EXAMPLE In order to measure the vibration of a test part the output of an air coupled transducer is sent into a spectrum analyzer The spectrum analyzer display is shown in the figure to the right Note that the spectrum analyzer displays the magnitude of the signal versus frequency an output analogous to the FFT of a time domain signal The measurement is made using a transducer second order system with a resonance frequency of 15kHz a damping ratio of 0 8 and a static sensitivity of 10V mm You note that the part seems to be vibrating strongly at two resonance frequencies and you are interested in quantifying the magnitude of the displacement at each frequency a Estimate the dynamic error at each frequency b Estimate the actual magnitude of the displacement in mm of the part at each frequency c Estimate the time delay at the lowest frequency WOW WE NEED TO DECIDE HOW TO ATTACK THIS PROBLEM y t A sin t System 1 K1 M 1 Frequencies 1 2 f1 2 4 9 x103 30 78 x103 rad s 2 2 f 2 2 20 x103 125 66 x103 rad s System Parameters n 2 f n 2 15 x103 94 24 x103 rad s 0 8 K 10V m m y t AK1M 1 sin t NOW WE CAN CALCULATE THE DYNAMIC ERROR The magnitude responses are M 1 M 2 1 2 3 2 3 30 78 x10 2 0 8 30 78 x10 1 3 94 24 x103 2 94 24 x10 1 2 2 3 2 3 125 66 x10 2 0 8 125 66 x10 1 3 3 2 94 24 x 10 94 24 x10 2 What is the dynamic error at each frequency 1 M 1 1 0 034 2 M 2 1 0 562 1 0 966 0 7980 0 2730 1 0 438 0 6502 4 551 HOW ABOUT THE INPUT AMPLITUDE y t A sin t System 1 K1 M 1 AK1M 1 0 32V 1 0 25V 2 A 1 0 32V 1 0 033mm 10V mm 0 96 A 2 0 25V 1 0 058mm 10V mm 0 43 y t AK1M 1 sin t FINALLY WE HAVE THE TIME DELAY AT THE LOWEST FREQUENCY What do we need first 3 2 0 8 30 78 x10 2 3 94 24 x 10 5225 n Arc tan Arc tan 0 5292rad Arc tan 2 2 8933 1 30 78 x103 1 2 3 2 n 94 24 x10 What is the time delay 5292 17 2m s 3 30 78 x10 Using computerized data acquisition measurements are taken at discrete set of times The time interval between measurements is quantified by the sample rate 1 t How do we choose the sample rate and what effect will it have on our results Consider the following sampled data What is the frequency of the analog signal What is the sample rate What will we measure if the sample rate is equal to the frequency of the signal The same effect happens if the sample rate is any integer fraction of the base frequency fm fm fm f m 2 3 Let s see what discrete data looks like for different sample rates Original Signal fm 10Hz Sampled at fs 100Hz Sampled at fs 27Hz Sampled at fs 12Hz We see a phantom signal at 2Hz The 2Hz signal that appeared is the result of the sampling process and can lead to misleading results in processing experimental data It occurred at f s f m 2 Hz These incorrect frequencies are known as aliases artifacts in the measurement process For any sampling rate greater than twice fm the lowest apparent frequency will be the same as the actual frequency SAMPLING RATE THEOREM SAMPLING RATE MUST BE GREATER THAN TWICE THE HIGHEST FREQUENCY COMPONENT IN THE SIGNAL IN ORDER TO RECONSTRUCT THE ORIGINAL WAVEFORM CORRECTLY Nyquist Theorem fs 2 fm fs 1 2 2 t Any frequencies that occur in the signal at frequencies greater than fN will occur as alias frequencies at less than fN in the sampled signal The Nyquist frequency is given by f N The alias frequencies can be predicted using the folding diagram fs f N 6 Hz 2 f m 10 Hz 1 666 f N Aliasing frequency f A 0 333 f N 2Hz Determine the alias frequency that results from sampling fm with a sample rate of fs a fm 1 2 KHz fs 2kHz b fm 16 Hz fs 8Hz a b fs 1kHz f m 10Hz 1 2 f N 2 f f N s 4 Hz f m 14 Hz 4 f N 2 fN f A 0 8 f N 0 8kHz f A 0 f N 0Hz How do we avoid aliasing for a given sample rate find fN and filter all components greater than fN anti aliasing filter Choose your sample rate to be so high that you are confident that there are no frequencies above fN Digital Signals are discrete in both time and amplitude Digital systems use some variation of binary numbering system to represent or transmit signal information Binary Digit bit can be 0 or 1 Numerical word ordered sequence of bits Byte specific sequence of 8 bits M bits can represent 2M different words For example 2 bits have 4 possible arrangements We can convert each of these arrangements to decimal base 10 integers 0 1 2 3 multiply Example 0101 0 x 23 1 x 22 0 x 21 1 x 20 From right to left 1 0 4 0 5 Convert the following binary numbers to positive integer base 10 numbers a 11111 b 1100001 Converts an analog signal to binary number quantization Specifications include full scale voltage range EFSR total number of bits in register An M bit A D converter can represent 2M different binary numbers Example an 8 bit A D converter with a EFSR of 10V would be able to represent the voltage between 0 10V unipolar or 5V bipolar with 28 different values 256 Resolution smallest voltage increment that will cause a bit change EFSR Q M 2 Quantization Error the finite resolution leads to an error if input signal falls between two bits noise on the signal bit noise 0 4 V 2bit A D Quantization error 1 1 eQ resolution Q 2 2 Bias voltage added internally such that the error is Q 2 Additional errors associated with A D Conversion saturation error outside range of A D hysteresis linearity sensitivity zero repeatability A D converter resolution is also specified in terms of signal to noise ratio Ps SNR dB 20log PN E2 R SNR dB 20 log 2 E M R2 Signal power Power resolved by quantization SNR dB 20log 2M The output voltage from a thermocouple is being used to measure temperatures from 50 70 C The output voltages will vary linearly over this range from 2 585 mV to 3 649 mV a If the thermocouple voltage is input to a 12bit A D converter having a 5V range estimate the percent quantization error in the digital …