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I have a system which is sampling at 100Hz. my input is sinusodial. The output is similar to cosine waveforms with varying frequency. I have no clue how to find out the exact formula to put into the cosine function to generate the exact output as the system

input is sinusoidal graph with 1 as amplitude, and output is cosine plot with the following local max

[1,476,1101,1730,2358,2986,3615,4243,4871,5500,6128,6756,7385,8013,8641,9270,9898]

from here How to find out the control function of a cosine wave?

i calculated my b to be 1*10e-6, but i can't multiply it out with thw input sinusodial plot.. matlab wont let me.

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oluwatola imoleayo
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1 Answers1

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There seems to be a typo in your data; $550$ should be $5500$. The $1$ is probably spurious since it doesn't fit with the rest of the pattern. The remaining maxima are evenly spaced (here's a plot), so your output is just sinusoidal; there's no reason to apply the answers of that other question that dealt with maxima spaced in a particular way, not evenly.

To describe your output, note that the maxima occur roughly every $628{\frac13}$ samples. The one at $476$ is slightly displaced, but the remaining ones have distances $629,628,628,629,\dotsc$, so to get the highest precision you can choose one that's symmetrically located between two differences of $628$, say, $2358$; then the corresponding function would be $\cos(2\pi(t-2358)/628{\frac13})$.

joriki
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  • @oluwatola: I was assuming amplitude $1$. If the minimum is at $-8$, you'll have to multiply the function I gave by $8$. I just noticed I forgot the factor of $2\pi$; I fixed that. Note that $\cos(2\pi(0-2358)/628{\frac13})\approx0$ comes out about right. – joriki May 04 '12 at 07:56
  • @oluwatola: You'd need to say more about the context, what you're using how as a control function for what. All I've done here is show you how to find the functional form given evenly spaced local maxima of a function presumed to be have sinusoidal form. – joriki May 04 '12 at 08:03
  • @oluwatola: the method to find these parameters is linear regression (in the previous thread the function was quadratic like $x_2\mapsto y_2$ and not linear). To find yourself the parameters you may try Alpha getting $y=628.339\cdot x+472.952=628.339\cdot (x+0.7527)$. This way you'll get the $628\frac 13$ proposed by joriki. See his fine explications for the remainder. – Raymond Manzoni May 04 '12 at 11:51
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    @oluwatola: Ignoring my responses and then just deleting your comments without explanation doesn't exactly increase my motivation to interact further. – joriki May 04 '12 at 11:55