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enter image description here Hello, this is one of my homework questions and i have already solved the first question but im having trouble gettin a relation that helps me solve the second one. From the question i understand that R(z)=H(z).H(1/z) but i dont know how to develop this. This is my response to the 1st question enter image description here

Thanks

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  • $\begingroup$ Estou a ver que foi a partir daqui que fizeste o primeiro trabalho. Tenho muita pena mas se não te retirar alguma cotação não é justo para os teus colegas que fizeram tudo por si próprios. $\endgroup$ – AJF Apr 30 '17 at 15:34
  • $\begingroup$ Eu entendo professor, mas não foi mesmo com a intenção de copiar descaradamente sem entender, embora tenha toda a razão em não dar toda a cotação. $\endgroup$ – Luce Sky Walker May 1 '17 at 12:23
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Since this looks like a homework problem I'll help you to solve it on your own, rather than just present you the solution.

The problem here is that you have a mistake in the first part of the solution. The evaluation of the first sum (from $n=-\infty$ to $n=-1$) is wrong. Redo the calculation and you should end up with

$$R(z)=\frac{1}{(1-az^{-1})(1-az)}\tag{1}$$

with some appropriate value for $a$. From $(1)$ it's immediately clear how you can factor $R(z)$ as

$$R(z)=H(z)H(z^{-1})\tag{2}$$

and how you have to choose $H(z)$ to get a minimum-phase systems.

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  • $\begingroup$ Thanks Matt, i did the calculation again and i was in fact wrong. With some trickery and variable changes i got to the expression u mentioned. Thanks again. $\endgroup$ – Luce Sky Walker Mar 13 '17 at 19:04
  • $\begingroup$ Just one more question, i got the expression that expression with a=0.5, now im doing the ROC of the whole signal. is it |z| > |0.5| && |z| < |2| like i had before? I having doubt on the signal that u told me to redo. Thanks in advance $\endgroup$ – Luce Sky Walker Mar 13 '17 at 20:06
  • $\begingroup$ @LuceSkyWalker: Yes, the ROC is a ring between the two poles. This ring includes the unit circle $|z|=1$, which means that also the Fourier transform of $r[n]$ exists and can be obtained by setting $z=e^{j\omega}$. $\endgroup$ – Matt L. Mar 13 '17 at 20:09

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