# Basic transfer function question

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.

• 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. – percusse Dec 21 '16 at 2:37
• @percusse Not sure I understand your hint about positive powers of z... – James HJ Dec 21 '16 at 7:12
• multiply both numerator and denominator with z and it becomes a low pass filter. unity on the forward path -a on the negative path – percusse Dec 21 '16 at 7:59

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