Fourier Transform Fast Fourier Transform discovered by Carl Gauss ~1805 re- re invented by Cooley & Tukey in 1965

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1 ourier Transform ast ourier Transform discovered by Carl Gauss ~85 re-invented by Cooley & Tukey in 965 wikipedia

initial phase Vector representation ti of sinusoidal id

2 Next concepts Shannon s Theorem ourier analysis Complex notation Rotating vectors Angular and circular frequency, initial phase Vector representation ti of sinusoidal id signal A sinusoid sampled with the sample rate twice its frequency

requency aliasing and Shannon s Theorem A sinusoid sampled with the sample rate less than twice its frequency: A false frequency is detected In order to

frequency detectable without aliasing is equal to half of the sample rate. Half of the sample rate is called the Nyquist frequency.

3 requency aliasing and Shannon s Theorem A sinusoid sampled with the sample rate less than twice its frequency: A false frequency is detected In order to prevent the frequency aliasing the signal cannot contain components with frequencies above half of the sample rate (they must be filtered out) The highest frequency detectable without aliasing is equal to half of the sample rate. Half of the sample rate is called the Nyquist frequency. Two-dimensional vector Complex notation a + j b a cos θ b sin θ Imag jb a + b θ atan (b/a) θ a (cos θ +jsin θ ) Euler s relationship e jθ (cos θ + j sin θ ) e -jθ (cos θ -j sin θ ) e jθ Real

4 Complex conjugate Imaginary part has the opposite sign Imag jb * (cos θ -j sin θ ) * (a - j b ) θ a Real * e -jθ * A few tricks Let s rotate the axes so the Real points up Imag jb θ a Real *

5 A few tricks Let s rotate the axes so the Real points up A few tricks Let s rotate the axes so the Real points up

6 A few tricks Let s flip the labels A few tricks Let s flip the labels Real * θ a Imag jb -jb

7 Next concepts Negative frequency Discrete ourier Transform olded spectrum requency resolution Discussion on parameters for frequency domain analysis Two counter-rotating rotating vectors Real Angular frequency ω (radians/s) Circular frequency * f(hz) or (revs/s) a ω π f Initial phase (at time) φ (radians) ( t ) ( t) ( t) + j e e ( ω t +φ ) j( ω t +φ ) ( t) Imag j( ω t + φ ) j( ω t + φ ) ( e + e ) jb θ -jb θ ω t + φ [ cos( ωt + sin j( ωt + cos( ωt sin j( ωt ] cos( ωt θ πf t+ φ

8 Two counter-rotating rotating vectors ( t) cos( ωt + cos( ωt j( ωt+ φo ) j( ωt+ φ ) ( e + e ) [ j sin( ωt + cos( ωt j sin( ωt ] Two counter-rotating rotating vectors (negfreq.m) With half amplitude Being a complex conjugate of each other red-positive freq..5.5 Real Imaginary Time

9 How to make a vector rotating with frequency ω stationary At a given time the vector is ( t ) e e j( π f t+ φ ) j ( ω t+ φ ) Multiply by j t e ω j( ωt+ φ ) jωt e e e jφ After the multiplication the result is a vector whose phase angle does not change, i.e. it is stationary. Normally the signal contains more than one sinusoid (rotating vectors). If the signal contains another vector rotating with frequency ω, initial phase φ and magnitude, after the same multiplication by exp(-jωt) the resulting vector will still rotate with the frequency (ω ω) j( ω t+ φ ) j( ω t+ φ ) jωt jφ j( ω ω) t+ φ ) ( e + e ) e e + e By making this multiplication for all k data points sampled with the sampling interval Δt and adding the terms together th we obtain a stationary ti vector with the magnitude k * and, due to the cancelling effect, a near zero small vector for the second vector that still rotates

10 In general k n g( t n ) e jω t n G( ω) Discrete ourier Transform G(ω) is the magnitude of the harmonic component of the signal at frequency ω, multiplied k times. ast ourier Transform is a fast DT algorithm applicable to data samples with the size being a power of. 3-D representation of an element of the frequency spectrum (negfreq.m) red-positive freq..5.5 Real Re Imaginary Im req

11 G DC value (offset) (mean value) k jωtn ( ω) g( tn) e When ω G( ω n ) k n g( t ) e n k n g( t ) n T (ast ourier Transform) T algorithm is only AST if the number of points is a power of, e.g. 5, 4, 48 etc In a spectrum the DC value is at zero frequency (at the origin) In order to produce correct values the result of the T algorithm must be divided by the number of points When the negative frequency components are eliminated, the remaining matching positive frequency components must be doubled to compensate Example signal.m fftexample.m fftexample.m

