Question 2 of 5: Transfer function from a pole/zero plot and multiplier-free realisations

Question 2: Transfer function from a pole/zero plot and multiplier-free realisations (25 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. $H(z)$ as a ratio of polynomials in $z^{-1}$, its Direct Form I structure, and at least two alternative structures that use fewer multipliers.

[Figure not reproduced: Figure 1 redrawn: a conjugate pole pair at 0.25 +/- j0.25 (radius 0.354, angle 45 degrees) and a single zero at z = -1 on the unit circle. See the official exam paper.]

Approach. Build $H(z)$ from its factors, expand the conjugate pair into real coefficients, then exploit the two special features of this constellation — a unity numerator coefficient and pole coefficients that are exact negative powers of two — to remove multipliers.

1. Write the factored transfer function. A rational $H(z)$ with a zero at $z = z_1$ and poles at $z = p, p^{*}$ is $$H(z) = K\,\frac{1 - z_1 z^{-1}}{(1 - p z^{-1})(1 - p^{*} z^{-1})} ,$$ with $z_1 = -1$, $p = 0.25 + j0.25$ and $p^{*} = 0.25 - j0.25$. The plot fixes only the pole and zero locations, so the overall gain is arbitrary; take $K = 1$.
2. Expand the conjugate pair into real coefficients. For a conjugate pair the product of the two factors is real: $$(1 - p z^{-1})(1 - p^{*} z^{-1}) = 1 - 2\,\mathrm{Re}\{p\}\,z^{-1} + |p|^{2} z^{-2}.$$ Here $2\,\mathrm{Re}\{p\} = 2(0.25) = 0.5$ and $|p|^{2} = 0.25^{2} + 0.25^{2} = 0.125$, giving the denominator $1 - 0.5 z^{-1} + 0.125 z^{-2}$. The numerator is simply $1 - (-1)z^{-1} = 1 + z^{-1}$.
3. State the transfer function (part a). $$\boxed{\;H(z) = \frac{1 + z^{-1}}{1 - 0.5\,z^{-1} + 0.125\,z^{-2}}\;}$$ equivalently the difference equation $$y[n] = x[n] + x[n-1] + 0.5\,y[n-1] - 0.125\,y[n-2].$$
4. Check the result before drawing it. The pole radius is $|p| = 0.25\sqrt{2} = 0.354 \lt 1$, so the filter is stable, and the pole angle is $\theta = 45^{\circ}$, i.e. a resonance near $\omega = \pi/4$. At $z = 1$ (dc) the gain is $H(1) = 2/(1 - 0.5 + 0.125) = 3.2$; at $z = -1$ (Nyquist) the zero forces $H(-1) = 0$. A low-pass character with a null at the folding frequency is exactly what the constellation promises.
5. Draw Direct Form I (part b). Direct Form I realises the numerator and denominator as two separate delay chains: the input feeds a one-sample delay whose tap is added straight in (numerator coefficient $+1$ needs no multiplier), while the output feeds a two-deep delay line whose taps are scaled by $+0.5$ and $-0.125$ and summed back. Two delay elements are used for the feed-forward path and two more for the feedback path — the non-canonic price Direct Form I pays for its simplicity.

Part (c) asks for structures that remove one of those two multipliers. Two independent routes exist, and both exploit the arithmetic of this particular pole pair.

6. Structure 1 — the coupled (Gold–Rader) form. Write the poles in polar form, $p = r e^{\,j\theta}$ with $$r = |p| = 0.25\sqrt{2} = 0.3536, \qquad \theta = 45^{\circ}.$$ A coupled resonator propagates the state vector by the rotation-and-scaling matrix $$\begin{bmatrix} s_1[n] \\ s_2[n] \end{bmatrix} = \begin{bmatrix} r\cos\theta & -r\sin\theta \\ r\sin\theta & \phantom{-}r\cos\theta \end{bmatrix} \begin{bmatrix} s_1[n-1] \\ s_2[n-1] \end{bmatrix} + \begin{bmatrix} 1 \\ 0 \end{bmatrix} x[n],$$ whose characteristic polynomial is the required $1 - 2r\cos\theta\,z^{-1} + r^{2} z^{-2}$. Because $\theta = 45^{\circ}$ exactly, $$\boxed{\,r\cos\theta = r\sin\theta = 0.25\,}$$ so the two distinct Direct Form I coefficients $0.5$ and $-0.125$ are replaced by a single coefficient value, $0.25$, used on every arm.
7. Take the all-pole output off the coupled section. State $s_1$ realises $(1 - r\cos\theta\, z^{-1})/D(z)$ and $s_2$ realises $(r\sin\theta\, z^{-1})/D(z)$, so the pure all-pole signal $1/D(z)$ is the combination $s_1 + \cot\theta \cdot s_2$. At $\theta = 45^{\circ}$, $\cot\theta = 1$, so that combination is a plain adder: $$w[n] = s_1[n] + s_2[n], \qquad y[n] = w[n] + w[n-1].$$ The numerator $1 + z^{-1}$ is one delay and one adder, again multiplier-free. This structure is also the numerically best-behaved of the three: its coefficient sensitivity is uniform over the whole unit disc, whereas Direct Form I loses pole-placement resolution near $z = \pm 1$.

8. Structure 2 — binary-shift (multiplier-free) realisation. The two feedback coefficients are exact negative powers of two, $$0.5 = 2^{-1}, \qquad 0.125 = 2^{-3},$$ so in a fixed-point implementation each product is a hard-wired arithmetic right shift, and no general multiplier is instantiated at all: $$y[n] = x[n] + x[n-1] + \bigl(y[n-1] \gg 1\bigr) - \bigl(y[n-2] \gg 3\bigr).$$ Shifting is exact — unlike a rounded coefficient it introduces no pole-placement error — and the structure below removes both multipliers, not merely one.

9. Structure 3 (bonus) — nested, or Horner, feedback. The denominator factors by inspection as $$1 - 0.5 z^{-1} + 0.125 z^{-2} = 1 - 0.5 z^{-1}\bigl(1 - 0.25 z^{-1}\bigr),$$ so the recursion can be nested, $$y[n] = x[n] + x[n-1] + 0.5\Bigl(y[n-1] - 0.25\,y[n-2]\Bigr).$$ Both surviving constants, $0.5 = 2^{-1}$ and $0.25 = 2^{-2}$, are again shifts, and the same physical scaler can be time-shared between them. This form is useful when a single hardware multiplier must serve the whole section.