Let $x[n]\$ be a real periodic sequence with fundamental period $N_0\$ and Fourier coefficients $c_k = a_k+jb_k\$, where $a_k\$ and $b_k\$ are both real.

Show that $a_{-k} = {a}_{k}\$ and $b_{-k}=-b_k\$.

If $x[n]\$ is real we have (equation for Fourier coefficients):

$c_{-k} = \frac{1}{N_0} \sum_{n=0}^{N_0-1} x[n]{e}^{jk \omega_0 n}$

and further:

$= \left ( \frac{1}{N_0} \sum_{n=0}^{N_0-1} x[n]e^{-jk \omega_0 n} \right ) ^{*} = c^{*}_k$

Therefore:

$c_{-k} = a_{-k} + jb_{-k} =(a_k + b_k)^{*} = a_k - jb_k\$

So now we can see that:

$a_{-k} = a_k\$ and $b_{-k} = -b_k\$

## Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood