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Abstract of Master’s Thesis Author: Kulvik Aleksanteri Name of thethesis: ibrational Study of Cryo I Helsinki, and Testing...
Master’s ThesisKaarle Aleksanteri Kulvik52664T24th April 2008Vibrational Study of Cryo I Helsinki, and Testing of Homogene...
Contents1 Preface 22 Introduction 33 Fourier Methods 34 Piezoelectric Accelerator 125 Low-pass filter and measurement setup...
1 PrefaceDedicated to Alexiel, Pulla and those who do not dwell amongst us anymore...Albert, Kustaa, and Rosiel.Solitary t...
2 IntroductionThis master thesis will study the vibrational, and rotational aspects of Cryo IHelsinki, which has now been ...
bn =2TpTp2−Tp2f(t) sin(nωt)dt. (4)The frequencies nω are known as the nth harmonics of ω. The series may bewritten in expo...
The series approach has to be abandoned when the waveform is not pe-riodic for example when Tp becomes infinite. As Tp incr...
∞−∞f′(x)e−i2πxsdx =∞−∞limf(x + ∆x) − f(x)∆xe−i2πxsdx = lim∞−∞f(x + ∆x)∆e−i2πxsdx − lim∞−∞With these interesting properties...
Figure 1: Sampling a continuous function at regular intervals.ak =1TT0x(t) cos(2πktTdtbk =1TT0x(t) sin(2πktTdt. (17)The pr...
Figure 2: Calculating Fourier coefficients from a discrete series using approximation.t = r∆, and ∆ = T/N, then the integral...
This can be seenN−1k=0Xke2iπkr/N=N−1k=01NN−1s=0xse−2iπks/Ne2iπkr/N=N−1k=0N−1s=01Nxse(−2iπk/N)(s−r)(24)and by interchanging...
ωk =2πkN∆=π∆(29)where k is in the range k = 0, 1, . . ., N/2. The frequencies above π/∆rad/s, which are present in the ori...
Firstly we separate the odd and the even terms in {xr} gettingXk =1NN−1r=0xre−2iπkr/N=1NN/2−1r=0x2re−i2π(2r)kN +N/2−1r=0x2...
N2/(N log2 N) = 52428, 8. Also one should make the sampling time intervalso small that the largest frequency that can be m...
Figure 3: Crystalline structure of a piezoelectric crystal, before and after polarization.along the direction of polarizat...
Figure 4: Poling process: (i) Before polarization; (ii) Polarization is gained using a verylarge DC electric field; (iii) T...
Figure 5: Axis for linear piezoelectric material describing the electromechanicalequations.Dm = dmiσi + ξσikEk, (38)where ...
S23 = S32, S12 = S21, S44 = S55, S66 = 2(S11 −S12), and others are zero. Thenon-zero piezo-electric strain constants are d...
circuit charge per unit area flowing between connected electrodes perpendic-ular j to the stress applied along i. A force F...
Ability of a piezoceramic to transform electrical energy to mechanicalenergy and vice verca is denoted by kij. The stress ...
The charge generated can be determinedq = D1 D2 D3dA1dA2dA3 , (55)where dA1, dA2 and dA3 are the differential electrode...
Figure 6: A typical configuration of a piezoelectric accelerometer.Figure 7: Applied force and displacements.20
The resonance sought is the lowest value of ω at which a maximum of(x1 − x2) occurs by varying ω. When the system is in dy...
From this we can get the upper limit of the usable frequency. Usually thisvalue is given, and sometimes the relative frequ...
5 Low-pass filter and measurement setupTaking into account the different frequency aspects namely the Nyquist fre-quency, an...
Figure 8: An active low-pass filter.Figure 9: Butterworth low-pass filter in a circuit used to obtain vibration spectra [19]...
other noisy vibrations, and other is resulting from external vibrations e.g.electronic devices on the cryostat, the pumps ...
When springs are in series the total spring constant can be calculated asthe deflection at the free end, δ, experienced app...
Figure 10: Beam deflection.From beam theory [21], M/I = E/R, where R is the radius of curvatureand EI is the flexural rigidi...
damping, and the viscous damping force is proportional to the first power ofthe velocity across the damper, and it is alway...
When ζ = 1 the damping is critical and equation x = (A + Bt)e−ωtisvalid. Finally damping greater than critical ζ > 1 gives...
From thisF2= (kX − mXν2)2+ (cXν)2, (94)orX = F/ ((k − mν2)2 + (cν)2), (95)andtan(φ) = cXν/(kX − mXν2). (96)The steady stat...
Figure 11: Isolating vibrating machine.X/Xs is known as the dynamic magnification factor, where Xs is staticdeflection of th...
The force transmitted to the foundation is the sum of the spring forceand the damper force. Thus the transmitted force is ...
from 5 Hz upwards can be achieved. Cork is one of the oldest materialsused for vibration isolation. It is usually used in ...
in natural frequency can simply be affected by change in the volume of thebellows. Greater attenuation of the exciting forc...
Figure 12: System with undamped vibration absorber.It can be easily seen thatX0 =F(k − mν2)∆, (112)andx0 =Fk∆, (113)where ...
amplitude. If correctly tuned ω2= K/M = k/m, and if the mass ratioµ = m/M, the frequency equation ∆ = 0 is ( [21], p.196)ν...
Figure 13: Model of a pump.mp ¨x1 + kpx1 = F sin νt, (120)gettinx1 = X1Fk1 − mpν2, (121)where k1 can estimated or deviced....
Figure 14: Adding the absorber on the pump.Taking x1,2 = X1,2 sin νt givingX1(k1 + k2 − m1ν2) − X2k2 = F (124)and−X1k2 + X...
From this and smallest absorber mass ν1/Ω = 0.8 as then ν2/Ω = 1.25,which is acceptable. Thus µ = 0.2 and hencem2 = µmp = ...
Figure 15: System with damped vibration absorber.X0 =FK − Mν2, (135)and when c is very large,X0 =FK − (M + m)ν2. (136)7 To...
the allowable magnitude of the torsional vibration. In our case the determin-ing factor is the heat leak into the nuclear ...
Figure 16: Circular shaft subject to torsional loading.Figure 17: Free-body diagram of the differential element of shaft at...
Mext= Meff(139)givesT(x, t)dx − Tr(x, t) + Tr(x, t) +δTr(x, t)δxdx = ρJdxδ2θ(x, t)δt2(140)orT(x, t) +δTr(x, t)δx= ρJδ2θδt2...
Considerδ2θδx2=δ2θδt2. (147)Let us look at some cases to analyse the cryostat with the above analysis.Firstly let us make ...
Now infinity of solutions arise corresponding toXk(x) = Dk sin (2k − 1)π2x (154)for any Dk. The modes are orthogonal giving...
dX(1)dx= βλX(1). (161)The solution isX(x) = D sin√λx (162)givingtan√λ =1β√λ. (163)There are countable but infinite values o...
Integrating the second equation, after the multiplication by Xi(x), from0 to 1:βλjXj(1)Xi(1) −10dXidxdXjdxdx + λj10XiXjdx ...
The torsional oscillations, in terms of nondimensional variables, withθ(0, t) = 0 areδθδx(1, t) = −βδ2θδt2(1, t) +T0LJGsin...
8 Balancing rotationThe unbalance of rotating machinery is the most common malfunction, evenso that any lateral vibrations...
first mode. The average unbalance angular force location will be referredto as a heavy spot. Rotors are usually similarly c...
Figure 18: Rotor lateral motion measured by two displacement proximity transducers inorthogonal orientation and the Keypha...
Figure 19: Time-base waveforms of rotor response to unbalance inertia force. Note thatresponse lags the force by the phase...
F = mrΩ2(182)Force phase that is the angular orientation δ mesured in degrees or radiansfrom a reference angle zero marked...
Figure 20: Unbalance force vector and rotor fundamental response vector in polarcoordinate format. Note conventional direc...
due to anisotropy of the rotor support system, which is the most commoncase in machinery. One Keyphasor dot on the orbit i...
Figure 21: Angular positions of unbalance force vector (heavy spot) and response vectorsat two rotational speeds and two d...
Figure 22: Balancing in one-plane.−→F−→H =−→B . (189)In this relationship, there are two unknown vectors,−→H and−→F . Ther...
problem and calculate the unknown parameters. The unkonwn vector −−→Fand the corrective mass, mc are calculated, therefore...
from wheremcrc(cos δc + i sin δc) = mτ rτ (cos δτ + i sin δτ )BB − B1(cos (β1 − β) + i sin (β1 − β))= mτ rτ (cos δτ + iNow...
Figure 23: One plane balancing of a rotor using polar plot.−→B −−→B1. Since the response is proportional to the input forc...
done with all the cryostat’s electronics in place, and other set of measure-ments were done when all the electronics were ...
0 5 10 15 20 25 30 35 40 45 50 5500.511.522.533.5x 104V(10−3)Tangential Noise with Electronics at 250 mrad/s0 5 10 15 20 2...
0 5 10 15 20 25 30 35 40 45 50012345678x 104HzVHz(10−3)Tangential Additive Noise Integral with Electronics250100020003000F...
0 5 10 15 20 25 30 35 40 45 5000.511.522.5x 104HzVHz(10−3)Normal Noise Additive Integral Plot with Electronics250100020003...
0 5 10 15 20 25 30 35 40 45 50 5502000400060008000V(10−3)Tangential Noise without Electronics at 250 mrad/s0 5 10 15 20 25...
0 5 10 15 20 25 30 35 40 45 50 5501234x 104V(10−3)Tangential Noise With Electronics at 3000 mrad/s0 5 10 15 20 25 30 35 40...
0 5 10 15 20 25 30 35 40 45 5002000400060008000100001200014000HzVHz(10−3)Tangential Additive Noise Integration without Ele...
case is helium is to be measured in a cylindrical cell. Solenoid magnet isthus a very good choice for this experiment, and...
Figure 29: Number of turns on each layer, and the dimensions of the magnet.dividual electrons, but it consists of Cooper p...
where the volume term has been neglected. The associated magneticenergy in the presence of an applied magnetic field ¯Ba is...
phonon energy from oscillating atoms in the lattice, then the electron pairwill stick together, thus not experiencing resi...
distributed disturbances are usually caused by heat leak to the cryogenicenvironment. Now transient disturbances are sudde...
13 Nuclear Magnetic Resonance, and Imag-ingQuantum mechanical magnetic proterties of atom’s nucleus can be studiedwith the...
zero for a net absorption. Now in a similar fashion using a alternating fieldwe get for the absorption energy [57]dEdt= n ω...
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION
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Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION

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Master Thesis on Rotating Cryostats and FFT, DRAFT VERSION

  1. 1. Abstract of Master’s Thesis Author: Kulvik Aleksanteri Name of thethesis: ibrational Study of Cryo I Helsinki, and Testing of Homogeneity of aSuperconducting Magnet Date: 23.4.2008 Number of pages: 60Department: Department of Engineering Physics and Mathematics Pro-fessorship: Tfy-3. Material PhysicsSupervisor & Instructor: Matti KrusiusIn the thesis, rotational aspects and vibrational analysis of Cryo I Helsinkicryostat have been studied.These studies were done extensively using Fourier methods. The mathe-matics of Fourier methods have been presented, and also mathematical mod-els of vibrations have been studied.The main goal is to lower the noise due to various aspects that affect theoperations of the cryostat.The second part of the work considers a superconducting magnet, andthe testing of how homogenous it was. There was a great preparation madeinto building of the testing equipment.1
  2. 2. Master’s ThesisKaarle Aleksanteri Kulvik52664T24th April 2008Vibrational Study of Cryo I Helsinki, and Testing of Homogeneity of aSuperconducting MagnetSupervisor M. Krusius2
  3. 3. Contents1 Preface 22 Introduction 33 Fourier Methods 34 Piezoelectric Accelerator 125 Low-pass filter and measurement setup 236 Vibrations 237 Torsional vibrations 408 Balancing rotation 499 Studies of the Noise of Old Rota I Cryostat 6010 Superconducting high-homogeneity magnet for NMR mea-surements 6711 Superconductivity 6812 Superconductor quenching 7113 Nuclear Magnetic Resonance, and Imaging 7314 Cooling of the Cryostat 7515 Preparing the NMR measurements 7916 Conclusions 801
  4. 4. 1 PrefaceDedicated to Alexiel, Pulla and those who do not dwell amongst us anymore...Albert, Kustaa, and Rosiel.Solitary trees, if they grow at all, grow strong. -Winston Churchill2
  5. 5. 2 IntroductionThis master thesis will study the vibrational, and rotational aspects of Cryo IHelsinki, which has now been dismantled. Suggestions for bettering rotaionalcryostat have been given in length, and the methods for study have beenmade clear for even a student reading this work. The second part consistof testing a superconducting magnet. The preparations are also included, asquite a lot of work needed to be done in order to test the magnet.3 Fourier MethodsFourier transformation can be seen as a tool that converts a ”signal”, or as inthis case, an output voltage into a sum of sinusoids of different frequencies,amplitudes and phases. In general, both input and output of the fouriertransform are complex vectors, which have the same length. A frequentdifficulty in understanding Fourier transformation lies in the comprehensionof the physical meaning of the results.The voltage versus time representation becomes magnitude versus fre-quency in the Fourier transform.The one dimensional Fourier series is given by the following formulaf(t) = a0 +∞n=1an cos(nωt) +∞n=1bn sin(nωt), (1)where t is an independent variable which in our case represents time, and Tpis the repetition period of the waveform. ω = 2π/Tp is the angular frequencyrelated to the fundamental frequency ff , by ω = 2πff . The constant a0 isgiven by the formulaa0 =1TpTp2−Tp2f(t)dt, (2)an byan =2TpTp2−Tp2f(t) cos(nωt)dt, (3)and bn by3
  6. 6. bn =2TpTp2−Tp2f(t) sin(nωt)dt. (4)The frequencies nω are known as the nth harmonics of ω. The series may bewritten in exponential formf(t) =∞n=−∞dneinωt(5)in whichdn =1TpTp2−Tp2f(t)e(−inωt)dt. (6)is complex and |dn| has the units of voltage in our case. Negative frequenciesdo not have any physical meaning rather being purely mathematical.The two conditions for f(t) are:1. The integral of |f(t)| from −∞ to ∞ exists 2. Any discontinuities inf(t) are finite.The squared modulus of a transform is referred as the energy spectrum.|F(ω)|2is the energy spectrum of f(t). Usually the graphs are given as theenergy spectrum versus ω.The complex and trigonometric forms are related by the followingdn = |dn|eiφn, (7)where|dn| = (a2n + b2n)1/2(8)andφn = − tan(bn/an), (9)where φn is the phase angle of the nth harmonic component.4
  7. 7. The series approach has to be abandoned when the waveform is not pe-riodic for example when Tp becomes infinite. As Tp increases the spacingbetween 1/Tp = ω/2π decreases to dω/2π eventually becoming zero. Thediscrete variable nω becomes continuos ω, and the amplitude and phasespectra become continuos. This means that dn → d(ω) and Tp → ∞. Withthese modifications we get the normalized formula [1],dn = F(iω) =12π−∞∞f(t)e(−inωt)dt. (10)F(iω) is the complex Fourier integral,F(iω) = Re(iω) + iIm(iω) = |F(iω)|eiφ(ω), (11)where the amplitude is given by,|F(iω)| = (Re(iω)2+ Im(iω)2)12(12)and the phase by,φ(w) = arctan[Im(iω)/Re(iω)]. (13)|F(iω)| has the units of V Hz−1.The Fourier transform (FT) has very useful properties [2]. If f(x) has theFourier transform F(s), then f(ax) has the Fourier transform |a|−1F(s/a).Its application to waveforms and spectra is well known as compression ofthe time scale corrensponds to expansion of the frequency scale. If f(x)and g(x) have the Fourier transforms F(s) and G(s), then f(x) + g(x) hasF(s) + G(s) as the FT. The FT of f(x) is F(s), then f(x − a) has the FTe−2πiasF(s). If f(x) has the FT F(s), then f(x) cos ωx has the FT 12F(s −ω/2π) + 12F(s + ω/2π). If f(x) and g(x) have FTs F(s) and G(s), thenconvolution f(x) ∗ g(x) has the FT F(s)G(s). The squarred modulus of afunction versus the squarred modulus of a spectrum yields;∞−∞|f(x)|2dx =∞−∞|F(s)|2ds. (14)If f(x) has the FT F(s) then f′(x) has the FT i2πF(s).It can be seen from:5
  8. 8. ∞−∞f′(x)e−i2πxsdx =∞−∞limf(x + ∆x) − f(x)∆xe−i2πxsdx = lim∞−∞f(x + ∆x)∆e−i2πxsdx − lim∞−∞With these interesting properties let us turn to the discrete Fourier trans-form (DFT) and the fast Fourier transform (FFT).The digitalization of the analogue data requires the Fourier transformsto be discrete. The analogue values are sampled at regular intervals andthen converted to binary representation. The operational viewpoint is thatit is irrelevant to talk about existence of values other than those given, andthose computed namely the input, and the output. Therefore we need themathematical theory to manipulate the actual quantified measurements.Discreteness arises in connection with periodic functions. Discrete in-tervals describing a periodic function may be viewed as a special case ofcontinuous frequency. This transform is thus regarded as equally spaceddeltafunctions multlipied by coeffients to determine their strengths.A typical fuction x(t) of the measurement is fed through an analogue todigital converter. It samples x(t) at a series of regularly spaced times asseen in Figure 1. Taking the sampling interval as ∆, then the discrete valueof x(t) = xr at time t is t = r∆, and can be written as a discrete timesequence {xr}, r = . . . , −1, 0, 1, 2, 3, . . .. We are interested in the frequencycomposition of sequence {xr} by analysis obtained from a finite length ofsamples.The historical method to estimate spectra from measured data was toestimate an appropriate correlation function first and then to FT this func-tion to obtain the required spectrum. This approach was until late 1960’s,and practical calculations follewed the mathematical route of spectra definedas FTs of correlation functions. The classical methods are studied in greatdetail, and there is extensive literature ( [3], [4] and [5] on this subject.The position was changed when fast Fourier transforms (FFT) camealong. This way of calculating the FT is much more efficient and faster.Instead of determining a correlation function, and then calculating the FT,FFT directly estimates the original FT of the time series.If x(t) is a periodic function with period T, then it can be written:x(t) = a0 + 2αk=1ak cos(2πktT) + bk sin(2πktT) (16)where k ≥ 0 is an integer, and6
  9. 9. Figure 1: Sampling a continuous function at regular intervals.ak =1TT0x(t) cos(2πktTdtbk =1TT0x(t) sin(2πktTdt. (17)The previous can be combined into a single equation:Xk = ak − ibk (18)and puttinge−i2πkt/T= cos2πktT− i sin2πktT(19)from we getXk =1TT0x(t)e−i2πkt/Tdt. (20)Knowing only the equally spaced samples of the continuous time seriesx(t) represented by the discrete series {xr}, r = 0, 1, . . ., (N − 1), where7
  10. 10. Figure 2: Calculating Fourier coefficients from a discrete series using approximation.t = r∆, and ∆ = T/N, then the integral may be replaced approzimately bythe summationXk =1TN−1r=0xre(−i2πk/T)(r∆)∆. (21)This is just assuming the total area under the curve in Figure 2. PuttingT = N∆ givesXk =1N r=0N − 1xre−i2πkr/N. (22)This may be regarded as approximation for calculating the Fourier series.The inverse formula for the series {xr} isxr =N−1k=0Xke2iπkr/N. (23)8
  11. 11. This can be seenN−1k=0Xke2iπkr/N=N−1k=01NN−1s=0xse−2iπks/Ne2iπkr/N=N−1k=0N−1s=01Nxse(−2iπk/N)(s−r)(24)and by interchanging the summation=N−1s=0N−1k=0e−2i(πk/N)(s−r) 1Nxs (25)and the exponentials all sum to zero unless s = r when the summationequals N and henceN−1s=0N−1k=0e−2i(πk/N)(s−r) 1Nxs = xr. (26)The components Xk are limited to k = 0 to N − 1 corresponding tofrequencies ωk = 2πk/T = 2πk/N∆.DFT of the series {xr}, r = 0, 1, . . ., N − 1 is defined asXk =1NN−1r=0xre−2iπkr/N(27)for k = 0, 1, . . ., N − 1. Calculating values of Xk for k is greater thanN − 1. Letting k = N + l thenXN+l =1NN−1r=0xre−(2iπr/N)(N+l)=1NN−1r=0xre−2iπr/Ne−2iπr= Xl. (28)The coefficients Xk just repeat for k > N −1, so plotting | Xk | along thefrequency axis ωk = 2πk/N∆, the graph repeats periodically. It is also easyto see that X−l = Xl (the complex conjugate) and hence | X−l |=| Xl | issymmetrical about the zero frequency position. The unique frequency rangeis | ω |≤ π/∆ rad/s. The higher frequencies are repetitions of those whichapply below π/∆ rad/s. The coefficients Xk calculated by the DFT arecorrect for frequencies up to9
