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Let's take a periodic wave function defined by f(t) = (1 - t) with a period of 1.

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The Fourier coefficients can be calculated: $$ \omega_0 = 2\pi/T=2\pi \\ F(n) = \int_{0}^{1}(1 - t)e^{-jn 2\pi t}dt=-\frac{j}{2\pi n} \\ F(0) = 1/2 $$

There is a property of Fourier series which states:

If f(t) is real and can be developed into a Fourier series: F(n) is purely imaginary and odd if and only if f(t) is odd.

My question is, why is the F(n) expression I get is purely imaginary if my f(t) function isn't odd (nor even in fact).

I might just be blind, I might not see something, in either case I would appreciate you point it to me, but I get that:

  • f(t) is not odd, nor even (\$f(-0.3) = 0.3, f(0.3) = 0.7\$)
  • F(n) is purely imaginary as \$ F(n) = -j/2\pi n \$ (no real part)

So does this example contradicts the stated property of Fourier transforms, or is it due to something I don't see? I expected to get a complex F(n), with both a real and an imaginary part. Might turn out to be a dumb question, oh well...

Thanks

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    \$\begingroup\$ F(0) is real and non-zero so F(n) in neither imaginary nor odd \$\endgroup\$ – Olof Oct 13 '16 at 5:24
  • \$\begingroup\$ It is asserted that F(n) = ... = -j/2pin. My head hurts when trying to do that stuff, so let's assume that's right. The second part is '... and can be developed into a Fourrier series'. Now that integral only defines f(t) in the 0 to 1 interval. If f(t) in other intervals is equal to f(mod(t,1)), as is plotted, then I don't see any development of F(n) for one interval into the full transform. Similarly if f(t) for the interval -1 to 0 is -f(abs(t)), we have our odd function. Is it the development into the full Fourrier series that's the issue, or is the f(t) = -j/2pin wrong? \$\endgroup\$ – Neil_UK Oct 13 '16 at 8:20
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Let me rewrite that statement for you, I believe I know where your misunderstanding lies:

\$F(n) \space \forall \space (n \in \mathbb{Z}) \$ is purely imaginary and odd if and only if f(t) is odd.

In plain English: \$F(n)\$ (for all n in integers including 0) is purely imaginary and odd [...]. In your example, \$F(0)\$ is not imaginary. Therefore the statement remains correct.

Now think of a new waveform that looks exactly the same as your original "sawtooth" function, but that is now odd: \$g(t) = f(t) - F(0) = f(t) - 1/2\$. If you redo the exercise, you will find \$G(0) = 0\$, confirming the statement.

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  • \$\begingroup\$ I kind of didn't consider zero for some strange reason, even though I had this all calculated by myself. Much thanks, small mistake but important one. \$\endgroup\$ – Yannick Oct 13 '16 at 16:29
  • \$\begingroup\$ Hey, quick side question. Say we had an hypothetical function h(t) with the same \$ H(n) = -j/2\pi n\$ and \$H(0) = -j\$, that is, with a H(0) purely imaginary but non zero, would H(n) be considered odd and thus conforming to the statement? Thanks \$\endgroup\$ – Yannick Oct 13 '16 at 16:52
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    \$\begingroup\$ No, for a function to be considered odd, its value at (n = 0) must be 0 (to satisfy -f(x) = f(-x)). If its value at 0 is not specified (like a sawtooth waveform) the average of limits from the right and from the left is considered. Also, H(0) = -j can be thought as a DC component with a (90º) phase shift... which doesn't make sense for real signals. The condition "If f(t) is real" still applies. \$\endgroup\$ – Vicente Cunha Oct 13 '16 at 18:10
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    \$\begingroup\$ Ninja edit: to satisfy -H(n) = H(-n). The explanation was a bit ambiguous using f(x). @Yannick \$\endgroup\$ – Vicente Cunha Oct 13 '16 at 18:17
  • \$\begingroup\$ So basically an odd function always have f(t = 0) = 0, while an even function may have f(t = 0) != 0. Thank you very much. \$\endgroup\$ – Yannick Oct 13 '16 at 21:29

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