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2/1/2010 Analog Measurements & Signal Conditioning • Module goals… – – – – Obtain a basic understanding of bridge circuits Learn concepts related to sampling frequency Review basics of filters and amplifiers Explore use of filters in Labview Motivation… • Many sensors respond to the sensed variable via a change in electrical resistance (e.g., strain gauges, RTDs, thermistors) • Computer data acquisition systems are able to sample voltages of incoming signals. • Thus Thus, the electrical resistance signal of sensors must be converted into a corresponding voltage signal. 1 2/1/2010 Voltage Dividers… a review • Ohm’s law: V=iR • Current through each resistor is the same • Ei= i(R1+R2) • V1=i(R1) • V2=i(R2) • But i= Ei/(R1+R2) • So… V1 Ei R1 R2 V1 V2 Ei R1 R1 R2 V2 Ei R2 R1 R2 Prof. Sailor Wheatstone Bridges • Useful for measuring small changes in resistance – Strain gages have a nominal resistance of 120 or 350 ohms. This resistance changes under loading, but not by much… R1 iG R2 G R3 R4 Ei Prof. Sailor 2 2/1/2010 Wheatstone Bridges • Under a “balanced” condition iG = 0 R2 R4 R1 R3 R1 iG R2 G R3 R4 Ei Prof. Sailor Wheatstone Bridges – Null Method • Suppose R1 is a sensor with varying resistance • Balance bridge by changing a variable resistor (R2) R2 R4 R1 R3 R 1 R2R R4 so... R1 3 Requires active balancing of circuit to obtain no current flow through G. But, the input voltage does not need to be constant. iG R2 G R3 R4 Ei 3 2/1/2010 Wheatstone Bridges – Deflection Method • Suppose R1 is a sensor with varying resistance • Balance bridge by changing a variable resistor (R2) R1 R2 V R3 R4 Measure voltage across this leg of the bridge bridge. Use eqn. 6.15… or (if resistances are initially equal… 6.16 to determine the change in R1 Ei Wheatstone bridge – deflection method (general) R1R4 R3 R2 Eo Eo Ei R R R R 2 3 4 1 (6.15) R1 R2 This equation can be simplified if we set all resistances initially equal… Eo R3 R4 Ei 4 2/1/2010 Wheatstone bridge – deflection method (all resistances initially equal) Eo Ei R / R 4 2(R / R ) ((6.16)) R R A useful approach as the response (voltage) can track high frequency variations in the sensor resistance. This approach does require accurate control of the input voltage as well as accurate measurement of the output voltage. Eo R R Ei Why use a Wheatstone Bridge rather than a simple Voltage Divider? • Consider the case of a 120 Ohm strain gage used to measure strain. If a small applied strain results in a 0.1 Ohm increase in resistance what would be the voltage signal that one would get from each circuit? Voltage Divider Wheatstone Bridge Eo Ei R1 R1 R2 Eo Eo Ei and, R1 R1 R1 R1 R2 Eo Ei R / R 4 2(R / R ) • If R1 = R2 we can show that the voltage divider sensitivity is the same as the Wheatstone bridge. 5 2/1/2010 Why use a Wheatstone Bridge rather than a simple Voltage Divider? • You can set up a bridge circuit so that the voltage is ZERO at the nominal conditions. • Consider a strain gage installed on a beam. Under no load conditions you’d like the output to read 0.0. You’d also like the sign of the output voltage to correspond to whether the gage is in tension or compression. • The output of a bridge circuit can be centered on 0.0 • The output of a voltage divider circuit will necessarily read between 0.0 and the supply voltage. Sampling and Data Acquisition (Ch 7) • Sampling theorem – to reconstruct the frequency content of a measured signal the sample rate must be more than