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This source states that the unit step function in the Z-domain is \$\frac{z}{z-1}\$. However, in its derivation it states \$\sum_{k=0}^{\infty} z^{-k} = \frac{z}{z-1}\$.

But doesn't that last relation only hold true for \$z>1\$? I don't see how that condition is met. I know that the poles have to be in the unit circle for the system to be stable, but I don't see if and how that connects to the condition above.

What am I missing?

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  • \$\begingroup\$ This question is better asked on the Mathematics StackExchange as you're asking about a mathematical concept and nothing electrical engineering related... or at least you haven't told us exactly what your system is... \$\endgroup\$ – KingDuken May 31 '18 at 17:51
  • \$\begingroup\$ Maybe you're right, I'll post it there. \$\endgroup\$ – Runge Kutta May 31 '18 at 17:54
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    \$\begingroup\$ You may want to view this from 3blue1brown on the Riemann zeta function. \$\endgroup\$ – jonk May 31 '18 at 20:44
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    \$\begingroup\$ A pole on the unit circle indicates critical stability, and a unit step sequence is critically stable, in the sense that it neither increases nor decreases. \$\endgroup\$ – Chu May 31 '18 at 21:25
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    \$\begingroup\$ Yes, I know that, but that still doesn't the answer my question. I think the answer is closer to a comment on Math.sx which basically says the function can be continued outside the range where it is defined, similarly to what is described in the @jonk 's link. \$\endgroup\$ – Runge Kutta Jun 19 '18 at 14:03
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Here is one way to look at it.

If you consider your derived expression as a transfer function in z $$\frac{Y(z)}{X(z)}=\frac{z}{z-1}$$

you can rewrite in terms of the delay operator (1/z) as

$$\frac{Y(z)}{X(z)}=\frac{1}{1-z^{-1}}$$

then cross multiply to get the difference equation $$y(k)-y(k-1)=x(k)$$ or $$y(k)=x(k)+y(k-1)$$ To get the impulse response x(k)=1 for k=0 but zero for k >0

from which y(k)=1 for k>=0, zero elesewhere, aka the step function.

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  • \$\begingroup\$ This is certainly conceptually helpful in general, but still I don't think this answers the question of why the convergence can be used in the proof if it only holds for if z>1. \$\endgroup\$ – Runge Kutta Jun 19 '18 at 12:43

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