- #1
neelakash
- 491
- 1
I want to solve this 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?
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?
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