Standard and ladder programming block diagrams

The difference equation as printed has a typo. It is read as the standard second-order filter

$$y(k) + a_1y(k-1) + a_2y(k-2) = b_0x(k) + b_1x(k-1) + b_2x(k-2)$$ $$G(z) = \frac{Y(z)}{X(z)} = \frac{b_0 + b_1z^{-1} + b_2z^{-2}}{1 + a_1z^{-1} + a_2z^{-2}} = \frac{b_0z^2 + b_1z + b_2}{z^2 + a_1z + a_2}$$

(i) Standard programming

Let $W(z) = X(z)/(1 + a_1z^{-1} + a_2z^{-2})$:

$$\begin{aligned} w(k) &= x(k) - a_1w(k-1) - a_2w(k-2) \\ y(k) &= b_0w(k) + b_1w(k-1) + b_2w(k-2) \end{aligned}$$

x(k)-->(S)------------o--[b0]------->(S)--> y(k)
        ^             |               ^
        |          [z^-1]             |
        |             | w(k-1)        |
       (S)<[-a1]------o--[b1]------->(S)
        ^             |               ^
        |          [z^-1]             |
        |             | w(k-2)        |
        +<[-a2]-------o--[b2]---------+

Only 2 delays are used, the minimum for second order.

(ii) Ladder programming

Expand as a continued fraction:

$$G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2}}}}$$

Define $c_1 = b_1 - a_1b_0$, $c_2 = b_2 - a_2b_0$, $d = a_1c_1 - c_2$ and $e = c_2 - \dfrac{a_2c_1^2}{d}$. Then:

$$\begin{aligned} G(z) &= b_0 + \frac{c_1z + c_2}{z^2 + a_1z + a_2} \\ A_0 &= b_0, \quad B_1 = \frac{1}{c_1}, \quad A_1 = \frac{c_1^2}{d}, \quad B_2 = \frac{d}{c_1e}, \quad A_2 = \frac{e}{a_2} \end{aligned}$$

X(z)--o-----------------[A0]--------------->(S)--> Y(z)
      |                                      ^
      v +                                    | Y1
     (S1)-->[z^-1/B1]-------------o----------+
      ^ -                         | +
      |                           v
      o<--E1--[1/A1]<----------- (S2)
      |                           ^ -
      v +                         |
     (S3)-->[z^-1/B2]-------------o Y2
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+

$$\begin{aligned} y_1(k) &= \tfrac{1}{B_1}[x(k-1) - e_1(k-1)], & e_1(k) &= \tfrac{1}{A_1}[y_1(k) - y_2(k)] \\ y_2(k) &= \tfrac{1}{B_2}[e_1(k-1) - e_2(k-1)], & e_2(k) &= \tfrac{1}{A_2}y_2(k) \\ y(k) &= A_0x(k) + y_1(k) \end{aligned}$$

Example: For $b_0 = 2$, $b_1 = 2.2$, $b_2 = 0.2$, $a_1 = 0.4$, $a_2 = -0.12$: $c_1 = 1.4$, $c_2 = 0.44$, $d = 0.12$, $e = 2.4$. This gives $A_0 = 2$, $B_1 = 0.7143$, $A_1 = 16.33$, $B_2 = 0.0357$, $A_2 = -20$.