“But if a finite width has noninfinitesimal “amounts” at every frequency, then the total amount is infinite, which makes no sense”. The problem with this statement is that the width is finite and therefore there are only a finite number of finite amounts and the sum is therefore finite! If I recall my history,this is precisely the polemic that raged for years regarging whether in the limit, the differential becomes vanishingly small on the one hand or 0 on the other… The point was that as the slice becomes smaller and smaller it’s contribution also becomes smaller, and in the limit, the sum converges on the integral. The argument is whether an infinite number of infintely small pieces totals up to a finite value. Certainly we know that when we integrate a simple function that the answer almost always gives us the area bounded by the curve, not an infinite value, and in this regard, for all practical considerations it does not seem to matter if we consider that the differential truly reaches 0 in the limit or not.I suppose that we could argue about this for years, just as mathematicians did in the 19th century, However, it appears that Paul Dirac must have finally resolved this debate with the invention of the Delta Function. In the interest of all the other readers of this posting, I think we should put the whole matter to rest. At least that’s what I am going to do!
Yes, what you are saying is correct, but the “amounts” have to tend toward zero as the “pieces” become smaller to give a finite result. I am just explaining why setting the amount at each piece to a constant (non infinitesimal) would give trouble.
The reason the Fourier transform of 1+sin(t) seems to be blowing up is because it does… The Fourier transform of a pure Fourier mode will always just be a delta function centered around the appropriate frequency. In the case of the zero frequency component, we expect zero anywhere away from zero, but an infinitely thin spike around zero.
Glenn,
I should have mentioned the 1 cycle issue. Here’s the explanation you are looking for (I hope). Since Sin and Cos are periodic, the integral over one cycle is exactly the same as the integral over an infinite number of cycles. Also, since the integral 1+sin(t)dt evaluates to t-cos(t), then divide by t to get 1-cos(t)/t. In the limit of infinity, this clearly becomes exactly equal to 1 - not to mention that cos(t) can never exceed one anyway, so this integral can never blow up - at least not with real valued t. If t is complex, I’m not sure what may happen. Why Maxima can’t cope with this probably has something to do with how it deals with the definition of infinity. Also, if I remember the deatils, the 1/p issue has to do with changing the limits of the integration to cover exactly one cycle. It is essentially the same as integrating from -2Pi to +2Pi. You might also consider forgetting about the negative frequency part of the spectrum. For all real valued data (that is, all real data!) it is exacly the mirror image of the positive spectrum anyway,
Daniel,
Are you sure about this? I’ve done a lot of fourier analysis on single frequency sine waves with and without a DC offset and indeed, the result is always a spike at the fundamental frequency of the sine wave and if there is a DC offset, another at the origin.However, these are not of infinite amplitude - quite the contrary, the heights are always exactly equal to the amplitude of the sine wave and it’s DC component. If this were not to be the case, the Fourier transform would not really be very useful for AC signal analysis. I suspect that you are thinking about some other interesting property of the transform and I would like to have you clarify what you are thinking about - perhaps the transform of the delta function?
Math is as easy as pie! but needs someone to bake it as good as you did!
Great!!
Congra!
I’d be happy to explain myself. I say the Fourier Transform of $$1+\sin(x)$$ has three spikes, one at $$s=0,$$ and one at both $$s=-\frac{1}{2\pi}$$ and $$s=\frac{1}{2\pi}$$ respectively. Namely, for $$f(x)=1+\sin(x):$$
$$\hat{f}(s)=\delta(s)+\frac{1}{2i}\left[\delta(s-\frac{1}{2\pi})-\delta(s+\frac{1}{2\pi})\right].$$
If you don’t buy this, well, let’s just check it. According to the definition given at the beginning of the post of the Inverse Fourier Transform, we have
$$\int_{-\infty}^{\infty}\delta(s)+\frac{1}{2i}\left[\delta(s-\frac{1}{2\pi})-\delta(s+\frac{1}{2\pi})\right] e^{-2\pi i s x} ;ds.$$
Then when we perform the integration, the delta functions yield the integrand evaluated at the value of s that makes the argument of that delta function zero. This gives
$$e^0+\frac{1}{2i}\left[e^{i x}-e^{-i x}\right]=\boxed{1+\sin(x)},$$
as desired.
The old Yamaha DX synthesizers series used frequency modulation to create, from 4 to 6 sine waveforms, quite complex sounds.
I didn’t enjoy this article as much as the others. This is because the recipe analogy didn’t feel intuitive. If anyone is having difficulties processing this, I highly recommend reading James’ http://practicalcryptography.com/miscellaneous/machine-learning/intuitive-guide-discrete-fourier-transform/. It’s way more verbose and more intuitive (if you understand simple statistic concepts like correlation and variance).
This is awesome! Nicely intuitive.
@Matija: re Q4, I’m not sure if this helps, but the above examples/annotations are all based on an overall time interval of 1 second, and 1Hz, 2Hz, etc. Probably in a real application the overall time interval would not be 1 second, and therefore the frequencies would change accordingly. For example (as I understand it), if the time interval was 1/10th second then the waves would have frequencies of 0Hz, 10Hz, 20Hz, etc, for as many samples as you have in that interval. If the time interval were 2 seconds then you would actually have 0Hz, 0.5Hz, 1Hz, 1.5Hz, etc, which might be of some use in your case. Also bear in mind the article uses the unit “Hz” (and “seconds”) a little loosely in places, which it explains. Hope this helps.
I had asked Kalid about the Fourier Transform a while back and he had emailed a great brief explanation. I just returned to this site to see whether there were any updates and I’m happy to see that the Fourier Transform is on here. This is a fantastic website!
Thanks Claude! 
"One of my giant confusions was separating circles from sinusoids."
This was one of my most notable Aha!-moments during the Analysis class as well. Glad to see it explained clearly here. You might find the website below very interesting, especially ‘phasor phactory’ (in case you haven’t encountered it yet).
http://www.jhu.edu/signals/index.html
Keep up this useful work,
M.
You ask how to easily prove “When every cycle has equal power and 0 phase, we start aligned and cancel afterwards.”
This is because the roots of x^n = 1 sum to 0 (Vieta’s formulas):
x^n - 1 = (x-r1)(x-r2)…(x-rn) = x^n - (r1 + r2 + … + rn)x^(n-1) + …
so r1 + r2 + … + rn = 0.
@Glukk: The Fourier Transform cannot stay in the hands of the math elite!
@Garth: Awesome, I love it when math gets addictive! Thanks for the kind words
@Gary: Thanks! I’ll have to check that out. I’m looking to do a more math-focused follow-up.
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I really like your presentation!
I recently wrote a tutorial on the DFT as well, though I came at it from a different point of view. My understanding of it is based on correlation between the time domain signal and a series of sinusoids of increasing frequency: http://practicalcryptography.com/miscellaneous/machine-learning/intuitive-guide-discrete-fourier-transform/
Where were you when I got my BSEE!! Great explanation.
@Steve: Glad you enjoyed it! I don’t have an article on convolution yet, but it’d be a great follow-up.
@Marnee: Looks like it didn’t work for you! The key is realizing whether you’re looking at “ingredients” (inputs) or the “cooked meal” (output). The transform lets you switch between the two. Smoothies are nice because there’s just blending/separating, no “cooking” that is difficult to undo.