Difference between revisions of "Chapter 5 OldKiwi" - Rhea

Revision as of 16:02, 30 March 2008

File:Lecture5 OldKiwi.pdf

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}