What is the Fourier transform of the signal x(n)=u(n)?(a) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω+π)}\)(b) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)}\)(c) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω+π)/2}\)(d) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)This question was addressed to me in unit test.This question is from Frequency Analysis of Discrete Time Signal in section Frequency Analysis of Signals and Systems of Digital Signal Processing

What is the Fourier transform of the signal x(n)=u(n)?(a) \(\frac{1}{2

1.	What is the Fourier transform of the signal x(n)=u(n)?(a) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω+π)}\)(b) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)}\)(c) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω+π)/2}\)(d) \(\frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)This question was addressed to me in unit test.This question is from Frequency Analysis of Discrete Time Signal in section Frequency Analysis of Signals and Systems of Digital Signal Processing
Answer» 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) = \(\frac{e^{jω/2}}{2jsin(ω/2)} = \frac{1}{2sin⁡(ω/2)} e^{j(ω-π)/2}\)

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