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ZTransform and DTFT 
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#1
Apr412, 05:05 AM

P: 200

1. The problem statement, all variables and given/known data
You are given the following pieces of information about a real, stable, discretetime 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)=\pmX(\omega[/tex]. a) x is a finitelength 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? 


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