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Libniz
 3 years ago
y(n)5y(n1)+6y(n2)=0
Libniz
 3 years ago
y(n)5y(n1)+6y(n2)=0

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Libniz
 3 years ago
Best ResponseYou've already chosen the best response.0we have to solve for y(n)

Libniz
 3 years ago
Best ResponseYou've already chosen the best response.0so I took fourier transform \[y(e^{jw})5y(e^{jw})e^{jw}+6y(e^{jw})e^{2jw}=0\] \[y(e^{jw})(15e^{jw}+6e^{2jw})=0\]

Libniz
 3 years ago
Best ResponseYou've already chosen the best response.0\[y(e^{jw})(15e^{jw}+6e^{2jw})=0\] let s=e^jw \[y(s)(15/s +6/s^2)=0\]

Libniz
 3 years ago
Best ResponseYou've already chosen the best response.0\[(15/s +6/s^2)=0\] \[(13/s)(12/s)=0\]

Libniz
 3 years ago
Best ResponseYou've already chosen the best response.0\[(13s^{1})(12s^{1})=0\]

phi
 3 years ago
Best ResponseYou've already chosen the best response.1I think the roots are +3 and +2 (I would have solved s^2 5s +6=0 to get (s2)(s3)=0

Libniz
 3 years ago
Best ResponseYou've already chosen the best response.0then how would we go back to fourier form?

phi
 3 years ago
Best ResponseYou've already chosen the best response.1this reminds me of laplace transforms, but I have not seen it done using fourier

Libniz
 3 years ago
Best ResponseYou've already chosen the best response.0z transform since it is discreet

phi
 3 years ago
Best ResponseYou've already chosen the best response.1I would have to read up on it, because I don't know this off the top of my head.

phi
 3 years ago
Best ResponseYou've already chosen the best response.1btw, what kind of problem is this?

Libniz
 3 years ago
Best ResponseYou've already chosen the best response.0this is from discrete signal processing

phi
 3 years ago
Best ResponseYou've already chosen the best response.1The best I can come up with is that you are solving a homogenous equation for the nonforced 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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