# Homework Help: WHAT IS THIS Phase Spectrum of this function.

1. Jan 5, 2012

### jose_peeter

dear all,

after doing a complex fourier series on a function. i am asked to find and graph the magnitude and phase spectrum of a fucntion.

now to cut the long story short. let's say we arrived a fourier series like this.

= $\sum_{i=-∞}^{∞}$ $\frac{1}{npi}$sin $\frac{npi}{2}$exp(jnpit)

so $c_{n}$ = $\frac{1}{npi}$sin $\frac{npi}{2}$

= the phase spectrum of a function is the phase of $c_{n}$, but notice that this is PURELY REAL like this.

=$\frac{1}{npi}$sin $\frac{npi}{2}$ + j *(0)

= phase is found as $tan^{-1}$ ( $\frac{0}{$\frac{1}{npi}$sin $\frac{npi}{2}$}$ )

= now is it not supposed to be zero for all n. but this function has a phase plot

= please explain to me how they did this,

my idea is that when we look at the complex plane.POSITIVE REAL part has angle 0 while NEGATIVE REAL part has 180 OR -180(I AM NOT SURE WHICH).

= please explain to me how $tan^{-1}$ 0/anything is accepted. AND which angle to choose 180 or -180

THANKS A LOTTTTTTTTTTTTTTTTTTTTTTT.....!!!

Last edited: Jan 5, 2012
2. Jan 5, 2012

### LCKurtz

I'm not sure about the term "phase spectrum". If it means frequency spectrum, isn't that just a plot of $|c_n|$ against $\omega_n$? Or is it something else?

3. Jan 10, 2012

### jose_peeter

I mean the graph of the PHASE against the frequency what you have suggested is the magnitude against frequency.

when you do it. please don't just show the graph, i know the answer but don't understand how to get it.
my doubts and questions i have asked in previous post above

thanks