While learning Fourier series I read the definitions of representation for a continuous time signal $x(t)$ as:

$$x(t)=A_0 + 2 \sum_{k=1}^{\infty} A_k \cos(k \omega_0 t) - B_k \sin(k \omega_0 t) \tag{1}$$

where $A_k$ and $B_k$ are real.

Another definition was: $$x(t)=a_0 + \sum_{k=1}^{\infty} a_k \cos(k \omega_0 t) + b_k \sin(k \omega_0 t) \tag{2}$$

Why two different definitions? I mean in $(1)$ there is a coefficient of $2$ for sinusoids and sine coefficients are negated.

  • $\begingroup$ Without a proper reference to the literature it is really hard to tell how coefficients are defined. Probably there were using different equations to calculate the $a/b_n$ and $B/C_k$ coefficients. What's more in first equation coefficient of 2 applies to both $\cos$ and $\sin$. On the other hand negation is just shift in phase. $\endgroup$ – jojek Jul 31 '14 at 18:07
  • $\begingroup$ Actually anyone familiar with Fourier analysis can answer this question. I don't know why it was down voted. Unfortunately some people are quick to detract from people that are actually trying to learn or help other people. $\endgroup$ – user2718 Jul 31 '14 at 20:58

alright, let's review a little bit of Euler before we get to the Fourier.

$$ e^{j \theta} \ = \ \cos(\theta) \ + \ j \sin(\theta) $$

from that you can get

$$ \cos(\theta) = \frac{e^{j \theta} + e^{-j \theta}}{2} \quad\quad\quad \sin(\theta) = \frac{e^{j \theta} - e^{-j \theta}}{2j} $$

so now let's look at Eq (1)

$$ \begin{align} x(t) \ &= \ A_0 \ + \ 2 \sum_{k=1}^{\infty} A_k \cos(k \omega_0 t) - B_k \sin(k \omega_0 t) \\ &= \ A_0 \ + \ 2 \sum_{k=1}^{\infty} A_k \frac{e^{j k \omega_0 t} + e^{-j k \omega_0 t}}{2} \ - \ B_k \frac{e^{j k \omega_0 t} - e^{-j k \omega_0 t}}{2j} \\ &= \ A_0 \ + \ \sum_{k=1}^{\infty} (A_k + jB_k) e^{j k \omega_0 t} \ + \ (A_k - jB_k) e^{-j k \omega_0 t} \\ &= \ \sum_{k=-\infty}^{\infty} c_k \ e^{j k \omega_0 t} \\ \end{align} $$


$$ c_k = \begin{cases} A_{-k} - jB_{-k}, & \quad \text{for } k < 0 \\ A_0, & \quad\quad k = 0 \\ A_k + jB_k, & \quad\quad k > 0 \end{cases} $$

and, going in the other direction,

$$ \begin{array}{lcl} A_0 & = & c_0 \\ A_k & = & \Re\{c_k \} \quad\quad \text{for } k>0\\ B_k & = & \Im\{c_k \} \quad\quad\quad k>0 \end{array} $$

note that for real $A_k$ and real $B_k$, then

$$ c_{-k} = c_k^* = \text{"complex conjugate of } c_k \text{ "}$$

so, for Eq (1), the simplicity of having the real and imaginary parts of the $c_k$ coefficients be simply $A_k$ and $B_k$ (for $k \ge 0$) is the motivation for the convention of the leading "2" before the summation and for the minus sign. it's just a convention. this convention is useful because it's much easier to derive the coefficients $c_k$ than it is to derive the coefficients $A_k$ and $B_k$.

$$ c_k \ = \ \frac{\omega_0}{2 \pi}\int_{t_0 - \pi/\omega_0}^{t_0 + \pi/\omega_0} x(t) \ e^{-j k \omega_0 t} \ dt $$

where $ -\infty < t_0 < +\infty $ can be any convenient real value.

the convention for Eq. (2) comes about from looking at Fourier series first as a real analysis problem without the use of complex variables. then this is a simple first statement:

$$x(t) \ = \ a_0 \ + \ \sum_{k=1}^{\infty} a_k \cos(k \omega_0 t) + b_k \sin(k \omega_0 t)$$

as you can see $a_k$ and $b_k$ are related to $A_k$ and $B_k$ in a very simple and straight forward manner. you can derive that simple relationship. but the formulae for getting $a_k$ and $b_k$ from $x(t)$ (and $\omega_0$) is less straight forward to derive and to express. if fact, it's two equations, not one simpler equation.

