--- title: "Ladder programming of a third-order filter" subject: "Digital Control System" chapter: 4 key: "semester_6/digital_control_system:ladder-third-order-128" source: https://www.bctengineer.com/q/semester_6/digital_control_system/ladder-third-order-128 site: BCT Engineering --- # Ladder programming of a third-order filter Digital Control System · Chapter 4: Design and Compensation of Discrete Time Control System - Asked in: 2075 Baisakh, 8 marks - Video: [Ladder Structure | DSP Ladder structure | Routh array formation | Realization of Ladder structure](https://www.youtube.com/watch?v=6zCP30VUoFU) — Grow with ashish ( Vlog & YouTube) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/digital_control_system/ladder-third-order-128 - Whole chapter: https://www.bctengineer.com/md/semester_6/digital_control_system/chapter-4 ## Question Realize the digital filter by ladder programming: G(z)=(128z⁻³+224z⁻²+106z⁻¹+11)/(128z⁻³+160z⁻²+34z⁻¹+1). ## Short answer Ladder realization expands G(z) in continued fraction: A0=1, B1=2, A1=4, B2=1, A2=2, B3=2, A3=4, giving G(z)=1+1/(2z+1/(4+1/(z+1/(2+1/(2z+1/4))))). The block diagram alternates constant gains A_i and delay branches z^-1/B_i, requiring three delays. ## Answer **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 $A_3 = 4$. $$ G(z) = 1 + \cfrac{1}{2z + \cfrac{1}{4 + \cfrac{1}{z + \cfrac{1}{2 + \cfrac{1}{2z + \cfrac{1}{4}}}}}} $$ | $A_0$ | $B_1$ | $A_1$ | $B_2$ | $A_2$ | $B_3$ | $A_3$ | |---|---|---|---|---|---|---| | 1 | 2 | 4 | 1 | 2 | 2 | 4 | #### Ladder block diagram ```text 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 $G(z)$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A third-order filter in ladder programming has how many constants A and B, and how many delays? A. A0..A2 and B1..B2; 2 delays B. A0..A3 and B1..B3; 6 delays C. A0..A3 and B1..B3; 3 delays D. A1..A3 only; 3 delays Answer: C. The continued fraction has A0, B1, A1, B2, A2, B3, A3 and each B term has one delay, so 3 delays. (Key point: Ladder structure for third order) 2. The answer assumes G(z) is written in positive powers of z. What does this give? A. A filter with unit gain B. A first-order filter C. A parallel structure D. Integer ladder constants and a stable filter Answer: D. With positive powers the poles are inside the unit circle and the continued fraction constants come out as integers. (Key point: Assumption about form) 3. For G(z) = (128z^3+224z^2+106z+11)/(128z^3+160z^2+34z+1), A0 equals: A. 128 B. 224/160 C. 1 D. 0 Answer: C. Equal degrees, so A0 is the ratio of the leading coefficients 128/128 = 1. (Key point: A0 from division) 4. The continued-fraction constants (A0,B1,A1,B2,A2,B3,A3) obtained are: A. 1, 2, 4, 1, 2, 2, 4 B. 2, 1, 4, 2, 1, 4, 2 C. 1, 4, 2, 1, 2, 4, 2 D. 1, 2, 2, 1, 4, 2, 4 Answer: A. Successive divisions give A0=1, B1=2, A1=4, B2=1, A2=2, B3=2, A3=4. 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