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From: Bret Cahill on 6 Jun 2010 11:23 > > Several functions in an equation have ac and dc components. If all > > the dc is filtered out of the functions for ac only and then all the > > ac out of the dc, will the equation hold for just the dc and just the > > ac? > > > The equation doesn't do anything more than add, multiply and divide > > the functions. > > > Bret Cahill > > Consider > > sin t + cos t = (\sqrt 2) sin (t + Pi / 4) > > as a test case. Supposing the functions were all always in phase, i.e., all either sin or all cos in the example above? Bret Cahill
From: Bret Cahill on 6 Jun 2010 11:38 > > Several functions in an equation have ac and dc components. If all > > the dc is filtered out of the functions for ac only and then all the > > ac out of the dc, will the equation hold for just the dc and just > > the ac? > > > The equation doesn't do anything more than add, multiply and divide > > the functions. > > No. Almost any nontrivial set of functions and equations would > provide a counterexample. > > For almost the simplest possible nontrivial example, you could try > f(t) = sin(t) + 1, g(t) = f(t) * f(t). I mis stated the problem. Both side of an equation have ac and dc terms. Will all the ac terms on one side equal all the terms on the other side, and all dc on one side = all dc on the other? Bret Cahill
From: hagman on 6 Jun 2010 12:44 On 6 Jun., 17:38, Bret Cahill <BretCah...(a)peoplepc.com> wrote: > > > Several functions in an equation have ac and dc components. If all > > > the dc is filtered out of the functions for ac only and then all the > > > ac out of the dc, will the equation hold for just the dc and just > > > the ac? > > > > The equation doesn't do anything more than add, multiply and divide > > > the functions. > > > No. Almost any nontrivial set of functions and equations would > > provide a counterexample. > > > For almost the simplest possible nontrivial example, you could try > > f(t) = sin(t) + 1, g(t) = f(t) * f(t). > > I mis stated the problem. > > Both side of an equation have ac and dc terms. > > Will all the ac terms on one side equal all the terms on the other > side, and all dc on one side = all dc on the other? > > Bret Cahill Of course, if g = f*f, then AC(g) = AC(f*f) and DC(g) = DC(f*f) and in general foobar(g) = foobar(f*f), however we do not necessarily have AC(g) = AC(f)*AC(f) or DC(g) = DC(f)*DC(f)
From: Bret Cahill on 6 Jun 2010 14:06 > > > > Several functions in an equation have ac and dc components. If all > > > > the dc is filtered out of the functions for ac only and then all the > > > > ac out of the dc, will the equation hold for just the dc and just > > > > the ac? > > > > > The equation doesn't do anything more than add, multiply and divide > > > > the functions. > > > > No. Almost any nontrivial set of functions and equations would > > > provide a counterexample. > > > > For almost the simplest possible nontrivial example, you could try > > > f(t) = sin(t) + 1, g(t) = f(t) * f(t). > > > I mis stated the problem. > > > Both side of an equation have ac and dc terms. > > > Will all the ac terms on one side equal all the terms on the other > > side, and all dc on one side = all dc on the other? > > > Bret Cahill > > Of course, if g = f*f, then AC(g) = AC(f*f) and DC(g) = DC(f*f) and in > general > foobar(g) = foobar(f*f) Singling out specific frequencies should work as well. Bret Cahill
From: mecej4 on 7 Jun 2010 18:26
Bret Cahill wrote: >> > Several functions in an equation have ac and dc components. If all >> > the dc is filtered out of the functions for ac only and then all the >> > ac out of the dc, will the equation hold for just the dc and just the >> > ac? >> >> > The equation doesn't do anything more than add, multiply and divide >> > the functions. >> >> > Bret Cahill >> >> Consider >> >> sin t + cos t = (\sqrt 2) sin (t + Pi / 4) >> >> as a test case. > > Supposing the functions were all always in phase, i.e., all either sin > or all cos in the example above? Please define 'in phase' for functions with different frequencies. For example, are sin (t) and sin (2t) 'in phase' ? > > > Bret Cahill |