Question

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

Answer 1

we have to solve for y(n)

Answer 2

so I took fourier transform

\[y(e^{jw})-5y(e^{jw})e^{-jw}+6y(e^{jw})e^{-2jw}=0\]

\[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\]

Answer 3

\[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\]

let s=e^jw

\[y(s)(1-5/s +6/s^2)=0\]

Answer 4

\[(1-5/s +6/s^2)=0\]
\[(1-3/s)(1-2/s)=0\]

Answer 5

\[(1-3s^{-1})(1-2s^{-1})=0\]

Answer 6

@phi

Answer 7

s=-3,-2

Answer 8

I think the roots are +3 and +2
(I would have solved s^2 -5s +6=0 to get (s-2)(s-3)=0

Answer 9

yes, s=3 and s=2

Answer 10

then how would we go back to fourier form?

Answer 11

this reminds me of laplace transforms, but I have not seen it done using fourier

Answer 12

z transform since it is discreet

Answer 13

I would have to read up on it, because I don't know this off the top of my head.

Answer 14

ok, thanks

Answer 15

btw, what kind of problem is this?

Answer 16

this is from discrete signal processing

Answer 17

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
\[y= c_0 2^{n} +c_1 3^{n} \]