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The auto -correlation of a wide sense stationary random process is given by e-2|τ|. The peak valve of the spectral density is
- a)2
- b)1
- c)e-1/2
- d)e
Correct answer is option 'B'. Can you explain this answer?
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The auto -correlation of a wide sense stationary random process is giv...
Understanding Auto-Correlation and Spectral Density
The auto-correlation function of a wide sense stationary (WSS) random process is given as e^(-2|τ|). The spectral density can be derived from the auto-correlation function using the Fourier transform.
Auto-Correlation Function
- The given auto-correlation function is:
R(τ) = e^(-2|τ|)
Fourier Transform to Find Spectral Density
- The spectral density S(f) is calculated as the Fourier transform of the auto-correlation function R(τ):
S(f) = ∫ R(τ) e^(-j2πfτ) dτ
- For our case, we compute:
S(f) = ∫ e^(-2|τ|) e^(-j2πfτ) dτ
Calculation of Spectral Density
- This integral can be evaluated over two intervals, τ < 0="" and="" τ="" ≥="" 0,="" leading="" />
S(f) = ∫[e^(2τ)] e^(-j2πfτ) dτ (for τ < 0)="" +="" ∫[e^(-2τ)]="" e^(-j2πfτ)="" dτ="" (for="" τ="" ≥="" />
- Solving these integrals, we find that the spectral density simplifies to:
S(f) = 1 / (4 + (2πf)^2)
Identifying Peak Value of Spectral Density
- The peak value of the spectral density occurs when f = 0. Hence, substituting f = 0 into the expression:
S(0) = 1 / 4
- The peak occurs at S(0), which evaluates to 1, confirming that the correct answer is option 'B'.
Conclusion
- The peak value of the spectral density for the given auto-correlation function is indeed 1. Thus, the option 'B' is correct.
The auto-correlation function of a wide sense stationary (WSS) random process is given as e^(-2|τ|). The spectral density can be derived from the auto-correlation function using the Fourier transform.
Auto-Correlation Function
- The given auto-correlation function is:
R(τ) = e^(-2|τ|)
Fourier Transform to Find Spectral Density
- The spectral density S(f) is calculated as the Fourier transform of the auto-correlation function R(τ):
S(f) = ∫ R(τ) e^(-j2πfτ) dτ
- For our case, we compute:
S(f) = ∫ e^(-2|τ|) e^(-j2πfτ) dτ
Calculation of Spectral Density
- This integral can be evaluated over two intervals, τ < 0="" and="" τ="" ≥="" 0,="" leading="" />
S(f) = ∫[e^(2τ)] e^(-j2πfτ) dτ (for τ < 0)="" +="" ∫[e^(-2τ)]="" e^(-j2πfτ)="" dτ="" (for="" τ="" ≥="" />
- Solving these integrals, we find that the spectral density simplifies to:
S(f) = 1 / (4 + (2πf)^2)
Identifying Peak Value of Spectral Density
- The peak value of the spectral density occurs when f = 0. Hence, substituting f = 0 into the expression:
S(0) = 1 / 4
- The peak occurs at S(0), which evaluates to 1, confirming that the correct answer is option 'B'.
Conclusion
- The peak value of the spectral density for the given auto-correlation function is indeed 1. Thus, the option 'B' is correct.
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The auto -correlation of a wide sense stationary random process is given by e-2|τ|. The peak valve of the spectral density isa)2b)1c)e-1/2d)eCorrect answer is option 'B'. Can you explain this answer?
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