Why decompose higher-order filters; ladder realization

Why higher-order pulse transfer functions are decomposed

A high-order $G(z)$ realized directly (direct or standard programming) is very sensitive to coefficient errors. The controller coefficients are stored with finite word length. A small rounding error in one coefficient of a high-order denominator can move all the poles a lot, even outside the unit circle, and make the filter unstable. So the transfer function is split into first- and second-order blocks (series, parallel or ladder programming):

• A coefficient error then affects only the one or two poles of that block.
• The structure is less sensitive to quantization, and the computations are better conditioned.
• Low-order blocks are easier to program, test and scale.

Ladder realization

Write the filter in positive powers of $z$:

$$G(z) = \frac{2 + 2.2z^{-1} + 0.2z^{-2}}{1 + 0.4z^{-1} - 0.12z^{-2}} = \frac{2z^2 + 2.2z + 0.2}{z^2 + 0.4z - 0.12}$$

In ladder programming, $G(z)$ is expanded as a continued fraction:

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

Each $1/(B_iz)$ is realized by a delay $z^{-1}/B_i$, and each $A_i$ by a constant.

Step 1: $A_0$.

$$G(z) = 2 + \frac{1.4z + 0.44}{z^2 + 0.4z - 0.12}$$

So $A_0 = 2$.

Step 2: $B_1$.

$$\frac{z^2 + 0.4z - 0.12}{1.4z + 0.44} = 0.7143z + \frac{0.0857z - 0.12}{1.4z + 0.44}$$

So $B_1 = 1/1.4 = 0.7143$.

Step 3: $A_1$.

$$\frac{1.4z + 0.44}{0.0857z - 0.12} = 16.333 + \frac{2.4}{0.0857z - 0.12}$$

So $A_1 = 49/3 = 16.333$.

Step 4: $B_2$ and $A_2$.

$$\frac{0.0857z - 0.12}{2.4} = 0.0357z - 0.05$$

So $B_2 = 1/28 = 0.0357$ and $A_2 = -1/0.05 = -20$.

$$G(z) = 2 + \cfrac{1}{0.7143z + \cfrac{1}{16.333 + \cfrac{1}{0.0357z - 0.05}}}$$

Ladder 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
      ^ -                         |
      |                           |
      +<--E2--[1/A2]<-------------+

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

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

With the values: $1/B_1 = 1.4$, $1/A_1 = 0.06122$, $1/B_2 = 28$, $1/A_2 = -0.05$.

The ladder needs only two delays. Because of its continued-fraction structure, it has low sensitivity to coefficient rounding.