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From: Clay on 3 May 2010 18:25 On May 3, 5:20 pm, "fisico32" <marcoscipioni1(a)n_o_s_p_a_m.gmail.com> wrote: > hello forum, > > Two sinusoids of different frequency w1 and w2 are orthogonal if integral > of their product (one time the complex conj of the other) gives zero. > That is only true if the integral of integration is T, and and the period > of w1 and the period of w2 fit an integer number of times in T... > > That means that two sinusoids are orthogonal only if their frequencies are > different and one an integer multiple of the other.... > Correct? > > In all other cases where the two frequencies are different and not multiple > of each other the integral is nonzero....or is there some interval of > integration over which they are actually orthogonal? > > For instance, > w1=(4pi) rad/s > w2=(3.2pi) rad/s > > y(t)=sin(w1*t+phi) > x(t)=sin(w2*t+theta) > Is the integral zero for certain limits of integration? How do I find the > limits? Clearly the phases theta and phi matter.... > > thanks > fisico32 The phases don't matter. Basically for orthogonality find the ratio of w1 to w2. If it is irrational you need infinite limits. Otherwise express as ratio of integers in the form of a/b. Then for an arbitrary lower limit of "p", your upper limit of "q" is simply q=p+n*a*2*pi/w1 where n is any integer. you may also write q=p+m*b*2*pi/w2 where m is any integer For your example w1=4pi and w2=3.2pi thus a/b=1.25=5/4 q=p+n*5*(0.5)=p+2.5*n IHTH, Clay
From: Tim Wescott on 3 May 2010 18:58 fisico32 wrote: > hello forum, > > Two sinusoids of different frequency w1 and w2 are orthogonal if integral > of their product (one time the complex conj of the other) gives zero. The definition I have seen uses the _average_ of the integral of their product. This makes no difference for a finite interval -- more on that later. > That is only true if the integral of integration is T, and and the period > of w1 and the period of w2 fit an integer number of times in T... > > That means that two sinusoids are orthogonal only if their frequencies are > different and one an integer multiple of the other.... > Correct? Incorrect, even by your definition of orthogonality. > In all other cases where the two frequencies are different and not multiple > of each other the integral is nonzero....or is there some interval of > integration over which they are actually orthogonal? > > For instance, > w1=(4pi) rad/s > w2=(3.2pi) rad/s > > y(t)=sin(w1*t+phi) > x(t)=sin(w2*t+theta) > Is the integral zero for certain limits of integration? How do I find the > limits? Clearly the phases theta and phi matter.... Take the integral of the _average_ of your two sinusoids over the interval -T/2 to +T/2, and take the limit of the average as T goes to infinity. What conditions must be met for the two sinusoids to _not_ be orthogonal? -- Tim Wescott Control system and signal processing consulting www.wescottdesign.com
From: glen herrmannsfeldt on 3 May 2010 19:07 fisico32 <marcoscipioni1(a)n_o_s_p_a_m.gmail.com> wrote: > Two sinusoids of different frequency w1 and w2 are orthogonal if integral > of their product (one time the complex conj of the other) gives zero. > That is only true if the integral of integration is T, and and the period > of w1 and the period of w2 fit an integer number of times in T... Look at the Wikipedia pages for Orthogonal_functions and Inner_product. > That means that two sinusoids are orthogonal only if their > frequencies are different and one an integer multiple of the other.... > Correct? Well, you can generalize a little more than that. You can have finite or infinite limits on the integral. Also, some inner products have a weight function in addition to the two functions contributing to the product. With infinite limits, sinusoids are orthogonal when the frequency is different. With finite limits, it is usual that only sinusoids with an integer number of periods over the range of integration are used. (Depending on boundary conditions, sometimes a half integer also works.) Another possibility is a Gaussian weighting function of a specified width. The integration limits would be infinite, though the main contribution comes from a finite width region. -- glen
From: glen herrmannsfeldt on 3 May 2010 22:10 Clay <clay(a)claysturner.com> wrote: > On May 3, 5:20?pm, "fisico32" wrote: >> Two sinusoids of different frequency w1 and w2 are orthogonal if integral >> of their product (one time the complex conj of the other) gives zero. >> That is only true if the integral of integration is T, and and the period >> of w1 and the period of w2 fit an integer number of times in T... (snip) > Basically for orthogonality find the ratio of w1 to w2. If it is > irrational you need infinite limits. Otherwise express as ratio of > integers in the form of a/b. > Then for an arbitrary lower limit of "p", your upper limit of "q" is > simply > q=p+n*a*2*pi/w1 where n is any integer. I suppose it could work that way, but it is usual for the limits to come first, along with the boundary conditions and any weighting function, and then find which basis functions satisfy the boundary conditions. Two common boundary conditions are for the function or its derivative to go to zero at the boundary, which allows for a whole number of periods over the region of interest. A little less common, is to have the function go to zero at one end, and the derivative at the other, which defines the modes for a cylindrical tube closed at one end, such that an odd number of half periods of sine or cosine fit. (And not counting end effects for real tubes.) Even more rare are mixed boundary conditions, allowing for some linear combination of the function and its derivative. And all that with the assumption that the basis functions are sinusoids. For a circular drum head with uniform tension the radial basis functions are Bessel functions, which have their own orthogonality relationship. -- glen
From: Clay on 4 May 2010 10:07
On May 3, 10:10 pm, glen herrmannsfeldt <g...(a)ugcs.caltech.edu> wrote: > Clay <c...(a)claysturner.com> wrote: > > On May 3, 5:20?pm, "fisico32" wrote: > >> Two sinusoids of different frequency w1 and w2 are orthogonal if integral > >> of their product (one time the complex conj of the other) gives zero. > >> That is only true if the integral of integration is T, and and the period > >> of w1 and the period of w2 fit an integer number of times in T... > > (snip) > > > Basically for orthogonality find the ratio of w1 to w2. If it is > > irrational you need infinite limits. Otherwise express as ratio of > > integers in the form of a/b. > > Then for an arbitrary lower limit of "p", your upper limit of "q" is > > simply > > q=p+n*a*2*pi/w1 where n is any integer. > > I suppose it could work that way, but it is usual for the limits > to come first, along with the boundary conditions and any > weighting function, and then find which basis functions satisfy > the boundary conditions. > Glen, I constrained my response to the OP's particular example. Since the OP was having trouble with it, I didn't even want to venture into weight functions, Hilbert spaces, and Sturm-Liouville systems. 'cause we know one can delve deeply into orthogonality and its various generalizations. Clay .. > > -- glen |