Let |four| be a real periodic sequence with fundamental period |zero| and Fourier coefficients |one|, where ak and bk are both real.
$ tex:{N}_{0} $
.. |one| image:: tex
:alt: tex:{c}_{k}={a}_{k}+j{b}_{k}
Show that |two| and |three|.
.. |two| image:: tex
:alt: tex:{a}_{-k}={a}_{k}
.. |three| image:: tex
:alt: tex:{b}_{-k}=-{b}_{k}
If |four| is real we have (equation for Fourier coefficients):
.. |four| image:: tex
:alt: tex:x[n]
.. image:: tex
:alt: tex:{c}_{-k}=\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{jk{\omega}_{0}n}
and further:
.. image:: tex
:alt: tex:={\left(\frac{1}{{N}_{0}}\sum_{n=0}^{{N}_{0}-1}x[n]{e}^{-jk{\omega}_{0}n} \right)}^{*}={{c}^{*}}_{k}
Therefore:
.. image:: tex
:alt: tex:{c}_{-k}={a}_{-k}+j{b}_{-k}={({a}_{k}+{b}_{k})}^{*}={a}_{k}-j{b}_{k}
So now we can see that:
|five| and |six|
.. |five| image:: tex
:alt: tex:{a}_{-k}={a}_{k}
.. |six| image:: tex
:alt: tex:{b}_{-k}=-{b}_{k}
