I want to solve this MCQ by observation: Consider a step function: F(t)=1 for 0<t< \(\frac{\pi}{\omega}\) F(t)=-1 for -\(\frac{\pi}{\omega}\)<t<0 and the same pattern is repeated over time. What should be the Fourier Series: (A) \(\frac{4}{\pi}\) \(\sum^\infty_1\frac{sin(n\omega\ t )}{n}\) (B) \(\frac{4}{\pi}\) \(\sum^\infty_0\frac{sin(2n+1)\omega\ t }{2n+1}\) It is easy to check that the actual answer is (B);but I want to see if this can be seen by observation only...(A) and (B) differe in that (A) contains the even terms as well where (B) lacks them... Can anyone say something if it could be done by using observation?
Well your function is symmetric around \(t = \frac{ \pi }{ 2 \omega }\). You can use this to rule out (A) since the even sine terms are antisymmetric around this axis.
OK,realize that the given function is symmetric about \(\ t=\)\(\frac{\pi} {2\omega}\) Then,try to sketch the graphs of even terms of sine (like n=2,n=4...) adjusting the scale on the t axis.(A) will show zeros at \(\frac{\pi}{2\omega}\)
What I meant by a function being "antisymmetric" about a point t[sub]0[/sub] was the property: \(f( t - t_0 ) = - f( -t - t_0 ), \; \forall t \in \mathbb{R}\) In the case of the even sines we have: \(\sin \left[ n \omega \left( t - \frac{\pi}{2 \omega} \right) \right] = - \sin \left[ n \omega \left( -t - \frac{\pi}{2 \omega} \right) \right]\) if \(n \in 2 \mathbb{Z}\) In general, if a \(\frac{2 \pi}{\omega}\)-periodic function F(t) is symmetric about a point t[sub]0[/sub], and a \(\cos(n \omega t)\) or \(\sin(n \omega t)\) is antisymmetric about that point (or vice versa), then the corresponding Fourier coefficient (a[sub]n[/sub] or b[sub]n[/sub]) is zero. The fact that the Fourier series of even functions only contain cosines, and the series of odd functions only contain sines is a special case of this (with t[sub]0[/sub] = 0).
