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Direct form II of y(n)=−0.1y(n−1)+0.72y(n−2)…
Short answer
H(z)=(0.7-0.252z^{-2})/(1+0.1z^{-1}-0.72z^{-2}). Direct Form II uses a shared delay line: w(n)=x(n)-0.1w(n-1)+0.72w(n-2), y(n)=0.7w(n)-0.252w(n-2). Only two delays are needed.
Determine the Direct Form II realization of the following system: y(n) = −0.1y(n−1) + 0.72y(n−2) + 0.7x(n) − 0.252x(n−2).
System function
y(n) = −0.1y(n−1) + 0.72y(n−2) + 0.7x(n) − 0.252x(n−2)
Taking the z-transform: Y(z)[1 + 0.1z⁻¹ − 0.72z⁻²] = X(z)[0.7 − 0.252z⁻²]
H(z) = (0.7 − 0.252z⁻²)/(1 + 0.1z⁻¹ − 0.72z⁻²) = 0.7(1 − 0.36z⁻²)/((1 − 0.8z⁻¹)(1 + 0.9z⁻¹))
Poles: 0.8 and −0.9 (inside the unit circle, so the system is stable). Zeros: ±0.6.
Direct form II (canonical)
Split H(z) into an all-pole part followed by an all-zero part, sharing one delay line:
w(n) = x(n) − 0.1·w(n−1) + 0.72·w(n−2)
y(n) = 0.7·w(n) − 0.252·w(n−2)
(Here the feedback coefficients are −a₁ = −0.1 and −a₂ = +0.72.) Only two delays are needed (order N = 2).
x(n) ──►(+)───────────●──── ×0.7 ──────►(+)──► y(n)
▲ │ w(n) ▲
│ [z⁻¹] │
├─ ×(−0.1) ◄─● w(n−1) │
│ │ │
│ [z⁻¹] │
└─ ×0.72 ◄───● w(n−2) ×(−0.252) ┘
Reading the diagram: the left adder forms w(n) = x(n) + (−0.1)w(n−1) + 0.72w(n−2); the right adder forms y(n) = 0.7w(n) + (−0.252)w(n−2). The feedback and feed-forward taps are both taken from the same two delays. There is no w(n−1) feed-forward tap, because the x(n−1) coefficient is 0.
| Quantity | Direct form I | Direct form II |
|---|---|---|
| Delays | 4 | 2 |
| Multipliers | 4 | 4 |
| Adders (two-input) | 3 | 3 |
Direct form II saves memory (2 delays instead of 4) and is called canonic.
✓ Answer checked by Claude Opus 5.5 (high effort)
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