Principal frequencies of a Waveform

[rude man]

here's my updated worksheet with the A and B coefficients calculated (cos and sin resp.)

 

[his_tonyness]

Re- Uploaded the file showing the actual An and Bn values. You forgot to multiply An by 2 and Bn by - 2.

 

[rude man]

I think there are more problems.
If the data is samples of I², which gives the graph you posted, then there can be no real and imaginary components.
So this time I will assume that the samples are I. I'll work on this and send it for what it's worth.

 

[Electest]

I'm wondering if the coefficients calculated by excel (Nzn) should be divided by N to give us zn, before entering into column C. What you think?

 

[rude man]

I'm wondering if the coefficients calculated by excel (Nzn) should be divided by N to give us zn, before entering into column C. What you think?

Good question. I intend to run a short example for which I know the answers already (N=4 only) to see what excel really does. It's too bad they don't reveal their algorithm, since there are innumerable versions of DFT's floating around.

 

[Electest]

How did you get on RM? I've looked at various websites to establish how things work with excel, but they all have different answers, so am curious how your assessment went.

 

[rude man]

How did you get on RM? I've looked at various websites to establish how things work with excel, but they all have different answers, so am curious how your assessment went.

Hang on, I think I finally have the goods, will send an updated worksheet today and also will try to go thru the theory. I've had to do a bit of re-education here myself!

 

[Electest]

Brilliant, look forward to it

 

[rude man]

Ok, here is (I hope) my final effort.
I strongly urge referring to the attached dft1.pdf file. I have used their nomenclature for most of my column headings.
First thing is that I had to consider the 1024 data points as current, not current-squared, since some of the numbers are negative.
My way of looking at it, what the OP's original graph showed was the rms current component divided by √2 (column J), which I think is weird since that divides each component amplitude by 2 instead of √2. Still, I decided to chart the J column to recall the OP's chart. You can change the J data to I data or whatever of course for a new chart.

I hope that most of my worksheet is self-explanatory. I apologize for my lousy chart, I never could figure out how to run my x-axis the way I want, the way the OP did (nice 100Hz segments running 0 to 500 Hz).
If the power density spectrum is desired my belief is that it has to be my column I which would give way bigger y-axis values, peaking at 504 A² at f = 52.73 Hz.

Again, if you don't get my worksheet, or a strange one, pls. let me know. I do seem to have some problems sometimes with .xlsx files since I am running office 2003. (I almost quit my job when I found out what microsoft did to excel after 2003!).
BTW I ran the 4-data example in dft1.pdf using excel's data analysis package Fourier analysis and it agreed with the example's numbers for F[n] so that does give me confidence in using the excel program.

 

[rude man]

BTW disregard the A and B columns far to the right. They were part of the OP's original worksheet. I did not compute these as I explained earlier.
I added column L which performs the inverse transform. Happily, all the f[k] values in columns B and L seem to agree.

 

[Electest]

Hi RM, away for the bank holiday, but will take a look on my return. It's quite an interesting question.

 

[M P]

Hi RM, away for the bank holiday, but will take a look on my return. It's quite an interesting question.

Sorry but I draw it in paint...

 

[earthloop]

Going back to the calculation of the THD -

But your n=2 peak is at 8.77 A² and the n=3 is at 6.25 A² so why are you again taking rms of those two numbers? You didn't take the rms of 16, and rightly so!

Why wouldn't you take the RMS of 16? The question states that $I_1$ is the RMS value of the fundamental current. Is the $16^2$ a peak value or is it already an RMS value? I would have assumed it was peak. The same with $I_2$ and $I_3$.

I am getting :

$\displaystyle\frac{1}{\sqrt{16}*\sqrt{2}} * \sqrt{\sqrt{2}*8.77+\sqrt{2}*6.25} * 100 = 81.43\%$

What have I missed here? Can anyone explain?

Thanks

 

[rude man]

Going back to the calculation of the THD -
View attachment 84470
Why wouldn't you take the RMS of 16? The question states that $I_1$ is the RMS value of the fundamental current. Is the $16^2$ a peak value or is it already an RMS value? I would have assumed it was peak. The same with $I_2$ and $I_3$.
I am getting :
$\displaystyle\frac{1}{\sqrt{16}*\sqrt{2}} * \sqrt{\sqrt{2}*8.77+\sqrt{2}*6.25} * 100 = 81.43\%$
What have I missed here? Can anyone explain?
Thanks

Please see post #44. I again strongly advise scrutiny of the attached pdf file for the DFT theory.
Also, the attached excel file in that post gives the correct rms values (column G).
Keep in mind that the frequencies are the bin frequencies and the rms values the corresponding bin current values. If the data sampled three discrete frequencies the dft frequencies will be approximate of course.

To answer your question, you do take the rms of I1, I2 and I3 in the THD formula. (Actually, you could use the amplitudes too. Works either way).

I don't understand why you multiplied your I² terms by √2 but the I² values are incorrect anyway. I get THD = (1/22.46)*√(12.40² + 8.84²) = 67.8%.

 

[earthloop]

Ok thanks RM, I'll give the pdf a thorough read and try again.

 

[rude man]

Ok thanks RM, I'll give the pdf a thorough read and try again.

Good. I would like to see the thread continue until everyone's happy!

