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Why is the transfer function from node 2 to 3 $H_{23}=\frac{z^{-1}}{1+az^{-1}}$? I don't understand the $z^{-1}$ in the numerator.

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  • $\begingroup$ Forget about the figure and do it manually from (18). It's a horrible illustration. Hint : write the $H_{23}$ in terms of the positive powers of z. $\endgroup$ – percusse Dec 21 '16 at 2:37
  • $\begingroup$ @percusse Not sure I understand your hint about positive powers of z... $\endgroup$ – James HJ Dec 21 '16 at 7:12
  • $\begingroup$ multiply both numerator and denominator with z and it becomes a low pass filter. unity on the forward path -a on the negative path $\endgroup$ – percusse Dec 21 '16 at 7:59
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Considering:

$H_{0, 1} = z^{-1} H_{2, 3} = \frac{z^{-1}}{1 + az^{-1}}$

then:

$H_{2, 3} = \frac{1}{1 + a z^{-1}}$

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  • $\begingroup$ At first I didn't understand your comment but after some time I realized that I have completely misunderstood (19). The statement does not provide expressions for 2 transfer functions! but write one transfer function in terms of the other. Now everything makes sense. $\endgroup$ – James HJ Dec 21 '16 at 7:11
  • $\begingroup$ Yes, actually (19) is a single function. I think the previous sentence, that is "By inspection of Figure 1, they are:", may confuse the reader due to its plural subject referring both the transfer functions. $\endgroup$ – rudicangiotti Dec 21 '16 at 12:13

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