Z-Transform and DTFT Homework: Poles, Zeros, and Endpoints

[zcd]

Homework Statement

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?

 

[mighty2000]

Based on the given information, we can make the following observations about the signal x and its DTFT X:

1. The signal x is finite-length, which means it has a finite number of non-zero values. This also implies that x(n) = 0 for n outside the range of the signal.

2. The DTFT X has exactly two poles at z=0, which means that the signal x has two terms in its transfer function that have the form z^(-n). This also means that x(n) = 0 for n < 0.

3. X(\omega)\Big|_{\omega=\pi}=2 tells us that x(n) has a non-zero value at n=0. This can be seen from the definition of the DTFT, where X(\omega) is the sum of the values of x(n)e^(-i\omega n) over all values of n. Since X(\pi) = 2, this means that x(n)e^(-i\pi n) = 2 for n=0, which implies that x(0) = 2.

4. The given expression for \theta_x(\omega) tells us that the phase of X(\omega) is a piecewise linear function with a slope of 1/2 and a discontinuity at \omega = 0. This means that the signal x has a linear phase characteristic, with a slope of 1/2, and a jump of \pi/2 at n=0.

5. The given expression for \hat{X}(\omega) also tells us that the magnitude of X(\omega) is always positive or zero, and it is equal to 2 for \omega = \pi. This means that x(n) is a real signal, and its absolute value is equal to 2 at n=0.

6. The given expression for the integral of e^(2i\omega)\frac{d}{d\omega}X(\omega)d\omega tells us that the signal x has a constant value of 2 for all values of n, except for n=0 where it has a jump of 4\pi i. This means that x(n) = 2 for n \neq 0, and x(0) = 2 + 4\pi i.

7. Finally, the expression for the DTFT of V(\omega) = Re(X(\omega)) tells us