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From: Jeroen Belleman <jeroen@nospam.please>
Newsgroups: sci.electronics.design
Subject: Re: Distorted Sine Wave
Date: Sun, 2 Jun 2024 23:45:43 +0200
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On 6/2/24 21:44, Joe Gwinn wrote:
> On Sun, 2 Jun 2024 20:18:50 +0200, Jeroen Belleman
> <jeroen@nospam.please> wrote:
> 
>> On 6/2/24 18:19, Joe Gwinn wrote:
>>> On Sun, 2 Jun 2024 11:31:33 -0000 (UTC), Cursitor Doom
>>> <cd999666@notformail.com> wrote:
>>>
>>>> On Sun, 2 Jun 2024 11:17:58 -0000 (UTC), piglet wrote:
>>>>
>>>>> Cursitor Doom <cd999666@notformail.com> wrote:
>>>>>> On Sat, 1 Jun 2024 22:00:58 -0000 (UTC), piglet wrote:
>>>>>>
>>>>>>> Cursitor Doom <cd999666@notformail.com> wrote:
>>>>>>>> On Sat, 1 Jun 2024 15:44:17 +0200, Jeroen Belleman wrote:
>>>>>>>>
>>>>>>>>> On 6/1/24 14:07, Cursitor Doom wrote:
>>>>>>>>
>>>>>>>>>> I've taken a shot of the waveform into the 50 ohm input. It's
>>>>>>>>>> around 850mV peak-peak. Hopefully the slight distortion I spoke
>>>>>>>>>> about is visible; the slightly more leisurely negative-going
>>>>>>>>>> excursions WRT their positive-going counterparts. So it's not a
>>>>>>>>>> pure sine wave as one would expect. Does it matter? I don't know!
>>>>>>>>>>
>>>>>>>>>> https://disk.yandex.com/i/7cuuBimDbOIBZw
>>>>>>>>>
>>>>>>>>> The shape looks perfectly acceptable to me. This is +3dBm into 50
>>>>>>>>> Ohms.
>>>>>>>>> Is that what it's supposed to be? Canned reference oscillators most
>>>>>>>>> often deliver +13dBm, sometimes +10dBm.
>>>>>>>>
>>>>>>>> Is it? I only make it about half your figure: +1.65dBm.
>>>>>>>> I admit I'm frequently prone to careless errors, so stand to be
>>>>>>>> corrected,
>>>>>>>> but here's my method:
>>>>>>>> 850mV peak to peak is 425mV peak voltage. Average of that is
>>>>>>>> 0.425x0.636 =
>>>>>>>> 0.27V. Average power is average volts squared divided by the load
>>>>>>>> impedance of 50 ohms = 1.46mW = +1.65dBm.
>>>>>>>>
>>>>>>>> I shall consult the manual to see what it ought to be - if I can find
>>>>>>>> it, that is, as PDF manuals are a nightmare to navigate IME.
>>>>>>>>
>>>>>>>>
>>>>>>>>
>>>>>>> Use 0.71 for RMS instead of 0.636 ! I make that about 1.8mW or +2.6dBm
>>>>>>> ?
>>>>>>
>>>>>> Thanks, Erich. But there's no such thing as "RMS power" strictly
>>>>>> speaking IIRC, so that's why I took the average figure; not that it
>>>>>> makes much difference in practice. it does seem a bit on the low side,
>>>>>> but despite reading through the most likely sources (the service manual
>>>>>> and the trouble-shooting/repair manual) I can find nothing stated for
>>>>>> what that signal level should be! This may be due to the
>>>>>> user-unfriendliness of very large PDF manuals; I just don't know.
>>>>>> Anyway, not very satisfactory! Later today I plan to do a direct power
>>>>>> meter measurement of the ref osc (since none of us here seem to agree
>>>>>> on what 850mV vs 50 ohms equates to!!)
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>>>
>>>>> 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.
>>>
>>>
>>> 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:
>>>
>>> .<https://www.amazon.com/MWRF-Source-Male-Female-Attenuator/dp/B07MP9D9GC?th=1>
>>>
>>> Just buying a few of these and doing some experiments will be far
>>> cheaper and faster than the various alternatives discussed.
>>>
>>> Joe Gwinn
>>
>> What scope picture are you looking at? I see only 0.88Vpp.
> 
> This one, posted by CD on 1 June '24:
> 
> .< https://disk.yandex.com/i/7cuuBimDbOIBZw>
> 
> 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

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