is easy to derive from Shannon's work. That relationship is:

PWR ratio in dB = 10*log(B2/B1)
---where log is log base 10, B2, is the bandwidth of channel 2 [ssb ~2500] and B1 is the bandwidth of channel 1 [say, psk31~31.25].

3.125 Hz is interesting as many waterfalls I see show distinct 3rd IMD and even higher. I am assuming the transmitter is properly adjusted and linear.

To get the power invert the dB back to Power:

P2 = P1 * 10^(dB/10)

---where P2 will be the equivalent power to achieve the same S/N on channel 2 as P1 achieved on channel 1. dB is the dB you calculated in the first formula.

You do get figures similar to what are discussed in this email thread. In fairness to ssb, there is the potential for a lot more information in its bandwidth and there is also a relationship between signal levels and required bandwidth. It is these relationships and their consequences that were the core of Shannon's work.

It is important to remember that what Shannon was driving at was channel capacity in relationship to noise and your results may vary based upon band conditions. He assumed white [random] noise and that 1.5 KW station next door 3 Hz away produces different issues.

Hope this helps.

Rob WW6G
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