twice the highest frequency (fm) within the signal. • Alias frequencies – apparent (lower) frequency response of a signal measured at less than 2fm • Nyquist frequency – if fs is the sampling frequency, fN = fs/2 is the frequency level above which any frequency components will show up as alias frequencies. Prof. Sailor 6 2/1/2010 DATA ACQUISITION AND SAMPLING FREQUENCY 1.2 1 0.8 0.6 0.4 0.2 0 -0.2 0 0.5 1 1.5 2 2.5 3 -0.4 -0.6 -0.8 -1 Prof. Sailor Original Signal to Sample (4 Hz) Original Signal: y=sin(2Pi*f*t) (frequency= 4hz; Period = 0.25sec) 1.5 1 0.5 0 0 0.5 1 1.5 2 -0.5 -1 -1.5 1 second 2 seconds 7 2/1/2010 Sampling theorem requires that we sample at more than 2* 4hz = 8hz in order to reconstruct the signal… Alias Signal (under-sampled) Original Signal: y=sin(2Pi*f*t) (frequency= 4hz; Period = 0.25sec) 1.5 1 0.5 0 0 0.5 1 1.5 2 -0.5 -1 -1.5 1 second 2 seconds 8 2/1/2010 Sample at ~2 Hz Original Signal: y=sin(2Pi*f*t) (frequency= 4hz; Period = 0.25sec) 1.5 1 0.5 0 0 0.5 1 1.5 2 -0.5 -1 -1.5 Sample at 4 Hz Original Signal: y=sin(2Pi*f*t) (frequency= 4hz; Period = 0.25sec) 1.5 1 0.5 0 0 0.5 1 1.5 2 -0.5 -1 -1.5 9 2/1/2010 Sample at 4 Hz (offset) Original Signal: y=sin(2Pi*f*t) (frequency= 4hz; Period = 0.25sec) 1.5 1 0.5 0 0 0.5 1 1.5 2 -0.5 -1 -1.5 Sample at 8 Hz Original Signal: y=sin(2Pi*f*t) (frequency= 4hz; Period = 0.25sec) 1.5 1 0.5 0 0 0.5 1 1.5 2 -0.5 -1 -1.5 Now Sample Frequency = 2* Fm and we get the right frequency content Of the signal. Of course, we miss the amplitude… 10 2/1/2010 Sample at 12 Hz Original Signal: y=sin(2Pi*f*t) (frequency= 4hz; Period = 0.25sec) 1.5 1 0.5 0 0 0.5 1 1.5 2 -0.5 -1 -1.5 Now Sample Frequency > 2* Fm and we get the right frequency content, And most of the amplitude content… Estimating Alias Frequencies (folding diagram of Figliola/Beasley Fig 7.3) Input frequency= f Nyquist frequency = fN 32 3.2 3.0 fN 2.6 2.2 2.8 2.4 2 0 fN 2.0 1.8 1.6 1.4 1.2 0 0.2 0.4 0.6 0.8 1.0 fN 11 2/1/2010 Estimating Alias Frequency from the example • We are sampling at 6 Hz so the Nyquist frequency is 3 Hz. • We expect an alias frequency based on the folding diagram… • The signal is at 4 Hz or f/fN = 4/3 = 1.33. Thus, the signal frequency is too high to resolve and will result in an alias frequency. • According to the folding diagram the alias frequency is 0.67*fN or 2 Hz. ME 411/511 Prof. Sailor Avoiding Alias Frequencies • If you are sampling at a particular rate fs you can avoid aliasing, simply by filtering out all frequencies at or above the Nyquist frequency. • For example, if your sampling rate is 100 Hz, the Nyquist frequency is 50 Hz. • You should apply a filter to remove all frequencies in the signal that are above 50 Hz. ME 411/511 Prof. Sailor 12 2/1/2010 Basics of Filters • Filters remove undesired frequency content from a dynamic signal. This “undesired” frequency content is generally noise in the signal that can arise from a variety of influences, including the 60 Hz “hum” of AC circuits. • A filter can be thought of as a selective amplifier in which the Gain is frequency dependent. • Passband – frequency range where signal is passed through the filter • Stopband – frequency range where signal is stopped • Cut-off frequency (fc) – frequency where the filter magnitude ratio (M(f)) transitions from a value of 1.0 to 0.0 Ideal filters M(f) 1.0 fc • Low-pass f M(f) 1.0 fc • High-pass f M(f) 1.0 • Band-pass fc2 fc1 f M(f) • Notch filter 1.0 fc1 fc2 f 13 2/1/2010 Real filters • Butterworth Low-pass RC filter • Cut-off frequency q