Harmonic waves - Fundamental voltages

[Jason-Li]

Homework Statement

An ac voltage, comprises of a fundamental voltage of 100 Vrms at a frequency of 120 Hz, 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 components.
(iii) Determine the voltage at 20 ms.
(iv) Given an ideal V = 100 V rms, what is the percentage error at 20 ms?

The Attempt at a Solution

(i)
A 1Hz = 2π rads^-1 so:
Fundamental = 120Hz = 240π, 3rd = 360Hz = 720ω & 5th = 600Hz = 1200π lagging by 1.2 radians
Vrms = 100V so Vmax or A = 100 * √2 = 141.421V
3rd = 0.2 * 141.421 = 28.2842V
5th = 0.1 * 141.421 = 14.1421V

so written as one expression:
141.421sin(240πt)+28.284*sin(720πt)+14.1421*sin(1200πt-1.2)

(ii) Attached at bottom

(iii) where t = 20ms or 0.02s

141.421sin(240πt)+28.284*sin(720πt)+14.1421*sin(1200πt-1.2)

141.421sin(240π*0.02)+28.284*sin(720π*0.02)+14.1421*sin(1200*π*0.02-1.2)
=96.844V

(iv) Ideal Vrms = 100V, V at 20ms = 96.844V
As 96.844V was found from 'max' value,
Percentage error = (1-(V@20ms/V)*100
= (1-(96.844/141.421)*100
=(1-0.6848)*100
= 31.5%

Can anyone confirm if I am along the right lines? just seems like an excessive percentage error and unsure if I should use max values or RMS values for (iv)

 

[gneill]

You can use either peak or rms values when discussing percentage error (as long as you're consistent) as rms is essentially a scaling factor.

For part (iv) you need to find the voltages at 20 ms for the ideal waveform and the the actual waveform. The ideal waveform will not be 100 Vrms at t = 20 ms. 100 Vrms is its maximum value which occurs at other times along its waveform. So compute the values for the ideal and actual waveforms for t = 20 ms, then do your %error calculation.

 

[Jason-Li]

Ahh I see, so does the following look more suitable:

For instantaneous ideal V at 0.02s:
V=141.421sin(240π*0.02) = 83.125V
Percentage error = (1- 83.125/ 96.844 ) * 100 = 14.166%

Also do the other parts look okay? More specifically the sin wave drawing? Thanks as always!

 

[gneill]

Ahh I see, so does the following look more suitable:

For instantaneous ideal V at 0.02s:
V=141.421sin(240π*0.02) = 83.125V
Percentage error = (1- 83.125/ 96.844 ) * 100 = 14.166%

I think you may have swapped the "actual" and "expected" values there. Percent error (pe) is given by

$pe = \frac{expected - actual}{expected} \times 100$

$pe = \left(1 - \frac{actual}{expected}\right) \times 100$

Also do the other parts look okay? More specifically the sin wave drawing? Thanks as always!

With a cursory look your plots look okay.

 

[Jason-Li]

Ah yeah, I wrote them the other way round so that the percentage error would be positive If I write it properly as you state:

Percentage error = ((96.844-83.125)/ 96.844 )) * 100 = 14.166%
or
Percentage error = (1- 83.125/ 96.844 ) * 100 = 14.166%

My main quandary is that I've seen other workings through this such as this:
Ideal = 100Vrms
Error at 20ms = 100V - 96.844 = 3.156, hence 3.155%
I assume this is incorrect as they have used an instantaneous and an RMS value together?

 

[gneill]

Hmm. I think that the expected value would be the 83.125 V (from the fundamental voltage waveform), and the actual value 96.844 V. The fundamental represents the expected value, while the 96.844 V is the actual value with the harmonics taken into account.

 

[gneill]

My main quandary is that I've seen other workings through this such as this:
Ideal = 100Vrms
Error at 20ms = 100V - 96.844 = 3.156, hence 3.155%
I assume this is incorrect as they have used an instantaneous and an RMS value together?

The instantaneous versus peak argument doesn't matter here. It is clearly incorrect as they have not taken into account the time. A 100 V (peak or rms) sinewave at some frequency will not be 100 V at 20 ms from t = 0.

 

[Jason-Li]

Hmm. I think that the expected value would be the 83.125 V (from the fundamental voltage waveform), and the actual value 96.844 V. The fundamental represents the expected value, while the 96.844 V is the actual value with the harmonics taken into account.

Makes sense,
Percentage error = ((83.125-96.844)/83.125 )) = -0.165 so expressed as a percentage = -16.5%.

You're post seems to make sense but when I read other physics papers I found this (below) - being experimental as actual and theoretical as expected? I have found quite a few different sources that have actual and expected different ways around, just unsure which way I should use them as it returns different values.

 

[gneill]

The difference between the formula that you quoted above and the one I presented in post #4 is that the positions of the "experimental" and "theoretical" values are swapped in the numerator. That will change only the sign of the result, not its magnitude. The interpretation of what a % error is meant to express (in terms of its sign) sometimes requires a bit of thought involving the context of the measurement. The formula that you quoted is fine.

However, the important thing to note is that the denominator should always be the "theoretical" (or "expected") value.