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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!
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!