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gneill
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Big Jock said:At 20ms I get - 47.87 that can't be right though
Nope. What value do you get for the fundamental alone? Write out the calculation.
Big Jock said:At 20ms I get - 47.87 that can't be right though
If you are going to express it in the form of a time-varying function A.sin ωt then A must be the peak value.gneill said:Looking back at the original question I just noticed that it stated: "An A.C. voltage, V comprises of a fundamental voltage of 100V rms at a frequency of ..." [my emphasis in red]. There should be no reason to convert to peak values unless requested to do so. So your waveform function would look like:
v(t) = (100sin(ωt) + 20sin(3ωt) + 10sin(5ωt - 1.2) ) V
where ##\omega = 2\;\pi\;120Hz##.
Now what values do you get for the actual and ideal waveforms at t = 20 ms?
NascentOxygen said:If you are going to express it in the form of a time-varying function A.sin ωt then A must be the peak value.
Big Jock said:100sin(240pi x 0.02) = 58.78
This is the way I have been trying to calculate the point for the graph but them look wrong...
Big Jock said:So the voltage waveform should be 141.4sin(240∏t)) + (28.3sin(720∏t)) + (14.1sin(1200∏t-1.2) or this
100sin(ωt) + 20sin(3ωt) + 10sin(5ωt - 1.2) ? Getting very confused. Ill clear this up first then ask about my second question...Try and keep things simple
Big Jock said:ok so think I have that.
Now this part I have sitting working at for days and I just can't get it. Sketch the waveforms of the harmonic components. Now this I know must include the fundamental,3rd and 5th harmonics. My problem is when I use this formula 141.4sin(240∏t)) + (28.3sin(720∏t)) + (14.1sin(1200∏t-1.2). For either the fundamental or the 3rd and 5th harmonics I don't get my points to look right. I use time values from 0.001 to 0.25 but to no avail. I may add that is all I do I just replace t with these values but I must be doing this wrong. Thanks for your patience so far and hopefully help with this part also...
The figure in post #9 is the sort of thing you're aiming for.Big Jock said:Trying to make a table with all values first then make the graph. 5w=1200pi then multiply by time but they still don't look like the graph in post#9.
Would it be possible to show an example then I know what it is I am trying to achieve?
Big Jock said:So you think If I follow this procedure to 20ms and calculate my values for the 3rd and 5th harmonics the same way I will be correct?
Big Jock said:So just calculate all point to 10ms you reckon then make the graph from there? Sorry for being a bit thick been at this for days and my head feels like mush!
Big Jock said:On thing gneil for the 5th harmonic would it be 14.14sin(1200pi x 0.001-1.2) then just change the time value as I go along till I reach 10ms for my various points?
Big Jock said:those values are from 0 to 10ms the same as I used for the fundamental and third harmonic. Its only this fifth one which doesn't look correct when plotted...
The root 2 needs to trail along. It doesn't make sense to speak of a waveform having an instantaneous value of .. RMS. Whether neater or not!gneill said:The RMS conversion for a sinusoid is just a scale factor. The function of time using the RMS instead of peak for the constants will be a scaled version of the actual voltage waveform. One can always multiply results by √2 to obtain the actual voltage. It's just that the numbers are a bit easier to work with when they're nice multiples of ten![]()
NascentOxygen said:The root 2 needs to trail along. It doesn't make sense to speak of a waveform having an instantaneous value of .. RMS. Whether neater or not!
Ebies said:Can I ask, what program do you guys use to plot the graphs with?
Ebies said:For some reason i do not get the same values if i try to calculate the values of voltage for the fundamental wave, 3rd harmonic or 5th harmonic... For instance if i type 141.4sin(240pi*0.01) into my calculator i get 18.553 as an answer... When from previous posts i know it should be more... Also calculating the voltage at 20ms i get a different answer for each waveform thus giving me an overall incorrect answer... Any ideas...?
gneill said:Looks like you've got your calculator set for degrees rather than radians.
oxon88 said:iv) Given an ideal V = 100V rms, what is the percentage error at 20ms
Show us your own attempt. Helpers cannot simply provide answers to problems, but are more than willing to help you arrive at the correct solution by guiding your demonstrated efforts.grinder76 said:Does anyone have a confirmed answer for this question?