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

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

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

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

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

phi Group TitleBest 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
 2 years ago

Libniz Group TitleBest ResponseYou've already chosen the best response.0
yes, s=3 and s=2
 2 years ago

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

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

Libniz Group TitleBest ResponseYou've already chosen the best response.0
z transform since it is discreet
 2 years ago

phi Group TitleBest 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.
 2 years ago

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

Libniz Group TitleBest ResponseYou've already chosen the best response.0
this is from discrete signal processing
 2 years ago

phi Group TitleBest 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 \[y= c_0 2^{n} +c_1 3^{n} \]
 2 years ago
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