Ladder programming of a third-order filter

Assumption: the textbook form of this filter has positive powers of $z$:

$$G(z) = \frac{128z^3 + 224z^2 + 106z + 11}{128z^3 + 160z^2 + 34z + 1}$$

This gives a stable filter (poles $-0.9895$, $-0.2254$, $-0.0350$) and integer ladder constants. If the $z^{-k}$ form is used literally, the same continued-fraction steps apply but the constants are not integers.

For a third-order filter, ladder programming expands $G(z)$ as:

$$G(z) = A_0 + \cfrac{1}{B_1z + \cfrac{1}{A_1 + \cfrac{1}{B_2z + \cfrac{1}{A_2 + \cfrac{1}{B_3z + \cfrac{1}{A_3}}}}}}$$

Step 1: $A_0$

$$G(z) = 1 + \frac{64z^2 + 72z + 10}{128z^3 + 160z^2 + 34z + 1}$$

So $A_0 = 1$.

Step 2: $B_1$

$$\frac{128z^3 + 160z^2 + 34z + 1}{64z^2 + 72z + 10} = 2z + \frac{16z^2 + 14z + 1}{64z^2 + 72z + 10}$$

So $B_1 = 2$.

Step 3: $A_1$

$$\frac{64z^2 + 72z + 10}{16z^2 + 14z + 1} = 4 + \frac{16z + 6}{16z^2 + 14z + 1}$$

So $A_1 = 4$.

Step 4: $B_2$

$$\frac{16z^2 + 14z + 1}{16z + 6} = z + \frac{8z + 1}{16z + 6}$$

So $B_2 = 1$.

Step 5: $A_2$

$$\frac{16z + 6}{8z + 1} = 2 + \frac{4}{8z + 1}$$

So $A_2 = 2$.

Step 6: $B_3$ and $A_3$

$$\frac{8z + 1}{4} = 2z + \frac{1}{4}$$

So $B_3 = 2$ and $A_3 = 4$.

$$G(z) = 1 + \cfrac{1}{2z + \cfrac{1}{4 + \cfrac{1}{z + \cfrac{1}{2 + \cfrac{1}{2z + \cfrac{1}{4}}}}}}$$

Ladder block diagram

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
      ^ -                         | +
      |                           v
      o<--E2--[1/A2]<----------- (S4)
      |                           ^ -
      v +                         |
     (S5)-->[z^-1/B3]-------------o Y3
      ^ -                         |
      |                           |
      +<--E3--[1/A3]<-------------+

Signal equations of the ladder ($o$ = take-off point, (S) = summer):

$$\begin{aligned} Y_1 &= \tfrac{z^{-1}}{B_1}(X - E_1), & E_1 &= \tfrac{1}{A_1}(Y_1 - Y_2) \\ Y_2 &= \tfrac{z^{-1}}{B_2}(E_1 - E_2), & E_2 &= \tfrac{1}{A_2}(Y_2 - Y_3) \\ Y_3 &= \tfrac{z^{-1}}{B_3}(E_2 - E_3), & E_3 &= \tfrac{1}{A_3}Y_3 \\ Y &= A_0X + Y_1 \end{aligned}$$

With the values: $1/B_1 = 0.5$, $1/A_1 = 0.25$, $1/B_2 = 1$, $1/A_2 = 0.5$, $1/B_3 = 0.5$, $1/A_3 = 0.25$. Three delays are used.

Simulating these equations gives the same impulse response as $G(z)$.