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Re: Quick Questions



Original poster: Jim Lux <jimlux@xxxxxxxxxxxxx>

At 08:07 PM 11/9/2005, Tesla list wrote:
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?

RMS value is the square root of the mean squared value. A square wave (if symmetric around zero volts), when you square it, is constant, so the squared value is constant, so the mean of the squared value is constant, etc...



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

A DC offset increases the RMS value and does make the RMS value sensitive to the duty cycle.


In any given cycle, with duty cycle "D"...(from 0 to 1)

For D amount of time the DC value is Offset+PeakToPeakValue/2
For 1-D amount of time, the DC value is Offset - PeaktoPeakValue/2

the squared value for the first one is (offset+PP/2)^2 or offset^2 + offset*PP + PP^2/4
for the second, offset^2 - offset*PP + PP^2/4


The mean of the squared values is then:
(D+ (1-D)) * offset^2
+ D * offset*PP
+ (1-D) * -offset*PP
(D+(1-D)) * PP^2/4

collecting and cancelling:
offset^2 + PP^2/4 + (D - (1-D)) * offset*PP
=offset^2 + PP^2/4 - (2*D - 1) * offset*PP




Thank you,

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