## anonymous 3 years ago y(n)-5y(n-1)+6y(n-2)=0

1. anonymous

we have to solve for y(n)

2. anonymous

so I took fourier transform $y(e^{jw})-5y(e^{jw})e^{-jw}+6y(e^{jw})e^{-2jw}=0$ $y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0$

3. anonymous

$y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0$ let s=e^jw $y(s)(1-5/s +6/s^2)=0$

4. anonymous

$(1-5/s +6/s^2)=0$ $(1-3/s)(1-2/s)=0$

5. anonymous

$(1-3s^{-1})(1-2s^{-1})=0$

6. anonymous

@phi

7. anonymous

s=-3,-2

8. phi

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

9. anonymous

yes, s=3 and s=2

10. anonymous

then how would we go back to fourier form?

11. phi

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

12. anonymous

z transform since it is discreet

13. phi

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

14. anonymous

ok, thanks

15. phi

btw, what kind of problem is this?

16. anonymous

this is from discrete signal processing

17. phi

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 $y= c_0 2^{n} +c_1 3^{n}$