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

  • Thread starter FrogPad
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In summary, the Z-transform for x[n] is X(z) = (z(z-cos Ω_0))/(z^2-2cos Ω_0 z+1), with a region of convergence |z| > 1 and poles at e^jΩ_0 and e^-jΩ_0, and zeros at 0 and cos Ω_0. To sketch the pole/zero plot, draw a unit circle with its center at the origin and mark the points where the poles and zeros lie. The region of convergence is the area outside of the unit circle.
  • #1
FrogPad
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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.

[tex] x[n] = \cos( \Omega_0 n) u[n] [/tex]

Where [tex] u[n] [/tex] is the step function.

"Answer"

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

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

with a region of convergence: [itex] |z|>1 [/itex]

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

Thus: zeros = [itex] \{ 0, \cos \Omega_0 \} [/itex]

Using the quadratic formula yields:

poles = [tex] 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} [/tex]

This can be further simplified to: [tex] 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} \}[/tex]Now if I assume I am right (which I'm not sure about), then:

ROC = [tex] |z| > 1 [/tex]
POL = [tex] \{ e^{j\Omega_0}, \,\,\, e^{-j\Omega_0} \}[/tex]
ZER = [tex] \{ 0, \,\,\, \cos \Omega_0 \}[/tex]

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!
 
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  • #2
Yes, the in-between steps are necessary in order to sketch the pole/zero plot. You can think of it as a “map” of the z-plane that shows where the poles and zeros of the transfer function (in this case, X(z)) lie. A pole is represented by an “X” and a zero is represented by an “O”. To plot the pole/zero plot, you will need to draw a unit circle with its center at the origin. Then, mark the points where the two poles lie (in this case, e^jΩ_0 and e^-jΩ_0) with an “X”. Lastly, mark the point 0 and the point cos Ω_0 with an “O”. The region of convergence (ROC) will be the area of the z-plane outside of the unit circle.
 

1. What is a z-transform?

A z-transform is a mathematical tool used to convert a discrete-time signal into a complex frequency domain representation, similar to the way a Fourier transform converts a continuous signal into a frequency domain representation.

2. Why would someone use a z-transform?

A z-transform is useful for analyzing and designing discrete-time systems, such as digital filters, in the frequency domain. It allows for easier manipulation and understanding of the system's characteristics.

3. How is a z-transform different from a Fourier transform?

A z-transform is specifically designed for discrete-time signals, while a Fourier transform is used for continuous signals. Additionally, a z-transform produces a complex frequency domain representation, while a Fourier transform produces a real-valued frequency spectrum.

4. Can a z-transform be reversed or undone?

Yes, a z-transform can be inverted, known as an inverse z-transform, to convert the complex frequency domain representation back into the original discrete-time signal.

5. Are there any limitations to using a z-transform?

One limitation is that it assumes the signal is causal, meaning it only depends on past values and not future values. Additionally, it can only be applied to signals with a finite number of values.

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