# What is the phase for the two poles shown?

#### u-will-neva-no

Joined Mar 22, 2011
230
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

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

#### t_n_k

Joined Mar 6, 2009
5,455
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.