  12. 12. ωk =2πkN∆=π∆(29)where k is in the range k = 0, 1, . . ., N/2. The frequencies above π/∆rad/s, which are present in the original signal, introduce a distortion calledaliasing. The high frequencies contribute to {xr}, and therefore distorts theDFT coefficients for frequencies below π/∆ rad/s. When ω0 is the maximunfrequency present in x(t), then the problem can be avoided by taking thesampling ∆ small enough soπ∆> ω0 (30)or f0 = ω0/2π giving12∆> f0. (31)This 1/2∆ Hz is called the Nyquist frequency, which is also the maximumfrequency that can be detected with particular time spacing ∆ (seconds).Aliasing is most important when analysing measured data, and to ensurethat DFT is good the sampling frequency 1/2∆ must be high enough tocover the full frequency range that the continuous time series operates in. Ifthis is not satisfied the the spectrum from equally spaced samples will differfrom the true spectrum because of aliasing. One way to ensure this is to filterall frequencies above the frequency components above 1/2∆ before makingthe DFT.FFT is an algorithm for calculating the DFTs. For working out values ofXk by directly calculating from the basic DFT definition for each N valuesrequires N2multiplications. The aim of the FFT is to reduce the number ofoperations to the order of N log2 N. The FFT therefore offers great amountof reduction in the prosessing time, and accuracy increases as fewer round-offerrors is reduced.FFT partitions the sequence {xr} in shorter sequences, and then combinesthese to yield the full DFT. Suppose {xr}, r = 0, 1, . . ., N − 1 is a sequencewhere N is an even number and this is partitioned to two shorter sequences{yr} and {zr} where yr = x2r and zr = x2r+1, r = 0, 1, . . ., (N/2 − 1). TheDFT’s of these areYk =1N/2N/2−1r=0yre−i2πkrN/2 Zk =1N/2N/2−1r=0zre−i2πkrN/2 , k = 0, 1, . . ., N/2 − 1.(32)10
  13. 13. Firstly we separate the odd and the even terms in {xr} gettingXk =1NN−1r=0xre−2iπkr/N=1NN/2−1r=0x2re−i2π(2r)kN +N/2−1r=0x2r+1e−i2π(2r+1)kN =1NN/2−1r=0yre−i2πrkN/2 + efrom whichXk =12Yk + e−i2πk/NZk , k = 0, 1, . . ., N/2 − 1. (34)The original DFT can be obtained from Yk and Zk. If the original numberof samples is a power of 2, then the half-sequences {yr} and {zr} can bepartitioned into quarter-sequences, and so on, until the last sub-sequenceshave only one term each. Yk and Zk are periodic and repeat themselves withperiod N/2 so that Yk−N/2 = Yk and Zk−N/2 = Zk.CalculatingXk =12Yk + e−2iπk/NZk , k = 0, 1, . . ., N/2 − 1Xk =12Yk−N/2 + e−2iπk/NZk−N/2 , k = N/2, N/2 + 1orXk =12Yk + e−2iπk/NZk Xk+N/2 =12Yk + e−2iπ(k+N/2)/NZk =12Yk − e−2iπk/NZk , k = 0, 1, . . ., N/These are the formulas occuring in most FFT programs, and definingW = e−2iπ/Nwe obtain what is called coputational butterfly [6]. The FFTchanged the approach to digital spectral analysis when it was implementedin 1965 ( [7] and [8]).For general purposes Matlab’s FFT is used. It is based upon FFTW-libraries [9]. FFTW uses several combinations of algorithms, including vari-ation of the Cooley-Tukey algorithm, a prime factor algorithm [10], and asplit-radix algorithm [11]. The split-radix FFT requires N to be a power of2 so the original sequence can be partitioned into two half-sequences of equallength, and so on.With these methods one is able to study the frequency depedence of theinput data. One should make the number of samples taken to be in theform 2n, where n is an integer. Even with number of samples the FFTworks quite some faster, for example a sequence that has N = 1048576 =220samples calculated directly with DFT compared to FFT has the ratio11
  14. 14. N2/(N log2 N) = 52428, 8. Also one should make the sampling time intervalso small that the largest frequency that can be measured is well withing theNyquist frequency to avoid distortion due to aliasing. Otherwise if this isnot possible then filtering should be used in the experimental setup to cutthe frequency components above the Nyquist frequency to avoid aliasing.4 Piezoelectric AcceleratorPiezoelectric effect was found in 1880 by Jaques and Pierre Curie in crys-talline minerals, when subjected to a mechanical force the crystal becameelectrically polarized. Compression and tension generated oppositely po-larized voltages in proportion to the force. In converse if voltage-genaratingcrystal was exposed to a electric field it contracted or expanded in accordancewith the polarity and field strength. These effects were called piezoelectriceffect and inverse piezoelectric effect. Quartz and other natural crystalsare widely used today in microphones, accelerometers, and ultrasonic trans-ducers. Their applications include smart materials for vibration control,aerospace, and astronautical applications of flexible surfaces, and vibrationreduction in sports equipment.Consistent with the IEEE standards of piezoelectricity [12], the transduc-ers are made of piezoelectric materials that are linear devices whose prop-erties are governed by a set of tensor equations. To better understand theworkings of piezoelectricity we firstly turn to making of a piezoelectric ce-ramic crystal. A piezoelectric ceramic is a perovskite crystal composed ofa small tetravalent metal ion placed inside a larger lattice of divalent metalions and O2 (see Figure 3). Preparing such a ceramic, fine powders of thecomponent metal oxides are mixed in very specific proportions, and heatedto form a uniform powder, which is then mixed with an organic binder. Thepowder turns into dense crystalline structure via specific process of heating,and cooling.Above the Curie temperature, each perovskite crystal exibit no dipolemoment (see Figure 4). Just below the Curie temperature each crystal hastetragonal symmetry, and a dipole moment. Alaining these dipoles usingelectrodes on the appropriate surfaces to create a strong DC electric field,gives a net polarization. This is called poling process. After removing theelectric field most of the dipoles remain in a locked place, creating permanentpolarization and permanent elongation. The length increase of the elementis usually within the micrometer range. Tension or mechanical compressionchanges the dipole moment associated with the particular element creatinga voltage. Tension perpendicular to direction of polarization or compression12
  15. 15. Figure 3: Crystalline structure of a piezoelectric crystal, before and after polarization.along the direction of polarization generates voltage of the same polarityas the poling voltalge. Tension along the polarization or compression per-pendicular to the direction or polarization generates an opposite voltage tothat of the poling voltage. The voltage and the compressive stress generatedapplying stress to the piezoelectric crystal are linearly proportional up to aspecific stress. In this way the crystal works as a sensor. The piezoelec-tric crystal expands and contracts when poling voltages are applied, and inthis way the use is an actuator. This way electric energy is converted intomechanical energy. When electric fields are low, and small mechanical stressthe piezoelectric materials have a linear profile. Under high stresses and elec-tric fields this breaks into very nonlinear behavior. Straining mechanically apoled piezoelectric crystal makes it electrically polarized, producing an elec-tric charge on the surface of the material. This is the direct piezoelectriceffect and it is the basis of sensory use.The electromechanical equations for a linear piezoelectric crystal are ([12], [13]):εi = SEij σj + dmiEm (37)13
  16. 16. Figure 4: Poling process: (i) Before polarization; (ii) Polarization is gained using a verylarge DC electric field; (iii) The remnant polarization after removing the field.14
  17. 17. Figure 5: Axis for linear piezoelectric material describing the electromechanicalequations.Dm = dmiσi + ξσikEk, (38)where i, j = 1, 2, . . ., 6 and m, k = 1, 2, 3 are different directions in thecoordinate system shown in Figure 5. The equations are usually written inanother form when the appications involve sensory actions:εi = §Dij σj + gmiDm (39)Ei = gmiσi + βσikDk (40)where σ . . . stress vector (N/m2) ε . . . strain vector (m/m) ξ . . . (F/m) E. . . vector of applied electric field (V/m) d . . . matrix of piezoelectric strainconstants (m/V ) S . . . matrix of compliance coefficients (m2/N) g . . . matrixof piezoelectric constants (m2/C) β . . . impermitivity component (m/F) D. . . vector of electric displacement (C/m2)The asumption here is that measurements of D, E, and σ are taken atconstant electric displacement, constant stress and constant electric field.Usually the crystal is poled along axis 3, and piezoelectric crystals aretransversely isotropic. Thus the equations simplify as S11 = S22, S13 = S31 =15
  18. 18. S23 = S32, S12 = S21, S44 = S55, S66 = 2(S11 −S12), and others are zero. Thenon-zero piezo-electric strain constants are d31 = d32, and d15 = d24. Alsothe non-zero dielectric coefficients are eσ11 = eσ22, and eσ33. One can write thesein matrix form to give:ε1ε2ε3ε4ε5ε6=S11 S12 S13 0 0 0S12 S11 S13 0 0 0S13 S13 S33 0 0 00 0 0 S440 0 0 0 S44 00 0 0 0 0 2(S11 − S12)σ1σ2σ3σ4σ5σ6+0 0 d310 0 d310 0 d330 d15 0d15 0 00 0 0E1E2E3(41)andD1D2D3 =0 0 0 0 d15 00 0 0 d15 0 0d31 d31 d33 0 0 0σ1σ2σ3σ4σ5σ6. (42)dij is the ratio of the strain in the j-axis to the electric field along thei-axis, taking external stresses constant. Voltage V is apllied as Figure 6 withthe crystal beign polarized in the z-direction, generates electric field:E3 =Vt. (43)This strains the transducer, e.g.ε1 =∆ll(44)and∆l =d31V lt. (45)Constant d31 is usually negative as the positive electric field generates apositive strain in z-direction. dij can also be interpreted as the ratio of short16
  19. 19. circuit charge per unit area flowing between connected electrodes perpendic-ular j to the stress applied along i. A force Fk is applied to the transducergenerates the stressσ3 =Flw(46)resulting in the elecric chargeq = d33F (47)flowing through the short circuit. The constant gij denotes the electricfield along i when the material is stressed along j. Force F applied in thepositive i, resulting in the voltage:V =g31Fw. (48)The other way to intepret gij is to take the ratio of strain along j to thecharge per unit area deposited on electrodes perpendicular to i. Placing anelectric charge Q on the surface electrodes (plates are on top and bottomperpendicular to k) changes the thickness by:∆l =g31Qw. (49)Constant Sij is the ratio of the strain in i-direction to the stress in j-direction, given that there’s no stress along the other two directions. Directstrains and stresses have indeces 1 to 3, and shear stresses and strains haveindeces 4 to 6. E is used to mark elastic compliace SEij measured with theelectrodes short-circuited, and similarly D denotes that the measurementswere done with electrods left open-circuited. SEij is maller than SDij as me-chanical stress results in an electrical responce that can increase the resultantstrain meaning that a short circuited piezo has a smaller Young’s modulusof elasticity than open-ciruited.The dielectric coefficient eij is the charge per unit area along x-axis due toapplied field applied in the y-axis. Relative dielectric constant K is definedas the ratio of the absolute permitivity of the material by the permittivity offree space. σ in eσ11 is the permittivity for a field applied along x-axis, whenthe material is not restrained.17
