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I have a set of samples that represents a waveform. This waveform resembles a frequency modulated sinusoidal wave (only it is not).

I would like to invert this waveform or shift it by $2\pi$ shift it by $\pi$. of course taking the cosine of samples as they are without preprocessing is wrong.

What should I do to achieve this?

Thank you.

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    $\begingroup$ Shifting by 2π is the same as doing nothing. I guess it's not what you want. Can you clarify the question ? $\endgroup$
    – Frédéric Grosshans
    Commented Nov 23, 2010 at 13:08
  • $\begingroup$ you are right, it wouldn't do a thing! what I would like to get is a mirror reflection of the waveform over the x-axis but with the same dc-component as the original waveform (I don't want to shift it on the Y-axis). $\endgroup$
    – mbadawi23
    Commented Nov 23, 2010 at 13:26
  • $\begingroup$ Do you mean you have a Fourier spectrum of the waveform? $\endgroup$
    – ptomato
    Commented Nov 23, 2010 at 13:31
  • $\begingroup$ I think my problem that I was shifting by $2\pi% instead of just pi, Dah! And thanks Frédéric Grosshans. I'll share my answer with you guys. $\endgroup$
    – mbadawi23
    Commented Nov 23, 2010 at 13:36
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    $\begingroup$ What does this have to do with physics? $\endgroup$
    – endolith
    Commented Nov 24, 2010 at 1:08

2 Answers 2

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Based on what you write in the comments, perhaps you can just calculate the DC component by taking the average, subtract that, flip it over the X axis, and add the DC component back.

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  • $\begingroup$ Yes, but how exactly do you suggest to flip it over the x-axis? $\endgroup$
    – mbadawi23
    Commented Nov 23, 2010 at 15:37
  • $\begingroup$ Multiply by -1? $\endgroup$
    – Marton Trencseni
    Commented Nov 23, 2010 at 17:22
  • $\begingroup$ @mbadawi23 : another way to apply @mtrencseni solution. Compute the average $a$, and replace each sample $s_i$ by $2a - s_i$ $\endgroup$
    – Frédéric Grosshans
    Commented Nov 26, 2010 at 15:42
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  • For each sample I calculated the angle $\theta = i * 2\pi$
  • Then I added $\pi$ to $\theta$ while calculating sine and cosine components.
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  • $\begingroup$ This solution is way more complex than @mtrencseni 's solution, at least for an angla as simple than $\pi$ $\endgroup$
    – Frédéric Grosshans
    Commented Nov 26, 2010 at 15:44

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