Understanding Z-Plane Poles and Zeros: A Sketching Guide

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I have been asked to sketch the Z planes poles and zeros for the follwoing:

H(z) = ((z-1)^2 (z+1)^2) /(z-0.7)(z+0.9+0.95j)(z+0.9-0.95j)


I have the denominator points sorted. Would the numerator points be:

(z-1)^2 =0
z-1 =0
z=1

and

(z+1)^2 = 0
z+1 = 0
z = -1
 
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Yes, 1 and -1 are the zeros of this function.
 
Ok so now to sketching these values.
Do I need to get:

z = -0.9 +0.95j
and
z=-0.9 -0.95 j

in the form r< theta

So r would be:

SQRT ((-0.9)^2 + (0.95)^2))
r = 1.309

and find theta with:

tan theta = b/a

giving me theata = -46.548 and 46.548 degress?
 
Well, if you want to plot the values, you COULD get them into polar form. It seems you have done this incorrectly: note that the real parts of each of the numbers is negative, but you are planning to plot them in the first and fourth quadrants.

I would just plot them straight from the x+yj form; the x-axis is the real axis, and the y-axis is the imaginary axis. If you do this and compare to the polar form, you'll see that they're the same, but it seems more straightforward to me to just plot the points (-0.9,0.95) and (-0.9,-0.95).
 
Ah I see its just the programme I have been provided to sketch the zplane poles and zeros requires values for r and theta. is this correct:

so r would be:

SQRT ((-0.9)^2 + (0.95)^2))
r = 1.309

and find theta with:

tan theta = b/a

(tan^(-1) 0.95/0.9)-pi

theta = -133.452

this would plot the two points in the second and third quadrants as required
 
Yes, seems right.
 
Thanks I just have a final question. On my work sheet it says what is the significence that the order of numerator is greater than that of denominator??
 
Well, it might tell you something about the behavior of the function as |z| goes to infinity...
 
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