We know that a Fourier series for signal $x(t)$ is given as

$$\frac {a_0} 2 + \sum \limits _{m=1} ^\infty (a_m \cos \frac {2 \pi m t} T + b_m \sin \frac {2 \pi m t} T)$$

So my question is

what do $a_0$,$a_m$ and $b_m$ terms mean in the Fourier series formula?

How are they important in Fourier analysis and synthesis ?

  • $\begingroup$ It means how heavy this term contributes to the entire solution. $\endgroup$ – MonkeyKing May 8 '15 at 18:27
  • $\begingroup$ @MonkeyKing yes sir you are right $\endgroup$ – pandu May 8 '15 at 18:32
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    $\begingroup$ You can think of the Fourier modes in terms of sound instead of sight. You can reconstruct any sound over $[0,T]$ by combining a fundamental tone and all of its harmonics if you get the amplitudes right. Once you get the amplitudes of the harmonics right, the sound will repeat itself over and over with periodicity $T$ seconds. The term 'harmonic analysis' is related. $\endgroup$ – DisintegratingByParts May 8 '15 at 20:21
  • $\begingroup$ @T.A.E. sir but what is $a_0$ ? $\endgroup$ – pandu May 25 '15 at 17:16
  • $\begingroup$ @pandu : an offset. $\endgroup$ – DisintegratingByParts May 25 '15 at 17:54

The $a_i$'s and $b_i$'s are respectfully the real and imaginary parts of the complex number in the $i$th position of the vector $F_nx$, where $F_n \in \mathbb{C}^{n \times n}$ is the Fourier transform matrix. For example, $F_4$ is $$\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & \omega & \omega^2 & \omega^3 \\ 1 & \omega^2 & \omega^4 & \omega^6 \\ 1 & \omega^3 & \omega^6 & \omega^9 \end{bmatrix}$$ and with $n=4$, $\omega=\exp(\frac{-2\pi i}{n})=\exp(\frac{-\pi i}{2})$.

Now if $$x=\begin{bmatrix} 1 \\ 0 \\ 1 \\ 2 \end{bmatrix}$$

Then $F_4x$ is $$\begin{bmatrix} 4 \\ 1+i \\ 0 \\ -2i \end{bmatrix}$$

So that $a_0=4, b_0=0, a_1=1, b_1=1, a_2=b_2=0, a_3=0$ and $b_3=-2$

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    $\begingroup$ The original question concerned the real Fourier transform, and its quite likely that its really about explaining sigma notation. Your answer uses complex numbers, matrices, and a quite abstract view of the Fourier Transform. Looking at the question, I very much doubt that the original questioner has the slightest idea of what you are talking about. Anybody who could understand your answer would never ask the original question. $\endgroup$ – Peter Webb May 9 '15 at 5:30
  • $\begingroup$ This only covers the case when the function to develop is a trigonometric polynomial. In general one has an infinite number of modes. $\endgroup$ – Giuseppe Negro May 9 '15 at 12:27
  • $\begingroup$ Frankly, I have no idea what this answer has to do with the Fourier transform. Is it maybe related to the discrete Fourier transform? $\endgroup$ – Alex M. May 9 '15 at 14:49

One is studying periodic functions with given period $T>0$. Among these there are very special ones, the pure harmonic functions $$t\mapsto\cos{2\pi k t\over T}\quad(k\geq0),\qquad t\mapsto\sin{2\pi k t\over T}\quad(k\geq1)\ .$$ These functions have very useful algebraic and analytic properties, and we understand them very well.

It is a miracle that any reasonable $T$-periodic function $f$ can be written as a linear combination (or superposition) of these special functions: $$f(t)={a_0\over2}+\sum_{k=1}^\infty\left(a_k\cos{2\pi k t\over T}+b_k\sin{2\pi k t\over T}\right)\ .$$ The "physical" interpretation of the coefficient pair $(a_k,b_k)$ is the following: It tells you with which intensity and phase the pure harmonic $t\mapsto\cos{2\pi k t\over T}$ is present in $f$. An example: If $a_7=10$, $b_7=3$, and all other $a_k$, $b_k$ are $\ll1$, and altogether they converge to $0$ when $k\to\infty$ then the function $f$ is not far away from a pure harmonic with $7$ peaks per period.

Concerning light spectra: The frequency $\omega$ of light waves can have any real value in a certain interval $[\omega_\min,\omega_\max]$. The Fourier representation of an arbitrary (steady) light signal would then look as follows: $$f(t)=\int_{\omega_\min}^{\omega_\max} \bigl(a(\omega)\cos(\omega t)+b(\omega)\sin(\omega t)\bigr)\>d\omega\ .$$ A physical device that can separate the various frequencies, like a prism or a spectrometer, will show bright bands or dots along an $\omega$-scale at places where the corresponding values $a(\omega)$, $b(\omega)$ are large. From white light you get a broad band (which even shows the colors, but this is secondary), since all frequencies are equally present; whereas from a methane flame you only get dots at places characteristic for methane.

  • $\begingroup$ sir,thank you very much for your answer but what is $a_0$ specifically ? $\endgroup$ – pandu May 25 '15 at 17:18
  • $\begingroup$ because suppose a signal $x(t)$ is made of only 3 sinusoidal functions,2 are sine functions and 1 cosine functions .then where is the $a_0$ i.e. dc signal term comes ? $\endgroup$ – pandu May 25 '15 at 17:58
  • $\begingroup$ The constant term ${a_0\over2}$ in the Fourier expansion of a periodic function $f$ is the average of $f$ over one period. $\endgroup$ – Christian Blatter May 25 '15 at 18:10
  • $\begingroup$ suppose these 3 signals are $2*sin(wt)$ ,$sin(4*wt)$ and $cos(wt)$. Then, will you write the equation for $x(t)$ using Fourier series formula including the $a_0$ term? $\endgroup$ – pandu May 25 '15 at 18:17
  • $\begingroup$ If $\omega$ is a constant then the Fourier expansion of your combined signal is just $$x(t)=2\sin(\omega t)+\sin(4\omega t)+\cos(\omega t)\ .$$ The whole machinery is only needed when the signal is just an old periodic function, and is not already given as a superposition of specific harmonics. $\endgroup$ – Christian Blatter May 25 '15 at 18:50

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