# z-transform

#### jut

Joined Aug 25, 2007
224
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?

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#### vvkannan

Joined Aug 9, 2008
138
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 .