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LibnizBest ResponseYou've already chosen the best response.0
we have to solve for y(n)
 one year ago

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

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

LibnizBest ResponseYou've already chosen the best response.0
\[(15/s +6/s^2)=0\] \[(13/s)(12/s)=0\]
 one year ago

LibnizBest ResponseYou've already chosen the best response.0
\[(13s^{1})(12s^{1})=0\]
 one year ago

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

LibnizBest ResponseYou've already chosen the best response.0
then how would we go back to fourier form?
 one year ago

phiBest ResponseYou've already chosen the best response.1
this reminds me of laplace transforms, but I have not seen it done using fourier
 one year ago

LibnizBest ResponseYou've already chosen the best response.0
z transform since it is discreet
 one year ago

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

phiBest ResponseYou've already chosen the best response.1
btw, what kind of problem is this?
 one year ago

LibnizBest ResponseYou've already chosen the best response.0
this is from discrete signal processing
 one year ago

phiBest ResponseYou've already chosen the best response.1
The 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 \[y= c_0 2^{n} +c_1 3^{n} \]
 one year ago
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