Trigonometric and Exponential Fourier Series of a Multitone Signal: 17-Phys-B4 Signals and Communications, December 2013, Q1

Question 1: Trigonometric and Exponential Fourier Series of a Multitone 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. $x(t)$ is handed to us already written as a finite trigonometric Fourier series $x(t)=C_0+\sum_n C_n\cos(n\omega_0 t+\theta_n)$, with fundamental $\omega_0=1\ \text{rad/s}$ (the greatest common divisor of the three angular frequencies present, $1,3,8$):

Given data — harmonic content of $x(t)$
Term	$n$	$C_n$	$\theta_n$
$2\cos t$	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. (a) the trigonometric-FS amplitude and phase spectra; (b) the exponential-FS amplitude and phase spectra, obtained purely by inspection of (a); (c) the exponential Fourier series for $x(t)$.

Trigonometric Fourier series: amplitude spectrum $|C_n|$ (top, nonzero only at $n=1,3,8$) and phase spectrum $\theta_n$ (bottom).

Approach. No integration is required — $x(t)$ is already a sum of harmonically related cosines, so the trigonometric-FS coefficients are read off directly, and the exponential coefficients follow from the standard identity $C_n\cos(n\omega_0 t+\theta_n)=\tfrac{C_n}{2}e^{j\theta_n}e^{jn\omega_0 t}+\tfrac{C_n}{2}e^{-j\theta_n}e^{-jn\omega_0 t}$.

Part (a) — trigonometric amplitude and phase spectra. Matching each term of $x(t)$ to $C_n\cos(n\omega_0t+\theta_n)$ with $\omega_0=1$: the amplitude spectrum $|C_n|$ has three nonzero lines, at $n=1$ ($C_1=2$), $n=3$ ($C_3=1$) and $n=8$ ($C_8=2$); the phase spectrum $\theta_n$ has the corresponding values $$\theta_1=0,\qquad \theta_3=-\frac{2\pi}{3}=\boxed{-120^{\circ}},\qquad \theta_8=+\frac{2\pi}{3}=\boxed{+120^{\circ}}.$$ Both spectra are plotted above; every $n$ not equal to $1,3,8$ is identically zero.
Part (b) — exponential spectra by inspection. Applying $X_n=\tfrac{C_n}{2}e^{j\theta_n}$ for $n>0$ and the conjugate-symmetry rule $X_{-n}=X_n^{*}$ (valid because $x(t)$ is real) to each line found in (a) gives six exponential-series lines: $$X_1=1\angle 0^{\circ},\ \ X_{-1}=1\angle 0^{\circ},\qquad X_3=0.5\angle{-120^{\circ}},\ \ X_{-3}=0.5\angle{+120^{\circ}},\qquad X_8=1\angle{+120^{\circ}},\ \ X_{-8}=1\angle{-120^{\circ}}.$$ Each trigonometric line of height $C_n$ splits into two exponential lines of half that height, one at $+n$ carrying phase $\theta_n$ and its mirror image at $-n$ carrying $-\theta_n$ — exactly the "by inspection" step the question asks for, with no new computation.

Exponential Fourier series: two-sided amplitude spectrum $|X_n|$ (top) and phase spectrum $\angle X_n$ (bottom), obtained from Figure 1 by inspection.

Part (c) — exponential Fourier series. Summing the six exponential terms $X_n e^{jn t}$ found in part (b), $$x(t)=\boxed{e^{j2\pi/3}e^{j8t}+0.5e^{-j2\pi/3}e^{j3t}+e^{jt}+e^{-jt}+0.5e^{j2\pi/3}e^{-j3t}+e^{-j2\pi/3}e^{-j8t}}.$$

Final results
Quantity	Value
Fundamental $\omega_0$	1 rad/s
Trig-FS lines $(n,C_n,\theta_n)$	$(1,2,0)$, $(3,1,-120^{\circ})$, $(8,2,+120^{\circ})$
Exp-FS lines $(n,X_n)$	$(\pm1,1\angle0^{\circ})$, $(\pm3,0.5\angle\mp120^{\circ})$, $(\pm8,1\angle\pm120^{\circ})$
Exponential series	$x(t)=\sum X_n e^{jnt}$, six terms above

Topic: Trigonometric vs. exponential (complex) Fourier series representations of a periodic signal.

Key relations: $x(t)=C_0+\sum_{n=1}^{\infty}C_n\cos(n\omega_0t+\theta_n)$; $X_n=\tfrac{C_n}{2}e^{j\theta_n}$ for $n>0$; $X_{-n}=X_n^{*}$ for real $x(t)$; $x(t)=\sum_{n=-\infty}^{\infty}X_n e^{jn\omega_0 t}$.

Why this works: Euler's formula $\cos\phi=\tfrac12(e^{j\phi}+e^{-j\phi})$ turns every real cosine harmonic into a conjugate pair of counter-rotating phasors; the trigonometric form collects each pair back into one real-valued amplitude and phase, so the two representations carry identical information and either is obtained from the other purely by bookkeeping — no new Fourier analysis (integration) is needed once one form is known.

Common pitfall: Taking $\omega_0$ to be the frequency of the lowest term present (here mistakenly $1$ vs. reading it as $3$ if the $2\cos t$ term were overlooked) instead of the largest angular frequency that evenly divides every harmonic present; and dropping the minus sign in $X_{-n}=X_n^{*}$, which flips the sign of every negative-index phase.

Source: Oppenheim & Willsky, Signals and Systems, 2nd ed., Ch. 3 (Fourier series representation of periodic signals).

Question 1 of 6: Trigonometric and Exponential Fourier Series of a Multitone Signal

Question 1: Trigonometric and Exponential Fourier Series of a Multitone Signal (20 marks)