  20. 20. Ability of a piezoceramic to transform electrical energy to mechanicalenergy and vice verca is denoted by kij. The stress or strain is j-orientedand electrodes are perpendicular to i. Applying a force F to the crystal,while leaving the terminals open circuited makes the device deflect like aspring. This deflection is ∆z, and the mechanical work is:WM =F∆z2. (50)Electric charges accumulate on the electrodes due to piezoelectric effectamounting to the elecrical energy in the piezoelectric capacitor:WE =Q22CP. (51)From this we getk33 =WEWM=QF∆zCp. (52)The coupling can be written otherwise ask2ij =d2ijSijEeσij = gijdijEp, (53)where Ep is the Young’s modulus of elasticity. Now we turn to see how thepiezoelectric sensor works on these basis, as described above. Piezoelectricsensors offer superior signal to noise ratio, and better high-frequency noiserejection, thus they are quite suitable for applications that involve measur-ing low strain levels. When a piezoelectric crystal sensor is subjected to astress field, assuming the applied electric field is zero, the resultind electricaldisplacement vector is:D1D2D3 =0 0 0 0 d1500 0 0 d15 0 0d31 d31 d33 0 0 0σ1σ2σ3σ4σ5σ6(54)18
  21. 21. The charge generated can be determinedq = D1 D2 D3dA1dA2dA3 , (55)where dA1, dA2 and dA3 are the differential electrode areas in the y − z,x − z and x − y planes. The voltage generated VP is related to chargeVP =qCP, (56)where CP is the capacitator of the sensor. By measuring VP , the strain canbe calculated from the integral above. A calibrated piezoelectric accelerom-eter is a sensor, and the voltage measured can then be used as measure oracceleration thus this is very useful for very precise frequency analysis. Fig-ure 6 shows a typical configuration of a piezoelectric accelerometer mounted.Small size, and rigidness generally means that stucture’s vibrational charac-teristics will be minimal, but the structure does often affect the vibrationalcharacteristics of the attached accelerometer. Accelerometer’s sensitivity isdefined as the ratio of the output signal (voltage in our case) to the acceler-ation of its base. The major resonant frequency of the accelerometer is thelowest frequency for which the sensitivity has a maximum. The frequencyrange of use is generally taken as that region in which sensitivity does notchange significantly from the value found near 100 Hz [14] when calibratedon a conventional shaker table. The upper limit of an accelerometer is lowerthan the determined by resonance of the accelerometer alone, when the massof accelerometer affects the motion of the structure.To estimate the largest frequency to be measured can be calculated usingthe following model. We assume the system consisting of masses (m1, m2 andm3), and springs (assumed massless, and their spring constants ka and ks)as in Figure 6. Let us suppose that a sinusoidally varying mechanical forceF cos ωt, is imposed on m3 from outside the system taking the position x3 isA3 cos ωt. The resultant motion of m1, and m2 or the varying positions x1and x2 (Figure 7) are considered as the frequency of the drive force is varied.After transient effects die away, the equations describing the motion of m1and m2 under the dynamic forces are:m1 ¨x1 + ka(x1 − x2) = 0 (57)m2 ¨x2 + ka(x2 − x1) + ks(x2 − x3) = 0. (58)19
  22. 22. Figure 6: A typical configuration of a piezoelectric accelerometer.Figure 7: Applied force and displacements.20
  23. 23. The resonance sought is the lowest value of ω at which a maximum of(x1 − x2) occurs by varying ω. When the system is in dynamic equilibriumand resonance is approached from below, the motions will be at the drivefrequency and in phase.x1 = A1 cos ωt (59)x2 = A2 cos ωt (60)¨x1 = −A1ω2cos ωt (61)From this we obtain(ka − m1ω2)A1 − kaA2 = 0 (62)−kaA1 + (ks + ka − m2ω2)A2 = ksA3 (63)Resonance occurs when A1 − A2 has a maximum value. There’s no prob-lem due to phase considerations because resonance is approached from belowand, with no damping, x1 and x2 can be considered to be in phase with themotion of the driving element i.e. with x3. The solution for A1 and A2 eachhas the determinant of its coefficients in the denominator. Thus maximumvalues of A1 and A2 occur when this determinant vanishes. An equation inthe resonant frequency ω results:(ka − m1ω2)(ka + ks − m2ω2) − k2a = 0. (64)This can be written as quadratic in ω2,ω4− [ka(1m1+1m2) + ks1m2]ω2+kaksm1m2= 0 (65)Now we simplify the calculations by taking new constants a = m2/m1,r = ks/ka, and ω20 = ka/m1. Substituting these into the above equation andcalculating the frequency:ω = ω20[1 + 1a(1 + r)] − [1 + 1a(1 + r)]2 − 4ra2. (66)21
  24. 24. From this we can get the upper limit of the usable frequency. Usually thisvalue is given, and sometimes the relative frequency is given as ω/ω0. If thisvalue is not given, then one can estimate. Now in our case the measurementson the cryostat imply that the mass of the accelerometer does not alterthe results, and the accelerometers base m2 is rigidly attached (in our casea very strong magnet and straps) to a very large mass m3 (cryostat’s orthe supporting frame’s mass) so that m1 and ka are the only resonance-determined parameters. These parameters are usually given, but if not itcan be relatively easy to approximate those. In this case of rigid attachmentmasses m2 and m3 are combined, and taking m3 = ∞ we getω = ka/m1. (67)The Nyquist frequency should be set little below this frequncy, and the allfrequencies above the Nyquist frequency should be filtered out (see next sec-tion). Or if it is know what frequencies are to be looked for then the highestfrequency should be set according to that. Accelerometers are also subjectedto thermal-transient stimuli from stronger vibrations. Certain propertiesof piezoelectric accelerometers can cause them to generate spurious outputsignals in response to such thermal transients, leading to significant measure-ment errors. Many piezoelectric crystalline materials are also pyroelectric [15]that is, a change of temperature causes a change in the polarization chargesin the material. Pyroelectric output signals can result from a uniform ornon-uniform distribution of thermal charges within the material. In addi-tion, mechanical strain within the piezoelectric element, resulting differentialthermal expansion of the components of an accelerometer subjected to ther-mal transients, may generate spurious output signals. In conditions wherethe accelerometer is exposed to blasts, non-uniform heating is propable. Theresultant output signal will thus include pyroelectrically generated chargesand charges produced by changes in the mechanical loading of the crystalresulting from differential expansion of accelerometer components. This isthe reason why after rigid attaching of the piezoelectric accelerometer on thecryostat or its frame one should wait for some time for normal conditions toreappear. Usually the pyroelectric effect under normal stated conditions formost piezoelectric crystals are not significant under frequencies of 3000 Hzand amplitudes of 5 g [16]. So there should not be any problem with normalmeasurements with the setup of the cryostat. However care should be takenas to where the accelerometer is placed, e.g. it should not be placed in closeproximity of electronic devices that give strong electric fields or directly leakheat into the surroundings.22
  25. 25. 5 Low-pass filter and measurement setupTaking into account the different frequency aspects namely the Nyquist fre-quency, and the maximum frequency of the piezoelectric accelerometer underwhich it operates, one needs to have a low-pass filter. Using this wisely willeliminate almost all coputed aliasing, and bad signals from the crystal. Usu-ally the window of interest lies somewhere in the region of 0 to 100 Hz, whichis easily attained by the machinery used. Ideally low-pass filters completelyeliminate all frequencies above the cut-off frequency while passing those be-low unchanged. Real time filters approximate the ideal filter by windowingthe infinite impulse response to make a finete impulse response. Digital filter-ing in our case is not the best solution, better is to use an electronic low-passfilter. A second-order filter does a better job of attenuating higher frequen-cies. There are many different types of filter circuits, with different responsesto changing frequency. A first-order filter will reduce the signal amplitude byhalf every time the frequency doubles (goes up one octave). As the frequencyreach of the equipment used is so large, the low-pass filter can be relativelysimple one. One could use or build easily an active low-pass filter. In theoperational amplifier shown in the Figure 8, the cutoff frequency is definedasfc =12πR2C(68)or equivalently in radians per secondωc =1R2C. (69)The gain in the passband is −R2/R1, and the stopband drops off at −6dBper octave, as it is a first-order filter.If this doesn’t work as wished one can easily build a second order (orhigher) Butterworth filter (see Figure 9 [19]), which decreases −12dB peroctave. Also the frequency responce of the Butterworth filter is maximallyflat [18] in the passband compared to Chebyschev Type I / Type II or anelliptic filter [17].6 VibrationsThe types of vibrations in our case can be divided into two main categories:the unbalanced rotation of the cryostat creates harmonic oscillations and23
  26. 26. Figure 8: An active low-pass filter.Figure 9: Butterworth low-pass filter in a circuit used to obtain vibration spectra [19].24
