Quick fundamental question about fourier analysis

[Plecto]

Hi. The general theory states that "any periodic signal can be synthesized with a sum of harmonically related sinusoids". What if I have a signal f(x)=cos(2*pi*110x)+cos(2*pi*210x)? Clearly this signals contains the frequencies 110hz and 210hz, but they are not harmonically related? Is the fundamental frequency much much lower? What am I missing?

 

[Deleted member 4993]

Hi. The general theory states that "any periodic signal can be synthesized with a sum of harmonically related sinusoids". What if I have a signal f(x)=cos(2*pi*110x)+cos(2*pi*210x)? Clearly this signals contains the frequencies 110hz and 210hz, but they are not harmonically related? Is the fundamental frequency much much lower? What am I missing?

GCF of 110 and 210 is 10

 

[Plecto]

So the fundamental frequency in my example is 10hz, but it's amplitude is zero thus not making any contribution? How do I find the fundamental frequency? I'm trying to do an assignment where I'm asked to find f₀ and all the complex amplitudes with the use of Euler's relation, I'm not supposed to need to use integration. Here's the original signal:

10+20cos(2*pi*100t+1/4*pi)+10*cos(2*pi*250t). I could convert them to complex exponentials, but I don't see how that would help me

 

[Ishuda]

Hi. The general theory states that "any periodic signal can be synthesized with a sum of harmonically related sinusoids". What if I have a signal f(x)=cos(2*pi*110x)+cos(2*pi*210x)? Clearly this signals contains the frequencies 110hz and 210hz, but they are not harmonically related? Is the fundamental frequency much much lower? What am I missing?

The arguments of the cosines and sines of a Fourier series are n (a x) where a is related to the periodicity of the function and n is the n^th harmonic. So, find an n₁ and n₂ such that
a = \(\displaystyle \dfrac{220\, \pi}{n_1}\, = \, \dfrac{420\, \pi}{n_2}\)
or
\(\displaystyle 220\, \pi n_2\, = \, 420\, \pi\, n_1\)
or
\(\displaystyle 11\, n_2\, = \, 21\, n_1\)
and we have n₁ is a multiple of 11 and n₂ is a multiple of 21 and we have
f(x) = \(\displaystyle cos(11\, (20 \pi x))+cos(21\, (20 \pi x))\)

Since a is sometimes written as \(\displaystyle \frac{2\pi}{L}\) where L is the length of the periodicity, you can see why Subhotosh Khan said "GCF of 110 and 210 is 10", i.e. if L is \(\displaystyle \frac{1}{10}\) then
f(x) = \(\displaystyle cos(11\, (\frac{2 \pi x}{L}))\, +\, cos(21\, (\frac{2 \pi x}{L}))\)

EDIT: Also note that these are the 11th and 21st harmonics for the f(x) given above. Had L been equal to 1, it would be the 110th and 210th harmonics. Thus the length of the periodicity is undetermined in the problem.