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0132130084_ism09

# 0132130084_ism09 - CHAPTER 9 Exercises E9.1 The equivalent...

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347 CHAPTER 9 Exercises E9.1 The equivalent circuit for the sensor and the input resistance of the amplifier is shown i Figure 9.2 in the book. Thus the input voltage is in sensor in sensor in R R R v v + = We want the input voltage with an internal sensor resistance of 10 k to be at least 0.995 times the input voltage with an internal sensor resistance of 5 k . Thus with resistances in k , we have in in sensor in in sensor R R v R R v + + 5 995 . 0 10 Solving, we determine that R in is required to be greater than 990 k . E9.2 (a) A very precise instrument can be very inaccurate because precision implies that the measurements are repeatable, however they could have large bias errors. (b) A very accurate instrument cannot be very imprecise. If repeated measurements vary a great deal under apparently identical conditions, some of the measurements must have large errors and therefore are inaccurate. E9.3 V 2 . 0 5 . 5 7 . 5 2 1 = = = v v v d V 6 . 5 ) 7 . 5 5 . 5 ( ) ( 2 1 2 1 2 1 = + = + = v v v cm E9.4 The range of input voltages is from -5 V to +5 V or 10 V in all. We have 256 2 2 8 = = = k N zones. Thus the width of each zone is 1 . 39 10 = = N mV. The quantization noise is approximately 3 . 11 3 2 rms = q N mV. E9.5 Look at Figure 9.14 in the book. In this case, we have f s = 30 kHz and f = 25 kHz. Thus, the alias frequency is f alias = f s - f = 5 kHz. E9.6 The file containing the vi is named Figure 9.17.vi and can be found on the CD that accompanies this book.

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348 Problems P9.1 The elements of a computer-based instrumentation system include sensors, signal conditioners, analog-to-digital converters, a computer, and suitable software. P9.2* The equivalent circuit of a sensor is shown in Figure 9.2 in the book. Loading effects are caused by the voltage drop across R sensor that occurs when the input resistance of the amplifier draws current from the sensor. Then the input voltage to the amplifier (and therefore overall sensitivity) depends on the resistances as well as the internal voltage of the sensor. To avoid loading effects, we need to have R in much greater than R sensor . P9.3 The equivalent circuit for a sensor for which the short-circuit current is proportional to the measurand is In this case we want the short-circuit current to flow into the amplifier (which is often a current-to-voltage converter). Thus we need to have R in much smaller than R sensor . P9.4* The equivalent circuit for the sensor and the input resistance of the amplifier is shown in Figure 9.2 in the book. Thus the input voltage is in sensor in sensor in R R R v v + = We want the overall sensitivity to be affected by less than 1% by loading. Thus we require 99 . 0 + in sensor in R R R Using the fact that R sensor = 1 k and solving we determine that we need k 99 in R .
349 P9.5* We need a current-to-voltage converter that converts the short- circuit current into a voltage that can be applied to an analog-to- digital converter. The equivalent circuit is: The input current to the converter is in sensor sensor sensor in R R R I i + = Furthermore we want the overall sensitivity to change by no more than 1% as R sensor

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