Direct programming of a third-order controller

Direct programming realizes the difference equation exactly as written. It uses one chain of delays for past inputs and another for past outputs.

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

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

$$\frac{Y(z)}{X(z)} = \frac{0.6z^{-1} + 0.7z^{-2}}{1 - 0.6z^{-1} + 0.7z^{-2} + 0.5z^{-3}}$$

Step 2: Difference equation

$$y(k) = 0.6y(k-1) - 0.7y(k-2) - 0.5y(k-3) + 0.6x(k-1) + 0.7x(k-2)$$

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)<[0.6]--------o
  |               ^               |
[z^-1]            |             [z^-1]
  | x(k-2)        |               | y(k-2)
  o--[0.7]------>(S)<[-0.7]-------o
                  ^               |
                  |             [z^-1]
                  |               | y(k-3)
                  +<[-0.5]--------o

• The left delay chain stores $x(k-1)$ and $x(k-2)$. The right chain stores $y(k-1)$, $y(k-2)$ and $y(k-3)$.
• Feed-forward gains are 0.6 and 0.7. Feedback gains are $+0.6$, $-0.7$ and $-0.5$ (the negatives of the denominator coefficients).
• There is no direct path from $x(k)$ to $y(k)$, because the $z^0$ term of the numerator is zero.
• Direct programming needs $m + n = 2 + 3 = 5$ delays. Standard programming would need only 3.