Direct programming of a first-order filter

Step 1: Difference equation

$$\frac{Y(z)}{X(z)} = \frac{2 - 0.6z^{-1}}{1 + 0.5z^{-1}} \;\Rightarrow\; Y(z)(1 + 0.5z^{-1}) = X(z)(2 - 0.6z^{-1})$$ $$y(k) = -0.5y(k-1) + 2x(k) - 0.6x(k-1)$$

Step 2: Direct programming block diagram

Each delayed input and output term is formed with its own delay element and added in a summer:

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

• Feed-forward: $x(k)$ with gain 2, and $x(k-1)$ with gain $-0.6$.
• Feedback: $y(k-1)$ with gain $-0.5$.
• Direct programming uses $1 + 1 = 2$ delays.

For comparison, standard programming needs only one delay: $w(k) = x(k) - 0.5w(k-1)$ and $y(k) = 2w(k) - 0.6w(k-1)$.

The pole at $z = -0.5$ is inside the unit circle, so the filter is stable.