The question already has a zero and a pole in the complex plane. It is asking to add additional zero and pole so LTI becomes zero valued.

zeros=-1/2 + 1/2 i , poles = -1/3 -1/3 i .

What I understand from it is that we have to make sure that in the end transfer function, there are no complex parts (there is no iota).. that would be real valued... but then why are these zeros and poles not in conjugate pairs? If they are in conjugate then it is already real valued .... so is the question there to confuse us or what? If I am missing something very basic please mention it so I can learn it.

Thank you. !

  • 2
    $\begingroup$ So you have to add poles and zeros to the transfer function to make sure that the corresponding system is real-valued. And you know that poles and zeros must occur in complex-conjugate pairs (unless they are real-valued). So why don't you just add the complex conjugates of the given poles and zeros? It seems you know everything to solve the problem but somehow you still don't know how to solve it ... $\endgroup$ – Matt L. Jul 5 '20 at 16:12
  • $\begingroup$ Thank you so much Matt. Just didn't understand why the question mentioned a zero and a pole having a complex part but not in conjugate pairs. It was possibly there to add confusion to the question. Thank you for your reply ! $\endgroup$ – Abdul Jul 5 '20 at 20:37

Your Answer

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

Browse other questions tagged or ask your own question.