## Libniz Group Title y(n)-5y(n-1)+6y(n-2)=0 one year ago one year ago

1. Libniz Group Title

we have to solve for y(n)

2. Libniz Group Title

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. Libniz Group Title

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

4. Libniz Group Title

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

5. Libniz Group Title

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

6. Libniz Group Title

@phi

7. Libniz Group Title

s=-3,-2

8. phi Group Title

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. Libniz Group Title

yes, s=3 and s=2

10. Libniz Group Title

then how would we go back to fourier form?

11. phi Group Title

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

12. Libniz Group Title

z transform since it is discreet

13. phi Group Title

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

14. Libniz Group Title

ok, thanks

15. phi Group Title

btw, what kind of problem is this?

16. Libniz Group Title

this is from discrete signal processing

17. phi Group Title

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