Fourier Problem by observation

[neelakash]

I want to solve this by observation:

Consider a step function:

F(t)=1 for 0<t< [tex]\frac{\pi}{\omega}[/tex]

F(t)=-1 for -[tex]\frac{\pi}{\omega}[/tex]<t<0

and the same pattern is repeated over time.

What should be the Fourier Series:

(A) [tex]\frac{4}{\pi}[/tex] [tex]\sum^\infty_1\frac{sin(n\omega\ t )}{n}[/tex]


(B) [tex]\frac{4}{\pi}[/tex] [tex]\sum^\infty_0\frac{sin(2n+1)\omega\ t }{2n+1}[/tex]



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?

 

[neelakash]

OK,realize that the given function is symmetric about
[tex]\ t= [/tex][tex]\frac{\pi}{2\omega}[/tex]

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 [tex]\frac{\pi}{2\omega}[/tex]

 

[tacman]

By observation, we can see that the given step function has a period of 2π/ω. This means that the Fourier series should only contain odd terms, since the even terms would cancel out due to the symmetry of the function. This observation leads us to option (B), which only contains odd terms. Additionally, we can also observe that the amplitude of the function is 1, which is reflected in the coefficient of 4/π in both options. Therefore, by analyzing the period and amplitude of the function, we can deduce that the correct Fourier series is (B).