Data Acquisition Systems. Signal DAQ System The Answer?
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1
2 Outline Analysis of Waveforms and Transforms How many Samples to Take Aliasing Negative Spectrum Frequency Resolution Synchronizing Sampling Non-repetitive Waveforms Picket Fencing A Sampled Data System
3 Data Acquisition Systems Signal DAQ System The Answer?
4 Fourier Series for Repetitive Waves N N pn cos( nω t) q qn sin( n t) n= n= x( t) = p ω Amplitude H H3 SUM Time (sample number) P s & Q s determine the resulting wave
5 Finding the Frequency Components How much sinewave at frequency h.f is in signal s(t) S( h) = Corrolation[ s( t), sin(2π. f. h)] The correlation is carried out as follows: S ( h) = s( t). sin(2π. f. h) dt Similarly (to account for phase angles) C ( h) = s( t). cos(2π. f. h) dt C( h) Using complex number operations j.2π. f. h. t + j S( h) = s( t). e dt
6 Fourier Transforms = = 2 ) ( ) ( N n N n f j e n s N f S π The component in waveform s(t) at frequency f is found by a Fourier Transform: For sampled data, the discrete Fourier transform (DFT) of repetitive waveform: The Fast Fourier Transform (FFT) is a fast calculation method for the DFT = dt e t s f S t f j π 2 ) ( ) (
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8 How Many Samples Are Needed? Sampling Theorem: The Sampling Frequency must be At Least Twice the Bandwidth of the System! So for a 5 Hz signal will Hz be OK?
9 Sampling a 5 Hz Signal Signal at 5Hz ADC To DSP Noise at.4 khz ( 28*5) Sampling Frequency =.6 khz, 32 samples/cycle
10 5 Hz Signal with Noise 5Hz Sine with 4Hz "Noise" Amplitude f n n Time Now Sample at.6 khz, (32 times 5 Hz).
11 Sampling Too Slowly 32 Samples Per Cycle of Noisey Sinewave Amplitude Time
12 Aliasing of.4 khz Noise in.6khz Sampling System True Spectrum.4 khz Noise Aliasing Line H.6 khz / 2 H28 H H6 H28 Wrapped Spectrum H H4 H6 Wrapped Around
13 Sampling Too Slowly 32 Samples Per Cycle of Noisey Sinewave Amplitude Time
14 Use of an Anti-Aliasing Aliasing Filter Signal at 5Hz Low Pass Filter, -8db at Fs/2 ADC To DSP Noise at any frequency Sampling Frequency = Fs
15 Negative Spectrum Amplitude Sample Number 6 Point FFT a+jb a-jb Negative frequencies Use Hilbert Transform to Cancel ve Spectrum
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17 Negative Frequencies Inverse FFT k = x ( t) = X exp( jkωt) k For k negative, we get Negative Frequencies As the negatives are conjugates of the positives, they cancel the imaginary parts and give a real series x(t)
18 Inverse FFT IFFT Time Series is Complex!
19 Inverse FFT Time Series is Real IFFT Complete the ve half of the spectrum (Use complex conjugate)
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21 Repetitive Signals and Windowing The Fourier Transform is Defined over Infinite Time. In practice only a finite duration signal is available. (we can t wait forever!) Repetitive Signals -Take a snap shot of the signal Use an Integer Number of Cycles Ti m e
22 FFT Frequency Resolution Time Domain Frequency Dom ain Resolution Cycle Windows Resolution = /T Hz 75Hz Frequency (Hz) Time Domain 2 Cycle Windows Resolution = /2T Frequency Domain Resolution Hz 62.5Hz Frequency (Hz)
23 Cycle Windows and Interharmonics Time Domain Frequency Domain Resolution Harmonic Frequency (Hz) Interharmonics
24 Smearing or Gibb s Phenomenon Ti m e Smeared Spectrum Normalised Modulus Value Harm onic Num ber Synchronise the Samples to the Window!
25 Shaped Windows Tim e Smeared Spectrum Normalised Modulus Value Amplitude Harmonic Number
26 Non-Repetitive Signals Made of H, H2 and H3 But My Next Talk..
27 A.M. and Sidebands Carrier Modulator Time Domain Frequency Domain
28 Only Certain Frequencies Can Be seen
29 Picket Fencing: 5 Hz + 25 Hz Two Cycle FT.2 Amplitude Time Normalised Amplitude Fourier Frequency 5Hz 25Hz One Cycle FT.2 Amplitude Time Normalised Amplitude Fourier Frequency 5Hz 25Hz
30 Summary A Sampled Data System Sync Sample Timing Sample Clock Signal Anti-Aliasing Filter Analog to Digital Converter Transformed Data Data Storage and Processing
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