22-Elec-A3 Signals and Communications · December 2016

Question 2 of 6: Discrete-Time System — Realisation, Impulse Response and Stability

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

Notes on this paper

Paper format. Professional Engineers Ontario / Engineers Canada national examination 07-Elec-A3 — Signals and Communications, December 2016. Three hours, closed book; a standard non-programmable calculator is the only permitted aid. Six questions of equal value (20 marks each); any five constitute a complete paper. All six are solved below, because this set is a study resource rather than an examination script.

Reference texts. B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. (Ch. 2–4 Fourier analysis, Ch. 4 amplitude modulation, Ch. 5 angle modulation, Ch. 6 sampling and pulse-code modulation); A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. (Ch. 3 Fourier series, Ch. 10 the z-transform); J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. (Ch. 3 z-transform, Ch. 7 realisation structures); S. Haykin and M. Moher, Communication Systems, 5th ed. (Ch. 3 amplitude modulation, Ch. 4 angle modulation, Ch. 7 digital transmission).

Notation used throughout. The normalised sinc function is $\operatorname{sinc}(x)=\dfrac{\sin \pi x}{\pi x}$, so that $\operatorname{sinc}(0)=1$ and the zeros sit at the non-zero integers. Average power of a periodic signal is $P=\frac{1}{T_0}\int_{T_0}|x(t)|^{2}\,dt$, and the exponential Fourier series is $x(t)=\sum_{n=-\infty}^{\infty} c_n e^{j n \omega_0 t}$ with $P=\sum_n |c_n|^{2}$ (Parseval).

Question 2: Discrete-Time System — Realisation, Impulse Response and Stability (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. The causal discrete-time transfer function $H(z) = \dfrac{1 + 2z^{-1}}{1 - \frac14 z^{-2}}$.

Find. A minimum-delay realisation, the impulse response $h(n)$, the input–output difference equation, and a stability verdict.

Approach. Realise the system in direct form II (which shares one delay chain between numerator and denominator and therefore uses the minimum number of delays), read the difference equation off the transfer function, and obtain $h(n)$ by partial fractions on the two real poles.

Minimum number of delay elements. The denominator has order 2 and the numerator order 1, so the system order is $N = 2$. Direct form II uses $\max(N, M) = 2$ delays — half of the four a direct-form-I structure would need. Defining the intermediate signal $$w(n) = x(n) + \tfrac14 w(n-2),\qquad y(n) = w(n) + 2\,w(n-1),$$ gives the realisation drawn below: one adder feeds a two-stage delay line, the tap $w(n-2)$ returns through a multiplier of $\tfrac14$, and the output adder takes $w(n)$ with unit gain plus $w(n-1)$ with gain 2.

Figure Q2.1 — direct-form-II realisation using the minimum of two delay elements; w(n) is the shared internal state.

With the structure fixed, the difference equation and the impulse response follow directly from the transfer function.

Difference equation. Cross-multiplying $Y(z)\left(1 - \tfrac14 z^{-2}\right) = X(z)\left(1 + 2z^{-1}\right)$ and inverting term by term, $$\boxed{\,y(n) - \tfrac14\,y(n-2) = x(n) + 2\,x(n-1)\,}$$ which answers part (c). Eliminating $w$ from the two realisation equations reproduces exactly this relation, confirming the block diagram.
Locate the poles. The denominator factors as $1 - \tfrac14 z^{-2} = \left(1 - \tfrac12 z^{-1}\right)\left(1 + \tfrac12 z^{-1}\right)$, so the poles are at $z = +\tfrac12$ and $z = -\tfrac12$, and there is one zero at $z = -2$.
Partial fractions. Write $H(z) = \dfrac{A}{1 - \frac12 z^{-1}} + \dfrac{B}{1 + \frac12 z^{-1}}$. Clearing denominators, $A\left(1 + \tfrac12 z^{-1}\right) + B\left(1 - \tfrac12 z^{-1}\right) = 1 + 2z^{-1}$, so $A + B = 1$ and $\tfrac12(A - B) = 2$. Hence $A = \tfrac52$ and $B = -\tfrac32$.
Invert to the time domain. Using the causal pair $\dfrac{1}{1 - az^{-1}} \leftrightarrow a^{n}u(n)$, $$\boxed{\,h(n) = \left[\tfrac52\left(\tfrac12\right)^{n} - \tfrac32\left(-\tfrac12\right)^{n}\right]u(n)\,}$$ The first few values are $h(0) = 1$, $h(1) = 2$, $h(2) = 0.25$, $h(3) = 0.5$, $h(4) = 0.0625$, which match the direct recursion $y(n) = \tfrac14 y(n-2) + x(n) + 2x(n-1)$ driven by a unit impulse. The response is a pair of interleaved geometric decays: the even samples decay as $\left(\tfrac12\right)^{n}$ with weight 1, the odd samples with weight 4.
Stability test. Both poles have $|z| = \tfrac12 \lt 1$, so for the causal (right-sided) system the region of convergence $|z| \gt \tfrac12$ contains the unit circle and the system is BIBO stable. Directly, $$\sum_{n=0}^{\infty}|h(n)| = H(z)\big|_{z=1} = \frac{1+2}{1-\frac14} = \boxed{4 \lt \infty}$$ (the absolute-value signs may be dropped because every $h(n)$ here is positive), so the answer to part (d) is yes, the system is stable.

Figure Q2.2 — impulse response h(n) = 2.5(0.5)ⁿ − 1.5(−0.5)ⁿ for n ≥ 0: interleaved geometric decays.

Quantity	Result
(a) Minimum delay elements	2 (direct form II)
(b) Impulse response	$h(n) = \left[2.5(0.5)^{n} - 1.5(-0.5)^{n}\right]u(n)$
(c) Difference equation	$y(n) - 0.25\,y(n-2) = x(n) + 2\,x(n-1)$
Poles / zero	$z = \pm 0.5$ / $z = -2$
(d) Stability	Stable; $\sum \|h(n)\| = 4$

Topic. Direct-form-II realisation, inverse z-transform by partial fractions, and the BIBO stability criterion for causal LTI systems.

Key relations. $H(z) = Y(z)/X(z)$; direct form II needs $\max(N,M)$ delays; $\dfrac{1}{1-az^{-1}} \leftrightarrow a^{n}u(n)$ for $|z| \gt |a|$; causal stability $\Leftrightarrow$ all poles inside the unit circle $\Leftrightarrow \sum_n |h(n)| \lt \infty$; and $\sum_n h(n) = H(1)$.

Why this works. Direct form II is the canonical minimum-memory structure because the poles and zeros act on the same internal state $w(n)$: the feedback path builds $w$ from $x$, and the feed-forward taps weight the same delayed copies to build $y$. Partial fractions work here because the two poles are real and distinct, so each term maps to a single geometric sequence. The zero at $z = -2$ lies outside the unit circle, which affects the phase response (the system is non-minimum-phase) but has no bearing whatsoever on stability — only pole locations matter.

Common pitfall. The most common error is expanding $H(z)$ rather than $H(z)/z$ when using the residue form, which shifts $h(n)$ by one sample; the safeguard used above is to check $h(0)$ and $h(1)$ against the difference equation driven by an impulse. A second error is declaring the system unstable because the zero at $z = -2$ sits outside the unit circle. A third is using four delays (direct form I) when the question explicitly asks for the smallest number.

Source. Proakis & Manolakis, Digital Signal Processing, 4th ed., §3.4 (inverse z-transform by partial fractions) and §7.2 (direct-form structures); Oppenheim & Willsky, Signals and Systems, 2nd ed., §10.7.

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