22-Elec-A3 Signals and Communications · December 2013

Question 1 of 6: Fourier Series Spectra of a Multi-Tone Signal

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

Notes on this paper

Paper format. National Exams, December 2013 — 07-Elec-A3, Signals and Communications. Three hours, closed book (one two-sided aid sheet and an approved calculator permitted). Six questions, all of equal value at 20 marks; five constitute a complete paper.

All six questions are solved here. The examination marks only the first five answers submitted, but this document is a study resource, so every question — and every sub-part — is worked in full.

Reference texts.

B. P. Lathi and Z. Ding, Modern Digital and Analog Communication Systems, 4th ed. — Ch. 3 (signal transmission), Ch. 4 (amplitude modulation), Ch. 5 (angle modulation), Ch. 6 (sampling).
B. P. Lathi, Linear Systems and Signals, 2nd ed. — Ch. 6 (Fourier series), Ch. 7 (Fourier transform).
A. V. Oppenheim and A. S. Willsky, Signals and Systems, 2nd ed. — Ch. 4 (Fourier transform properties), Ch. 7 (sampling), Ch. 10 (z-transform).
S. Haykin, Communication Systems, 5th ed. — Ch. 3 (amplitude modulation), Ch. 4 (angle modulation).
J. G. Proakis and D. G. Manolakis, Digital Signal Processing, 4th ed. — Ch. 3 (z-transform and inversion).

Question 1: Fourier Series Spectra of a Multi-Tone Signal (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 sum of three cosines with radian frequencies 1, 3 and 8 rad/s:

Term	Frequency $\omega$ (rad/s)	Amplitude	Phase (rad)
$2\cos t$	1	2	0
$\cos(3t-2\pi/3)$	3	1	$-2\pi/3$
$2\cos(8t+2\pi/3)$	8	2	$+2\pi/3$

Find. The one-sided trigonometric amplitude and phase spectra, the corresponding two-sided exponential spectra, and the exponential Fourier series written out explicitly.

Approach. Identify the fundamental frequency as the greatest common divisor of the three frequencies, read the trigonometric coefficients $C_n,\theta_n$ directly off the given cosines, then map each one-sided line of height $C_n$ into a conjugate pair of exponential lines of height $C_n/2$.

Establish the fundamental frequency and period. The signal is periodic only if all three frequencies are integer multiples of a common $\omega_0$. Since $\gcd(1,3,8)=1$, $$\omega_0 = 1\ \text{rad/s}, \qquad T_0 = \frac{2\pi}{\omega_0} = \boxed{2\pi\ \text{s}}$$ The three tones are therefore the harmonics $n = 1,\,3,\,8$; every other harmonic is absent, and there is no DC term.
Write the trigonometric series in standard form. Comparing with the compact form $x(t) = C_0 + \sum_{n=1}^{\infty} C_n\cos(n\omega_0 t + \theta_n)$, the coefficients are read off by inspection — no integration is required, because the signal is already expressed as a sum of harmonically related cosines: $$C_0 = 0,\quad C_1 = 2,\ \theta_1 = 0;\quad C_3 = 1,\ \theta_3 = -\tfrac{2\pi}{3};\quad C_8 = 2,\ \theta_8 = +\tfrac{2\pi}{3}$$ with $C_n = 0$ for all other $n$.
Sketch the one-sided spectra (part a). The amplitude spectrum is a set of lines at $\omega = 1, 3, 8$ of heights 2, 1, 2. The phase spectrum carries a line only where the amplitude is non-zero: 0 at $\omega=1$, $-2\pi/3$ at $\omega=3$ and $+2\pi/3$ at $\omega=8$.

Trigonometric Fourier-series spectra: amplitude C_n (top) and phase θ_n (bottom), one-sided, plotted against ω = nω₀.

Converting to the exponential form is now a purely mechanical splitting operation, which is exactly what the examiner means by “by inspection”.

Convert each trigonometric line into a conjugate pair (part b). Using Euler's identity, $C_n\cos(n\omega_0 t+\theta_n) = \tfrac{C_n}{2}e^{j\theta_n}e^{jn\omega_0 t} + \tfrac{C_n}{2}e^{-j\theta_n}e^{-jn\omega_0 t}$, so $$D_n = \frac{C_n}{2}e^{j\theta_n}, \qquad D_{-n} = D_n^{*} = \frac{C_n}{2}e^{-j\theta_n}$$ Numerically $|D_{\pm 1}| = 1$, $|D_{\pm 3}| = 0.5$, $|D_{\pm 8}| = 1$, and $D_0 = 0$.
Assign the phases. The magnitude spectrum $|D_n|$ is even and the phase spectrum $\angle D_n$ is odd, because $x(t)$ is real: $$\angle D_1 = \angle D_{-1} = 0,\qquad \angle D_3 = -\tfrac{2\pi}{3},\ \angle D_{-3} = +\tfrac{2\pi}{3},\qquad \angle D_8 = +\tfrac{2\pi}{3},\ \angle D_{-8} = -\tfrac{2\pi}{3}$$

Exponential Fourier-series spectra: |D_n| (top) is even, ∠D_n (bottom) is odd — each trig line of height C_n splits into a conjugate pair of height C_n/2.

