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I am confused by two questions on the region of convergence (ROC) when applying the $Z$-transform.

  1. Consider a discrete-time linear system $$x[n+1] = Ax[n]+Bu[n], \qquad y[n]=Cx[n]+D$$ whose transfer function is given by $$H(z) = C(zI - A)^{-1}B + D$$ Is there a region of convergence for $H(z)$? Is it supposed to be the whole complex plane?

    Also from the Woodbury matrix identity, $(zI-A)^{-1}$ can be written as $\sum_{k=0}^\infty \frac{1}{z^{k+1}}A^k$. How should I derive the ROC of the latter series? Does it matter if the ROC of series is different from the ROC of original transfer function H(z)?

  2. If only considering causal sequences, is it possible to have two sequence with same $Z$-transforms, but with different ROC? To be specific, consider two causal sequences $\left\{x_1(k)\right\}_{k=0}^{\infty}$, $\left\{x_2(k)\right\}_{k=0}^{\infty}$, is it possible denote the two $z$-transform as $H_1(z), H_2(z)$ respectively. If $H_1(z) = H_2(z), \forall z\in ROC_1 \cap ROC_2$, can I claim that $x_1(k) = x_2(k), \forall k$?

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  • $\begingroup$ What's this got to do with digital communications? $\endgroup$ – Dilip Sarwate Mar 1 at 22:02
  • $\begingroup$ The transfer function is a actually matrix of transfer functions, each of which should have a ROC. $\endgroup$ – Rodrigo de Azevedo Mar 2 at 20:54
  • $\begingroup$ @RodrigodeAzevedo Thanks for the reply! Is there any way to identify the ROC of each transfer function, in terms of the matrix $A$? $\endgroup$ – mw19930312 Mar 4 at 14:54
  • $\begingroup$ The set of poles should be a subset of the set of roots of the polynomial $\det (z I - A)$. Once one has the poles, one can find the ROC based on whether the system is causal or not. Be careful with pole-zero cancelations that may occur. $\endgroup$ – Rodrigo de Azevedo Mar 4 at 21:13

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