Direct programming of a third-order controller

Step 1: Write in powers of $z^{-1}$

Divide the numerator and denominator by $10z^3$:

$$\frac{Y(z)}{X(z)} = \frac{0.6z^{-1} + 0.7z^{-2} + 1.6z^{-3}}{1 + 0.8z^{-1} + 0.9z^{-2} + 1.5z^{-3}}$$

Step 2: Difference equation

$$y(k) = -0.8y(k-1) - 0.9y(k-2) - 1.5y(k-3) + 0.6x(k-1) + 0.7x(k-2) + 1.6x(k-3)$$

Step 3: Block diagram (direct programming)

x(k)
  o              (S)--------------o--> y(k)
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-1)        |               | y(k-1)
  o--[0.6]------>(S)<[-0.8]-------o
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-2)        |               | y(k-2)
  o--[0.7]------>(S)<[-0.9]-------o
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-3)        |               | y(k-3)
  o--[1.6]------>(S)<[-1.5]-------o

• The input delay chain gives $x(k-1)$, $x(k-2)$, $x(k-3)$, with gains 0.6, 0.7, 1.6.
• The output delay chain gives $y(k-1)$, $y(k-2)$, $y(k-3)$, fed back with gains $-0.8$, $-0.9$, $-1.5$.
• All products are added in the summer column to give $y(k)$. There is no direct $x(k)$ term.

Direct programming uses $3 + 3 = 6$ delays. (The denominator has roots $-1.1507$ and $0.1754 \pm j1.1282$, which are outside the unit circle, so this filter is unstable as given. The question asks only for its realization.)