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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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    Well, that is a function of $z$ in there, no?2012-02-03

2 Answers 2

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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