What is the phase for the two poles shown?

Thread Starter

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
5,447
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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