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'[EE]: Measuring the peak value of an AC squarewave'
2002\09\16@114603 by smerchock, Steve

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Friends,

I am having a problem that I really need help with.
I am trying to measure an AC squarewave at 800Hz with a
dutycycle of 50%. I am using a Keithley 2000 multimeter
set to read ACV. I need to measure the volts peak, the meter
gives me the volts RMS. I figure I could do the formula but
nothing matches.

I am measuring 5.6Vpp and 2.8Vrms. This doesn't make any sense to
me. When I measure the sinewave of an equivelant circuit everything
works out mathematically. Could someone point me to a possible solution.
I have a feeling I am missing / forgetting something trivial.

Thanks in advance!!

Best regards,
Steve

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2002\09\16@115322 by Jim

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One word (what we used in the 'old days'!):

   "Oscilloscope".

RF Jim

----- Original Message -----
From: "Kosmerchock, Steve" <spam_OUTSteve.KosmerchockTakeThisOuTspamRFSWORLD.COM>
To: <.....PICLISTKILLspamspam@spam@MITVMA.MIT.EDU>
Sent: Monday, September 16, 2002 10:38 AM
Subject: [EE]: Measuring the peak value of an AC squarewave.


> Friends,
>
> I am having a problem that I really need help with.
> I am trying to measure an AC squarewave at 800Hz with a
> dutycycle of 50%. I am using a Keithley 2000 multimeter

<snip>

> Steve
>

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2002\09\16@115326 by Jochen Feldhaar

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Hi Steve,

everything is OK!
You have 50% duty cycle, and pp Voltage is double the RMS value (what
would be dissipated as equivalent DC in a resistor).

I wonder why your results are so good in precision that I always get
suspicious that Mr. Murphy is lurking just around the corner.....hehehe

Greets Jochen

"Kosmerchock, Steve" schrieb:
{Quote hidden}

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2002\09\16@120333 by Jochen Feldhaar

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Jochen Feldhaar schrieb:
>
> Hi Steve,
>
> everything is OK!
> You have 50% duty cycle, and pp Voltage is double the RMS value (what
> would be dissipated as equivalent DC in a resistor).
>
> I wonder why your results are so good in precision that I always get
> suspicious that Mr. Murphy is lurking just around the corner.....hehehe
Oh God, my English.....
I wanted to say that results that tally so fine are a surefire indicator
of a Murphy situation...

Jochen

{Quote hidden}

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2002\09\16@120704 by Giles Honeycutt

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Steve,
The formula for the RMS of a AC square wave is Divide by 2
Or you can just flip the negative, and you have DC and DC is RMS

Best regards,
Giles

{Original Message removed}

2002\09\16@140833 by Herbert Graf

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Have you checked the input bandwidth of your meter? It is likely that it is
attenuating (and possibly doing some ugly non linear things to) the incoming
signal in such a way that it causes your mysterious readings. I'd suggest
dumping the meter and using a scope, one of the new digital scopes would be
best since they often have measurement capabilites (otherwise you can figure
it out with a "normal" scope using a calculator). TTYL

> {Original Message removed}

2002\09\16@144014 by Smith,Steven W

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There is this really great little book called "An Introduction to
Circuits and Electronics by J.R. Cogdell"  It explains it nicely (Hard
to get but an really nice intro to basic circuits, page 153).

The root mean square of a signal is really determined by its integral

Vrms = Square Root( 1/T Integral( V(t)^2 dt) over 1 period of the wave.

When a meter gives true RMS value, this is actually what it does, a
non-true RMS meter gives an approximation for a sin function - a factor
of 2 * square root of 2.  
So you take your square wave over one period, square it and find its
integral over the single period.  Then you divide by the period.  
Then you take the square root to get the final answer.

Since you're squaring the "square wave, 50% duty cycle", the areas below
0 Volts, become positive.  That means that if you have a 5 Volt peak
square wave, you have a section that's +5 and -5 each of 1/2 the period.
So when you square it you get (25V * 1/2 T) + (25V *1/2 T) = 25V T/T =
25, which has a square root of 5 V -- The peak to peak in this case is
10V -- or divide by 2 as stated before.  If you change your function,
then you change the factor.  A triangle is different than a square, a
sin is different than the others, etc.  
You need to check your meter - a Keithley 2000 is most likely a true RMS
meter.  The manuals of Keithley products often have this type of
explanation also.  
Steve Smith
Electrical Engineer

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2002\09\16@173644 by Peter L. Peres

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On Mon, 16 Sep 2002, Jochen Feldhaar wrote:

*>I wonder why your results are so good in precision that I always get
*>suspicious that Mr. Murphy is lurking just around the corner.....hehehe

Or he spent too much money on test equipment and Murphy took care of
other details instead ;-)

Peter

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2002\09\16@175941 by smerchock, Steve

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Jim,
This is for an automated test setup. I can not afford
to buy a GPIB o-scope. Thanks anyways for the response.

Steve

{Original Message removed}

2002\09\16@191618 by Peter L. Peres

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On Mon, 16 Sep 2002, Kosmerchock, Steve wrote:

*>Friends,
*>
*>I am having a problem that I really need help with.
*>I am trying to measure an AC squarewave at 800Hz with a
*>dutycycle of 50%. I am using a Keithley 2000 multimeter
*>set to read ACV. I need to measure the volts peak, the meter
*>gives me the volts RMS. I figure I could do the formula but
*>nothing matches.
*>
*>I am measuring 5.6Vpp and 2.8Vrms. This doesn't make any sense to
*>me. When I measure the sinewave of an equivelant circuit everything
*>works out mathematically. Could someone point me to a possible solution.
*>I have a feeling I am missing / forgetting something trivial.

I don't know if your dmm is a true rms one. It seems to be from the
following. rms means root (of) mean (of voltages measured over a second)
squared as you know.

You do not say if you measure 5.6Vpp with a scope. For a square wave with
5.6Vpp a true rms meter should indicate 2.8Vrms as yours does.

When in doubt picture the rms value as the area between the time axis and
the waveform, divided by time (this happens to be mathematically correct
too if you use the correct timescale and volts for the vertical scale).

For a sine wave the rms value is 0.707 of half of the pk-pk value.

For a symmetrical triangle wave the rms value is 0.5 of the half of the
pk-pk value (regardless of the shape of the triangle wave, as long as it
has no dc component).

Peter

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2002\09\16@215919 by smerchock, Steve

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Giles,

This is what I figured, but what would be the "peak" value?
In an AC sinewave the peak value is half of the peak to peak
value and the RMS is the peak x 0.707. Is their a math formula
for finding the peak value?

Thanks for your help!
Steve

{Original Message removed}

2002\09\16@215936 by smerchock, Steve

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Jochen,

This is for a military spec'd device. Precision is
a MUST. Mr. Murphy isn't allowed in the building! ;-)
Thanks for your help.

Steve

{Original Message removed}

2002\09\16@230148 by Jim

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Steve, if this really is a "mil spec'd"
device I would say you really need to
get one of your EE's to buy-off on your
test methodology ... no offense to the
list, but this is like getting "medical
advice" from a panel over the internet ...

What if, during your "testing" of this particular
device or box, something shifts or changes such
that the RMS value reads within tolerances on
the Keithley - yet the actual value/waveform
has shifted considerably from the 800 Hz 50%
duty cycle square wave?

Depending on the box you're checking/testing, this
may not be an issue - perhaps this is a simple
functional check an LRU (Line Replaceable Unit)
of some other non-critical parameter and this
isn't an issue.


RF Jim

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