Calculating Total Harmonic Distortion: Is My Result Accurate?

[Morgz129]

Homework Statement

estimate the total harmonic distortion in the current waveform using the formula below

Relevant Equations

see attached pdf

i have started by taking the rms values of the results from the spreadsheet making:

I1= 2.818 amps
I3=2.095 amps
I5=1.767 amps

i then added I3 and I5 to give me 3.863 amps which i then input into the formula to yield a result of 135.202% which seems way off to me, any help would be greatly appreciated

 

[BvU]

Hi,

i have started by taking the rms values of the results from the spreadsheet making:
I1= 2.818 amps
I3=2.095 amps
I5=1.767 amps

I can't see these come out: how did you do that ?
And: why do you call them like that when they are at 50, 250 and 350 Hz ?

i then added I3 and I5 to give me 3.863 amps which i then input into the formula

well, the formula does not have $(\sum I)^2\$ but $\sum (I^2)\$

I'm pretty sure the THD is huge when I look at the signal (did you ? It looks horrible ! ) ,
but not 135.202%.

Several remarks & 1 question :

So the whole sample represents 57 ms --- not a nice multiple of the main signal period of 20 ms which means the Fourier peaks are smeared out

THD is approximate, there is no justification for 6 digit accuracy

Looking at the sample one would expect a 300 Hz peak, but it gets split up in a 250 and a 350 Hz peak. My trigonometric math is rusty, but I suspect it's because of a multiplication

 

[BvU]

Because I like the subject, I played around a bit more and found a much lower THD than you did.

With a bit of handwork the sample in column B can be reconstructed almost exactly by adjusting $a, b, c$ in $\ \ a\sin(2\pi\, 50t) + b\sin(2\pi\, 250t) + c\sin(2\pi\, 350t)$

This bypasses the distortion from extending the 1024 point sample as if it were exactly one period of the original signal, which is clearly is not (it jumps from $-$53 to 0 when going from 1023 to 0 again)

I don't want to hijack your exercise, but the amplitude ratios I found that way are nowhere near your I1 : I2 : I3
$\$

Oh, and in the preceding post forget my remark on trig multiplication.

 

[Morgz129]

Hi BVU thanks for the reply, i came to those figures as the current in the graph is measured in i².. so i took the square root of the current and then multiplied by √2 to give the rms value required in the question.

 

[Morgz129]

right i have just had another go and get a totally different figure:
I1= √15.8 = 3.97.. this is a peak value so multiply by 0.707 to get rms value of 2.8Amps
In = √8.7+6.3= 3.87 .. multiply by 0.707 = 2.73Amps

using the given formula: 1/2.8 * (2.73) =0.975 multiply by 100 = 97.5%

Is that looking better?

 

[Morgz129]

Hi could anyone please confirm I have this somewhere near as i am unsure on any other way of working it out? Thanks

 

[BvU]

I get that too. So I suppose that's what you should hand in.

Now, this is where the interesting part begins for me:

Your signal in column B is equal to $$\ \ 32.528\sin(2\pi\, 50t)
-19.5132\sin(2\pi\, 250t) + 11.709\sin(2\pi\, 350t) $$

Amplitude ratios are 1, 0.6, 0.36 which in the formula yields THD 70.0 %.

So -- due to the mismatch between signal period and sample size the peaks smear out and the THD estimate is 40% higher. A considerable effect, some of which may be recovered by taking the area of the peaks instead of the height.
As I said, interesting.

 

[Morgz129]

yeah its definitely a contentious question, interesting, thankyou for your help BVU