# z-transform

Discussion in 'Homework Help' started by jut, Oct 10, 2009.

1. ### jut Thread Starter Senior Member

Aug 25, 2007
224
2
I'm trying to take the (two-sided, aka defined for all n) Z-transform of

$x(n)=sin(Bn)$

What I tried to do was split x(n) into

$x(n)=sin(Bn)u(n)+sin(Bn)u(-n-1)$

But a problem arises now because the z-transform of the first equals the negative of z-transform of the second, which makes the z-transform = 0. Waaah? Could this be?

Last edited: Oct 10, 2009
2. ### vvkannan Active Member

Aug 9, 2008
138
11
I dont think the z transform would exist.The right sided sequence will converge only when magnitude of z is greater than 1 and left sided sequence will converge only when z magnitude is lesser than 1 and hence the whole series is not summable .