Fourier analysis - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: September 17, 2017, 09:13:26 pm
content: so next part of my assignment question: I have to develop the Fourier analysis of:f (wt) = 0 for 0 ? wt ? pi/2f (wt) = Vsin(wt) for pi/2 ? wt ? pif (wt) = 0 for pi ? wt ? 3/2 pif (wt) = Vsin(wt) for 3/2 pi ? wt ? 2piit's sin(x) with the first and third quarter removed and made into 0.now they have put "w" in, but w=2*pi*f and i don't know the frequency, well I assume based on it being part of a question that states 50Hz that it is 50Hz but this is not the same waveform i was just asked to calculate on so who knows what they mean (they do this often, just ask an ambiguous question)But assuming i don't know the frequency, do i need to ? can i just take w out and work on the basis that it is a function of 2pi ?I tried a first attempt at fourier analysis and got the attached which i think is a little more complex than the answer was supposed to be so should probably go back and try and work out what i should be doing.

author: bson
author_rank: Unknown
timestamp: September 18, 2017, 06:28:53 am
content: It sounds like they want you to create a graph with wt on the horizontal axis ranged 0-2pi, and f(wt)/V on the vertical axis ranged 0 to 1.But this would be a simple step function with f=1 at pi/2 to pi and 3/2pi to 2pi, and 0 everywhere else. Not sure if they're looking for something quite so simple.

author: IanB
author_rank: Super Contributor
timestamp: September 18, 2017, 06:40:35 am
content: Normally, "Fourier analysis" means to decompose a waveform into a sum of sine waves of different frequencies and amplitudes (the harmonic content) that all add up to the original waveform.I think they are trying to indicate that when you use a phase angle dimmer on a 50 Hz supply you introduce a lot of additional frequencies much higher than 50 Hz. In the real world this translates to unwanted electrical noise that can escape and cause problems, so dimmer circuits need to include filters to suppress it.Do you know how you are supposed to work this question? Are you intended to use some software to do the Fourier analysis? It doesn't seem like something you could do by hand.

author: Simon
author_rank: Unknown
timestamp: September 18, 2017, 06:59:41 am
content: We the RMS question was the first part of the question setting a firing angle at 60 degrees, the second part of the question asked what would happen to the harmonics is the firing angle was not symmetrical on both halves of the waveform. This question is the third part, it does not explicitly refer to the first two parts of the question. Unfortunately their questions often seem haphazard and with insufficient background and i don't like making assumptions, if i have to I state them in my response.I am asked to develop the analysis as far as I am able, perhaps it's the distinction part of the question.You say I can't do this by hand ? I am using microsoft mathematics to perform calculations as i don't trust myself but it won't do any Fourier work. They have explained how to use the FFT of excel, but that is for random wave-forms of which I have a few discrete points. This function seems to be described mathematically with the sine function. Perhaps if i put in enough points in excel it would yield a decent result ? What I have attempted so far in ms mathematics has yielded some unwieldy results.

author: IanB
author_rank: Super Contributor
timestamp: September 18, 2017, 07:28:52 am
content: I'm afraid this question is going beyond my knowledge and experience.From what little I know, to do the Fourier analysis by hand, analytically, given a stated mathematical function, would require some fairly advanced mathematics (degree level stuff). I doubt they are expecting that of you. Most people do Fourier analysis by computer using numerical algorithms. So maybe, yes, putting lots of points into Excel would be a way to do it.Since a Fourier analysis really requires an "infinite" wave without beginning and end, if you do do it numerically you should put in data points for many repetitions of the basic wave rather than just one cycle.Another way to look at it is that the beginning point where the triac fires is like a step change from zero to the full voltage. Perhaps if you have already encountered square waves and step changes they may want you to refer back to that in some way?But as I say, without having gone through the course material up to this point, I am not really able to give more than vague suggestions.

author: Vtile
author_rank: Super Contributor
timestamp: September 18, 2017, 07:38:31 am
content: Those equations do look really cryptic with all those questionmarks.

author: IanB
author_rank: Super Contributor
timestamp: September 18, 2017, 07:54:55 am
content: Quote from: Vtile on September 18, 2017, 07:38:31 amTry this:\$f(\omega t) = 0 \text{ for } 0 \le \omega t \lt \pi / 2 \$\$f(\omega t) = V \sin(\omega t) \text{ for } \pi / 2 \le \omega t \lt \pi \$\$f(\omega t) = 0 \text{ for } \pi \le \omega t \lt 3 \pi / 2 \$\$f(\omega t) = V \sin(\omega t) \text{ for } 3 \pi / 2 \le \omega t \lt 2 \pi \$

author: Simon
author_rank: Unknown
timestamp: September 18, 2017, 10:28:51 am
content: Yes sorry, not a good copy and paste from the pdf

author: Simon
author_rank: Unknown
timestamp: September 18, 2017, 10:31:25 am
content: I have done square waves but I'm still not sure how that helps, this certainly will be a weird one but the if you are synthesizing with sine and cosine and the wave is a sine with bits missing that does rather make it a bit difficult I would say intuitively. I'll try exel and if it crashes give up.

author: sibeen
author_rank: Frequent Contributor
timestamp: September 18, 2017, 10:47:51 am
content: Quote from: Simon on September 18, 2017, 06:59:41 amI'd be interested to see what your response was.As to the Fourier, I'm afraid it was a long, long, long time age when I had to dig down into equations like that.

