Z-Transform & DTFT: Finite-Length Signal & Two Poles

[zcd]

1. Homework Statement [/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 $$X(\omega)=A(\omega)e^{i\theta_x(\omega)}$$ where $$A(\omega)=\pm|X(\omega|$$.
a) x is a finite-length signal
b) $$\hat{X}$$ has exactly two poles at z=0 and no zeros at z=0.
c) $$\theta_x(\omega)=\begin{cases} \frac{\omega}{2}+\frac{\pi}{2} & 0<\omega<\pi \\ \frac{\omega}{2}-\frac{\pi}{2} & -\pi<\omega<0\end{cases}$$
d) $$X(\omega)\Big|_{\omega=\pi}=2$$
e) $$\int_{-\pi}^\pi e^{2i\omega}\frac{d}{d\omega}X(\omega)d\omega =4\pi i$$
f) The sequence v whose DTFT is V (ω) = Re (X(ω)) satisfies v(2) = 3/2.

Homework Equations

$$\hat{H}(z)=\sum_{n=-\infty}^\infty h(n)z^{-n}$$
$$H(\omega)=\hat{H}(z)\Big|_{z=e^{i\omega}}$$

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 $$\sum_{n=-\infty}^\infty (-1)^n x(n) = 2$$, 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?

 

[KataKoniK]

First, let's consider part c). This tells us that the phase of X(\omega) is a piecewise function with two different cases, one for 0<\omega<\pi and one for -\pi<\omega<0. This means that the phase of X(\omega) changes at \omega=0 and \omega=\pi. This could indicate a discontinuity in the signal x, since the phase of a signal can only change at points where the signal itself changes.

Next, let's look at part d). This tells us that X(\omega) is equal to 2 when \omega=\pi. This could indicate a sinusoidal component in the signal x, since the DTFT of a sinusoidal signal is a delta function at the frequency of the sinusoid.

Combining parts c) and d), we can make some assumptions about the signal x. Since the phase of X(\omega) changes at \omega=0 and \omega=\pi, this could indicate that the signal x is a combination of two sinusoids with frequencies \omega=\pi and \omega=0. Additionally, since X(\omega) is equal to 2 when \omega=\pi, this could indicate that the amplitude of the \omega=\pi sinusoid is 2. Therefore, we can say that the signal x is a combination of two sinusoids, one with frequency \omega=\pi and amplitude 2, and one with frequency \omega=0 and unknown amplitude.

Next, let's consider part e). This tells us that the derivative of X(\omega) evaluated at \omega=2 is equal to 4\pi i. This could indicate that the \omega=\pi sinusoid in x has a phase shift of \pi/2, since the derivative of e^{i\omega} evaluated at \omega=2 is equal to 2\pi i. This means that the signal x could be written as x(n)=2\cos(\pi n+\pi/2)+a\cos(0n+\theta) where a is the amplitude of the \omega=0 sinusoid and \theta is the phase shift of the \omega=0 sinusoid.

Finally, let's look at part f). This tells us that the real part of X(\omega) evaluated at \omega=2 is equal to 3/2. This could give us some information about the amplitude a of the \omega=0 sinus