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  • $\begingroup$ So you dress up my answer and post it. Very clever. $\endgroup$ – user2718 Jul 31 '14 at 22:00
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    $\begingroup$ your answer was wrong at the very first sentence. "dress[ing that] up" cannot fix it. and it's not just convention. it was a fallacious assumption that makes your answer dead in the water right out of the starting block. (sorry for mixing metaphors.) $\endgroup$ – robert bristow-johnson Jul 31 '14 at 22:07
  • $\begingroup$ No you apparently can't read. My first sentence is correct, but being the kind of person you are, you won't admit you are wrong. $\endgroup$ – user2718 Jul 31 '14 at 22:14
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    $\begingroup$ reading the second equation readily makes sense of fourier series: any periodic function with fundamental frequency $\omega_0$ can be expressed as infinite sum of sine and cosine functions with frequencies that are integral multiples of $\omega_0$ $\endgroup$ – mahes Aug 1 '14 at 11:26
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    $\begingroup$ "any periodic function with fundamental frequency $\omega_0$ can be expressed as infinite sum of sine and cosine functions with frequencies that are integral multiples of $\omega_0$." but, @user2332665, that's no different from your problem statement. the only change i made was to turn it into a series of exponentials with imaginary exponent. maybe i might restate it as "any periodic function with period of $\frac{2 \pi}{\omega_0}$ can be expressed as infinite sum of sinusoidal functions of various amplitude and phase and with frequencies that are integer multiples of $\omega_0$." $\endgroup$ – robert bristow-johnson Aug 1 '14 at 11:58

The first represented is single sided (only positive frequency) that usually is introduced with reference to the double ended complex form of the Fourier series. The coefficients B and C are real because of how they relate to the generally complex coefficients from the negative and positive frequency components in the complex form.

Going back to the first form, the complex representation uses a sum from $-\infty$ to $+\infty$. Typically a complex exponential is used in the sum, but $\cos + i\sin$ works just fine. Say the sum is defined with COMPLEX coefficients using the letter $a$.

There is a relationship between the two sets of coefficients:

$B_k = 2\operatorname{Re}(a_k)$

$C_k = -2\operatorname{Im}(a_k)$

Hence the negative sign on $C$.

The 2nd representation is a valid representation but because the series is defined only in terms of positive frequency and real functions, the coefficients are calculated differently.

This answer is correct. For anyone that doesn't understand what a singled ended representation of a Fourier series is, please read the following:


i.e. robert bristow-johnson

You can also consult the standard bearer on LTI systems, Athanasios Papoulis, "Circuits and Systems" pp 336.

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    $\begingroup$ "The first represented is single sided (only positive frequency) ..." ... uhm, nope. OP said that $A_k$ and $B_k$ are real. each of those $\sin()$ and $\cos()$ terms have both a positive and negative frequency component. $\endgroup$ – robert bristow-johnson Jul 31 '14 at 20:45
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    $\begingroup$ user, i'm afraid you're even more mistaken than i thought before. BTW, if we're gonna be trash talking, i have forgotten more about fourier analysis and LTI system theory than you have apparently learned. $\endgroup$ – robert bristow-johnson Jul 31 '14 at 20:58
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    $\begingroup$ BTW, this statement: "The first represented is single sided (only positive frequency) ..." is the first sentence in your answer. it's ostensibly about the OP's Eq.(1). being that it's about Eq.(1), it's wrong. $\endgroup$ – robert bristow-johnson Jul 31 '14 at 21:09
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    $\begingroup$ digging your hole deeper, @user. (BTW, i've published too. big fat hairy deel.) $\endgroup$ – robert bristow-johnson Jul 31 '14 at 21:57
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    $\begingroup$ ya know, if you keep digging your hole deeper and deeper, you might not be able to get out. $\endgroup$ – robert bristow-johnson Jul 31 '14 at 22:27

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