  27. 27. other noisy vibrations, and other is resulting from external vibrations e.g.electronic devices on the cryostat, the pumps and vibrations from groundor foundation vibrations. Torsional vibration analysis is vital for ensuringreliable machine operation, especially as very precise measurements are madeon the large cryostat. If rotating component failures occur on the cryostatas a result of torsional oscillations, the consequences can be catastrophic.In the worst case, the entire machine can be wrecked as a result of largeunbalancing forces, and worse injury to human beigns might be inflicted.The foundation and electronics vibations are easier to allocate, but are bigenough to cause problems as vibrations could affect the nuclear stage andthe demagnetization solenoid creating a heat leak there as suggested by [19].The level of vibration in a structure can be attenuated by reducing either theexcitation or the response of the structure to that excitation or both. Thesecould be relocating equipment, or isolating the structure from the excitingforce. The torsional vibrations can be reduced by balancing the load on therotating machinery. Real structures consist of an infinite number of elesticallyconnected masses and have infinite number of degrees of freedom. In realitythe motion is often such that only a few coordinates is needed to describethe motion. The vibration of some structures can be analysed using a sigledegree of freedom. Other motions may occur, but in our case for instancefor analysing the electric and foundation vibrations other vibrations can bedimished, and electrical devices can be measured one at the time (to see morecomprehensive study [20]. A body of mass m is free to move along a fixedhorizontal surface attached to a spring k one end fixed. Displacement of themass is denoted by x, so giving this initial displacement x0, and letting gowe get:¨x +kmx = 0 (70)givingx = A cos ωt + B sin ωt, (71)where A and B are constants, and ω is the circular frequency. Now withx = x0 and t = 0 gives A = x0, and ˙x = 0 and t = 0 givesx = x0 coskmt. (72)25
  28. 28. When springs are in series the total spring constant can be calculated asthe deflection at the free end, δ, experienced applying the force F is to bethe same in both cases,δ = F/ke =iF/ki (73)so that1/ke =i1/ki. (74)Similarly parallel springs giveke =iki. (75)Let us consider a beam with m as the mass unit length, and y is theamplitude of the deflection curve (see Figure 10) thenTmax =12˙y2maxdm =12ω2y2dm, (76)where ω is the natural circular frequency of the beam, and Tmax is themaximum kinetic energy.The strain energy of the beam is the work done on the beam which isstored as elastic energy. If the bending moment is M, and the slope of theelastic curve is θ, the potential energy isV =12Mdθ. (77)Assuming the deflection of beams smallRdθ = dx, (78)thus1R=dθdx=d2ydx2. (79)26
  29. 29. Figure 10: Beam deflection.From beam theory [21], M/I = E/R, where R is the radius of curvatureand EI is the flexural rigidity:V =12MRdx =12EId2ydx22dx. (80)Now Tmax = Vmax;ω2=EI d2ydx22dxy2dm. (81)This expression gives the lowest natural frequency of transverse vibrationof a beam. It can be seen that to analyse the transverse vibration of a partic-ular beam by this method requires y to be known as a function of x. In thecase of the cryostat’s frame this method can prove to be quite cumbersome.Real structures dissipate vibration energy, so damping sometimes becomessignificant. Damping is difficult to model exactly because the mechanismsof the structures. Using simplified models usually gain quite good results,and can give insight to the problem. Viscous damping is a common form of27
  30. 30. damping, and the viscous damping force is proportional to the first power ofthe velocity across the damper, and it is always opposed to motion, so thatdamping force is linearly continuous function of the velocity. Simple modelcan be imagined taking a horizontal m mass attached to a spring k and adamper c (damping force is proportional to velocity), which are both fixed.As before we get for the equation of motion:m¨x + c ˙x + kx = 0. (82)Assuming solution of the form x = Xest= 0, and substituting for roots:s1,2 = −c2m±(c2 − 4mk)2m, (83)hencex = X1es1t+ X2es2t, (84)where X1 and X2 are arbitrary constants found from initial conditions.The dynamic behavious of the system depends opon the numerical value ofthe radical, so defining critical damping cc = 2√km making the radical zero,and undamped natural frequency ω = cc/2m. Defining damping ratio byζ = c/cc, (85)ands1,2 = (−ζ ± (ζ2 − 1))ω. (86)When damping is less critical ζ < 1s1,2 = −ζω ± i (1 − ζ2)ω (87)sox = Xe−ζωtsin ( (1 − ζ2)ωt + φ). (88)28
  31. 31. When ζ = 1 the damping is critical and equation x = (A + Bt)e−ωtisvalid. Finally damping greater than critical ζ > 1 gives two negative realvalues of s so that x = X1es1t+ X2es2t. Substituting for damping constanta constant friction force Fd that represents dry friction (Coulomb damping)applicaple in many mechanisms:m¨x + kx = Fd. (89)Getting a solutionx = A sin ωt + B cos ωt +Fdk. (90)Hencex = (x0 +Fdkcos ωt +Fdk, (91)where the oscillation ceases with | x |≤ Fd/k, and the zone x = ±Fd/k iscalled the dead zone. Many real structures have both viscous and Coulombdamping. The two damping actions are sometimes dependent of amplitude,and if the two cannot be separated a mixture of linear and exponential decayfunctions have to be found by trial and error. In most real structures separat-ing stiffness and damping effects is often not possible. This can be modeledusing complex stiffness k∗= k(1 + iη), where k is the static stiffness, andη is the hysteric damping loss factor. A range of values for η can be foundfor common engineering materials in basic literature ( [22]). The electronicdevices on the cryostat behave as external excitation forces usually periodic.From previous we construct a model as taking mass m connected to a fixedspring and viscous damper, whilst a harmonic force of circular frequency νand amplitude F:m¨x + c ˙x + kx = F sin(νt). (92)Solution can be taken as x = X sin(νt − φ), where motion lags the forceby vector φ, so substituting and using cos − sin relations we getmXν2sin(νt − φ + π) + cXν sin(νt − φ + π/2) + kX sin(νt − φ) = F sin(νt).(93)29
  32. 32. From thisF2= (kX − mXν2)2+ (cXν)2, (94)orX = F/ ((k − mν2)2 + (cν)2), (95)andtan(φ) = cXν/(kX − mXν2). (96)The steady state solutionx =F((k − mν2)2 + (cν)2)sin(νt − φ), (97)whereφ = tan−1 cνk − mν2. (98)The complete solution includes the transient motion given by the com-plementary function:x = Ae−ζωtsin(ω (1 − ζ2)t + α), (99)where ω = k/m and Xs = F/k so thatXXs=1(1 − (ν/ω)2)2 + (2ζν/ω)2, (100)andφ = tan−1 2ζ(ν/ω)1 − (ν/ω)2. (101)30
  33. 33. Figure 11: Isolating vibrating machine.X/Xs is known as the dynamic magnification factor, where Xs is staticdeflection of the system under a steady force F, and X is the dynamic ampli-tude. The mechanical vibration arises from the large values of X/Xs, whenν/ω has a value near unity, meaning that a small harmonic force can producea large amplitude of vibration. Resonance occurs when the forcing frequencyis equal to natural frequency e.g. ν/ω = 1. The max of X/Xs can be attainedfrom differentiating to get:(ν/ω)(X/Xs)max = 1 − 2ζ2) ≃ 1, ζ ≈ 0, (102)and(X/Xs) = 1/(2ζ 1 − ζ2). (103)For small ζ, (X/Xs)max ≃ 1/2ζ is a measure of the damping and is knownas the Q factor. The force transmitted to the foundation or supporting struc-ture can be reduced by using flexible mountings with the correct properties.Figure 11 shows a model of such a system.31
  34. 34. The force transmitted to the foundation is the sum of the spring forceand the damper force. Thus the transmitted force is given byFT = (kX)2 + (cνX)2. (104)The transmissibility is given byTR =FTF=X k2 + (cν)2F(105)sinceX =F/k(1 − νω2)2 + 2ζ νω)2, (106)TR =1 + (2ζ νω)21 − (νω)2 + 2ζ νω2. (107)Therefore the force and motion transmissibilities are the same. It can beseen that for good isolation ( [21]) ν/ω >√2, hence for a low value of ωis required which implies a low stiffness, that is a flexible mounting. In thecryostat it is particularly important to isolate vibration sources e.g. the elec-trical devices because vibrations transmitted to structure radiate well, andserious heat leak problems can occur. Theoretically low stiffness isolatorsare desirable to gice a low natural frequency. There are four types of springmaterial commonly used for resilient mountings and vibration isolation: air,metal, rubber, and cork. Air springs can be used for very low-frequencysuspensions: resonance frequencies as low as 1 Hz can be achieved whereasmetal springs can only be used for resonance frequencies greater than about1.3 Hz. Metal springs can transmit high frequencies, however, so rubberor felt pads are often used to prohibit metal-to-metal contact between thespring and the structure. Different forms of spring element can be used ascoil, torsion, cantilever and beam. Rubber can be used in shear or compres-sion but rarely in tension. It is important to determine the dynamic stiffnessof a rubber isolator because this is generally much greater than the staticstiffness. Rubber also possesses some inherent damping although this maybe sensitive to amplitude, frequency and temperature. Natural frequencies32
  35. 35. from 5 Hz upwards can be achieved. Cork is one of the oldest materialsused for vibration isolation. It is usually used in compression and naturalfrequencies of 25 Hz upwards are typical. For precise isolation systems, andmaterials please refer to [23]. The above analysis is more for analysing thevibrations due to electrical devices and the pumps, as they are consistentusually having some periodicity giving only specific frequency peaks in fre-quency spectrum. For external noise from the surroundings are more likelyto be random processes (as in [19]) possibly due to heavy traffic on a nearbyroad or other large machinery used nearby. Collection of sample functionsx1(t), x2(t), . . . , xn(t), which make up the ensemble x(t). Normal or Gaussianprocess is the most important of random processes because a wide range ofphysically observed random waveforms represented by Gaussian process:p(x) =1√2πσe− 12x−xσ2, (108)is the density function of x(t), where σ is the standard deviation of x,and x is the mean of x. The values of σ and x may vary with time for a non-stationary process but are independent of time if the process is stationary.x(t) lies between −λσ and λσ, where λǫR+taking x with probabilityProb{−λσ ≤ x(t) ≤ λσ} =λσ−λσ1√2πσe(− 12x2σ2 )dx. (109)Probabilities with varying λ can be found for example in [24]. We nowturn to actual methods of damping the unwanted vibrations. Some reductioncan be achieved by changing the machinery generating the vibration, for ex-ample removing the fans from electrical devices, and using static heat sinkscommercially available noting problems involved. It is desirable for the cryo-stat and the framework to possess sufficient damping so that the responseto the expected excitation is acceptable. If damping in the structure is in-creased the vibrations and noise, and the dynamic stresses will be reduceddirectly resulting in lowered heatleak. However increasing damping might beexpensive and may require big changes in already existing buildings. Goodvibration isolation can be achvieved by supporting the vibration generatoron a flexible low-frequency mounting. Air bags or bellows are sometimesused for very low-frequency mountings where some swaying of the supportedsystem is allowed. Approximate analysis shows that the natural frequency ofa body supported on bellows filled with air under pressure is inversely pro-portional to the square root of the volume of the bellows, so that a change33
