Let $F[z]=N^2\frac{z(z-(3+j\sqrt{7})/2)(z-(3-j\sqrt{7})/2)}{(z-1)^3}$

This has 3 poles at $z=1$; one zero at $z=0$; and a conjugate pair of zeros at $z=\frac{3\pm{}j\sqrt{7}}{2}$

Assuming a contour $H$ encircles these roots and zeros, a unique inverse is obtained from direct inversion:

$$f=\frac{1}{2{\pi}j}\oint_H Fz^{n-1} \,dz$$

What, if any, is the mathematical interpretation of taking a contour around a single pole or zero?


For the integral to be an inverse Z transform, the contour $H$ must enclose the origin, and it must be inside the region of convergence (ROC). In this way, the contour defines the ROC. E.g., if it encircles all poles, the corresponding sequence is causal. If all poles are outside the contour, then the resulting sequence is anti-causal. Finally, if there are poles inside and outside the contour, the resulting sequence is two-sided.

In your example, all poles are at $z=1$, so there are only two options: either the contour encloses these poles, then the sequence is causal. Otherwise, the sequence is anti-causal. Note that the location of the zeros is irrelevant here.

If you had two distinct poles, you would have three choices:

  1. both poles are outside $H$: the resulting sequence is anti-causal (left-sided)
  2. $H$ encloses only 1 pole: the resulting sequence is two-sided (non-causal)
  3. $H$ encloses both poles: the resulting sequence is causal (right-sided)
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