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Ladder programming of a third-order controller

Short answer

Ladder programming expands G(z) as a continued fraction. For G(z)=(8z^3+7z^2+10z+6)/(4z^3+6z^2+9z+10), the gains are A0=2, B1=-0.8, A1=12.5, B2=-0.02051, A2=-3.8604, B3=0.05031, A3=-10.04. Three delays are used.

Realize the given digital controller by ladder programming whose transfer function is (8z³+7z²+10z+6)/(4z³+6z²+9z+10).

Ladder programming expands G(z)G(z) as a continued fraction:

G(z)=A0+1B1z+1A1+1B2z+1A2+1B3z+1A3G(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}}}}}}

with G(z)=8z3+7z2+10z+64z3+6z2+9z+10G(z) = \dfrac{8z^3 + 7z^2 + 10z + 6}{4z^3 + 6z^2 + 9z + 10}.

Step 1: A0=8/4=2A_0 = 8/4 = 2

G(z)=2+−5z2−8z−144z3+6z2+9z+10G(z) = 2 + \frac{-5z^2 - 8z - 14}{4z^3 + 6z^2 + 9z + 10}

Step 2: B1B_1

4z3+6z2+9z+10−5z2−8z−14=−0.8z+−0.4z2−2.2z+10−5z2−8z−14\frac{4z^3 + 6z^2 + 9z + 10}{-5z^2 - 8z - 14} = -0.8z + \frac{-0.4z^2 - 2.2z + 10}{-5z^2 - 8z - 14}

So B1=−0.8B_1 = -0.8.

Step 3: A1A_1

−5z2−8z−14−0.4z2−2.2z+10=12.5+19.5z−139−0.4z2−2.2z+10\frac{-5z^2 - 8z - 14}{-0.4z^2 - 2.2z + 10} = 12.5 + \frac{19.5z - 139}{-0.4z^2 - 2.2z + 10}

So A1=12.5A_1 = 12.5.

Step 4: B2B_2

−0.4z2−2.2z+1019.5z−139=−0.02051z+−5.0513z+1019.5z−139\frac{-0.4z^2 - 2.2z + 10}{19.5z - 139} = -0.02051z + \frac{-5.0513z + 10}{19.5z - 139}

So B2=−0.02051B_2 = -0.02051.

Step 5: A2A_2

19.5z−139−5.0513z+10=−3.8604+−100.396−5.0513z+10\frac{19.5z - 139}{-5.0513z + 10} = -3.8604 + \frac{-100.396}{-5.0513z + 10}

So A2=−3.8604A_2 = -3.8604.

Step 6: B3B_3 and A3A_3

−5.0513z+10−100.396=0.05031z−0.09961\frac{-5.0513z + 10}{-100.396} = 0.05031z - 0.09961

So B3=0.05031B_3 = 0.05031 and A3=−1/0.09961=−10.04A_3 = -1/0.09961 = -10.04.

A0A_0B1B_1A1A_1B2B_2A2A_2B3B_3A3A_3
2−0.812.5−0.02051−3.86040.05031−10.04

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 (oo = take-off point, (S) = summer):

Y1=z−1B1(X−E1),E1=1A1(Y1−Y2)Y2=z−1B2(E1−E2),E2=1A2(Y2−Y3)Y3=z−1B3(E2−E3),E3=1A3Y3Y=A0X+Y1\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}

Numerical gains: 1/B1=−1.251/B_1 = -1.25, 1/A1=0.081/A_1 = 0.08, 1/B2=−48.751/B_2 = -48.75, 1/A2=−0.25901/A_2 = -0.2590, 1/B3=19.8751/B_3 = 19.875, 1/A3=−0.09961/A_3 = -0.0996.

The expansion was checked by substituting back. It reproduces G(z)G(z) exactly, and the ladder's impulse response matches the direct difference equation. Three delays are used.

✓ Answer checked by Claude Opus 5.5 (high effort)

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