RMS voltage of a square wave
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Posted by panfilero on May 23, 2008, 10:37 am
Please log in for more thread options wave, a noisy square wave, with some measurement device, and I want to know the regular current of the thing can I take the RMS current and divide it by the square root of 2 in order to get my regular current? I know you can do this for sinusoidal ac signals but don't know if the math still works out the same for other signals... PS - I got my RMS current value by dividing the RMS voltage value by a known resistance I have in my circuit Much Thanks for any responses I get | |||||||||||||||||||||||||||||||||||||||||||
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Posted by Tim Wescott on May 23, 2008, 11:06 am
Please log in for more thread options There is no formal definition of "regular current", so if you define it right, sure. What do _you_ mean by "regular current"? With sinusoidal waveforms, the peak quantity is the RMS quantity times the square root of two. With square waveforms, the peak and RMS quantities are equal. -- Tim Wescott Control systems and communications consulting http://www.wescottdesign.com Need to learn how to apply control theory in your embedded system? "Applied Control Theory for Embedded Systems" by Tim Wescott Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html | |||||||||||||||||||||||||||||||||||||||||||
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Posted by Bob Eld on May 23, 2008, 12:10 pm
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> I was wondering if I take the RMS voltage of a messy looking square
> wave, a noisy square wave, with some measurement device, and I want to > know the regular current of the thing can I take the RMS current and > divide it by the square root of 2 in order to get my regular current? > > I know you can do this for sinusoidal ac signals but don't know if the > math still works out the same for other signals... > > PS - I got my RMS current value by dividing the RMS voltage value by a > known resistance I have in my circuit > > Much Thanks for any responses I get Yes the RMS current is the RMS voltage devided by R. No, the "regualr curent" What ever that is, is not the RMS current divided by the square root of two. With a sign wave, the peak current divided by the the square root of two is the RMS current, but ONLY for a sign wave. With any other waveform, the the square root of two relation does NOT hold. It must be calculated from first principles or measured with a true RMS meter. The equation for the RMS current is: Irms = ((int i(t)^2 dt )/ T)^1/2 Where i(t) is the current as a function of time, T is the time over which the integral is taken, ( 0 to T ), usually one period of the wave. That is, in words, the RMS current is the square root of the average time integral of the current squared. For a square wave of current Ip going plus and minus over a period of T the RMS value is: Irms =( ( (Ip^2 *T/2 + (-Ip)^2 * T/2)) / T)^1/2 = ( (Ip ^2)/2 + (Ip^2)/2)^1/2 = (Ip ^2)^1/2 Irms = Ip, The RMS equals the peak value for a square wave. It very much more complicated if noise is present. | |||||||||||||||||||||||||||||||||||||||||||
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Posted by BobG on May 23, 2008, 2:04 pm
Please log in for more thread options The RMS of an on and off wave is the average. Try it with 1V and 0V.
Then you have to believe that it also works with 5v and 0v. | |||||||||||||||||||||||||||||||||||||||||||
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Posted by Bob Eld on May 23, 2008, 3:27 pm
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> The RMS of an on and off wave is the average. Try it with 1V and 0V.
> Then you have to believe that it also works with 5v and 0v. Thats right. In the example I gave the wave was plus and minus, not plus and zero. | |||||||||||||||||||||||||||||||||||||||||||
> wave, a noisy square wave, with some measurement device, and I want to
> know the regular current of the thing can I take the RMS current and
> divide it by the square root of 2 in order to get my regular current?
>
> I know you can do this for sinusoidal ac signals but don't know if the
> math still works out the same for other signals...
>
> PS - I got my RMS current value by dividing the RMS voltage value by a
> known resistance I have in my circuit
>
> Much Thanks for any responses I get