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Libniz Group Title

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

  • 2 years ago
  • 2 years ago

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  1. Libniz Group Title
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    we have to solve for y(n)

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

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

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

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

    • 2 years ago
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    @phi

    • 2 years ago
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    s=-3,-2

    • 2 years ago
  8. phi Group Title
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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

    • 2 years ago
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    yes, s=3 and s=2

    • 2 years ago
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    then how would we go back to fourier form?

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

    • 2 years ago
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    z transform since it is discreet

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

    • 2 years ago
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    ok, thanks

    • 2 years ago
  15. phi Group Title
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    btw, what kind of problem is this?

    • 2 years ago
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    this is from discrete signal processing

    • 2 years ago
  17. phi Group Title
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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 \[y= c_0 2^{n} +c_1 3^{n} \]

    • 2 years ago
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