I have been working with discrete-time systems and I am not yet able to understand the reason behind using powers of $z^{-1}$ in transfer functions.

I get that $z^{-1}$ means a delay, but when we write a transfer function like this, usually some zeros/poles disappear, I can't see what is the practical use of this notation.

For instance, take $$H(z) = \frac{1}{(1-\frac{1}{2}z^{-1})(1-\frac{3}{4}z^{-1})}$$

If we multiply by $\frac{z^2}{z^2}$ it becomes $$H(z)=\frac{z^2}{(z-\frac{1}{2})(z-\frac{3}{4})}$$ and we can clearly see that there are two zeros and two poles.

  • $\begingroup$ what you're doing appears to me to be exactly correct. there are two poles, one at $z=\frac12$ and the other at $z=\frac34$ and two zeros both at $z=0$. that is what you got. if there was a $z^{-2}$ in your numerator, there would be two poles (at the same two locations) and no zeros. turns out, if your $H(z)$ is for a causal system (one that only responds to input from the past and the present and does not respond to future input samples), then the number of zeros must be no greater than the number of poles. $\endgroup$ – robert bristow-johnson Jul 3 at 19:59
  • $\begingroup$ Exactly, so is it a matter of personal choice using $H(z^{-1})$ instead of $H(z)$? What would make someone prefer the first notation? (To me, the second notation, with positive powers of $z$, makes a lot more sense) $\endgroup$ – Vinícius Lopes Simões Jul 3 at 20:06
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    $\begingroup$ the first notation, with all of the negative powers, can first have its factors multiplied out: $$ H(z) = \frac{1}{1 - \frac54 z^{-1} + \frac38 z^{-2}}$$ and then be translated into a difference equation for the purpose of implementation: $$ y[n] = x[n] + \tfrac54 y[n-1] - \tfrac38 y[n-2] $$ this is because it only has delay elements depicted. both forms have a use, but different uses. $\endgroup$ – robert bristow-johnson Jul 3 at 20:31

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