I have been working on this problem for a few days and dont know if what I'm doing is right. I work the problem using different methods and get different answers at this point I think I'm just confusing myself. The problem is F(z)= (3) / (z^2 + (1/9))....find inverse z. I first divided both sides by z and used partial frac exp to get the constants and then just looked at table for transform... The book has a similar problem with the inverse transform being time shifted by k-2 so this is where I'm confused. Any help would be greatly appreciated.
Looks like a good approach to me... can you be more specific what it is that you don't understand? I'm guessing what exactly you mean here... but i'm guessing they just time shifted it just to make it causal...
what you can do is F(z)= 3/(z^2+1/9). since you are looking for the inverse you can use partial fractions to simplifies. but what i abserved is that the denominator can not be separated may be there is a book error, coz its suppose to be (z^2-1/9) so that you can express it in diference of two squares, as in (z - 1/3) (z +1/3). i hope you will check it
When I work the problem, F(z)= (3) / (z^2 + (1/9)) dividing both sides by z and solving constants I get... 27δ[k] + 9^k cos(k(pi/2)) u[k] The specific question in the book was (1)/ (z^2+4) with the answer given as...2^k-2 cos ((k-2)(pi/2)) u[k-2]...So I guess I just dont understand why this is time shifted? No this is correct, it expands to a complex number (z - j1/3) (z +j1/3)
Hmm can you show a bit more of your work? I'm guessing its a matter of whether you solved in terms of z^-2 or z^2. Also, you need a region of convergence to do this, really...
I have tried to post my solution hopefully someone can figure out if this is right or wrong. Thanks, Matt
Looks good to me (except for the C_2* ... but you take the abs anyway, so it doesnt matter haha) Here's what i think with regards to the time shift thing (note i said 'think' haha): If you go through and multiply your original transfer function by z^-2/z^-2 then you'll end up with the same result except there will be an extra factor of z^-2 on top. Applying the time shift property of the z-transform, you'll see where their time shift comes from. I BELIEVE that it's somewhat arbitrary which sample you call zero. Maybe if someone knows better, they can correct me, but i believe thats where it comes from.