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y(n)-5y(n-1)+6y(n-2)=0
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we have to solve for y(n)
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so I took fourier transform
\[y(e^{jw})-5y(e^{jw})e^{-jw}+6y(e^{jw})e^{-2jw}=0\]
\[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\]
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\[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\]
let s=e^jw
\[y(s)(1-5/s +6/s^2)=0\]
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\[(1-5/s +6/s^2)=0\]
\[(1-3/s)(1-2/s)=0\]
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\[(1-3s^{-1})(1-2s^{-1})=0\]
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@phi
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s=-3,-2
phi
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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
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yes, s=3 and s=2
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then how would we go back to fourier form?
phi
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this reminds me of laplace transforms, but I have not seen it done using fourier
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z transform since it is discreet
phi
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I would have to read up on it, because I don't know this off the top of my head.
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ok, thanks
phi
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btw, what kind of problem is this?
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this is from discrete signal processing
phi
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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
\[y= c_0 2^{n} +c_1 3^{n} \]