Finding Z-Transform of sin(Bn)

[jpm]

Homework Statement

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

[tex]x(n)=sin(Bn)[/tex]

Homework Equations

The Attempt at a Solution

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

[tex]x(n)=sin(Bn)u(n)+sin(Bn)u(-n-1)[/tex]

The Z transform of the first part of x(n) was found from a table. The z transform of the 2nd was found from the fact that the z transform of [tex]-sin(Bn)u(-n-1)[/tex] equals the z transform of [tex]sin(Bn)u(n)[/tex]

[tex]X(z)=\frac{z sin(B)}{z^2-2zcos(B)+1}-\frac{z sin(B)}{z^2-2zcos(B)+1}[/tex]

So the z transform is zero? Could this be?

 

[CEL]

Remember that [tex]sin x = \frac {e^{jx}-e^{-jx}}{2j}[/tex]

 

[Mene]

Yes, it is possible for the z-transform of sin(Bn) to be zero. This can occur when the poles of the z-transform expression cancel each other out, resulting in a zero value for the overall expression. In this case, it is likely that the poles of the two parts of the expression, sin(Bn)u(n) and sin(Bn)u(-n-1), are located at the same locations but with opposite signs. This would result in the cancellation of the two terms and a zero value for the z-transform.

However, it is important to note that the z-transform of a signal is not always finite and can also be infinite or undefined. This depends on the specific properties of the signal and the values of the poles and zeros of the z-transform expression. Therefore, it is important to carefully analyze the signal and its properties before drawing any conclusions about its z-transform.