Write the exponential Fourier series (part c). Summing $x(t) = \sum_n D_n e^{jn\omega_0 t}$ over the six non-zero coefficients: $$\boxed{\,x(t) = e^{jt} + e^{-jt} + \tfrac{1}{2}e^{-j2\pi/3}e^{j3t} + \tfrac{1}{2}e^{j2\pi/3}e^{-j3t} + e^{j2\pi/3}e^{j8t} + e^{-j2\pi/3}e^{-j8t}\,}$$
Check the result. Pairing the conjugate terms must return the original cosines, and Parseval's theorem for periodic signals gives the mean-square value two ways: $$P = \sum_n |D_n|^2 = 2(1)^2 + 2(0.5)^2 + 2(1)^2 = 4.5 \qquad\text{and}\qquad \sum_n \frac{C_n^2}{2} = \frac{4+1+4}{2} = 4.5\ \checkmark$$

Quantity	Result
Fundamental frequency / period	$\omega_0 = 1$ rad/s, $T_0 = 2\pi$ s
Trigonometric amplitudes $C_n$	$C_1 = 2,\ C_3 = 1,\ C_8 = 2$ (all others 0)
Trigonometric phases $\theta_n$	$0,\ -2\pi/3,\ +2\pi/3$
Exponential magnitudes $\|D_n\|$	$\|D_{\pm1}\| = 1,\ \|D_{\pm3}\| = 0.5,\ \|D_{\pm8}\| = 1$
Exponential phases $\angle D_n$	$0,\ \mp 2\pi/3,\ \pm 2\pi/3$ (odd)
Signal power	4.5 W (1 Ω)

Topic. Trigonometric and exponential Fourier series of a sum of harmonically related sinusoids.

Key relations. $x(t) = C_0 + \sum C_n\cos(n\omega_0 t+\theta_n) = \sum_{n=-\infty}^{\infty} D_n e^{jn\omega_0 t}$; $D_n = \tfrac{C_n}{2}e^{j\theta_n}$, $D_{-n}=D_n^{*}$, $D_0 = C_0$; Parseval $\overline{x^2} = \sum |D_n|^2$.

Why this works. A signal already written as a sum of cosines at integer multiples of one frequency is its own Fourier series — the analysis integrals would simply return the coefficients you can already see, because the harmonics are mutually orthogonal over $T_0$. The exponential form is a bookkeeping change: Euler's identity splits every real cosine into two counter-rotating phasors of half the amplitude and opposite phase, which is why the two-sided magnitude spectrum is even and the two-sided phase spectrum is odd for any real signal. That conjugate symmetry is not a coincidence of this problem; it is the frequency-domain fingerprint of realness.

Common pitfall. Taking $\omega_0$ to be the lowest frequency present rather than the greatest common divisor of the frequencies present — here the two happen to agree, but for tones at 2, 3 and 8 rad/s the fundamental would be 1 rad/s even though no 1 rad/s component exists. The second frequent error is forgetting to negate the phase on the negative-frequency lines, which destroys the odd symmetry and makes the reconstructed signal complex.

Source. Lathi, Linear Systems and Signals, 2nd ed., §6.1–6.4; Oppenheim & Willsky, Signals and Systems, 2nd ed., §3.3–3.5.

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Term	Frequency \(\omega\) (rad/s)	Amplitude	Phase (rad)
\(2\cos t\)	1	2	0
\(\cos(3t-2\pi/3)\)	3	1	\(-2\pi/3\)
\(2\cos(8t+2\pi/3)\)	8	2	\(+2\pi/3\)

Quantity	Result
Fundamental frequency / period	\(\omega_0 = 1\) rad/s, \(T_0 = 2\pi\) s
Trigonometric amplitudes \(C_n\)	\(C_1 = 2,\ C_3 = 1,\ C_8 = 2\) (all others 0)
Trigonometric phases \(\theta_n\)	\(0,\ -2\pi/3,\ +2\pi/3\)
Exponential magnitudes \(\|D_n\|\)	\(\|D_{\pm1}\| = 1,\ \|D_{\pm3}\| = 0.5,\ \|D_{\pm8}\| = 1\)
Exponential phases \(\angle D_n\)	\(0,\ \mp 2\pi/3,\ \pm 2\pi/3\) (odd)
Signal power	4.5 W (1 Ω)