author: Vtile
author_rank: Super Contributor
timestamp: September 18, 2017, 11:07:19 am
content: It have been a few years since I have done fourier and need to confess that this piecewise case doesn't ring any bells at which corner one should start with it by hand. You should get frequenzy distribution out of it with FFT, when there is a cas-system that will eat it.Have you tried something likehttps://www-fourier.ujf-grenoble.fr/~parisse/giac.htmlit should be rather good. 'I assume the following is the format that most CAS-system will eath a such function. It basicly uses the boolean operators to give multiplicator of zero in all ranges except where is is specified.\$f(\omega t) = ( V \sin(\omega t) *( \pi / 2 \le \omega t)*(\omega t \lt \pi ) + ( V \sin(\omega t) \text{ for } *( 3 \pi / 2 \le \omega t )*( \omega t \lt 2 \pi )) \$

author: Simon
author_rank: Unknown
timestamp: September 18, 2017, 11:35:03 am
content: Quote from: sibeen on September 18, 2017, 10:47:51 amI waffled:The harmonic content would certainly change, the waveform would no longer be symmetrical about the “x” axis. The waveform would also be neither odd or even. The average value of the waveform will also no longer be 0 but offset and so the Fourier series would now have an a0 term.The harmonic content will increase and could include frequencies above those expected if the waveform was symmetrical leading to the possibility of an increased risk of interference generated by the unit.

author: Assafl
author_rank: Frequent Contributor
timestamp: September 18, 2017, 11:55:34 am
content: Long time ago, but my approach would be to simplify the problem (in the time domain) - and rewrite it as the product of a sine wave (for which the transform is clear), and a square wave that is twice the frequency of the sine wave.The fourier transform of a product is the product of the two series (or transforms).

author: Simon
author_rank: Unknown
timestamp: September 18, 2017, 12:03:52 pm
content: Quote from: Assafl on September 18, 2017, 11:55:34 amI just spoke to someone through work (he writes heat transfer software for them) and he was saying something like that, yes of course, you are right, no calculations required, it's a sine plus the square wave series but the two "series" one of which is simply a sine fundamental needs to be at the right phase, or rather the square does.

author: sibeen
author_rank: Frequent Contributor
timestamp: September 18, 2017, 12:18:12 pm
content: Quote from: Simon on September 18, 2017, 11:35:03 amAhh. I suspect they are trying to get you to recognise that if a waveform is symmetrical above and below the line that there will be no even harmonics in the Fourier transform.

author: DJohn
author_rank: Regular Contributor
timestamp: September 18, 2017, 12:54:18 pm
content: Quote from: Assafl on September 18, 2017, 11:55:34 amNot quite: the Fourier series of a product is the convolution of the Fourier series of the two functions. It's messier than a simple product, but still a good short-cut.It's still not that difficult to do this one directly, by hand, if you have the right trig identities. To find the coefficients you need the integral from 0 to 2pi of the product of the function with either sin(nx) or cos(nx). Since the function is 0 in some places, just split it into two integrals from pi/2 to pi and from 3pi/2 to 2pi of either sin(x)sin(nx) or sin(x)cos(nx).That's where you need the trig identities: sin(ax)sin(bx) = 1/2( cos((a-b)x) - cos(a+b)x) ) and sin(ax)cos(bx) = 1/2( sin((a-b)x) + sin((a+b)x) ). The rest should be straight-forward.The coefficients will end up as functions of w. That's where the (unspecified) frequency comes in.

author: Assafl
author_rank: Frequent Contributor
timestamp: September 18, 2017, 01:03:05 pm
content: Quote from: DJohn on September 18, 2017, 12:54:18 pmOh dear I messed up! It the Convolution Theorem. At one side of the transform is a product <-> convolution in the other.

author: Simon
author_rank: Unknown
timestamp: September 19, 2017, 06:39:32 am
content: Uh yes I suppose I could start trying to develop the analysis of the square wave, as it is i just made it up in a graph program to look right and reported the resulting series. The attached goes as far as the 17th harmonic

author: orolo
author_rank: Frequent Contributor
timestamp: September 19, 2017, 03:04:09 pm
content: You can make the change of variable x = wt, and work for x in [0, 2*Pi], no problem.The Fourier series are a bit of a hassle to do. Since your waveform isn't even nor odd, you must consider sin and cos harmonics.The Fourier series I got is:\$ \displaystyle \frac{1}{2}\,\sin(x) \ - \frac{1}{\pi} \sum_{n=0}^{\infty}\, \frac{1}{2n+1}\left[\cos\left((4n+1)x\right)-\cos\left((4n+3)x\right)\right] \$This is a bit more complex than the typical exercise, so I checked it by plotting the Fourier series with n=30, getting the image attached below.The sine integrals are zero for n>1. The cosine integrals have modulo 4 periodicity, which is a bit ugly.Edit: Ok, corrected. The first time I got the quarters where the waveform was zero in reverse. Now it's corrected.

author: sibeen
author_rank: Frequent Contributor
timestamp: September 20, 2017, 08:56:16 am
content: Looks nothing like what we've been discussingGood job, orolo. I'm afraid my maths has sunken into an old age abyss.