Question 2 of 6: Discrete-Time IIR System — Realization, Impulse Response, Stability

Question 2: Discrete-Time IIR System — Realization, Impulse Response, Stability (1/6 of paper)

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. $H(z)=Y(z)/X(z)=\dfrac{1+2z^{-1}}{1-\tfrac14z^{-2}}$, a second-order IIR system (one zero, an all-pole quadratic denominator).

Find. (a) a minimal-delay block diagram; (b) $h[n]$; (c) the difference equation; (d) BIBO stability.

Approach. Factor the denominator to expose the two real poles, realize $H(z)$ in Direct Form II (the canonical structure that needs only $\max(\text{order of num.},\text{order of denom.})=2$ delays, shared between the pole and zero sections), invert by partial fractions for $h[n]$, read the difference equation straight off $H(z)$, and check the pole magnitudes for stability.

1. Part (a) — minimal-delay realization. Split $H(z)=H_p(z)H_z(z)$ with $H_p(z)=1/(1-\tfrac14z^{-2})$ (poles) processed first and $H_z(z)=1+2z^{-1}$ (the zero) tapped from the same delay chain (Direct Form II), using the intermediate node $w[n]$: $$w[n]=x[n]+\tfrac14w[n-2],\qquad y[n]=w[n]+2w[n-1]$$ This needs only $2$ unit-delay elements total (not 2 for the pole section plus 2 more for the zero section), which is the smallest possible count for a second-order system.
Direct Form II realization of H(z): two shared unit-delay elements, gains 1/4 (feedback) and 2 (feed-forward).
2. Part (b) — impulse response. Factor the denominator, $1-\tfrac14z^{-2}=(1-\tfrac12z^{-1})(1+\tfrac12z^{-1})$, and expand in partial fractions: $$H(z)=\frac{1+2z^{-1}}{(1-\tfrac12z^{-1})(1+\tfrac12z^{-1})} =\frac{A}{1-\tfrac12z^{-1}}+\frac{B}{1+\tfrac12z^{-1}}$$ Matching $1+2z^{-1}=A(1+\tfrac12z^{-1})+B(1-\tfrac12z^{-1})$ gives $A+B=1$ and $\tfrac12(A-B)=2$, so $A=2.5$, $B=-1.5$. Each term inverts as a one-sided geometric sequence: $$h[n]=\boxed{\big[2.5(0.5)^n-1.5(-0.5)^n\big]u[n]}$$ verified against a direct recursion of the difference equation in part (c) for $n=0,\dots,29$.
3. Part (c) — difference equation. Cross-multiplying $H(z)$, $$Y(z)\Big(1-\tfrac14z^{-2}\Big)=X(z)\Big(1+2z^{-1}\Big)$$ which is, term by term, $$\boxed{y[n]=x[n]+2x[n-1]+\tfrac14y[n-2]}$$
4. Part (d) — stability. The poles found in part (b) are $z=\pm0.5$, and $|\pm0.5|=0.5<1$ for both, so the system is $$\boxed{\text{BIBO stable}}$$ (the geometric terms $(0.5)^n$ and $(-0.5)^n$ in $h[n]$ both decay to zero, and $\sum_n|h[n]|<\infty$).