  36. 36. in natural frequency can simply be affected by change in the volume of thebellows. Greater attenuation of the exciting force at high frequencies can beachieved by using a two-stage mounting. In this arrangement the machineis set on flexible mountings on an inertia block, which is itself supported byflexible mountings. This may not be expensive to install since for examplethe cryostat can be used as the inertia block. Naturally, techniques used forisolating structures from exciting forces arising in machinery and plant canalso be used for isolating delicate equipment from vibrations in the struc-ture. Normal solution to vibrational problems is to place the cryostat on aheavy block supported by air springs, and rotating motors placed on bellowswrapped with isolating tape [25]. Of course increasing the mass of the blockincreases the resonant frequency decreases, this might be a problem with ro-tation. There are also active isolation systems in which the exiciting force ormoment is applied by an externally powered force or couple. The opposingforce or moment is applied by an externally powered force. The opposingforce can be produced by means such as hydraulic rams. All materials dis-sipate energy during cyclic deformation due to molecular dislocations andstress changes at grain boundaries. Such damping effects are non-linear andvariable within material. Some particular materials such as damping alloyshave a certain enhanced damping mechanisms. The load extension hysteresisloops for linear materials and structures are elliptical under sinusoidal load-ing, and increase in area according to the square of the extension. Althoughthe loss factor η of a material depends upon its composition, temperature,stress and the type of loading mechanism used, an approximate value for ηcan be given [26]. Pure aluminium has loss factor of 0.00002 − 0.002, andhard rubber has 1.0. In a single or multi degree of freedom system modeis excited into resonance, and the excitation frequency nor the natural fre-quency can be altereded then adding a single degree of freedom can be ofuse. One can consider this using a model such as in Figure 12, where K andM are the effective stiffness and mass of the primary system.The absorber is represented by the system with parameters k and m. Theequations of motion for the primary system:M ¨X = −KX − k(X − x) + F sin νt (110)and for the vibration absorberm¨x = k(X − x), (111)where X = X0 sin νt and x = x0 sin νt.34
  37. 37. Figure 12: System with undamped vibration absorber.It can be easily seen thatX0 =F(k − mν2)∆, (112)andx0 =Fk∆, (113)where ∆ = (k − mν2)(K + k − Mν2) − k2, and ∆ = 0 is the frequencyequation. Now the system possess two natural frequencies, Ω1 and Ω2, but byarranging k − mν2= 0, X0 can be made zero. Now if (k/m) = (K/M),the response of the primary system at its original resonance frequency canbe made zero. This is the usual tuning arrangement for undamped absorberbecause the resonance problem in the primary system is only severe whenν ⋍ K/M. When X0, x0 = −F/k, so that the force in the absorber spring,kx0 is −F thus the absorber applies a force to the primary system whichis equal and opposite to the exciting force. Hence the body in the primarysystem has a net zero exciting force acting on it and therefore zero vibration35
  38. 38. amplitude. If correctly tuned ω2= K/M = k/m, and if the mass ratioµ = m/M, the frequency equation ∆ = 0 is ( [21], p.196)νω4− (2 + µ)νω2+ 1 = 0, (114)henceΩ1,2ω= 1 +µ2± µ +µ241/2. (115)For a small µ, Ω1 and Ω2 are very close to each other, and near to ω,increasing µ gives better separation between Ω1 and Ω2. This is of impor-tance in systems where the excitation frequency may vary e.g. µ is small,resonances at Ω1 or Ω2 may be excited. Now:Ω1ω2= 1 +µ2− µ +µ24(116)andΩ2ω2= 1 +µ2+ µ +µ24(117)then multiplying givesΩ1Ω2 = ω2(118)andΩ1ω2+Ω2ω2= 2 + µ. (119)One can use these relations to desing an absorber, and can be used forinstance for a pump having mass of mp rotating at constant speed of ωprev/min, giving large unbalance vibrations. Fitting an undamped absorberso that the natural frequency of the system is removed by 20%. We modelthe pump as in Figure 13, so we get the equation of motion:36
  39. 39. Figure 13: Model of a pump.mp ¨x1 + kpx1 = F sin νt, (120)gettinx1 = X1Fk1 − mpν2, (121)where k1 can estimated or deviced. When X1 = ∞ then ν = k1/mp,that is resonance occurs when ν = ω = k1/m1. Assuming x2 > x1 (Figure14) we get:m2 ¨x2 = −k2(x2 − x1) (122)andmp ¨x1 = k2(x2 − x1) − k1x1 + F sin νt. (123)37
  40. 40. Figure 14: Adding the absorber on the pump.Taking x1,2 = X1,2 sin νt givingX1(k1 + k2 − m1ν2) − X2k2 = F (124)and−X1k2 + X2(k2 − m2ν2) = 0. (125)From thisX1 =F(k2 − m2ν2(k2 + k1 − mpν2)(k2 − mmν2) − k22. (126)If ν2= k2/m2, X1 = 0 then the frequency equation is (k1+k2−mpν2)(k2−m2ν2) − k22 = 0. Putting µ = m2/mp = k2/k1 and Ω = k2/m2 = k1/mp,givingνΩ2=2 + µ2±µ2 + 4µ4. (127)38
  41. 41. From this and smallest absorber mass ν1/Ω = 0.8 as then ν2/Ω = 1.25,which is acceptable. Thus µ = 0.2 and hencem2 = µmp = 0.2mp, (128)andk2 = 2πωp. (129)One good system is a ciscous damped absorber such as in Figure 15.Equations of motion are:M ¨X = F sin νt − KX − k(X − x) − c( ˙X − ˙x) (130)andm¨x = k(X − x) + c( ˙X − ˙x). (131)Substituting X = X0 sin νt and x = x0 sin (νt − φ) gives, after somemanipulation,X0 =F (k − mν2)2 + (cν)2((k − mν2)(K + k − Mν2) − k2)2 + (cν(K − Mν2 − mν2))2.(132)When c = 0 this reduces to the undamped vibration absorber. If c islarge thenX0 =FK − ν2(M + m). (133)Response of the primary system is minimized over a wide frequency rangeby choosing different c. If k = 0,X0 =F√m2ν4 + c2ν2((K − Mν2)mν2)2 + (cν(K − Mν2 − mν2)). (134)When c = 0,39
  42. 42. Figure 15: System with damped vibration absorber.X0 =FK − Mν2, (135)and when c is very large,X0 =FK − (M + m)ν2. (136)7 Torsional vibrationsHistorically torsional modes in machinery were always the first to considerand analyze, in order to avoid extreme stresses. Today torsional vibrationanalysis is routinely done throughout design of rotating machines. Their ex-istence can be discovered when using dedicated instruments [27] to measuretorsional vibrations. Torsional vibration is an oscillatory angular motioncausing twisting in the shaft of a system. Motion is rarely a concern withtorsional vibration unless it affects the function of a system. It is stressesthat affect the structural integrity and life of components and thus determine40
  43. 43. the allowable magnitude of the torsional vibration. In our case the determin-ing factor is the heat leak into the nuclear stage. The complicated system ofthe cryostat can be crudely modeled to gain insight into the problem. How-ever the torsional vibration is a complex vibration having many differentfrequency components. The cryostat can be crudely taken as cylinder ro-tating a perpendicular axis. The polar moment of inertia can be calculatedfrom the general fromula J = r2dm, where r is the instanteneous radius,and dm is the differential mass. The formula for the polar moment of inertiaof a cylinder rotating about a perpendicular axis isJ =πd4lγ32g, (137)where J is the polar moment of inertia, γ material density, d diameter ofcylinder, l is axial length of cylinder, and g acceleration due to gravity. Thetorsional stiffness is (πd4G)/(32l), where G is rigidity modulus, and substi-tuting d2−d1 for d gives you formula for an annulus with outer-inner diameterd2-d1. Taking the cryostat as a circular shaft (Figure 16) is made of materialof mass density ρ and shear modulus G and has a length L, cross-sectionalarea A, and polar moment of inertia as above. Let x be the coordinate alongthe axis of the shaft. The shaft is subject to a time-dependent torque perunit lenght, T(x, t). Let θ(x, t) measure the resulting torsional oscillationswhere θ is chosen positive clockwise. Figure 17 shows free-body diagramsof a differential element of the shaft at an arbitrary instant of time. Theelement is of infinitesimal thickness dx and its left face is a distance x fromthe left end of the shaft.The free-body diagram of the external forces shows the time-dependenttorque loading as well as the internal torques developed in the cross sections.The internal resisting torques are the resultant moments of the shear stressdistributions. If Tr(x, t) is the resisting torque acting on the left face of theelement, then a Taylor seris expansion truncated after the linear terms gives:Tr(x + dx, t) = Tr(x, t) +δTr(x, t)δtdx. (138)The directions of the torques shown on the free-body diagram are consis-tent with the choise of θ positive clockwise. Since the disk is infinitesimal,the angular acceleration is assumed constant across the thickness. Thus thefree-body diagram of the effective forces simply shows a moment equal to themass moment of inertia of the disk times its angular acceleration. Summationof moments about the center of the disk41
  44. 44. Figure 16: Circular shaft subject to torsional loading.Figure 17: Free-body diagram of the differential element of shaft at arbitrary instant.42
  45. 45. Mext= Meff(139)givesT(x, t)dx − Tr(x, t) + Tr(x, t) +δTr(x, t)δxdx = ρJdxδ2θ(x, t)δt2(140)orT(x, t) +δTr(x, t)δx= ρJδ2θδt2. (141)From mechanics of materials,Tr(x, t) = JGδθ(x, t)δx(142)which leads toT(x, t) + JGδ2θδx2= ρJδ2δδt2. (143)Using the following to simplify x∗= x/L, t∗= G/ρ(t/L), and T∗(x∗, t∗) =T(x, t)/Tm, where Tm is the maximum value of T. From theseL2TmJGT(x, t) +δ2θδx2=δ2θδt2, (144)where the ∗ has been dropped from nondimensional variables. The prob-lem formulation is completed by specifying appropriate initial conditions ofthe formθ(x, 0) = g1(x) (145)andδθ(x, 0)δt= g2(x). (146)43
