How Do You Calculate AC Waveform Components and Errors?

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SUMMARY

The discussion focuses on calculating the components of an AC voltage waveform, specifically a fundamental voltage of 100V rms at 120Hz, along with its 3rd harmonic (20% of the fundamental) and 5th harmonic (10% of the fundamental) with a phase angle of 1.2 radians lagging. The voltage waveform is expressed as v(t) = (141.4sin(240∏t)) + (28.3sin(720∏t)) + (14.1sin(1200∏t-1.2)). The voltage at 20ms is calculated to be 96.88V, leading to a percentage error of 16.5% when compared to the ideal instantaneous value. The discussion also emphasizes the importance of using the correct phase angles and RMS values in calculations.

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  • #31
I have used the various values and various different times but can't get anything to work out right. Think its something wrong in the (wt) of the equation for the graph Iam having trouble with...
 
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  • #32
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?
 
  • #33
So gneill take the separate components I.e 100sin(ωt) w=240pi and just plug in various time values then do the same with the others 20sin(3ωt) w=720pi and so on??
 
  • #34
Big Jock said:
So gneill take the separate components I.e 100sin(ωt) w=240pi and just plug in various time values then do the same with the others 20sin(3ωt) w=720pi and so on??

That's the idea. You want the sum of them for the "actual" waveform, while the first component by itself is that of the ideal case.

If you list the values that you obtain for each component at t = 20ms I can check them.
 
  • #35
At 20ms I get - 47.87 that can't be right though
 
  • #36
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.
 
  • #37
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?
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.
 
  • #38
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...
 
  • #39
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.

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 :smile:
 
  • #40
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...

That value looks okay. Multiply by √2 if you want the actual potential (see my post above in answer to NascentOxygen). The scale factor won't affect the % error calculation.

What about the other two components?
 
  • #41
So the voltage waveform expression 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
 
  • #42
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

The first one is the actual waveform that you'd measure on an oscilloscope. The second I guess you could call an "RMS scaled" version. If you like, you can write the peak version as:

##v_m(t) = \sqrt{2}\left(100 sin(\omega t) + 20 sin(3 \omega t) + 10 sin(5 \omega t - 1.2) \right)##

where the √2 is factored out in order to keep the constants looking nice (and minimizing truncation/rounding issues).

If you're looking for % errors then any scaling factor won't matter; it'll cancel out in the calculation of the % error.
 
  • #43
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...
 
  • #44
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...

What mechanism are you using to make your plots? Are you doing them by hand or using software? Try reducing your domain; the highest harmonic will be oscillating at 5x the rate of the fundamental. Try values between 0 and 0.01 seconds.
 
  • #45
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?
 
  • #46
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?
The figure in post #9 is the sort of thing you're aiming for.

You can sketch a sine wave easily enough by laying out the bounding lines specified by the magnitude, and marking off a period or two on the time axis --- If you know the frequency then you know the period. You can free-hand in a sine curve accordingly. For the phase-shifted one, convert the angular offset into a time offset and shift the curve accordingly (the phase angle represents a certain fraction of a period).

Since there's a 5x range of frequencies there'll be a similar 5x range for the periods. You need to be sure to take your points close enough together for the higher harmonics in order to pick up the shape.
 
  • #47
the frequencies are 120 hz 360 hz and 600hz. Now the w value for these are 240pi, 720pi and 1200pi my values for the fundamental are
0 = 0
1ms= 96.79
2ms= 141.12
3ms= 108.95
4ms= 17.72
5ms= -83.11
6ms= -138.9
7ms= -119.39
8ms= -35.16
9ms=68.12
10ms=134.48
Now to me those don't match the graph of #9 this is what I really am struggling with...
 
  • #48
The values look okay to me, and they lie on the plot of the fundamental. 10 ms covers just over one period of the fundamental wave.

Here are your points plotted on top of a sine curve matching the fundamental (peak).

attachment.php?attachmentid=66102&stc=1&d=1390930038.gif
 

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  • #49
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?
 
  • #50
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?

I think you only need to plot out to 10 ms at the most, really. That's just over 1 period for the fundamental. The period of the fundamental is 8.333 ms. The period of the fifth harmonic is only 1.667 ms. If you want to resolve the fifth harmonic (so its shape appears) you'll have to plot many points over each period.

It's easier to sketch by hand than it is to plot points!
 
  • #51
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!
 
  • #52
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!

You could do that but...

They asked for a sketch did they not? I wouldn't spend the time calculating points. I'd rough in the boundaries and the periods and free-hand the sine waves. You know how a sine wave looks, and if you have the amplitude and period it's easy enough to rough in.

For the one curve with a phase offset I might find the first zero crossing to fix the shift. After that it's handled the same way -- mark off periods from the zero crossing and lay down the sine waves to fit the within the periods and amplitude bounds.
 
  • #53
Almost got all the point values now so Ill finish that off then put them into open office and create a graph in there. This way I have all the working and graph so surely should tick all the boxes?
 
  • #54
Sure.
 
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  • #55
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?
 
  • #56
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?

0.001 corresponds to t = 1 ms, yes? Don't you want to start at t = 0?
 
  • #57
Yeah but checking was just as simple as plugging in the 1.2 as my figures for this are as follows
0
0.8
-13..18
13.67
-13.18
7.65
0.8
-13.18
13.67
-13.18
And am not sure those would be correct. Many thanks for your help again. Hopefully you can help clear up this final issue...
 
  • #58
I can't tell what the numbers "mean" without the corresponding time values associated with them.

However, as I cast my eye down the list I notice the large jumps in values between successive entries, such as -13.18 followed by 13.67 followed by -13.18 again. Looks like your time step is too coarse and you're losing shape of the curve. In digital terms, your sampling rate is too low :smile:
 
  • #59
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...
 
  • #60
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...

It's because your sampling rate is not high enough! You need about 10 or 12 samples for each millisecond to resolve the shape of the wave for the 5th harmonic. Try just the first ms of the plot using time steps of 0.0001 second.

If you want to see the shape of the wave, you need to have enough samples over every period of the wave to follow its outline.

This is why I suggested that sketching by hand would be quicker.
 

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