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?
\(x(n)=sin(Bn)\)
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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