Simple question about fourier representation
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From: Golabi Doon on 3 Jul 2010 04:03 Hello mathematicians, Consider a function f_n(x) defined on [-pi,pi] that is represented with n frequency components: f_n(x)=sum_{k=1}^n a_k cos(kx) + b_k sin(kx) The set {a_k,b_k} for k=1,...n is given. The function f_n may have multiple peaks (I think at most 2n). By a peak I mean a point that is a local minimum or local maximum. Now, does the fact that the basis functions have bounded frequency (due to finite n) imply a bound on how close two peaks of f_n can be? In other words, can we define a quantity d, that is a function of n, such that we can say "For any two peaks of f_n(x), whose location is denoted by x_1 and x_2, we always have |x_1-x_2| >= d(n)"? If so, what that d(n) is in terms of n? Your help would be greatly appreciated. Golabi
From: George Jefferson on 3 Jul 2010 05:27 "Golabi Doon" <golabidoon(a)gmail.com> wrote in message news:279cf805-79d9-4ebb-9f0a-b23b7d0628c2(a)z15g2000prn.googlegroups.com... > Hello mathematicians, > > Consider a function f_n(x) defined on [-pi,pi] that is represented > with n frequency components: > f_n(x)=sum_{k=1}^n a_k cos(kx) + b_k sin(kx) > > The set {a_k,b_k} for k=1,...n is given. > > The function f_n may have multiple peaks (I think at most 2n). By a > peak I mean a point that is a local minimum or local maximum. > > Now, does the fact that the basis functions have bounded frequency > (due to finite n) imply a bound on how close two peaks of f_n can be? > > In other words, can we define a quantity d, that is a function of n, > such that we can say "For any two peaks of f_n(x), whose location is > denoted by x_1 and x_2, we always have |x_1-x_2| >= d(n)"? > > If so, what that d(n) is in terms of n? > > Your help would be greatly appreciated. > If f is differentiable and bounded then yes. This is because the extrema should alternate in direction which creates a high frequency component. The closer the points the higher the frequency(just think of trying to map a sine wave to them). e.g., just look at the extrema. They may have different heights but we can map a modulated sine wave to them. As they get closer we would have to use a higher frequency. As long as the extrema alternate in direction this must be true. For any bounded differentiable function of the type given they must alternate(if it is not constant). Essentially the number of maxima must be +-1 the number of minima. d(n) = 1/n if x_1 and x_2 are adjacent and opposite direction. You have to realize that a bounded differentiable function is essentially a type of "jumbled" up sine wave. The extrema are not equally spaced and the same height. Effectively any differentiable(and some non-differentiable) function can be represented as A(t)*sin(w(t)*t). i.e., a modulated nonlinear frequency sinusoid. The amplitude doesn't come into play in the frequency. w(t) is the frequency and we can simply think of it as changing with time. When the function is slowly "moving" we would expect w(t) to be small. We can approximate w(t) by looking at the extrema of the function. Remember, this is basically like a distorted sine wave. To get the frequency of a sine wave we do what? Apply the same logic to sin(w(t)*t). Note that the highest frequency will depend on the closest extrema. If the extrema are far apart then it will be low. The main thing is to realize that it is independent of A(t)(for the differentiable case). Take sin(w(t)*t) and 1000*sin(w(t)*t). If w(t) is constant then we just have the standard sinusoids of the same frequency even though their amplitude is different. Once you get it in your head that such functions(bounded and differentiable) are "Frequency defined" only in the sense of their extrema(it matters not what happens in between). Two functions with exactly the same extrema will have the same "frequency" even though they may be drastically different. Square waves, Triangle Waves, sinusoids, etc can all have the same frequency and it is because their extrema all have the same relative distance(as you have defined d(n) to be the smallest period). The Fourier transform adds new "frequencies" because we are trying to represent a function in terms of sinusoids when it can't do it. So the Fourier transform needs extra, usually an infinite number, of frequencies to compensate for some functions not "fitting the mold". Take a square wave. It is a rather simple function but has an infinite number of Fourier components. This is because the Fourier series/transform does not really "understand" it. You can compare a sinusoid of the same frequency and it has one component but the Fourier transform gives an infinite spectrum. Yet they feel very similar to us. Of course this is because of it's discontinuities but it isn't the only reason. A(t)*sin(w(t)*t) can represent a square wave, sinusoid, triangle wave, saw tooth, etc... all by simply varying A(t) and setting w(t) constant. The real problem is that A(t)*sin(w(t)*t) doesn't form a basis and hence there is no straightforward and unambiguous way to "solve" for A(t) and w(t). Even though the Fourier transform/series generally obfuscates the problem it always works. Look up Emperical Mode Decomposition to learn more about how to decompose functions into those modulated non-linear frequency sinusoids. Essentially the methods I've seen define frequency as you are trying to understand it(by finding the evenlopes at different levels and using their extrema to define frequency) but in a continuous way(using the hilbert transform). If anyone ever finds a strong theoretical foundation for it then it will replace fourier theory overnight(as it offers much much more because it can work on nonlinear and non-stationary data while fourier theory only works on linear and stationary).
From: Golabi Doon on 3 Jul 2010 07:42 Thank you a lot George for your thorough respnse. Here are some comments. On Jul 3, 4:27 am, "George Jefferson" <phreon...(a)gmail.com> wrote: > d(n) = 1/n I think here you meant d(n)=pi/n. Right? I get your explanation about alternation of peaks and locally mapping a sine wave to the alternation region. It makes a lot of sense intuitively. However, it would be great if a bit more rigorous reasoning could be provided. Basically, I don't know how to transform the intuition that I originally had, even with aid of the intuition that you added, into a mathematical proof. Does any one have any idea about a rigorous proof? Regards Golabi > > if x_1 and x_2 are adjacent and opposite direction. > > You have to realize that a bounded differentiable function is essentially a > type of "jumbled" up sine wave. The extrema are not equally spaced and the > same height. Effectively any differentiable(and some non-differentiable) > function can be represented as A(t)*sin(w(t)*t). > > i.e., a modulated nonlinear frequency sinusoid. > > The amplitude doesn't come into play in the frequency. w(t) is the frequency > and we can simply think of it as changing with time. When the function is > slowly "moving" we would expect w(t) to be small. > > We can approximate w(t) by looking at the extrema of the function. Remember, > this is basically like a distorted sine wave. To get the frequency of a sine > wave we do what? Apply the same logic to sin(w(t)*t). Note that the highest > frequency will depend on the closest extrema. If the extrema are far apart > then it will be low. > > The main thing is to realize that it is independent of A(t)(for the > differentiable case). > > Take sin(w(t)*t) and 1000*sin(w(t)*t). If w(t) is constant then we just have > the standard sinusoids of the same frequency even though their amplitude is > different. > > Once you get it in your head that such functions(bounded and differentiable) > are "Frequency defined" only in the sense of their extrema(it matters not > what happens in between). Two functions with exactly the same extrema will > have the same "frequency" even though they may be drastically different. > Square waves, Triangle Waves, sinusoids, etc can all have the same frequency > and it is because their extrema all have the same relative distance(as you > have defined d(n) to be the smallest period). > > The Fourier transform adds new "frequencies" because we are trying to > represent a function in terms of sinusoids when it can't do it. So the > Fourier transform needs extra, usually an infinite number, of frequencies to > compensate for some functions not "fitting the mold". > > Take a square wave. It is a rather simple function but has an infinite > number of Fourier components. This is because the Fourier series/transform > does not really "understand" it. You can compare a sinusoid of the same > frequency and it has one component but the Fourier transform gives an > infinite spectrum. Yet they feel very similar to us. Of course this is > because of it's discontinuities but it isn't the only reason. > > A(t)*sin(w(t)*t) can represent a square wave, sinusoid, triangle wave, saw > tooth, etc... all by simply varying A(t) and setting w(t) constant. > > The real problem is that A(t)*sin(w(t)*t) doesn't form a basis and hence > there is no straightforward and unambiguous way to "solve" for A(t) and > w(t). > > Even though the Fourier transform/series generally obfuscates the problem it > always works. > > Look up Emperical Mode Decomposition to learn more about how to decompose > functions into those modulated non-linear frequency sinusoids. Essentially > the methods I've seen define frequency as you are trying to understand it(by > finding the evenlopes at different levels and using their extrema to define > frequency) but in a continuous way(using the hilbert transform). > > If anyone ever finds a strong theoretical foundation for it then it will > replace fourier theory overnight(as it offers much much more because it can > work on nonlinear and non-stationary data while fourier theory only works on > linear and stationary).- Hide quoted text - > > - Show quoted text -
From: Stephen Montgomery-Smith on 3 Jul 2010 11:01 Golabi Doon wrote: > Hello mathematicians, > > Consider a function f_n(x) defined on [-pi,pi] that is represented > with n frequency components: > f_n(x)=sum_{k=1}^n a_k cos(kx) + b_k sin(kx) > > The set {a_k,b_k} for k=1,...n is given. > > The function f_n may have multiple peaks (I think at most 2n). By a > peak I mean a point that is a local minimum or local maximum. > > Now, does the fact that the basis functions have bounded frequency > (due to finite n) imply a bound on how close two peaks of f_n can be? > > In other words, can we define a quantity d, that is a function of n, > such that we can say "For any two peaks of f_n(x), whose location is > denoted by x_1 and x_2, we always have |x_1-x_2|>= d(n)"? Think about sin(3x) + 8.9999 sin(x). It has two peaks very close to pi/2. If you replace 8.9999 be a number even closer to 9, you can make the peaks arbitrarily close to each other.
From: Rob Johnson on 3 Jul 2010 11:06
In article <279cf805-79d9-4ebb-9f0a-b23b7d0628c2(a)z15g2000prn.googlegroups.com>, Golabi Doon <golabidoon(a)gmail.com> wrote: >Consider a function f_n(x) defined on [-pi,pi] that is represented >with n frequency components: >f_n(x)=sum_{k=1}^n a_k cos(kx) + b_k sin(kx) > >The set {a_k,b_k} for k=1,...n is given. > >The function f_n may have multiple peaks (I think at most 2n). By a >peak I mean a point that is a local minimum or local maximum. > >Now, does the fact that the basis functions have bounded frequency >(due to finite n) imply a bound on how close two peaks of f_n can be? > >In other words, can we define a quantity d, that is a function of n, >such that we can say "For any two peaks of f_n(x), whose location is >denoted by x_1 and x_2, we always have |x_1-x_2| >= d(n)"? > >If so, what that d(n) is in terms of n? Consider the function for n = 2 f(x) = sin(2x) - 2(1-t) sin(x) Take the derivative of f f'(x) = 2 cos(2x) - 2(1-t) cos(x) = 4 cos^2(x) - 2(1-t) cos(x) - 2 Solving this equation for f'(x) = 0, we get 1-t + 3 sqrt(1 - 2t/9 + t^2/9) cos(x) = ------------------------------ 4 t + 3(1-sqrt(1 - 2t/9 + t^2/9)) = 1 - ------------------------------- 4 ~ 1 - t/3 Therefore, f'(x) = 0 when x^2 ~ 2t/3. By making t small, we can bring the two local extrema of f(x) near x = 0 as close as we want. Thus, there is no lower bound, based on n, for the distance between two local extrema of such a function. Rob Johnson <rob(a)trash.whim.org> take out the trash before replying to view any ASCII art, display article in a monospaced font |