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Old 26th May 2008, 07:30 AM   #11
awasson is offline awasson  Canada
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Chris are you sure that disk has a square wave 1k tone?

I have a number of test disks and I think they are all sine waves.

Andrew
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Old 26th May 2008, 03:10 PM   #12
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Hello,

Quote:
That's a pity about the camera. Isn't USB supposed to be plug and play, I think I would have a whinge about that !
Good news the camera is fine after replaced it with new batteries.

Quote:
Right --- I am still confused. The picture in post 7. Where has that squarewave come from. Is it the CD or the 'scope cal output.
Just confirm what we are looking at
The squarewave is from the scopes PROBE ADJUST connector / scope cal output

Quote:
When making any measurements you need all the "variable" controls on the 'scope ( Any pots in other words ) in the cal or calibrate position. If this isn't done the settings on the attenuator and the timebase are meaningless.
Yes, this is what the manual suggested procedure “OBTAIN BASELINE” before use each time.

Quote:
O.K. Amplitude of the sinewave on the screen. As an example if the 'scope is set to 1V/DIV and the sinewave is 4.25 squares from tip to tip this equals 4.25 Volts peak to peak. To find the RMS value ( only for sine waves ) divide the pk/pk by 2 to get the peak value. 2.125 volts peak in this case. Then divide this by the square root of 2. So the RMS value of 4.25 V pk/pk is 1.5 volts RMS. So working the other way a C.D. player with an output of 2Vrms will show on the scope as 2*1.414 *2 which is 5.65 volts pk/pk.
When scope is set at 1V/DIV
RMS in Volts = Number of squares (Pk-Pk) / 2 / 1.4142
1.5 = 4.25 / 2 / 1.4142
Does it matter what frequency? Or is it a normal practice that 1 kHz test frequency used in this case?

Experiment: try to find out what the C.D. players output at in RMS compare with what the manual stated in the specification section of a particular C.D. player.
1/ Reset Base Line
2/ Feed the scope with 1 kHz signal from C.D. player by connecting the probe to the audio output RCA center pin with probe ground to RCA outer jack.
3/ If the scope is set at 1V/DIV then count the number of squares peak to peak displaying on screen.
4/ use formula RMS in Volts = Number of squares (Pk-Pk) / 2 / 1.4142

Question:
a/ What if the scope is set at 2V/DIV then?
Will it become RMS in Volts = Number of squares (Pk-Pk) / 4 / 1.4142

b/ What if the scope is set at 0.5V/DIV then?
Will it become RMS in Volts = Number of squares (Pk-Pk) / 1 / 1.4142

c/ With this scope I should be able to measure from 2mV to 50V DC like any Volt meter am I correct?


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Regards Karl
Edit, just had another look at your pic, it's from the 'scope cal output isn't it. That looks fine, and when you get your X10 probe you adjust the trimmer in the probe so that the tops and bottoms of that squarewave are flat. When you try it it's obvious what you do.
It is from the cal output.


Thanks a lot Karl
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Old 26th May 2008, 03:13 PM   #13
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Quote:
Originally posted by awasson
Chris are you sure that disk has a square wave 1k tone?

I have a number of test disks and I think they are all sine waves.

Andrew

Hi Andrew,

I am not sure, but it is the CD's cover stated that it is supposed to be same as the other track of 100 Hz square wave but at 1 kHz. I tried the 100 Hz track as well but it only shown as sine wave as well. I guess it may have been misprinted on the CD cover.
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Old 26th May 2008, 03:44 PM   #14
Mooly is offline Mooly  United Kingdom
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Hi,
You always read the "number of squares" as accurately as you can. If your 'scope is set on 5Mv/div and you have a signal of 3.85 squares then that is 3.85 X 5Mv or 19.25 Mv. Same goes for the timebase, at say 20microseconds/div a signal that "repeats" every 3.65 squares along the horizontal axis is 3.65 X 20E-6 or 73 microseconds. The frequency is therefore F =1/T or 1/73E-6 or 13698 Hz.
A 'scope is not as accurate as meter and is only as accurate in any case to how well you can read it.
When measuring on the horizontal axis for determining frequency pick a point on the waveform (anywhere but usually easiest where it crosses the center line) and count the number of squares untill the same point is repeated on the waveform.
For amplitude you have to stick to sine waves for converting peak to rms values. For other repetitive waveforms there are various form factors that can be applied, but stick to sine waves for now.
Regards Karl
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Old 26th May 2008, 04:03 PM   #15
Mooly is offline Mooly  United Kingdom
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I see your back , The setting of the attenuator and timebase don't of course affect the actual reading, a signal of 4 squares at 1V/Div is 2 squares at 2V/div and so on. In either case it's still 4 volts pk/pk. Same for the timebase. A signal that repeats every 20 milliseconds will occupy 1 horizontal square at 20ms/div and 4 squares at 5 ms/div.
You asked about a 1 khz square wave from your test CD, well it just so happens I have an actual screen shot taken yesterday for another thread so I have attached it. All CD players will reproduce this identically.
Attached Images
File Type: jpg 1 khz off disc.jpg (37.3 KB, 315 views)
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Old 26th May 2008, 04:18 PM   #16
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What settings did you use when taking that 1KHz square wave screen shoot?

I could not make it look square no matter what settings I tried with my test CD.
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Old 26th May 2008, 04:33 PM   #17
Mooly is offline Mooly  United Kingdom
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Hi,
It would be 1 volt/div if I remember, but it was via an attenuator not direct from the line out. Now the timebase ---- try and work it out I will post the result in a mo.
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Old 26th May 2008, 04:44 PM   #18
Mooly is offline Mooly  United Kingdom
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How are you getting on, it's a 1000 hz signal so that is a time period of 1 millisecond. F=1/T so T=1/F. Now the signal is taking 5 horizontal squares so that is 1 / 5 or 0.2 milliseconds per div. (200microseconds/div)
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Old 26th May 2008, 04:47 PM   #19
Mooly is offline Mooly  United Kingdom
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Should say that the 'scope I use is a dual timebase type so the lower trace is part of the same waveform at a higher sweep speed to show detail--- sorry that would confuse
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Old 26th May 2008, 05:00 PM   #20
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Hi Karl,

I am at work and you've beat me to it

I wonder about that your screen shot was showing two time base. I will try again at home tonight. May be my CD mis-printed the sine wave instead of square wave.
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