hw2soln

hw2soln - ME 365 Homework Set 2 Out September 1 , 2011 Due...

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ME 365 Homework Set 2 Out September 1 , 2011 Due September 8, 2011 Problem No. 1 In this question two analog to digital converters (ADC) will be considered. The information about these ADC’s are given in the following table: ADC Number of bits Nominal Range ADC1: 8 bits 0 to 10 volts ADC2: 12 bits -10 to +10 volts (a) Determine the output code (decimal integer) for both of the ADCs for the given input voltages. i. -3.5 volts ii. 2.35 volts iii. 11.35 volts Solution: i. for ADC1, -3.5V is clipped to 0V and therefore it is coded as 0 for ADC2, Q 2 =20/2 12 =0.0049V code=round[(-3.5V-(-10V))/Q 2 ]=1327 ii. for ADC1, Q 1 =10/2 8 =0.0391V code=round[2.35V/Q 1 ]=60 for ADC2, code=round[(2.35V-(-10V))/Q 2 ]=2520 iii. for ADC1, 11.35 is clipped to V ADCmax =10V-Q 1 =9.9609V and it corresponds to the maximum positive code 2 8 -1=255 Similarly for ADC2, 11.35 is clipped to V ADCmax =10V-Q 2 =9.9951V and it corresponds to the maximum positive code 2 12 -1=4095 (b) ADC1 produced the follow output code (1110 0010) 2 . What is the range of input voltages that could produce this code, assuming positive codes only? Solution: (1110 0010) 2 =1*2 7 +1*2 6 +1*2 5 +1*2 1 =(226) 10 The estimated voltage from this output code is 226*Q 1 =8.8366V Therefore, the range of input voltage that could produce this code is (8.8366-Q 1 /2, 8.8366+Q 1 /2)=(8.8171V, 8.8562V) Note: If you did not use the rounded number of Q 1 , but use this (226*(10/2 8 )-(10/2 8 )/2, 226*(10/2 8 )+(10/2 8 )/2 ) You will get this range (8.8086V, 8.8477V)
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ME 365 Homework Set 2 Out September 1 , 2011
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hw2soln - ME 365 Homework Set 2 Out September 1 , 2011 Due...

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