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Euler theorem in Prony method

  1. Oct 17, 2011 #1
    I read in various articles about Prony analysis that

    [itex]{\displaystyle \sum_{i=1}^{L}}A_{i}e^{\sigma_{i}t}cos(2\pi f_{i}t+\phi_{i})[/itex]

    Using Euler theorem this equals:

    [itex]{\displaystyle \sum_{i=1}^{L}}A_{i}e^{\sigma_{i}t}\left(\frac{e^{j2\pi f_{i}t}e^{j\phi_{i}}}{2}+\frac{e^{-j2\pi f_{i}t}e^{-j\phi_{i}}}{2}\right)[/itex]

    [itex]={\displaystyle \sum_{i=1}^{L}}\left(\frac{A_{i}e^{(j2\pi f_{i}+\sigma_{i})t}e^{j\phi_{i}}}{2}+\frac{A_{i}e^{(-j2\pi f_{i}+\sigma_{i})t}e^{-j\phi_{i}}}{2}\right)[/itex]

    is supposed to be equal to

    {\displaystyle \sum_{i=1}^{N}}\frac{A_{i}}{2}e^{j\phi_{i}}e^{(σ_{i}+j2\pi f_{i})t}

  2. jcsd
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