mt_sol(2)

mt_sol(2) - Name Solution Real Time Computing Systems EE...

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Unformatted text preview: Name Solution Real Time Computing Systems EE 4770 Midterm Examination 12 March 1997, 8:40-9:30 CST Alias Problem 1 (33 pts) Problem 2 (34 pts) Problem 3 (33 pts) Exam Total (100 pts) Good Luck! Problem 1: A circuit is to be designed to convert temperature x ∈ [280K , 300K] to voltage H ( x ) = ( x- 280K) V 4K using a thermistor having model function H t2 ( x ) = R e β/x with β = 3000K and R = 0 . 059Ω. ( a ) Find a linear thermistor model appropriate to this problem using the method described in class. (In which the change in resistance with temperature (slope of model) matches H t2 at a particular point.) (8 pts) Linear model: H t4 ( x ) = R M (1 + α ( x- T M )) . As derived in class α =- β/T 2 M , where T M is the middle of the temperature range to be measured ([280K , 300K]) , so T M = 290K , and R M is the thermistor resistance at T M , R M = H t2 ( T M ) = R e β/T M = 1835Ω . Then α =- 3000K / (290K) 2 =- . 03567 /uK . The complete model function is H t4 ( x ) = 1835Ω ( 1- . 03567 K ( x- 290K) ) . ( b ) Find a linear thermistor model which matches H t2 at 280K and 300K. (7 pts) Let R 280 denote the thermistor resistance at 280 Kelvins, R 280 = H t2 (280K) , and similarly R 300 = H t2 (300K) . The linear model should pass through points (280K ,R 280 ) and (300K ,R 300 ) . The equation of such a line is H t5 ( x ) = R 280 + ( x- 280K) R 300- R 280 300 K- 280 K . Using R 280 = 2655Ω and R 300 = 1300Ω , H t5 ( x ) = 2655Ω + ( x- 280K)- 1355 Ω 20 K . ( c ) Draw the schematic of a circuit that could be used to generate the voltage. Show the sign (but not the value of) of voltage sources used. Show, but do not solve, an equation or equations that...
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This note was uploaded on 11/28/2011 for the course EE 4770 taught by Professor Staff during the Fall '99 term at LSU.

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mt_sol(2) - Name Solution Real Time Computing Systems EE...

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