--- title: "Cascade form realization of second order system" subject: "Digital Signal Analysis and Processing" chapter: 4 key: "semester_7/digital_signal_analysis_and_processing:cascade-form-3-4-1-8" source: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/cascade-form-3-4-1-8 site: BCT Engineering --- # Cascade form realization of second order system Digital Signal Analysis and Processing · Chapter 4: Discrete Filter Structures - Asked in: 2070 Chaitra, 4 marks - Video: [Digital Filters: Cascade Form Structure & Realization | EC Academy](https://www.youtube.com/watch?v=dQS17iEutmA) — EC Academy - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/digital_signal_analysis_and_processing/cascade-form-3-4-1-8 - Whole chapter: https://www.bctengineer.com/md/semester_7/digital_signal_analysis_and_processing/chapter-4 ## Question Determine the Cascade Form realization of the following system: y[n] − (3/4)y[n−1] + (1/8)y[n−2] − x[n] − 2x[n−1] = 0. ## Short answer Factor denominator (1−0.5z⁻¹)(1−0.25z⁻¹). Cascade two first-order DF-II sections: H₁(z)=1/(1−0.5z⁻¹) then H₂(z)=(1+2z⁻¹)/(1−0.25z⁻¹). Total 2 delays; feedback 0.5 and 0.25, feedforward tap 2 in the second section. ## Answer Rewrite the equation as y[n] = (3/4)y[n−1] − (1/8)y[n−2] + x[n] + 2x[n−1]. H(z) = (1 + 2z⁻¹) / (1 − (3/4)z⁻¹ + (1/8)z⁻²) #### Factorise the denominator 1 − 0.75z⁻¹ + 0.125z⁻² = (1 − 0.5z⁻¹)(1 − 0.25z⁻¹) (poles at z = 0.5 and 0.25, both inside the unit circle, so the system is stable). #### Split into first-order sections H(z) = H₁(z) · H₂(z), where - H₁(z) = 1 / (1 − 0.5z⁻¹) - H₂(z) = (1 + 2z⁻¹) / (1 − 0.25z⁻¹) (Another grouping, (1+2z⁻¹)/(1−0.5z⁻¹) times 1/(1−0.25z⁻¹), is equally valid.) Each section in Direct Form II uses one delay: - Section 1: w₁[n] = x[n] + 0.5w₁[n−1]; y₁[n] = w₁[n] - Section 2: w₂[n] = y₁[n] + 0.25w₂[n−1]; y[n] = w₂[n] + 2w₂[n−1] ```text x[n]──►(+)──────┬──────────►(+)──────┬──►(+)──► y[n] ▲ │ y1 ▲ │ ▲ │ ▼ z⁻¹ │ ▼ z⁻¹│ └─[0.5]─┘ └[0.25]┤ │ └[2]─┘ ◄──── section 1 ────► ◄──── section 2 ────► ``` Section 1: adder with feedback 0.5 through one delay. Section 2: adder with feedback 0.25 through one delay, and a feed-forward tap of 2 from the delayed value to the output adder. The cascade needs 2 delays, 3 multipliers and 3 adders; the overall transfer function is the product H₁H₂ = H(z). **Answer: H(z) = [1/(1 − 0.5z⁻¹)] · [(1 + 2z⁻¹)/(1 − 0.25z⁻¹)], two first-order Direct Form II sections in cascade.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does 1 - 0.75z^-1 + 0.125z^-2 factorise? A. (1 - 0.5z^-1)(1 - 0.25z^-1) B. (1 + 0.5z^-1)(1 + 0.25z^-1) C. (1 - 0.75z^-1)(1 - 0.125z^-1) D. (1 - 0.5z^-1)(1 + 0.25z^-1) Answer: A. Roots 0.5 and 0.25 have sum 0.75 and product 0.125. (Key point: Factorising denominator) 2. What defines a cascade-form realization? A. Parallel sum of first-order sections B. Product H(z) = H1(z)H2(z) of low-order sections in series C. A single section with all delays D. A section for each pole with no zeros Answer: B. The system function is split into a product of first- or second-order sections connected in series. (Key point: Cascade principle) 3. How many delays does the cascade of two first-order DF-II sections need? A. 1 B. 2 C. 3 D. 4 Answer: B. Each first-order DF-II section uses one delay, so 2 in total. (Key point: Cost of realization) 4. Why is the system stable? A. The zero at -2 is outside the circle B. Poles 0.5 and 0.25 are inside the unit circle C. The numerator has two terms D. The denominator has a constant term 1 Answer: B. Stability depends on poles only; both are inside the unit circle. 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