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Is there a way to understand visually the complex coefficient of Fourier series (shown as $C_n$ in many sources)? I mean something similar to this which represents $e^{j\omega_0 t}$: . phasor

$\omega_0 \triangleq 2 \pi f_0 .$

But when I read $C_n e^{jn\omega_0 t}$ my mind shuts down. Here is complex Fourier series: http://www.thefouriertransform.com/series/complexFourier.jpg. And its one element is $C_n e^{j n \omega_0 t}$. How can I visualize this element? And does $C_n$ have a meaning besides being a coefficient?

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if you think of $e^{j n \omega_0 t}$ as a "vector" (complex numbers are not quite the same as 2-dimensional vectors, but they look the same, and are both describable with two real numbers) with length 1 and starting out with an angle of 0 (pointing straight to the right, along the positive real axis, and this vector is rotating counter-clockwise at a rate of $n \omega_0$ radians per unit time. this vector is what you see looking down the spiral in the graphic referred to in the question, looking along the time axis.

then multiplying by $C_n$ changes the length of the vector to $|C_n|$ and the initial angle of the vector is the angle of $C_n$ or $\arg\{C_n\}$.

that's how i would visualize it.

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  • $\begingroup$ thanks if so i can visualize it in my brain. but they multiply by Cn which is a phasor not a scalar right? that confuses me. another thing how is Cn associated with power? $\endgroup$ – user16307 Nov 6 '15 at 1:10
  • $\begingroup$ a "phasor" is really just a complex number. both have magnitude (length) and angle. sometimes we think of phasors as the spinning vector (or $e^{j \omega t}$), but it's really the thing that mutliplies $e^{j \omega t}$. which is a complex-valued constant (w.r.t. time). $\endgroup$ – robert bristow-johnson Nov 6 '15 at 1:13
  • $\begingroup$ aha so i wanted to say vector not phasor. ok so Cn is a vector multiplied by a complex exponential. Cn is not scalar. so are we multiplying a vector with a phasor? $\endgroup$ – user16307 Nov 6 '15 at 1:15
  • $\begingroup$ NO $C_n$ is not a cartesian vector. it is a complex number. a vector multiplied by a complex exponential is a vector with elements that are complex instead of real. you are multiplying two complex numbers. one ($e^{j n \omega_0 t}$) might be changing in time but the other ($C_n$) is constant in time. but they are both complex numbers, which means when you multiply them, their lengths multiply and their angles add. $\endgroup$ – robert bristow-johnson Nov 6 '15 at 1:17
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    $\begingroup$ $C_n$ is the initial position (at $t=0$) of the product: $C_n \cdot e^{j n \omega_0 t}$. this is true because the length of $ e^{j n \omega_0 t}$ is always 1 and the initial angle of $ e^{j n \omega_0 t}$ (at $t=0$) is 0. $\endgroup$ – robert bristow-johnson Nov 6 '15 at 1:22
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An FFT complex coefficient is just a 2 dimensional vector that has a magnitude (length) and a phase angle (direction). Omega (w) is a frequency. So, roughly, the magnitude of each complex coefficient corresponds to the height or amount of energy of a sine wave at that frequency, and the phase angle suggests where the sine wave (zero crossing) at that frequency starts.

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The simple way I tend to think about it is to remember that what you are doing is approximating the signal with a series of sinusoids.

Technically you are treating your signal as a vector (function) and expressing it in terms of an orthonormal basis of complex exponential functions. That is to say the coefficients are the inner product of the signal with the basis vectors of the space. This is much the same process as you would use to get the coefficients of a vector in $\mathbb{R}^3$ . $$\vec{v}=v_i\vec{i}+v_j\vec{j}+v_k\vec{k}$$

From a practical perspective what you are doing for each term in your fourier expansion is taking a sinusoid and expanding it and sliding it along the axis until you have the best fit.

If you remember that any complex number can be written as $Re^{i\theta}$ where $R$ is the amplitude and $e^{i\theta}$ the phase. This means that your coefficient is really

                           "Amplitude x Phase"

or in simpler terms

              "the make it bigger bit" x "the slide it along bit"

Easy huh?

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  • $\begingroup$ great explanation! $\endgroup$ – user16307 Nov 6 '15 at 8:26

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