If x(n) is a real and odd sequence, then what is the expression for x(n)?(a) \(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω(b) –\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω(c) \(\frac{1}{π} \int_0^π\)[XI(ω) cosωn] dω(d) –\(\frac{1}{π} \int_0^π\)[XI(ω) cosωn] dωI got this question during an online interview.My query is from Properties of Fourier Transformfor Discrete Time Signals in chapter Frequency Analysis of Signals and Systems of Digital Signal Processing

If x(n) is a real and odd sequence, then what is the expression for x(

1.	If x(n) is a real and odd sequence, then what is the expression for x(n)?(a) \(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω(b) –\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω(c) \(\frac{1}{π} \int_0^π\)[XI(ω) cosωn] dω(d) –\(\frac{1}{π} \int_0^π\)[XI(ω) cosωn] dωI got this question during an online interview.My query is from Properties of Fourier Transformfor Discrete Time Signals in chapter Frequency Analysis of Signals and Systems of Digital Signal Processing
Answer» The CORRECT choice is (b) –\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω The EXPLANATION: If x(N) is REAL and odd then, x(n)cosωn is odd and x(n) sinωn is even. Consequently XR(ω)=0 XI(ω)=\(-2\sum_{n=1}^∞ x(n) sin⁡ωn\) =>x(n)=-\(\frac{1}{π} \int_0^π\)[XI(ω) sinωn] dω

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