Question 5 of 7: System Reconstruction from Poles, Zeros and d.c. Gain

Question 5: System Reconstruction from Poles, Zeros and d.c. Gain (20 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. $H(z)$ and its ROC; the difference equation; $h[n]$ in closed form; the total (zero-input plus zero-state) response to the given input with the given initial conditions; and a canonic realisation with its operation count.

Approach. Build $H(z)$ from its pole-zero form with an unknown gain, fix the gain from $H(1)$, then read the difference equation off the coefficients. Part (c) is a partial-fraction inversion, part (d) is the same inversion applied to the unilateral z-transform with the initial conditions carried through, and part (e) follows from the order.

1. (a) Assemble $H(z)$ and fix the gain. With the given zeros and poles,
   $$H(z) = K\,\frac{(z - 0)(z + 1)}{(z - 0.5)(z + 0.5)} = K\,\frac{z(z+1)}{z^{2} - 0.25}.$$

   Imposing the d.c. gain, $H(1) = K\,\frac{(1)(2)}{1 - 0.25} = \frac{8K}{3} = \frac{8}{3}$, so $K = 1$. Dividing numerator and denominator by $z^{2}$ puts it in the causal $z^{-1}$ form:

   $$\boxed{H(z) = \frac{1 + z^{-1}}{1 - \tfrac{1}{4} z^{-2}}, \qquad \text{ROC: } |z| \gt 0.5.}$$

   The ROC is the exterior of the outermost pole because the system is stated to be causal; since $0.5 \lt 1$ that region contains the unit circle, which is consistent with the stated stability. Both facts had to agree, and here they do.

2. (b) Read the difference equation from the coefficients. Cross-multiplying $Y(z)\left(1 - \tfrac{1}{4}z^{-2}\right) = X(z)\left(1 + z^{-1}\right)$ and inverting term by term,
   $$\boxed{y[n] = \tfrac{1}{4}\,y[n-2] + x[n] + x[n-1].}$$

   There is no $y[n-1]$ term because the two poles are symmetric about the origin and their sum vanishes.

3. (c) Invert $H(z)$ by partial fractions. Factoring the denominator as $(1 - \tfrac{1}{2}z^{-1})(1 + \tfrac{1}{2}z^{-1})$ and writing
   $$\frac{1 + z^{-1}}{(1 - \tfrac{1}{2}z^{-1})(1 + \tfrac{1}{2}z^{-1})} = \frac{A}{1 - \tfrac{1}{2}z^{-1}} + \frac{B}{1 + \tfrac{1}{2}z^{-1}},$$

   the residues are $A = \tfrac{3}{2}$ and $B = -\tfrac{1}{2}$. Because the ROC is the outer region, both terms invert to right-sided exponentials:

   $$\boxed{h[n] = \left[\tfrac{3}{2}\left(\tfrac{1}{2}\right)^{n} - \tfrac{1}{2}\left(-\tfrac{1}{2}\right)^{n}\right] u[n].}$$

   Checking the first few values against the difference equation: $h[0] = 1.5 - 0.5 = 1$, $h[1] = 0.75 + 0.25 = 1$, $h[2] = 0.375 - 0.125 = 0.25$ — and indeed $h[2] = \tfrac{1}{4}h[0] = 0.25$. The alternating $(-1/2)^{n}$ term makes $h[n]$ ring slightly rather than decay monotonically.

4. (d) Total response by the unilateral z-transform. Taking the one-sided transform of the difference equation, the delayed output term contributes the initial conditions:
   $$Y(z)\left(1 - \tfrac{1}{4}z^{-2}\right) = \tfrac{1}{4}\left(y[-2] + z^{-1}y[-1]\right) + X(z)\left(1 + z^{-1}\right) = 2 + X(z)\left(1 + z^{-1}\right),$$

   since $y[-2] = 8$, $y[-1] = 0$ and the input is causal so $x[-1] = 0$. The two parts separate cleanly. The zero-input part is

   $$Y_{zi}(z) = \frac{2}{1 - \tfrac{1}{4}z^{-2}} \;\Rightarrow\; y_{zi}[n] = \left[\left(\tfrac{1}{2}\right)^{n} + \left(-\tfrac{1}{2}\right)^{n}\right] u[n],$$

   which correctly gives $y_{zi}[0] = 2 = \tfrac{1}{4}(8)$. With $X(z) = 1/(1 - \tfrac{1}{3}z^{-1})$ the zero-state part is

   $$Y_{zs}(z) = \frac{1 + z^{-1}}{\left(1 - \tfrac{1}{2}z^{-1}\right)\left(1 + \tfrac{1}{2}z^{-1}\right)\left(1 - \tfrac{1}{3}z^{-1}\right)},$$

   whose residues are $4.5$, $-0.3$ and $-3.2$ at the poles $0.5$, $-0.5$ and $1/3$ respectively, so

   $$y_{zs}[n] = \left[4.5\left(\tfrac{1}{2}\right)^{n} - 0.3\left(-\tfrac{1}{2}\right)^{n} - 3.2\left(\tfrac{1}{3}\right)^{n}\right]u[n].$$

   Adding the two contributions and collecting like modes gives the total response:

   $$\boxed{y[n] = \left[5.5\left(\tfrac{1}{2}\right)^{n} + 0.7\left(-\tfrac{1}{2}\right)^{n} - 3.2\left(\tfrac{1}{3}\right)^{n}\right] u[n].}$$

   Verification against the recursion: $y[0] = \tfrac{1}{4}(8) + 1 + 0 = 3$, and the closed form gives $5.5 + 0.7 - 3.2 = 3$; $y[1] = \tfrac{1}{4}(0) + \tfrac{1}{3} + 1 = \tfrac{4}{3}$, and the closed form gives $2.75 - 0.35 - 1.0\overline{6} = 1.3\overline{3}$. All three modes have magnitude less than one, so the response decays to zero — as it must for a stable system with a decaying input.

5. (e) Canonic realisation. The system is second order, so the minimum possible number of delay elements is two. Direct Form II achieves it by placing the single delay chain between the feedback and feedforward arms:
   $$w[n] = x[n] + \tfrac{1}{4}w[n-2], \qquad y[n] = w[n] + w[n-1].$$
   Canonic Direct-Form II realisation. The two delays are shared by the recursive and non-recursive halves, which is what makes the structure canonic.

   Counting: one multiplication (by $\tfrac{1}{4}$), two additions (one at each summing node) and two delay elements. Both feedforward coefficients are unity, so they cost nothing. Since $\tfrac{1}{4} = 2^{-2}$, even that one multiplier reduces to a two-bit arithmetic right shift in a fixed-point implementation, giving a genuinely multiplier-free realisation.

   $$\boxed{2 \text{ delays}, \quad 1 \text{ multiplication}, \quad 2 \text{ additions.}}$$