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Unformatted text preview: that 1eλ b T min = 0 . 99 ⇐⇒ T min = 1 λ b ln[1 / (0 . 01)] = 4 . 6 hours 1eλ b T max = 0 . 01 ⇐⇒ T max = 1 λ g ln[1 / (0 . 99)] = 6 . 03 hours. (d) Using the result of (c) we have that 1 λ b ln(1 / (0 . 01)) ≤ T ≤ 1 λ g ln(1 / (0 . 99)) , which is only possible when γ = λ b λ g ≥ γ min = ln[1 / (0 . 01)] ln[1 / (0 . 99)] = 458 . 2 . For example, if company’s values are o± by about 15% each so that λ b = 0 . 85 and λ g = 1 / 525, then the test cannot be designed so that the detection and FA requirements are met. 0.5 1 1.5 2 2.5 20 40 60 80 100 EE464: Chugg, Spring ’94  Midterm 1 Score out of 130 Average = 57.7 0.5 1 1.5 2 2.5 3 3.5 20 40 60 80 100 EE464: Chugg, Spring ’94  Midterm 1 Score out of 130 Average = 57.7 EE 464 Midterm Solution 3...
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This note was uploaded on 02/02/2010 for the course EE 464 taught by Professor Caire during the Spring '06 term at USC.
 Spring '06
 Caire

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