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

 

[gneill]

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.

 

[NascentOxygen]

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.

 

[Big Jock]

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...

 

[gneill]

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

 

[gneill]

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?

 

[Big Jock]

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

 

[gneill]

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.

 

[Big Jock]

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...

 

[gneill]

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.

 

[Big Jock]

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?

 

[gneill]

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.

 

[Big Jock]

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...

 

[gneill]

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).

 

[Big Jock]

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?

 

[gneill]

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!