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I would just want to prove $\cos(t) + \cos(\pi t)$ is non periodic.

I don't know where to start it. Also I know that individually these signals ie $\cos(t)$ and $\cos(\pi t)$ are periodic with frequencies $1/(2 \pi)$ Hz and $1/2$ Hz respectively. But since the superpositon of these signals is not periodic should we represent them using Fourier transform only? What is the harm in drawing the frequency components at $1/(2 \pi)$ Hz and $1/2$ Hz using fourier series?

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    $\begingroup$ Fourier series are for signals that are periodic and your signal is not periodic, as Aaron's answer points out. I don't know what "drawing" the frequency components means, but it is perfectly fine to write the signal as $\cos(t) + \cos(\pi t)$ which is not a Fourier series, and to "draw" the signal as having impulses (or vertical lines) at $f = \pm (1/2\pi)$ and $f = \pm (1/2)$ if that is what you are looking for. $\endgroup$ – Dilip Sarwate Jan 1 '14 at 7:13
  • $\begingroup$ The answer is as usual 42, sorry, no, 44. Try it out, use your favorite plotting program to plot $f(x)$ against $f(x+44)$ ;-) (This is an extension of the famous $\sin(22)$ test, resp. the pi approximation of Archimedes.) $\endgroup$ – Dr. Lutz Lehmann Jan 1 '14 at 12:04
  • $\begingroup$ @LutzL : Are you speaking about periodicity. The signal is not periodic. $\endgroup$ – dexterdev Jan 1 '14 at 14:48
  • $\begingroup$ Of course not. But the difference $f(x+44)-f(x)$ has size $2\sin(22-7\pi)<0.018$, which is mostly invisible in function plots. Even smaller is the difference in $f(x+710)-f(x)$, bounded by $2\sin(355-113\pi)<0.0000603$. $\endgroup$ – Dr. Lutz Lehmann Jan 1 '14 at 15:07
  • $\begingroup$ @LutzL : Actually I was trying to look the theoretical aspects and I trusted software called 'GRAPH' for that matter in plotting. Now that is lesson for me. $\endgroup$ – dexterdev Jan 2 '14 at 5:41
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$cos(t)$ and $cos(\pi t)$ are both periodic with periods $2\pi$ and $2$ respectively. To find the period of the sum we need to find an integers $n,m$ such that $\pi n = 2 m$ or $\pi / 2 = m /n $, which is not possible since $\pi$ is irrational.

I think the Fourier series uses specific frequencies to form the signal. You can't necessarily put a frequency component at irrational frequencies.

Fourier series work fine with irrational frequencies too, but the frequencies are all harmonics, that is, integer multiples, of the fundamental frequency.

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    $\begingroup$ The last sentence is incorrect. Fourier series work just fine with irrational frequencies, but the frequencies are all harmonics, that is, integer multiples, of the fundamental frequency $f_0$ (or $\omega_0$ depending on which flavor of Fourier series one prefers). But notice that since $\omega_0 = 2\pi f_0$ and $\pi$ is irrational, if one set of folks are using a rational fundamental frequency, the other set of folks are using irrational fundamental frequency! $\endgroup$ – Dilip Sarwate Jan 1 '14 at 7:08
  • $\begingroup$ @Aaron : I have edited your answer regarding the point which Dilip Sarwate mentioned. $\endgroup$ – dexterdev Jan 1 '14 at 8:23

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