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

Mathematics
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we have to solve for y(n)
so I took fourier transform \[y(e^{jw})-5y(e^{jw})e^{-jw}+6y(e^{jw})e^{-2jw}=0\] \[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\]
\[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\] let s=e^jw \[y(s)(1-5/s +6/s^2)=0\]

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Other answers:

\[(1-5/s +6/s^2)=0\] \[(1-3/s)(1-2/s)=0\]
\[(1-3s^{-1})(1-2s^{-1})=0\]
s=-3,-2
  • phi
I think the roots are +3 and +2 (I would have solved s^2 -5s +6=0 to get (s-2)(s-3)=0
yes, s=3 and s=2
then how would we go back to fourier form?
  • phi
this reminds me of laplace transforms, but I have not seen it done using fourier
z transform since it is discreet
  • phi
I would have to read up on it, because I don't know this off the top of my head.
ok, thanks
  • phi
btw, what kind of problem is this?
this is from discrete signal processing
  • phi
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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