  46. 46. Considerδ2θδx2=δ2θδt2. (147)Let us look at some cases to analyse the cryostat with the above analysis.Firstly let us make x the length of the cryostat a unilength so that freeend is x = 1 and taking the fixed end at x = 0. The boundary conditionis θ(0, t) = 0, and ˙θ(1, t) = 0 (the derivate is in terms of x). Applying amoment M is statically applied to the end of the shaft leading to the initialcondition θ(x, 0) = Mx/(JG) = γx. Since the shaft is released from resta second initial condition is ˙θ(x, 0) = 0 (the derivative is in terms of t). Aseparation of variables is assumed θ(x, t) = X(x)T(t), which gives1X(x)d2Xdx2=1T(t)d2Tdt2. (148)leading tod2Tdt2+ λT = 0, (149)andd2Xdx2+ λX = 0, (150)where λ is the separation constant. The solution isT(t) = A cos√λt + B sin√λt, (151)where A and B are arbitrary constants of integration. SimilarlyT(t) = C cos√λx + D sin√λx (152)The initial conditions give C = 0, B = 0 and the only reasonable solutionλk = [(2k − 1)π2]2k = 1, 2, . . . (153)44
  47. 47. Now infinity of solutions arise corresponding toXk(x) = Dk sin (2k − 1)π2x (154)for any Dk. The modes are orthogonal giving(Xk(x), Xj(x)) =10DjDk sin (2k − 1)π2x sin (2j − 1)π2xdx = 0 (155)for j = k, but when k = j we get1 = (Xk, Xk) =D2k2(156)leading toθ(x, t) =∞k=1√2 sin (2k − 1)π2x[Ak cos (2k − 1)π2t]. (157)From the initial conditions we get the lastAk =4γ√2(−1)k+1π2(2k − 1)2(158)yielding in totalθ(x, t) =8γπ2∞k=1(−1)k+1 1(2k − 1)2sin((2k − 1)π2x) cos((2k − 1)π2t). (159)Let us now consider a circular shaft fixed at x = 0 and has a thin diskof mass moment of inertia I, similar to the electronics above the cryosta,attached at x = 1. The partial differential equation governing [28]δθ(1, t)δx= −βδ2θ(1, t)δt2(160)where β = I/(ρJL). Separation of variables give:45
  48. 48. dX(1)dx= βλX(1). (161)The solution isX(x) = D sin√λx (162)givingtan√λ =1β√λ. (163)There are countable but infinite values of λ, and at large k, λk approaches((k + 1)π)2. Let λi and λj are distinct solutions with corresponding modeshape Xi(x) and Xj(x) respectively. The mode shapes satisfy the boundaryconditions Xi(0) = 0, Xj(0) = 0, ˙Xi(1) = βλiXi(1), and ˙Xj(1) = βλjXj(1):d2Xidx2+ λiXi = 0 (164)andd2Xjdx2+ λjXj = 0. (165)Multplying the first of these by Xj(x) and integrating from 0 to 1 gives10d2Xidx2Xjdx + λi10XiXjdx = 0 (166)and integrating by parts leads to:Xj(1)dXidx(1) − Xj(0)dXidx(0) −10dXidxdXjdxdx + λi10XiXjdx = 0. (167)so thatβλiXi(1)Xj(1) −10dXidxdXjdxdx + λi10XjXidx = 0. (168)46
  49. 49. Integrating the second equation, after the multiplication by Xi(x), from0 to 1:βλjXj(1)Xi(1) −10dXidxdXjdxdx + λj10XiXjdx = 0 (169)and subtracting the last two equations leads to(λi − λj) βXi(1)Xj(1) +10XiXjdx = 0. (170)This implies since λi = λjβXi(1)Xj(1) +10XiXjdx = 0 (171)and defining scalar product of g and f by(f, g) =10f(x)g(x)dx + βf(1)g(1) (172)then (Xj, Xk) = 0.The mode shape is normalized1 = (Xk, Xk) =10D2k sin2( λkx)dx + D2kβ sin2λk = D2k1012(1 − cos(2 λkx))dx + β sin2λk =Little manipulation producesDk =√2(1 + β sin2√λ)−1/2(174)where λk is the kth solution.Let us see a forced vibration example. This is similar to the problemswith the motor and the belt rotating the cryostat. We use the above modelas the cryostat and subject the thin disk to harmonic torque,T(t) = T0 sin ωt. (175)47
  50. 50. The torsional oscillations, in terms of nondimensional variables, withθ(0, t) = 0 areδθδx(1, t) = −βδ2θδt2(1, t) +T0LJGsin ˜ωt (176)where ˜ω = L ρ/Gω. Since the external excitation is harmonic, thesteady state response is assumed asθ(x, t) = u(x) sin ˜ωt. (177)This leads tod2udx2sin ˜ωt = −˜ω2u sin ˜ωt (178)ord2dx2+ ˜ω2u = 0. (179)From the boundary conditions given in the previous example and u(0) = 0leads todudx(1) − β˜ω2u(1) =T0LJG. (180)The solution isu(x) =T0L(˜ω cos ˜ω − β˜ω2 sin ˜ω)JGsin ˜ωx. (181)Note that if ˜ω is equal to any of the system’s natural frequencies, thedenominator vanishes. The assumed for the solution must be modified toaccount for this resonance condition. The total solution is the steady-statesolution plus the homogeneous solution, which is a summation over all free-vibration modes. Initial conditions can then be applied to determine theconstants in the linear combination.48
  51. 51. 8 Balancing rotationThe unbalance of rotating machinery is the most common malfunction, evenso that any lateral vibrations are usually wrongly thought to be due to un-balances. In our case the unbalance of the cryostat is obvious as the electricdevices, and pumping systems mounted are not symmetric, and there arerestrictions in placing them. Quite frequently, balancing procedures per-formed on the machine, which another type of malfunction, worsens thesituation. These unbalances have been recognized for over 100 years. Bal-ancing procedures are equally old. However during the last 25 years theyhave experienced substantial improvements due to implementation of vibra-tion measuring electronic instruments and application of computers for dataacquisition and processing. For over a century researchers have publishedhundreds of papers on how to balance machines. For more advanced methodsone should consult [29]. The problem due to unbalance is easiest to identifyand correct. The unbalance causes vibrations and alternating or variablestress in the cryostat it self and the supporting structure elements. Thesevibrations are directly linked to heat leak, and thus again should be minizedas possible. The balancing problem is solved by either, relocating electronicequipment or adding masses. After proper balancing, rotating vibrationsshould be reduced in the entire range of rotational speeds, including the op-timal operating speeds, as well as the resonance speed range. The latter isespecially important when the cryostat is operated in hihg speeds exceed-ing the first, the second, or even higher natural modes of resocance. As theunbalance force is proportional to the rotational frequency squared [30], theunbalance-related grows considerably with increasing rotational speed. Theplane that is rigidly attached on the cryostat carrying the electronics can bethought of as beign a symmetric rotor with the axis of rotation directly inthe middle. The unblance condition changes the rotor mass centerline not tocoincide with the axis of rotation. Unbalance is due to the restricted placingof the electronic devices mounted on the plane. During rotation, the rotorunbalance generates a centrifugal force perpendicular to the axis of rotation.This force excites, rotor lateral vibration e.g. rotor fundamental response.In the following presentation, the modal approach to the rotor system as amechanical structure, has been adopted. At the beginning, the first lateralmode of the rotor is considered only. The lateral mode can either be rotorbending mode or susceptibility mode. Conventionally fundamental the vibra-tion response of the rotor at its lateral mode is due to the inertial centrifugalexciting force, generated by unbalance. In the modal approach, limited tothe first lateral mode, the unbalance-retaled exciting force is discrete, i.e. anaverage integral, lumped effect of the axially distributed unbalance in the49
  52. 52. first mode. The average unbalance angular force location will be referredto as a heavy spot. Rotors are usually similarly constrained in all lateraldirections. Therefore, they exhibit lateral vibrations in space, with two in-separable components of motion at each specific axial section of the rotor.These two components result in a two-dimensional orbiting motion of eachaxial section. Typically, two displacement proximity transducers, mounted inXY orthogonal configuration, will measure the lateral vibrations of the rotorin one axial section plane. The isotropic rotor lateral synchronous motion,as seen by the displacement transducers 90 degrees apart, will differ by 90degrees phase angle. The rotor lateral vibrations can be observed on an oscil-loscope in the time-base mode, and in orbital mode. The latter represents amagnified image of the actual rotor centerline path in this section. Figure 18illustrates the waveforms and an orbit of a slightly anisotropic rotor funda-mental response. The angular position of the force and response vectors arevital parameters for the balancing procedure. In practical applications, theresponse phase is measured by the Keyphasor transducer ( [31], and [32]).Keyphasor is a transducer generating a signal used in rotating for observinga once-per-revolution event. A notch is made on the rotor, which duringrotor rotation causes the Keyphasor displacement transducer to produce anoutput impulse, every time the Keyphasor notch passes under the trans-ducer. The one-per-turn impulse signal is simultaneously received, togetherwith the signals from the rotor lateral displacement-observing transducers.The Keyphasor signal is usually superimposed on the rotor lateral vibrationresponse time-base waveform presentation and on rotor orbits. On the os-cilloscope display, the Keyphasor pulse is connected to the beam intensityinput (the z-axis of the oscilloscope; while the screen displays x and y axis).The Keyphasor pulse causes modulation of the beam intensity, displayinga bright dot, followed by a blank spot on the time-base and/or orbit plots.The sequence bright/blank may vary for different oscilloscopes and for rotornotch/projection routine, but is always consistent and constant for a partic-ular oscilloscope and rotor configuration; this sequence should be checked onrotor waveform time-base responses when the oscilloscope is first used.The unbalance force at a constant rotational speed, Ω as seen in Figure19 can be characterized in the following way. There’s a fixed relation to therotating system. The nature of the rotating period is strictly harmonic time-base, expressed by sin Ωt, cos Ωt or eiΩt, where t is time. When the frequencyis equal to the actual rotational speed the unbalance is rotating at the samerate in sync with the rotor rotation. The force F is proportional to threephysical parameters namely: unbalance average, modal mass m, and squareof the rotational speed.50
  53. 53. Figure 18: Rotor lateral motion measured by two displacement proximity transducers inorthogonal orientation and the Keyphasor phase reference transducer.51
  54. 54. Figure 19: Time-base waveforms of rotor response to unbalance inertia force. Note thatresponse lags the force by the phase difference.52
