0
$\begingroup$

Got a couple of questions here guys

  1. I have a unit circle with just 1 pole and a lot of zeros. What condition could result to this and does it have any effect on the dsp system

  2. Why do poles appear as real/complex conjugate pairs.

Thanks

$\endgroup$
1
$\begingroup$

to have more zeros than poles require that the system is not causal. something that can exist in theory (and we do that with non-realtime DSP processing which can look ahead to "future" samples when it computes and outputs the "present" sample) but not in practice (those "future" samples referred to were really recorded long ago and saved in a file or buffer). perhaps you might think there are more zeros than poles but that might be because there are poles located at $z=0$ that you're not counting. FIR filters have all of their poles at the origin.

$\endgroup$
  • $\begingroup$ and the poles and zeros are either real or paired as complex conjuagates of each other because the transfer function (a ratio of two polynomials of $z$, both can be factored, the roots of the numerator are the zeros and the roots of the denominator are the poles) has real coefficients. if the coefficients are not real (and have a non-zero imaginary component), then at least some pole or zero is not purely real and also does not have a complex-conjugate. $\endgroup$ – robert bristow-johnson Apr 18 '15 at 22:21
  • $\begingroup$ Going by your answer to the first question, it means the system is causal if it has more poles than zeros. thanks for the explanation. $\endgroup$ – Shittu Olalekan Apr 19 '15 at 2:46
  • $\begingroup$ if the system is causal (that is $h[n]=0$ for all $n<0$) then the number of zeros cannot exceed the number of poles. (the number of zeros can equal the number of poles in a causal system.) if the number of zeros exceed the number of poles, you can do long division of the numerator polynomial by the denominator polynomial (that has lower order) and some non-zero terms of positive powers of $z$ will pop out. those terms are the opposite of delay (which are $z^{-1}$ terms). $\endgroup$ – robert bristow-johnson Apr 19 '15 at 2:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.