 

[Gremlin]

Homework Statement

Identify the principal frequencies in the current waveform

Homework Equations

The Attempt at a Solution

From the excel file, I have identified 3 principle frequencies at 52Hz, 246Hz and 351Hz. Is this correct? What does the term principle frequency actually mean?

Part 1 of this questions asks you to obtain the Fourier Transform for the data using the Fourier Analysis Tool of Excel. The transformed data should commence in cell D2.

I have done the analysis (attached spreadsheet), but am unsure exactly what data/figures i should put in my write up - can anyone advise?

 

[Gremlin]

Or is it inferred that you've done it from subsequent answers?

 

[Gremlin]

Thats the darndest thing, the sample interval does not state the time units. "The supply current was sampled 1024 times over a very short time interval" that's all it gives me with regards to time, I assume we can use any sensible small time period. If you hover the mouse cursor over cell I3, it does state that the sampling frequency is 360/20ms=18000. I re-ran the simulation at 20ms:

View attachment 83707
View attachment 83708

Can i ask how you did this? Did you feed it the figures from the spreadsheet?

 

[rude man]

I'm leaving this in the hands of earthloop and electest.
r m

 

[Gremlin]

Or is it inferred that you've done it from subsequent answers?

To answer my own question i can only assume it's inferred from subsequent answers that you've done it.

 

[Gremlin]

The final part of this question is

Attempt to synthesise the shape of the original waveform from its principal harmonics [e.g. sketch the waveforms of the harmonics on the same time axis and add them together].

The principle waveform is at f = 52.7Hz and the harmonics are at 246.4Hz & 351.6Hz - what axis should i use for the waveforms?

 

[Gremlin]

Can anyone offer the any assistance as to what axis to use when plotting the fundamental and the 2 harmonics?

I'm unsure how to get them to 'overlap' so i can subtract them.

 

[MattSiemens]

Please see post #44. I again strongly advise scrutiny of the attached pdf file for the DFT theory.
Also, the attached excel file in that post gives the correct rms values (column G).
Keep in mind that the frequencies are the bin frequencies and the rms values the corresponding bin current values. If the data sampled three discrete frequencies the dft frequencies will be approximate of course.

To answer your question, you do take the rms of I1, I2 and I3 in the THD formula. (Actually, you could use the amplitudes too. Works either way).

I don't understand why you multiplied your I² terms by √2 but the I² values are incorrect anyway. I get THD = (1/22.46)*√(12.40² + 8.84²) = 67.8%.

I'm so confused with this question different answers and calcs everywhere!

As the question asks for Rms, I found the square root of the I^2 values (which I assume are peak?), multiplied this by sqrt 2 to get RMS then plugged the values in??

1/5.635 Sqrt (4.1822^2 + 3.5364^2) then multiplied the whole thing by 100 to find the percentage.

Am I way off?

Thanks :-(

 

[HDG]

Sorry to go all the way back to the beginning but...

The spreadsheet shows principal frequency as being the highest peak (I believe this is both I1 and n=1). This is 52hz with a magnitude of 15.878. I'm confused though as to Why 15.878 is just rounded up to 16 yet the 2 distortion frequencies are not rounded and stay at 8.77 and 6.25 respectively.

I've used data analysis on excel to produce the same graph but am in 2 minds what number to use for 1/(I1) when using the THD formula. Do I use 1/square root 15.878 etc etc or stick
with 1/square root 16 etc etc detailed above.

Thanks

 

[HDG]

Using 16 = 96.89%
Using 15.878 = 97.26%

Both obviously equate to 97% but Just wondering if there was something I'd missed about needing to round up a fundamental or something

 

[HDG]

Grappled with this for a few days and I'm happy I've got the answer. It's done using 15.878.

For part iv) I changed the 3 magnitudes to rms values ((square root of magnitude)x0.707). I then used the formula:

Amplitude(rms)*sin(2pi x f x t)

in 3 separate columns in excel (for each frequency and it's associated rms magnitude) and just dragged the formulas down to complete the 1024 entries. I then summed all of the 3 entries, created another column for time(s) and then inserted a scatter graph to synthesis the original waveform. Only did it over a short time period though as the graph wouldn't accept all 1024 entries.

(To get the formula to drag down I had to lock some of the fields when writing the formula with the $ sign) e.g. $b$2*sin(2pi*$a$3*a8)).

 

[Jason-Li]

Grappled with this for a few days and I'm happy I've got the answer. It's done using 15.878.

For part iv) I changed the 3 magnitudes to rms values ((square root of magnitude)x0.707). I then used the formula:

Amplitude(rms)*sin(2pi x f x t)

in 3 separate columns in excel (for each frequency and it's associated rms magnitude) and just dragged the formulas down to complete the 1024 entries. I then summed all of the 3 entries, created another column for time(s) and then inserted a scatter graph to synthesis the original waveform. Only did it over a short time period though as the graph wouldn't accept all 1024 entries.

(To get the formula to drag down I had to lock some of the fields when writing the formula with the $ sign) e.g. $b$2*sin(2pi*$a$3*a8)).

Hi HDG,

Did you get the question correct? I got a final answer also of 97.271%

How did you select the fields to create the scattergraph?

Thanks for any help!