22-Elec-A3 Signals and Communications · May 2017

Question 2 of 5: Discrete-Time LTI System — Convolution, Structure 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 16-Elec-A3 — Signals and Communications, May 2017. Closed book (one non-programmable calculator permitted), 3 hours, 4 pages. Five questions, all compulsory, all of equal value (20 marks each). All five are solved in full below.

Reference texts. B. P. Lathi & Z. Ding, Modern Digital and Analog Communication Systems, 5th ed. (AM/DSB power and efficiency, PCM, FM/Carson); B. P. Lathi, Linear Systems and Signals, 2nd ed. (Fourier series, filtering, average power); A. V. Oppenheim & A. S. Willsky, Signals and Systems, 2nd ed. (LTI convolution, BIBO stability); J. G. Proakis & D. G. Manolakis, Digital Signal Processing, 4th ed. (FIR structures, transfer functions); C. K. Alexander & M. N. O. Sadiku, Fundamentals of Electric Circuits, 7th ed. (first-order RC response).

Question 2: Discrete-Time LTI System — Convolution, Structure 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. $h(n)=\{1,1,1,1,\tfrac12\}$ for $n=0,\ldots,4$ and zero elsewhere; $x(n)=1-\dfrac{|n-6|}{6}$ for $0\le n\le 12$, zero elsewhere — a symmetric triangle rising from $x(0)=0$ to $x(6)=1$ and back to $x(12)=0$ in steps of 1/6.

Find. The output $y(n)=x(n)*h(n)$, the input–output difference equation, a block diagram, the transfer function $H(z)$, and a justified BIBO-stability verdict.

Figure Q2.1 - Impulse response h(n).

Approach. The system is FIR, so every part follows from the five impulse-response samples: convolve for (a), read the taps as coefficients for (b) and (c), take the $z$-transform for (d), and test absolute summability for (e).

(a) Convolve input and impulse response. The output length is $13+5-1=17$ samples, $0\le n\le 16$, and $$y(n)=\sum_{k=0}^{4}h(k)\,x(n-k)=x(n)+x(n-1)+x(n-2)+x(n-3)+\tfrac12 x(n-4)$$ Each output sample is a sliding sum of four consecutive input samples plus half of the fifth. Evaluating with $x=\{0,\tfrac16,\tfrac26,\ldots,1,\ldots,\tfrac16,0\}$: $$\boxed{y(n)=\left\{0,\ \tfrac16,\ \tfrac12,\ 1,\ \tfrac53,\ \tfrac{29}{12},\ \tfrac{19}{6},\ \tfrac{43}{12},\ \tfrac{11}{3},\ \tfrac{41}{12},\ \tfrac{17}{6},\ \tfrac{25}{12},\ \tfrac43,\ \tfrac34,\ \tfrac13,\ \tfrac1{12},\ 0\right\}}$$ for $n=0,1,\ldots,16$. In decimals: 0, 0.1667, 0.5, 1, 1.6667, 2.4167, 3.1667, 3.5833, 3.6667, 3.4167, 2.8333, 2.0833, 1.3333, 0.75, 0.3333, 0.0833, 0. Two independent checks confirm the arithmetic: the peak $y(8)=11/3$ occurs where the sliding window is centred on the triangle's apex, and the total areas multiply, $\sum y = (\sum x)(\sum h) = 6\times 4.5 = 27$.

Figure Q2.2 - Output y(n) = x(n) * h(n), 17 samples.

(b) Time-domain input–output relation. Because the impulse response has finite length, the convolution sum is the difference equation — there is no recursive (feedback) term: $$\boxed{y(n)=x(n)+x(n-1)+x(n-2)+x(n-3)+\tfrac{1}{2}\,x(n-4)}$$ This is a five-tap moving-average (FIR) filter with a half-weighted last tap.
(c) Block diagram. The direct-form (transversal) realisation follows straight from the difference equation: a chain of four unit delays $z^{-1}$, taps of 1, 1, 1, 1 and 1/2 taken from the chain, and a single summing junction. No feedback path exists, which is the visual signature of an FIR system.

Figure Q2.3 - Direct-form (transversal) FIR realisation.

(d) Transfer function. Taking the $z$-transform of the difference equation, or equivalently $H(z)=\sum_n h(n)z^{-n}$, $$\boxed{H(z)=1+z^{-1}+z^{-2}+z^{-3}+\tfrac12 z^{-4} =\frac{z^{4}+z^{3}+z^{2}+z+\tfrac12}{z^{4}}},\qquad \text{ROC: } |z|\gt 0$$ All four poles sit at the origin, which is the standard fingerprint of a causal FIR filter; the four zeros are the roots of the numerator polynomial.
(e) BIBO stability. A discrete LTI system is BIBO stable if and only if its impulse response is absolutely summable. Here $$\sum_{n=-\infty}^{\infty}|h(n)| = 1+1+1+1+\tfrac12 = \boxed{4.5 \lt \infty}$$ so the system is BIBO stable. Equivalently, every pole lies at $z=0$, well inside the unit circle. Quantitatively, if $|x(n)|\le M$ for all $n$ then $|y(n)|\le 4.5M$ — a bounded input can never produce an unbounded output. Any FIR filter with finite coefficients is stable for the same reason.

Result	Value
Output y(n), n = 0…16 (a)	0, 1/6, 1/2, 1, 5/3, 29/12, 19/6, 43/12, 11/3, 41/12, 17/6, 25/12, 4/3, 3/4, 1/3, 1/12, 0
Peak output	y(8) = 11/3 = 3.667
Difference equation (b)	y(n) = x(n)+x(n−1)+x(n−2)+x(n−3)+½x(n−4)
Structure (c)	Direct-form FIR: 4 delays, taps 1,1,1,1,½, one adder
Transfer function (d)	H(z) = 1+z⁻¹+z⁻²+z⁻³+½z⁻⁴
BIBO stability (e)	Stable; ∑\|h(n)\| = 4.5 < ∞

Topic. Finite-impulse-response systems: convolution, direct-form realisation and the absolute-summability stability test.

Key relations. $y(n)=\sum_k h(k)x(n-k)$; output length $=N_x+N_h-1$; $H(z)=\sum_n h(n)z^{-n}$; BIBO $\iff \sum_n|h(n)|\lt\infty$; area identity $\sum_n y(n)=\left(\sum_n x(n)\right)\left(\sum_n h(n)\right)$.

Why this works. Linearity and time invariance mean the response to any input is the superposition of shifted, scaled impulse responses, which is exactly the convolution sum. For an FIR system the sum has finitely many terms, so it doubles as the difference equation and translates directly into a tapped-delay-line diagram. Absolute summability bounds the worst case $|y(n)|\le \max|x|\cdot\sum|h|$, which is why it is both necessary and sufficient for BIBO stability.

Common pitfall. Getting the output length or the alignment wrong (starting the sum at $n=1$, or truncating at $n=12$ instead of $n=16$) and so losing the tail; and quoting "all poles inside the unit circle" without noticing that an FIR system's poles are all at the origin, which makes the test trivially satisfied. The area check $\sum y=27$ catches an alignment slip immediately.

Source. Oppenheim & Willsky, Signals and Systems, 2nd ed., §2.1 and §2.3 (discrete convolution, stability); Proakis & Manolakis, Digital Signal Processing, 4th ed., Ch. 7 (FIR structures).

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