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I have the transfer function$$ H(z) = \frac{z-.75}{.1 z+.15} $$

how do I find the Poles and Zeros?

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    $\begingroup$ Poles are the values of $z$ where the denominator becomes zero; zeros are the values of $z$ where the numerator becomes, well, zero. I presume you know how to find the zeros of a linear function? $\endgroup$ – J. M. is a poor mathematician Feb 3 '12 at 6:28
  • $\begingroup$ yes I do. So finding poles and zeros in the z domain is exactly like in the s-domain? $\endgroup$ – richardnixonthethird Feb 3 '12 at 6:29
  • $\begingroup$ Well, that is a function of $z$ in there, no? $\endgroup$ – J. M. is a poor mathematician Feb 3 '12 at 6:34
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First of all, I'd advice to get rid of those awful decimal numbers and use fractions: $$H(z):=\frac{z-\frac{3}{4}}{\frac{1}{10}z+\frac{3}{20}}=\frac{\frac{1}{4}}{\frac{1}{20}}\frac{4z-3}{2z+3}=5\,\frac{4z-3}{2z+3}$$ From here, we clearly see the only zero of the function (i.e., exactly where its numerator vanishes) is $\,z=3/4\,$ , and its pole (i.e., exactly where the denominator vanishes) is $\,z=-3/2\,$, both simple.

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roots of numerator are zeros and roots of denonumerator are poles

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