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Libniz

  • 3 years ago

y(n)-5y(n-1)+6y(n-2)=0

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  1. Libniz
    • 3 years ago
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    we have to solve for y(n)

  2. Libniz
    • 3 years ago
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    so I took fourier transform \[y(e^{jw})-5y(e^{jw})e^{-jw}+6y(e^{jw})e^{-2jw}=0\] \[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\]

  3. Libniz
    • 3 years ago
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    \[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\] let s=e^jw \[y(s)(1-5/s +6/s^2)=0\]

  4. Libniz
    • 3 years ago
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    \[(1-5/s +6/s^2)=0\] \[(1-3/s)(1-2/s)=0\]

  5. Libniz
    • 3 years ago
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    \[(1-3s^{-1})(1-2s^{-1})=0\]

  6. Libniz
    • 3 years ago
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    @phi

  7. Libniz
    • 3 years ago
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    s=-3,-2

  8. phi
    • 3 years ago
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    I think the roots are +3 and +2 (I would have solved s^2 -5s +6=0 to get (s-2)(s-3)=0

  9. Libniz
    • 3 years ago
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    yes, s=3 and s=2

  10. Libniz
    • 3 years ago
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    then how would we go back to fourier form?

  11. phi
    • 3 years ago
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    this reminds me of laplace transforms, but I have not seen it done using fourier

  12. Libniz
    • 3 years ago
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    z transform since it is discreet

  13. phi
    • 3 years ago
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    I would have to read up on it, because I don't know this off the top of my head.

  14. Libniz
    • 3 years ago
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    ok, thanks

  15. phi
    • 3 years ago
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    btw, what kind of problem is this?

  16. Libniz
    • 3 years ago
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    this is from discrete signal processing

  17. phi
    • 3 years ago
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    The best I can come up with is that you are solving a homogenous equation for the non-forced system response. The solution is a linear combination of exponentials. you found \[ e^{jw_0} = 2 \text{ and } e^{jw_1} =3 \] the solution would be \[y= c_0 e^{jw_0n} +c_1 e^{jw_1n} \] which becomes \[y= c_0 2^{n} +c_1 3^{n} \]

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