Correct choice is (d) \(\frac{1}{2sin(ω/2)} e^{j(ω-π)/2}\)
The best explanation: Given x(n)=u(n)
We know that the z-transform of the given signal is X(z)=\(\frac{1}{1-z^{-1}}\) ROC:|z|>1
X(z) has a pole p=1 on the unit circle, but converges for |z|>1.
If we evaluate X(z) on the unit circle except at z=1, we obtain
X(ω) = \(\frac{e^{jω/2}}{2jsin(ω/2)} = \frac{1}{2sin(ω/2)} e^{j(ω-π)/2}\)