22-Elec-A3 Signals and Communications · May 2015

Question 2 of 7: Second-Order Discrete-Time System — Transfer Function, Impulse Response and Canonical Realisation

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 07-Elec-A3 Signals and Communications, May 2015 — 3 hours, closed book, non-programmable calculator permitted. Seven questions, all of equal value (20 marks each); any five constitute a complete paper and only the first five in the answer book are marked. All seven are solved here, since the set is a study resource.

Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication Systems, 5th ed.; B. P. Lathi, Linear Systems and Signals, 2nd ed.; A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed.; J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed.; S. Haykin & M. Moher, Communication Systems, 5th ed.

Check: pulse definition. The pulse is p(t) = t/a — a ramp, not a constant — and every part of Question 1 is solved against that reading. A constant-height pulse would make parts (a)–(c) trivial and would not produce the odd-harmonic structure the question is clearly aiming at.

Question 2: Second-Order Discrete-Time System — Transfer Function, Impulse Response and Canonical Realisation (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. A causal linear constant-coefficient difference equation with feedback coefficients $a_1 = a_2 = 1/4$ and feed-forward coefficients $b_0 = b_1 = 1$, at rest before the input is applied.

Find. $H(z)$, the impulse response $h(n)$ in closed form, and the realisation using the fewest delay elements.

Approach. Take the $z$-transform term by term using the shift property, factor the denominator to locate the poles, expand in partial fractions to invert, then draw the Direct Form II (canonical) structure, which needs only as many delays as the system order.

Transfer function. Applying $\mathcal{Z}\{y(n-k)\} = z^{-k}Y(z)$ to every term and collecting: $$Y(z)\left(1 - \tfrac{1}{4}z^{-1} - \tfrac{1}{4}z^{-2}\right) = X(z)\left(1 + z^{-1}\right)$$ $$\boxed{\,H(z) = \frac{1 + z^{-1}}{1 - \tfrac{1}{4}z^{-1} - \tfrac{1}{4}z^{-2}} = \frac{z(z+1)}{z^2 - 0.25z - 0.25}\,}$$ The DC gain is $H(1) = 2/(1-0.5) = 4$, so the filter has a low-pass character with a transmission zero at $z = -1$ (that is, at the Nyquist frequency).
Locate the poles. Solving $z^2 - 0.25z - 0.25 = 0$, $$z = \frac{0.25 \pm \sqrt{0.0625 + 1}}{2} = \frac{0.25 \pm 1.030776}{2}$$ $$p_1 = 0.640388, \qquad p_2 = -0.390388$$ Both are real and lie strictly inside the unit circle ($|p_1| \lt 1$, $|p_2| \lt 1$), so the system is BIBO stable and the impulse response decays as a sum of two real geometric sequences — one monotone, one alternating in sign.
Partial-fraction expansion. Expanding $H(z)/z$ to keep the inverse transforms in the standard $p^n u(n)$ form, $$\frac{H(z)}{z} = \frac{z+1}{(z-p_1)(z-p_2)} = \frac{A}{z-p_1} + \frac{B}{z-p_2}$$ with residues $A = \dfrac{p_1+1}{p_1-p_2} = 1.59141$ and $B = \dfrac{p_2+1}{p_2-p_1} = -0.59141$.
Impulse response. Inverting term by term gives $$\boxed{\,h(n) = \left[1.5914\,(0.6404)^{n} - 0.5914\,(-0.3904)^{n}\right]u(n)\,}$$ The first samples are $h(0)=1$, $h(1)=1.25$, $h(2)=0.5625$, $h(3)=0.4531$, $h(4)=0.2539$. Driving the original difference equation with a unit impulse reproduces exactly these values, which confirms both the residues and the algebra; note in particular that $A+B = 1 = h(0)$, as it must be for a filter whose numerator and denominator have equal degree in $z^{-1}$.
Minimum-delay realisation. Direct Form I would need four delay elements (two for the input history, two for the output history). Factoring $H(z)$ as a cascade of the all-pole part and the all-zero part and swapping their order lets the two delay chains be shared, giving the Direct Form II canonical structure with an intermediate signal $w(n)$: $$w(n) = x(n) + \tfrac{1}{4}w(n-1) + \tfrac{1}{4}w(n-2), \qquad y(n) = w(n) + w(n-1)$$ This uses two delay elements, the theoretical minimum for a second-order system.
Figure Q2.1 — Direct Form II (canonical) realisation, two delay elements: w(n) = x(n) + ¼w(n-1) + ¼w(n-2), y(n) = w(n) + w(n-1).

Quantity	Result
Transfer function	$H(z) = (1+z^{-1})/(1-0.25z^{-1}-0.25z^{-2})$
Poles	$p_1 = 0.6404$, $p_2 = -0.3904$ (stable)
Zeros	$z = -1$ and $z = 0$
Impulse response	$h(n) = [1.5914(0.6404)^n - 0.5914(-0.3904)^n]u(n)$
First samples	1, 1.25, 0.5625, 0.4531, 0.2539
DC gain	$H(1) = 4$
Minimum realisation	Direct Form II, 2 delay elements

Topic. $z$-domain analysis of a causal IIR difference equation and canonical filter structures.

Key relations. $\mathcal{Z}\{y(n-k)\} = z^{-k}Y(z)$ for a system at rest; $H(z) = \sum b_k z^{-k}\big/\left(1-\sum a_k z^{-k}\right)$; $\mathcal{Z}^{-1}\{z/(z-p)\} = p^n u(n)$; stability for a causal system $\Leftrightarrow$ all poles inside $|z| = 1$.

Why this works. The difference equation is linear with constant coefficients, so the $z$-transform converts the time-shifts into algebraic factors and the whole input–output relation into a rational function. Partial fractions then decompose that rational function into first-order pieces whose inverse transforms are memorised geometric sequences. The Direct Form II structure exploits the fact that cascaded LTI blocks commute: putting the recursive (pole) section first means the delay line already holds the history the feed-forward (zero) section needs, so one chain serves both.

Common pitfall. Expanding $H(z)$ rather than $H(z)/z$ in partial fractions — the residues then belong to $1/(z-p)$, whose inverse is the shifted sequence $p^{n-1}u(n-1)$, and forgetting the shift produces an impulse response that is wrong at every sample. Also watch the sign convention: the coefficients $a_1, a_2$ appear on the right-hand side of the difference equation with plus signs, so they enter the denominator with minus signs.

Source. Proakis & Manolakis, Digital Signal Processing, Ch. 3 ($z$-transform, inversion by partial fractions) and Ch. 9 (Direct Form I/II realisations); Oppenheim & Willsky, Signals and Systems, Ch. 10.

← Question 1 Next: Question 3 →

Quantity	Result
Transfer function	\(H(z) = (1+z^{-1})/(1-0.25z^{-1}-0.25z^{-2})\)
Poles	\(p_1 = 0.6404\), \(p_2 = -0.3904\) (stable)
Zeros	\(z = -1\) and \(z = 0\)
Impulse response	\(h(n) = [1.5914(0.6404)^n - 0.5914(-0.3904)^n]u(n)\)
First samples	1, 1.25, 0.5625, 0.4531, 0.2539
DC gain	\(H(1) = 4\)
Minimum realisation	Direct Form II, 2 delay elements