# What is the phase for the two poles shown?

Discussion in 'Homework Help' started by u-will-neva-no, Jan 5, 2012.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hi all, let me explain what that title is on about.

My z-transform function is:
$\frac{1}{2j} \left(\frac{z}{z-exp(jw_0T)} - \frac{z}{z-exp(-jw_0T)}\right)$ (1)

Now, the question asks for the phase, where $w_s = \frac{2\pi}{T}$
for $w_s = 3w_0$

So I make the answer to be:
$w_0 = \frac{2\pi}{3T}$ Just by combining the two equations together and then substituting this into (1) gives:

$\frac{1}{2j} \left(\frac{z}{z-exp(\frac{j.2\pi}{3})} - \frac{z}{z-exp(\frac{-j.2\pi}{3})}\right)$

So $\phi = \pm e(\frac{j.2\pi}{3})$ ?

The answer that has been provided has

$\phi = \pm e(\frac{j.\pi}{3})$

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
I would think you would need to make the substitution

$z=exp(j\omega T$

in the transfer function.

This would give the frequency response traversing the unit circle as the angular frequency ω varies.

To find the phase limits one presumably evaluates the function at ω=0 and ωs.

Interestingly at w=w0 your function leads to a case of division by zero.

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