# I am having some trouble with this z-transform, and would like a

1. Apr 29, 2007

I am having some trouble with this z-transform, and would like a little help.

Question:
Find the Z-transform, and sketch the pole/zero plot with the region of convergence.

$$x[n] = \cos( \Omega_0 n) u[n]$$

Where $$u[n]$$ is the step function.

Directly from a table in my book, $x[n]$ transforms to:

$$X(z) = \frac{1-[\cos \Omega_0]z^{-1}}{1-[2\cos \Omega_0]z^{-1}+z^{-2}}$$

with a region of convergence: $|z|>1$

For the poles/zeros, a little bit of algebra yields:
$$X(z) = \frac{z(z-\cos \Omega_0)}{z^2-2\cos \Omega_0 z + 1}$$

Thus: zeros = $\{ 0, \cos \Omega_0 \}$

poles = $$p_{1,2} = \frac{2 \cos \Omega_0 \pm \sqrt{4 \cos^2 \Omega_0 - 4}}{2} = \cos \Omega_0 \pm \sqrt{\cos^2 \Omega_0 -1}$$

This can be further simplified to: $$p_{1,2} = \cos \Omega_0 \pm \sqrt{-\sin^2 \Omega_0} = \cos \Omega_0 \pm j\sin \Omega_0 = \{ e^{j\Omega_0},\,\,\,\, e^{-j\Omega_0} \}$$

Now if I assume I am right (which I'm not sure about), then:

ROC = $$|z| > 1$$
POL = $$\{ e^{j\Omega_0}, \,\,\, e^{-j\Omega_0} \}$$
ZER = $$\{ 0, \,\,\, \cos \Omega_0 \}$$

So how do I plot this?

Do I say:
A zero exists at the origin, and the other zero exists from -1 to 1 on the real line.

A pole exists on the unit circle with the other pole "opposite" it.

I just need a little help in understanding this. Please let me know, if the in-between steps are necessary. Thanks!