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Re: Quick Questions

Original poster: "Gerry  Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

Hi Ian,

Often, RMS is computed over a periodic cycle. For a square wave (50% duty cycle), the percent of time the waveform is high doesn't change with frequency so RMS doesn't change with frequency. However, duty cycle does affect the rms value for a given amplitude square wave. Example, RMS (root mean square) for a 1 volt pulsed signal where the dute cycle is 50% and the waveform is 1 volt for 1 sec and 0 volt for 1sec is as follows:

Vrms = sqrt( (1v^2*1sec + 0v^2*1sec)/2sec_peroid ) = 0.707v

For a wavefore that was 1volt for 1sec and 0volt for 3sec (25% duty cycle):

Vrms = sqrt( (1v^2*1sec +0v^2*3sec)/4sec_peroid ) = 0.5v

Note the voltages are time weighted so when you divide by the period you get the average of the squares.

Gerry R.


Original poster: Ian Lacy <ianlacy@xxxxxxxxx>

Hey all,

My name is Ian and I am currently an undergraduate at the University
of Kansas. In my secondary circuit analysis course we're using
function generators and oscilloscopes to analyze the RMS values and
effective voltages. The experiment was very straight forward and I
found myself wanting to know more as the material was not covered in
depth. I subscribe to this mailing list and read it often because I
find it educational and fascinating. I find myself hesitant to reply
to most of the posts because my level of knowledge on the subject is
very minuscule comparatively. This is why I've decided to present my
questions to this group. Perhaps some will be able to help?

Why exactly do frequency and the duty cycle of a square wave have no
effect on the V RMS value?

Also, how does the DC offset effect the power of the signal?

Thank you,

Ian Lacy
The University of Kansas
Theta Tau, Z'996
IEEE Student Member