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Z-Transform and DTFT

  1. Apr 4, 2012 #1

    zcd

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    1. The problem statement, all variables and given/known data[/b]
    You are given the following pieces of information about a real, stable, discrete-time signal x and its DTFT X, which can be written in the form [tex]X(\omega)=A(\omega)e^{i\theta_x(\omega)}[/tex] where [tex]A(\omega)=\pm|X(\omega|[/tex].
    a) x is a finite-length signal
    b) [tex]\hat{X}[/tex] has exactly two poles at z=0 and no zeros at z=0.
    c) [tex]\theta_x(\omega)=\begin{cases} \frac{\omega}{2}+\frac{\pi}{2} & 0<\omega<\pi \\
    \frac{\omega}{2}-\frac{\pi}{2} & -\pi<\omega<0\end{cases}[/tex]
    d) [tex]X(\omega)\Big|_{\omega=\pi}=2[/tex]
    e) [tex]\int_{-\pi}^\pi e^{2i\omega}\frac{d}{d\omega}X(\omega)d\omega =4\pi i[/tex]
    f) The sequence v whose DTFT is V (ω) = Re (X(ω)) satisfies v(2) = 3/2.

    2. Relevant equations
    [tex]\hat{H}(z)=\sum_{n=-\infty}^\infty h(n)z^{-n}[/tex]
    [tex]H(\omega)=\hat{H}(z)\Big|_{z=e^{i\omega}}[/tex]


    3. The attempt at a solution
    By parts e) and f) I've figured out that x(-2)=4 and x(2)=-1. With that and part a) and b), I've realized that the rightmost endpoint of x is x(2)=-1 and the leftmost endpoint of x is less than n=-2 (the transfer function can't converge at infinity since the system cannot be causal). Part d) gives me [tex]\sum_{n=-\infty}^\infty (-1)^n x(n) = 2[/tex], but I'm unsure how to use it at the moment. I do not know how to use part c) at all. Can someone help shed some light on the problem?
     
  2. jcsd
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