Does f(t)=1 Have a Fourier Series Expansion?

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naven8
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Does f(t)=1 have Fourier series expansion or not?
 
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Every (reasonably behaving) function has a Fourier series. Some are just more boring than others.
 


Fourier series is defined only for periodic signals.
f(t)=1 is not a periodic signal-->no Fourier series expansion??
 


Depends on how you define "periodic" (if you want, it is periodic with any period P).

Without arguing about definitions and semantics, you can note that indeed you can write f(t) as
[tex]f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \, [a_n \cos(nx) + b_n \sin(nx)][/tex]
(copied from Wikipedia), where [tex]a_n = b_n = 0[/tex] except for [itex]a_0 = 2[/itex].
 
therimalaya said:
May be simple, but I'm getting problem with doing Fourier series expansion of Sin(x) for -pi[tex]\leq[/tex]x[tex]\leq[/tex]pi
The point is that sin(x)= (1)sin(x)+ 0 cos(x)+ (0)sin(2x)+ (0)cos(2x)+ ... is a perfectly good Fourier series!

naven8 said:
Does f(t)=1 have Fourier series expansion or not?
The function defined as f(t)= 1 for [itex]-\pi< x\le \pi[/itex] and continued periodically has a Fourier series expansion.