How Do You Calculate AC Waveform Components and Errors?

[oxon88]

Homework Statement

An A.C. voltage, V comprises of a fundamental voltage of 100V rms at a frequency of 120Hz, a 3rd harmonic which is 20% of the fundamental, a 5th harmonic which is 10% of the fundamental and at a phase angle of 1.2 radians lagging.

i) Write down an expression for the voltage waveform.

ii) Sketch the waveforms of the harmonic compnents.

iii) Determin the voltage at 20ms.

iv) Given an ideal V = 100V rms, what is the percentage error at 20ms

The Attempt at a Solution

part i)

V = V_rms * sqrt2 = 100* 1.414 = 141.4V at 120Hz

3rd harmonic = 20% of 141.4 = 28.28V at 360Hz

5th harmonic = 10% of 141.4 = 14.14V at 600Hz


v = (141.4sin(240∏t)) + (28.3sin(720∏t)) + (14.1sin(1200∏t-1.2))


does this first part look correct?

 

[gneill]

v = (141.4sin(240∏t)) + (28.3sin(720∏t)) + (14.1sin(1200∏t-1.2))


does this first part look correct?

Yes, it looks okay.

 

[oxon88]

Ok thanks. So for the sketch I just need to plot the 3rd harmonic and 5th harmonic. What time period would be appropriate?

 

[NascentOxygen]

A couple of cycles (of the fundamental) should do. But at least up to 20ms. (Note: the fundamental can also be referred to as the first harmonic.)

 

[gneill]

Ok thanks. So for the sketch I just need to plot the 3rd harmonic and 5th harmonic. What time period would be appropriate?

The fundamental should be plotted too (it's the "first harmonic").

Choose a time period that will display at least one full cycle of the fundamental.

 

[oxon88]

should it look something like this?

 

[NascentOxygen]

You're sketch for fifth harmonic should show x5 the frequency!

 

[mahnoorbloch]

i think the ans is right...

 

[oxon88]

Ok so i have changed the time scale and got this...


does this look as it should?

 

[NascentOxygen]

That looks more like it.

 

[oxon88]

ok great.

iii) Determin the voltage at 20ms.

v = (141.4sin(240∏*0.02)) + (28.3sin(720∏*0.02)) + (14.1sin(1200∏*0.02-1.2))

v = 83.11 + 26.91 - 13.14 = 96.88 vdoes this look correct?

 

[NascentOxygen]

does this look correct?

It does.

Might be worth going back and checking to make sure that the problem did not specify a phase angle for the 3rd harmonic.

 

[oxon88]

Many thanks. I have checked the original question and there is no mention of a phase angle for the 3rd harmonic.

 

[oxon88]

ok so last part

iv) Given an ideal V = 100V rms, what is the percentage error at 20ms100v - 96.845v = 3.155v

so would it be 3.155% ?

 

[NascentOxygen]

ok so last part

iv) Given an ideal V = 100V rms, what is the percentage error at 20ms


100v - 96.845v = 3.155v

so would it be 3.155% ?

I'm sure you can't mix RMS and instantaneous values to get anything meaningful.

 

[oxon88]

ok so would i need to work out the RMS at 20ms?

96.845 / (√2) = 68.5v


then work this out as a % of the 100v rms?


100v - 68.5v = 31.5v = 31.5% error?

am i making any sense?

 

[NascentOxygen]

ok so would i need to work out the RMS at 20ms?

96.845 / (√2) = 68.5v


then work this out as a % of the 100v rms?


100v - 68.5v = 31.5v = 31.5% error?

am i making any sense?

The only calculation that would make sense to me would be based on:

ideal instantaneous value at that time - actual instantaneous value

 

[oxon88]

Actual instantaneous value at 20ms v = (141.4sin(240∏*0.02)) + (28.3sin(720∏*0.02)) + (14.1sin(1200∏*0.02-1.2))

v = 83.11 + 26.91 - 13.14 = 96.88 v


Ideal instantaneous value at 20ms = 141.42Sin(240∏*0.02) = 83.11v


error = [(83.11v - 96.845v)/100]*100 = -13.735%

 

[oxon88]

does the answer above look ok?


i have ploted a graph to show actual Vs. Ideal

 

[NascentOxygen]

error = [(83.11v - 96.845v)/100]*100 = -13.735%

I would not have divided by 100 when determining fractional error. Some other data value would seem more appropriate.

 

[oxon88]

ok but the question asks for a % error?

 

[NascentOxygen]

The only calculation that would make sense to me would be based on:

ideal instantaneous value at that time - actual instantaneous value

On second thought, it would be better to swap them:

actual instantaneous value - ideal instantaneous value at that time

So if the actual value were a few volts too high, the error would be +ve.

 

[NascentOxygen]

ok but the question asks for a % error?

I was referring to where you divided by 100.

 

[oxon88]

ah right ok. so...

error = [(96.845v - 83.11v)/83.11]*100 = 16.5%

 

[NascentOxygen]

How did you decide that 96.845 would be the appropriate denominator here?

EDIT Yes, 83.1V does seem the better choice.

 

[oxon88]

changed it now. I saw it was incorrect. It will be 83.11 because that's the ideal voltage at 20ms.

 

[Big Jock]

Oxon88 was the sketch of the waveforms for the harmonic components correct?? I have values for them all but not sure how to construct the sketch as I am unsure of the time intervals to use for the different waveforms.

 

[oxon88]

Oxon88 was the sketch of the waveforms for the harmonic components correct?? I have values for them all but not sure how to construct the sketch as I am unsure of the time intervals to use for the different waveforms.

Yes it was correct. I can't remember what time intervals i used. I can have a look for the spreadsheet and check later for you.

 

[Big Jock]

Perfect Ill try and work it all out now from your images in the meantime

 

[Big Jock]

NascentOxygen or Oxon88 how to do you calculate the various different values for your graph? I can't quite get my head round that part and would be very grateful of a little help with understanding it please...

 

[Big Jock]

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...

 

[gneill]

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?

 

[Big Jock]

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??

 

[gneill]

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.

 

[Big Jock]

At 20ms I get - 47.87 that can't be right though