Cascade form realization of second order system

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]

 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.