Distorted Sine Wave

May 29, 2024 Last reply: 2 years ago 67 Replies

Since you have a power meter, a signal source, and an oscilloscope why not measure the peak to peak voltage on the scope and power on the power meter and see which calculation 0.636 vs 0.707 gives the closest agreement?

It wouldn't prove anything one way or ther other, though, since that power meter hasn't been calibrated for "quite a while" so to speak. :) It'll give a 'good enough' reading for my purposes, but won't be accurate enough to meaningfully test your otherwise fine suggestion.

To CD:

The above is what I did. 30 + 10*log( (0.88/(2*sqrt(2)))^2 / 50) =

2.869 dBm. Rounded to 3dBm.

What's the issue with RMS vs. average?

I have an 8566B which is currently not working. Both the status leds on the front panel at the bottom are red. I haven't started to investigate yet. The fault developed slowly. At first it would sometimes work, then progressively less often and now never. However, if the signal being discussed is available on the rear panel I could measure mine and see what it looks like and what voltage is delivered. John

OK, thanks for that clarification. Anyway, I finally measured the power of that oscillator with my HP RF power meter and it comes out at 1.74mW (or about +2.5dBm off the top of my head). Seems a tad on the low side, but I can't find what it's supposed to be in the manual.

When you dig into it, you find that what people really mean when they talk about "RMS Watts" is actually *average* power. I found this on the web which attempts to explain it:

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Yes, that could be very helpful, John, since your fault is clearly totally different to mine. Peak to peak volts into 50 ohms on a scope will be fine if don't have access to an RF power meter.

It’s really not this hard.

“RMS” stands for “root mean square”, which is a shorthand description of how you calculate the power delivered by an arbitrary voltage waveform (or equivalently current) in a resistive circuit.

You square the instantaneous voltage, compute the mean (I. e. time average), and then take the square root.

All those fudge factors like 0.5, 0.636, 0.707, and so forth, can be useful for quick calculations, but they just summarize the results of the above procedure _for_specific_situations_. Without first doing the math, and understanding the situation, they’re worse than useless.

The ‘rms power’ thing came as a response to lying advertisements for stereo systems, starting in the 1970s iirc. Crappy stereos were advertised as producing “250 watts PMP”, for “peak music power”, as though that were a thing. That led to very optimistic numbers, even before actual lies were added, which they usually were.

People started pushing back by insisting on knowing what sine wave power the amp could put out continuously without distorting or overheating.

That’s a very conservative spec, since music waveforms have a high peak/rms ratio and the ear is most sensitive to transient distortion on the peaks. It does have some basis in reality, though, and is easy to measure unambiguously, which cuts through the Audio BS” (tm).

While saying “rms watts“ is indeed redundant, strictly speaking, nevertheless it’s a useful shorthand for describing audio amps, Chinese switchers, and (I suppose) power FETs.

Cheers

Phil Hobbs

My old HP RF power meter uses this principle. It has a thermistor sensor head which is remarkably sensitive down into the microwatt range. Not only that, but there isn't any noticeable thermal lag, even at very low power levels, so the meter needle instantly flicks over to give the power reading. I've often wondered how they do that. More modern meters use a different principle IIRC.

Building myown RF power meter to emulate what a commercial one can do is way above my capabilities, sadly, Jan.

Phil, I believe you also have an 8566B. Do you know what the 10Mhz reference oscillator output level should be? Is yours anything close to

+2.5dBm?

The 0 to +10 dBm range I mentioned came from the service manual.

Looking at your scope picture, it looks like a 3 Vpp signal, which is

+13 dBm, a very common distribution level, but one that exceeds the analyzer's allowed range. All that's needed to fix this is a 3dB inline attenuator. Here is one for SMA connectors:

.

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Just buying a few of these and doing some experiments will be far cheaper and faster than the various alternatives discussed.

Joe Gwinn

I think you're looking at the first picture with the signal into the scope's 1 Meg input. The 50 ohm trace is only 850mV peak-to-peak or thereabouts and when I measured it with an actual power meter, came out at about +2.5dBm so within the range you stated; no attenuation needed (thanks for the range, by the way; I needed to know that).

I've now measured the 100Mhz oscillator and that seems fine, although I only saw 0.61V p-p into 50 ohms, so somewhat less than the 10Mhz oscillator's output. So far, I've not measured anything which screams "the fault's here!" as all the expected signals are present - although admittedly I have many more to test. But certainly all the *major* signals within this complex beast are present. It's looking like it could be an issue with one of the phase detectors or LPFs. Sigh....

e I'm guessing there should have been a 50 ohm load screwed into the rear

10Mhz reference oscillator output BNC socket when the analyzer's in use. There wasn't one but there is now. Unfortunately it hasn't cleared the PLL unlock issue.

Average power is not the same as average voltage! Average power is proportional to the average of the voltage squared. It makes a difference!

Jeroen Belleman

What we don't know is exactly how you made the various measurements. If you are observing the signal from the 10 MHz reference where it enters the analyzer, I would expect that there is a T-connector with the scope (set to 1 Mohm) listening in to passing signals.

In this case, the load seen by the incoming reference is that provided by the input on the analyzer. Which input is +10 dBm max. If you set the observing scope input to 50 ohm, the reference will see a 25 ohm load, cutting the signal seen by the analyzer by 3 dB. Which will take +13 dBm down to +10 dBm, which is in range.

A 3dB attenuator in line will drop the signal to 10 dBm as well.

I've built lots of systems like that. The 10 MHz reference is delivered to everybody at +13 dBm, and it is the receivers' responsibility to attenuate it to whatever they need.

To my eye, it does scream.

Joe Gwinn

What scope picture are you looking at? I see only 0.88Vpp.

Jeroen Belleman

Sorry, but I don't recall anyone claiming average power and average voltage were the same thing!

This one, posted by CD on 1 June '24:

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This is the one with the funny stuff at the bottom. If you look at the upper waveshape, the peak amplitude to the inflection point near the bottom is about 1.5 Vp, which implies 3 Vpp, which is +13 dBm into

50 ohms. Why the inflection point? Because in a undistorted sine wave, the zero crossing is linear, and does not flair. The scope picture does not show where zero volts is, so had to use the inflection point.

Joe Gwinn

You did ask me this before and did post an answer. See Message-ID: <v3fsbp$2u0a6$ snipped-for-privacy@dont-email.me

You also still appear to think that the 10Mhz signal is going into the analyzer. It isn't. It's coming out. Again, see Message-ID: <v3fsbp$2u0a6$ snipped-for-privacy@dont-email.me

Earlier, you said, I cite, "Average power is average volts squared divided by the load impedance".

It isn't. It's RMS volts squared divided by load impedance.

Jeroen Belleman

I'm afraid you have lost me there... I see only a roughly sine-shaped wave framed with cursors along the peaks being

0.88V apart. I don't care about the DC level, only the 10MHz component matters. Its amplitude is only 0.44V.

Jeroen Belleman

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