y – fc is frequency at which M(fc) = 0.5 1.0 M(f) fc f R fc 1 RC C Ei(t) Eo(t) Better filters 1.0 • Multistage filters – RC and LC components • Active filters • Numerical filters M(f) fc f Prof. Sailor 14 2/1/2010 Amplifier Basics • An amplifier scales the magnitude of an analog input signal: Eo(t) = h { Ei(t) } • For a linear amplifier p Eo(t) = G * Ei(t) where G is the “gain” • Amplifiers have many flavors: – – – – Non inverting Inverting Differential Integrating and differentiating • One can build an amplifier circuit using an op-amp and several resistors from a local electronics supply store. Signal Conditioning – a broader perspective • Many data acquisition systems will have aspects of signal conditioning (filtering and amplification) built into the system • In some cases filtering can be taken care of within the data analysis phase. • Build it or buy it? 15 2/1/2010 Labview Filter Examples • Start a Blank VI in LabView • Use the “SIMULATE SIGNAL” sub-VI that can be found on the Input list under the Express category on the Functions palette to create 3 separate sine waves with differing frequencies (say 10, 50, and 150 Hz) and magnitudes (say 1.0, 2.0, and 5.0). • Use a few addition operators to add the signals and then pipe them into a Waveform Graph indicator. g Analysis y menu within the • Use a Filter from the Signal Express category. • Create another Waveform Graph to display the filtered signal. • Explore… 16 2/1/2010 Data Acquisition Overview Software (e.g. LabView) and “virtual instrument” Data Acquisition card Terminals and/or signal conditioning Experimental apparatus and sensors Prof. Sailor 17 2/1/2010 Data Acquisition Issues • Accuracy – – – – 8 bit (28 or 256 possible states) 8-bit 12-bit (4096 possible states) 16-bit (65536 possible states) 24-bit (16.8 Million possible states) • Speed – ~100,000 samples per second – ~10,000,000 10,000,000 samples per second • Digital and analog I/O, counters, etc. • Software for automated DAQ/analysis Digital representation of analog signals 255=“11111111” 0 =“00000000” “00000000” Discrete digital reading 10 volts 0 volts Continuous analog signal 18 2/1/2010 255=“11111111” 10 volts For an 8-bit card with a 10 Volt range the uncertainty in the voltage reading will be +/- (10v/256) * ½ … or +// 0.02 0 02 volts lt In this example the DAQ would represent a 0.5000 volt signal as “00001101”=13. This would be converted by software into: 0.51 +/- 0.02 volts “.508v” “.467v” 13=“00001101” 12=“00001100” 0 =“00000000” Discrete digital reading Signal = 0.5000 volts 0 volts Continuous analog signal Can we use a 16-bit DAQ card to read temperatures from a thermocouple? • Answer this question in terms of zero-order uncertainty. • Could be a good in-class assignment if time allows… ? 19 2/1/2010 Can we use a 16-bit DAQ card to read temperatures from a thermocouple? • Answer this question in terms of zero-order uncertainty. • The resolution of the 16 16-bit bit card is as follows: – Assume the 10 volt range is represented by 216 possible values. – The scale resolution of the DAQ is 10V/216 = 10V/(65536-1) or 0.153 mV – So, the zero-order uncertainty (in terms of voltage) is 0.075 mV – Chapter 8 in Figliola and Beasley has a table for J-type thermocouples. If we are interested in a temperature range around 20 to 30 C, the sensitivity of the thermocouple is 0.052 mV/deg C – So, the zero-order sensitivity of the DAQ for J-type thermocouple readings is about 0.075 mV / 0.052 mV/degC = 1.4 deg. C. ME 411/511 Prof. Sailor 20