I am computing the Fourier series expansion of the given signal $x(t)$. In that I am having difficulties in calculating range of function $ \frac{1}{n\pi}(1-\cos(n\pi))\sin(\frac{n\pi}{2})$ (or) I am not able to get how can I arrive at equation (2) from equation (1)

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I have done some back calculation (and would like to know the forward way of calculating it).

For example, for the function to have a value of $ \frac{2n}{\pi} $

$$ \sin(\frac{n\pi}{2})=1 \,\, \textrm{and} \,\, \cos(n\pi)=-1$$

$$=> \sin(\frac{n\pi}{2})=\sin((4k+1)\frac{\pi}{2}) \, \textrm{and} \, \cos(n\pi)=\cos(2k+1)\pi$$ $$=>n=(4k+1) \, \textrm{and} \, n =(2k+1)$$

Which is a contradiction and doesn't match with the book. (Is this the case that among $=>n=(4k+1)$, and $n =(2k+1)$, we have to choose $n=(4k+1)$?)

And how have we arrived at 2nd part of equation (2)?

Please help in this regard.

Also clarify another doubt. How have we arrived at equation (3) from (2). It is confusing because it involves determination of the terms based on n which in turn is dependent on $k$.

  • $\begingroup$ Have you actually tried to evaluate expression (1) for $n=0,1,2,3,\ldots$? Sit down and try it and I'm pretty confident that you'll see how to arrive at (2). There' absolutely no magic involved. As for Eq. (3), from (2) you see that the coefficients are zero for even $n$, and for odd $n$ they just alternate in sign; you will see this after having followed my first advice. $\endgroup$
    – Matt L.
    Feb 19 '17 at 9:52
  • $\begingroup$ @MattL. I have done as you said and readily obtained eq (3) from eq (1) $$x_n={...,\frac{-2}{11\pi},0,\frac{2}{9\pi},0,\frac{-2}{7\pi},0,\frac{2}{5\pi},0,\frac{-2}{3\pi},0,\frac{2}{\pi},0(origin),\frac{2}{\pi},0,\frac{-2}{3\pi},0,\frac{2}{5\pi},0,\frac{-2}{7\pi},0,\frac{2}{9\pi},0},\frac{-2}{11\pi},....$$ Thank you. $\endgroup$
    – Soumee
    Feb 21 '17 at 8:56
  • $\begingroup$ I find obtaining eq(1) from (2) is easier than (2) from (1) $$=\frac{1}{n\pi}(1-cos(4k+1)\pi)sin((4k+1)\frac{\pi}{2})$$ $$=\frac{1}{n\pi}(1-cos(4k\pi+\pi))sin((\frac{4k\pi}{2}+\frac{\pi}{2}))$$ $$=\frac{1}{n\pi}(1+cos(4k\pi))cos(\frac{4k\pi}{2})$$ $$=\frac{1}{n\pi}(2)(1)=\frac{2}{n\pi}$$ and $$=\frac{1}{n\pi}(1-cos(4k+3)\pi)sin((4k+3)\frac{\pi}{2})$$ $$=\frac{1}{n\pi}(1-cos(4k\pi+3\pi))sin((2k\pi+\frac{3\pi}{2}))$$ $$=\frac{1}{n\pi}(1+cos(3\pi))cos(\frac{3\pi}{2})$$ $$=\frac{1}{n\pi}(2)(-1)=-\frac{2}{n\pi}$$ $\endgroup$
    – Soumee
    Feb 21 '17 at 8:57

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