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What will be inverse $\mathcal Z$-transform for this function:

$$H(z) = \frac{\left(1+\beta z^{-1}\right)\left(1+\beta z\right)}{\left(1+\alpha z^{-1}\right)\left(1+\alpha z\right)}$$

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    $\begingroup$ uhm, you gotta poles at both $-\alpha$ and $\frac{-1}{\alpha}$. one of them will be outside the unit circle. or they will both be on the unit circle at the same place (double pole). either way, this sonova bitch ain't gonna be stable. $\endgroup$ – robert bristow-johnson May 9 '16 at 5:24
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    $\begingroup$ It will be stable, but not causal (unless $|\alpha|=1$). That's the type of Z- transform you get from an autocorrelation sequence (two-sided IIR). $\endgroup$ – Matt L. May 9 '16 at 12:02
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    $\begingroup$ Yes, see the second example here which shows how such a transfer function can have three different ROCs, which say different things about causality and stability. If one of the ROCs includes the unit circle, then the system is BIBO stable... but not causal. $\endgroup$ – Peter K. May 9 '16 at 12:45
  • $\begingroup$ As you now the Z transform H(z) alone is not enough to uniquely deduce the inverse transform h[n], which also depends on the Region of Convergence (ROC) for the Z-Transform H(z). So can you please state the ROC to find the inverse. $\endgroup$ – Fat32 May 9 '16 at 13:20
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Given the $\mathcal Z$-transform : $$H(z) = \frac{\left(1+\beta z^{-1}\right)\left(1+\beta z\right)}{\left(1+\alpha z^{-1}\right)\left(1+\alpha z\right)}$$

The inverse $\mathcal Z$-transform can be found by the method of partial fraction expansion guided with the Region of Convergence ROC associated with $H(z)$

I will proceed with the most fundamental method and apply necesary algebraic manipulations to get the required standard form as follows:

\begin{align} H(z) &= \frac {\left(1 + \beta z^{-1}\right)\left(1 + \beta z\right)}{\left(1 + \alpha z^{-1}\right)\left(1 + \alpha z\right)} &\scriptstyle{\text{convert to negative power of $z$}}\\ &= \frac{\left(1 + \beta z^{-1}\right)\left(z^{-1} + \beta\right)}{\left(1 + \alpha z^{-1}\right)\left(z^{-1} + \alpha\right)} &\scriptstyle{\text{convert to standard form of } (1-d_k z^{-1})}\\ &= \frac{\beta}{\alpha} \frac{\left(1 + \beta z^{-1}\right)\left(1 + \frac{1}{\beta} z^{-1}\right)}{\left(1 + \alpha z^{-1}\right)\left(1 + \frac{1}{\alpha}z^{-1}\right)} &\scriptstyle{\text{expand the brackets}}\\ &= \frac{\beta}{\alpha} \frac{\left(1 + \left(\beta + \frac{1}{\beta}\right) z^{-1} + z^{-2}\right) }{\left(1 + \left(\alpha + \frac{1}{\alpha}\right) z^{-1} + z^{-2}\right)} &\scriptstyle{\text{apply the long division}}\\ &= \frac{\beta}{\alpha} \left( 1 + \frac{ \left(\beta + \frac{1}{\beta} - \alpha - \frac{1}{\alpha}\right)z^{-1} }{\left(1 + \alpha z^{-1}\right)\left(1 + \frac{1}{\alpha}z^{-1}\right)} \right) &\scriptstyle{\text{go further...}}\\ &= \frac{\beta}{\alpha} \left( 1 + \frac{A}{1 + \alpha z^{-1}} + \frac{B}{1 + \frac{1}{\alpha}z^{-1}} \right) &\scriptstyle{\text{where $A$ and $B$ are found from the right side quotient of $H(z)$}}\\ \end{align}

Now assuming $A$ and $B$ are found as functions of $\alpha$ and $\beta$, we shall define all possible ROC of $H(z)$ to find the corresponding inverse transform $h[n]$.

  • Now If $0<|\alpha|<1<\frac{1}{|\alpha|}$ and ROC is such that $|z|<|\alpha|$ then: $h[n]$ will be unstable and anti-causal as $$h[n] = \frac{\beta}{\alpha} \delta[n] + \frac{\beta}{\alpha} \left(- A\left(-\alpha\right)^n \text{u}[-n-1] - B\left(- \frac{1}{\alpha}\right)^n \text{u}[-n-1] \right)$$

  • Or else if $0<|\alpha|<1<\frac{1}{|\alpha|}$ and ROC is such that $|\alpha| <|z|<\frac{1}{|\alpha|}$ then: $h[n]$ will be stable and two-sided as $$h[n] = \frac{\beta}{\alpha} \delta[n] + \frac{\beta}{\alpha} \left( A(-\alpha)^n \text{u}[n] - B\left(- \frac{1}{\alpha}\right)^n \text{u}[-n-1] \right)$$

  • Or else if $0<|\alpha|<1<\frac{1}{|\alpha|}$ and ROC is such that $|z|>\frac{1}{|\alpha|}$ then: $h[n]$ will be unstable and causal as $$h[n] = \frac{\beta}{\alpha} \delta[n] + \frac{\beta}{\alpha} \left( A(-\alpha)^n \text{u}[n] + B\left(- \frac{1}{\alpha}\right)^n \text{u}[n] \right)$$

  • The cases for $0<\frac 1\alpha<1<\alpha$ are simply the same, merely interchanging $\alpha$ with $\frac 1\alpha$

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    $\begingroup$ Computing the constants $A$ and $B$ will reveal that $A=-B$, which shows that the stable solution is symmetrical. This is also immediately clear from $H(z)$, which has the form $H(z)=F(z)F(1/z)$, i.e., on the unit circle it corresponds to the squared magnitude $|F(e^{j\omega})|^2$, and so its time domain function must be a valid autocorrelation function. $\endgroup$ – Matt L. May 9 '16 at 16:14
  • $\begingroup$ @MattL. Do you mean there is an typo/error in the final stage? Or is it just a clearup reminder because I think plugging A=−B should provide what you define as the symmetrical ACF solution (in the second case) stable-two sided becomes stable and symmteric $\endgroup$ – Fat32 May 9 '16 at 16:37
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    $\begingroup$ I don't think there's an error; I just wanted to point out and clarify that the (stable) solution must be symmetrical and how to see that. $\endgroup$ – Matt L. May 9 '16 at 19:05

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