Sujet : Re: Distorted Sine Wave
De : jeroen (at) *nospam* nospam.please (Jeroen Belleman)
Groupes : sci.electronics.designDate : 02. Jun 2024, 22:45:43
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <v3ip2b$3h6vm$2@dont-email.me>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14
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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:
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Cursitor Doom <cd999666@notformail.com> wrote:
On Sat, 1 Jun 2024 22:00:58 -0000 (UTC), piglet wrote:
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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:
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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.
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Use 0.71 for RMS instead of 0.636 ! I make that about 1.8mW or +2.6dBm
?
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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!!)
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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
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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