anonymous
  • anonymous
y(n)-5y(n-1)+6y(n-2)=0
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
we have to solve for y(n)
anonymous
  • anonymous
so I took fourier transform \[y(e^{jw})-5y(e^{jw})e^{-jw}+6y(e^{jw})e^{-2jw}=0\] \[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\]
anonymous
  • anonymous
\[y(e^{jw})(1-5e^{-jw}+6e^{-2jw})=0\] let s=e^jw \[y(s)(1-5/s +6/s^2)=0\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[(1-5/s +6/s^2)=0\] \[(1-3/s)(1-2/s)=0\]
anonymous
  • anonymous
\[(1-3s^{-1})(1-2s^{-1})=0\]
anonymous
  • anonymous
@phi
anonymous
  • anonymous
s=-3,-2
phi
  • 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
anonymous
  • anonymous
yes, s=3 and s=2
anonymous
  • anonymous
then how would we go back to fourier form?
phi
  • phi
this reminds me of laplace transforms, but I have not seen it done using fourier
anonymous
  • anonymous
z transform since it is discreet
phi
  • phi
I would have to read up on it, because I don't know this off the top of my head.
anonymous
  • anonymous
ok, thanks
phi
  • phi
btw, what kind of problem is this?
anonymous
  • anonymous
this is from discrete signal processing
phi
  • 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 \[y= c_0 2^{n} +c_1 3^{n} \]

Looking for something else?

Not the answer you are looking for? Search for more explanations.