If h(n) is the real valued impulse response sequence of an LTI system, then what is the phase of H(ω) in terms of HR(ω) and HI(ω)?(a) \(tan^{-1}\frac{H_R (ω)}{H_I (ω)}\)(b) –\(tan^{-1}\frac{H_R (ω)}{H_I (ω)}\)(c) \(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)(d) –\(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)I have been asked this question during a job interview.This is a very interesting question from Frequency Domain Characteristics of LTI System topic in division Frequency Analysis of Signals and Systems of Digital Signal Processing

If h(n) is the real valued impulse response sequence of an LTI system,

1.	If h(n) is the real valued impulse response sequence of an LTI system, then what is the phase of H(ω) in terms of HR(ω) and HI(ω)?(a) \(tan^{-1}\frac{H_R (ω)}{H_I (ω)}\)(b) –\(tan^{-1}\frac{H_R (ω)}{H_I (ω)}\)(c) \(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)(d) –\(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)I have been asked this question during a job interview.This is a very interesting question from Frequency Domain Characteristics of LTI System topic in division Frequency Analysis of Signals and Systems of Digital Signal Processing
Answer» The CORRECT choice is (c) \(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\) The best explanation: If h(n) is the real VALUED impulse response sequence of an LTI SYSTEM, then H(ω) can be represented as HR(ω)+j HI(ω). => tanθ=\(\frac{H_I (ω)}{H_R (ω)}\) => Phase of H(ω)=\(tan^{-1}\frac{H_I (ω)}{H_R (ω)}\)

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