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
    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
    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.
    u-will-neva-no likes this.