12 olded spectrum frequency axis requency spectrum can only detect frequency components from from Hz to half of the sampling frequency. This is related to the Shannon Sampling Theorem. The limits of frequency axis are from Hz (DC value) to sreq/ Corresponding Matlab indices (folded) are from (one) to npts/+ Half of the sampling frequency is termed the Nyquist frequency irst - ve frequency A / Two-sided spectra Amplitude A / A / A / A o Last + ve frequency (Nyquist) -f npts- npts Δf Δf...f f s / 3 npts/+ A /4 Amplitude A /4 POWER A /4 A o A /4 -f Δf Δf f

13 IMPLEMENTATION O THE OLDING ALGORITHM - EXAMPLE sreq 8; npts 48; g?????; %obtain the signal df sreq/npts; fnyquist sreq / ; spec fft(g, npts); spec spec(:npts/+); spec spec / npts; spec(:end-) * spec(:end-); mag abs(spec); phase atan(imag(spec)./real(spec)); %or phase angle(spec); freq linspace(,fnyquist,npts/+)'; %or freq [: npts/]' * df; plot(freq, mag) plot(freq, phase); POWER olding into one-sided spectrum A /4 A /4 Amplitude A o A /4 A /4 -f Δf Δf Amplitude A / -sided A / POWER spectrum A o f Δf Δf f

14 Amplitude A / -sided POWER spectrum A o A / RMS spectrum -sided RMS amplitude spectrum Δf Δf Amplitude A o A / A / Δf Δf f f Resolution in time domain The smaller the sampling interval Δt, the better the time domain resolution Since Δt /sreq The faster the sampling frequency, the better the time domain resolution

15 Resolution in frequency domain The smaller the spacing between the lines of the spectrum Δf, the better the resolution in frequency domain. Δf sreq / npts, then The faster the sampling frequency, the worse the frequency resolution (opposite effect to the time domain). The more points, the better the frequency resolution. Because sreq / Δt Δf /(Δt * npts) /T, then The frequency resolution is inversely proportional to the time of observation, ie. it improves with the increased time of observation (recording). Example T settings () The RPM of a shaft is to be determined from the acceleration signal which contains fluctuations caused by the imbalance of the shaft. The acceptable error of the estimate is +/- rpm. The RPM is in the vicinity of 4 rpm. Discuss the settings of data acquisition system necessary to accomplish this task. The imbalance will produce a significant peak in the T spectrum. Its frequency (Hz) is equal to RPS (vicinity of 4/64 4 Hz). The theoretical minimum sampling frequency is at least twice that, ie 8 Hz

16 Example T settings () The RPM of a shaft is to be determined from the acceleration signal which contains fluctuations caused by the imbalance of the shaft. The acceptable error of the estimate is +/- rpm. The RPM is in the vicinity of 4 rpm. Discuss the settings of data acquisition system necessary to accomplish this task. The acceptable error corresponds to the resolution of the spectrum. Δf per min /6 per second /6 Hz Since Δf /T, the signal should be measured for at least 6 seconds. Since Δf sreq / npts sreq Δf * npts Assume the T size of 48 sreq /6 * Hz The sampling rate (frequency) should be below 34 Hz to obtain the resolution better than rpm (but more than 8 Hz). The sreq of 6 Hz will make the peak of appr. 4 Hz appear in the middle of the spectrum. In such a case npts 4 would be sufficient. Δf sreq / npts 6/4.56 Hz.94 min - Exercise Dynamic characteristics of a bridge is to be tested for structural resonances. The frequency bandwidth of interest is from to Hz and the required frequency resolution is at least. Hz Determine: The setting of anti-aliasing filter The possible combinations of sampling frequency and T size

17 Dynamic characteristics of a bridge system is to be tested for structural resonances. The frequency bandwidth of interest is from to Hz and the required frequency resolution is at least. Hz Determine: The setting of anti-aliasing filter The possible combinations of sampling frequency and T size Bandwidth - Nyquist frequency Sampling frequency > * Nyquist requency resolution /T T size T settings ling frequency, Hz Samp sreq npts *Δf.. requency resolution, Hz

3 What You Should Know About Complex Numbers

3 What You Should Know About Complex Numbers Life is complex it has a real part, and an imaginary part Andrew Koenig. Complex numbers are an extension of the more familiar world of real numbers that make