On page 223, we are presented with the transfer function sans the a0 variable. A factored version is shown and it is said that the zeros are at Z1 and Z2. From looking at the equation it seems that in order for that to be true, Z1=Z2=z. But that is not in the case since Z1 and Z2 are said to be complex conjugates of each other and won't be equal to z depending on the coefficients. How can this be, I seem to not be understanding something here.Â
I got caught on the fact that a function of z is actually over the whole complex plane, while I was thinking it was limited to the unit circle. Now it makes sense.
I found there's also another way to find the zeros and poles of the 2nd order filters, which is to algebraically manipulate (there must be a word for this) the transfer function so there is a quadratic equation in the numerator and/or denominator, and use the quadratic equation.
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