  55. 55. F = mrΩ2(182)Force phase that is the angular orientation δ mesured in degrees or radiansfrom a reference angle zero marked on the rotor circumference. The unbal-ance force causes rotor response in a form of two-dimensional orbital motion.The harmonic time-base is expressed by a similar harmonic function as theunbalance force and the frequency is equal to the actual rotational speedΩ. Amplitude B is directly proportional to the amplitude of the unbalanceforce, and F is inversely proportional to the rotor synchronous dynamic stiff-ness [33]. Phase lag β represents the angle between the unbalance force vectorand response vector plus the original force phase, δ. The response always lagsthe force, thus the phase moves in the direction opposite to rotation. Bothunbalance force and rotor response are characterized by the single frequencyequal to the frequency of the rotational motion. The vibrational signal readby a pair of XY displacement transducers should, therefore, be filtered tofrequency Ω, or what is the same, to frequency describing synchronous fre-quency of the rotor response as a multiple of one. There may exist otherfrequency components in the rotor response. These possible components ofthe vibrational signals are not directly useful for rotor balancing. A vectorfilter can, for instance, be used for filtering of the measured signal to the firstcomponent only. In the characterization of both the force and response, theamplitude and phase were emphasized as two equally important parameters.Using the complex number formalism, these two parameters can be lumpedinto one the force vector and response vector correspondingly. The amplitudewill represent the length of the vector, the phase its angular orientation inthe polar plot, coordinate format. The unbalance force and rotor responseare therefore, described in a very simple way. Unbalance force is:Fei(Ωt+δ)= mrΩ2ei(Ωt+δ)(183)and rotor fundamental responseRF = Bei(Ωt+β). (184)The corresponding vectors are obtained when the periodic function oftime, eiΩtis eliminated:−→F = Feiδ= mrΩ2ejδ(185)53
  56. 56. Figure 20: Unbalance force vector and rotor fundamental response vector in polarcoordinate format. Note conventional direction of response angle, β, lagging the forcevector in direction opposite to rotation.and (Figure 20)−→RF =−→B = Beiβ. (186)The keyphasor transducer provides a very important measurement of therotor response phase. Since the keyphasor notch is attached to the rotor,the keyphasor signal dot superimposed on the response waveform, representsthe meaningful angular reference system. A useful convention of coordinatesdescribes the angles. When the notch on the rotor is exactly under thekeyphasor transducer, the rotor section under the chosen lateral transducerhas the angle zero. In order to locate the heavy spot, looking from the drivingend of the rotor, rotate the rotor in the direction of the rotation by the angleβ. The heavy spot will then be found under the chosen lateral transducer.This way there is no angular ambiguity, independently of the lateral probepositions.The rotor orbit displayed on the oscilloscope is a magnified picture ofthe rotor centerline motion. The rotor fundamental response orbit, as canbe observed on the oscilloscope screen in orbital mode. Elliptical orbits are54
  57. 57. due to anisotropy of the rotor support system, which is the most commoncase in machinery. One Keyphasor dot on the orbit is at a constant posi-tion, when the rotational speed is constant. It means that during its onerotation cycle the rotor makes exactly one lateral vibration orbiting cycle.Direction of orbiting is the same as direction of rotation called forward or-biting. For a constant rotational speed, the orbit exhibits a stable shapeand the Keyphasor dot appears on the orbit at the same constant angularposition. The phase of the rotor fundamental response is often referred toas the high spot. It corresponds to the location, on the rotor circumference,which experiences the largest deflections and stretching deformations at aspecific rotational speed. Although just One-plane balancing does not havemany practical applications in machinery, it provides a meaningful generalscheme for balancing procedures. Basic equation for one plane balancing ofthe rotor at any rotational speed is represented by the one mode isotropicrotor relationship between input force vector, Feiδ, rotor response vector,Beiβand complex dynamic stiffness [34],−→k (Ω):−→k (Ω)Beiβ= Feiδ(187)where−→k (Ω) = K − MΩ2+ iDSΩ. (188)The complex dynamic Stiffness represents a vector with the direct part,kD = K − MΩ2and quadrature part kQ = DΩ.The rotor in Figure 18 is now defined more precisely it contains the rotorTransfer Function [35], which is an inverse matrix of the Complex DynamicStiffness,−→k . The inverse of the complex dynamic dtiffness is also knownreceptance. The objective of balancing is to introduce to the rotor a correctiveweight of mass, mc, which would create the inertia centrifugal force vectorequal in magnitude and opposite in phase to the initial unbalance force vector.This way, the rotor input theoretically becomes nullified and the vibrationaloutput results also as a zero. In practical balancing procedures, the inputvector force of the initial unbalance has therefore to be identified. Usingagain the block diagram formalism, the one-plane balancing at a constantrotational speed is illustrated in Figure 22.Introduce the vectorial notation:−→F = Feiδand−→B = Beiβ, for theunbalance force vector and response respectively, as well as,−→H = 1/−→k forthe rotor transfer function vector. The original unbalance response at aconstant speed, Ω is:55
  58. 58. Figure 21: Angular positions of unbalance force vector (heavy spot) and response vectorsat two rotational speeds and two directions of rotor rotation (a), (c), and (b), (d) and fortwo locations of lateral transducers (a), (b), and (c), (d). Note that minus signs for theangles are most often omitted and replaced by lag.56
  59. 59. Figure 22: Balancing in one-plane.−→F−→H =−→B . (189)In this relationship, there are two unknown vectors,−→H and−→F . Theresponse vector,−→B is measured thus its amplitude and phase are known. Inorder to identify the initial unbalance force vector, it is sufficient to stop therotor and introduce a calibration weight of a known mass mτ at a knownradial rτ and angular δτ position into the balancing plane. When the rotor isrun again at the same constant speed, Ω, the mass mτ generates an additionalinput force vector,−→F = mτ rτ Ω2eiδτ. This run is called a calibration run.The measured rotor response vector is now−→B 1 = B1eiβ1, which is differentfrom the response vector. For this second run, the following input/outputrelationship holds true:(−→F +−→F τ )−→H =−→B1. (190)In the above the unknown vectors are−→H and−→F , and the others areknown. The last two equations are suffiecient to solve the one-place balancing57
  60. 60. problem and calculate the unknown parameters. The unkonwn vector −−→Fand the corrective mass, mc are calculated, therefore, as follows:−−→F = mcrcΩ2eiδc=−→F τ−→B−→B −−→B1(191)or−FΩ2= mcrceiδc=−→F τ−→B(−→B −−→B 1)Ω2(192)where rc and δc are radial and angular positions of the corrective weightwith mass mc. Note that the corrective weight is supposed to be insertedat the same axial location on the rotor, as the calibration weight. Note alsothat if the radii for the calibration and corrective weights are equal (rτ = rc)and the original and calibration run measurements are taken at the samerotational speed giving:mceiδc= mτ eiδτ−→B−→B −−→B1. (193)Finally, note that the balancing procedure does not require calculation ofthe second unknown parameter, the rotor transfer function vector−→H providesthis vector as well:−→k =1−→H=−→F τ−→B 1 −−→B. (194)This synchronous dynamic stiffness vector, totally overlooked in balancingprocedures, represents a meaningful characteristic of the rotor. It should becalculated, stored and reused, if balancing is required in the future. Duringthe next balancing, the old and new rotor dynamic stiffness vectors should becompared. For a constant speed balancing, the synchronous dynamic stiffnessvector is often used in the form−→k /Ω2and is known as the sensitivity vector.The analytical solution for the corrective mass and its radius can be obtainedby splitting it into real and imaginary parts:mcrceiδc= mτ rτ eiδτBeiβBeiβ − B1eiβ1= mτ rτ eiδτBB − B1ei(β1−β)(195)58
  61. 61. from wheremcrc(cos δc + i sin δc) = mτ rτ (cos δτ + i sin δτ )BB − B1(cos (β1 − β) + i sin (β1 − β))= mτ rτ (cos δτ + iNow the real and imaginary parts are:mcrc cos δc = mτ rτ B(B − B1 cos (β − β1)) cos δτ − B1 sin (β − β1) sin δτB2 + B21 − 2BB1 cos (β − β1)(197)andmcrc sin δc = mτ rτ B(B − B1 cos (β − β1)) sin δτ + B1 sin (β − β1) cos δτB2 + B21 − BB1 cos (β − β1).(198)From these we get:mcrc = mτ rτBB2 + B21 − 2BB1 cos (β − β1)(199)andδc = δτ + arctanB1 sin (β − β1)B − B1 cos(β − β1). (200)These represent the analytic result for the one plane balancing. This canbe done first choosing a rotational speed Ω for balancing. Next is to run therotor and measure its original synchronous response vector,−→B = Beiβ, at therotational speed Ω. After this stop the rotor and choose a radial and angularscale for plotting vectors. Draw the vector−→B = Beiβin the polar plot.Now introduce a known calibration weight into the rotor at a convenient,known axial, radial and angular position. Convenience consists in installingthe calibration weight in the rotor in the opposite half-plane to the originalunbalance. On the polar plot draw the corresponding calibration force vector−→F τ = Fτ eiβτ= mτ rτ Ω2eiδτ. Run the rotor at the same speed Ω. Measurethe new rotor synchronous response vector−→B1 = B1eiβ1and draw it in thepolar plot using the same scale, and then stop the rotor. Subtract vectorially−→B1 from−→B in the plot; draw a vector−→B −−→B1. Find the corrective weightangular position as δc = δτ + θ. The angle θ is between the vectors−→B and59
  62. 62. Figure 23: One plane balancing of a rotor using polar plot.−→B −−→B1. Since the response is proportional to the input force, the triangles−→B ,−→B1 ,−→B1 −−→B , and−→F ,−→F +−→F τ ,−→F τ are similar; they havethe same angles. Now measure on the plot the length of the vector−→B −−→B1using the assigned scale. Next calculate the corrective mass, mc, applyingthe formula:mc = mτrτ Brc−→B −−→B 1. (201)Finally introduce the correct weight with mass mc at the angle δc, andradius rc to the same plane rotor as the calibration weight. This procedurecan be iterated when new equipment is added to the plane carrying theelectronics.9 Studies of the Noise of Old Rota I CryostatAs Rota I Cryostat was to be modified, and also partly rebuilt, rotationalcharacteristics were measured focusing on noise. The measurements were60
  63. 63. done with all the cryostat’s electronics in place, and other set of measure-ments were done when all the electronics were removed. The measurementswere done in the same way as previous vibrational noise measurements [36].The measurements were done in the normal, and tangential directions fromthe axis of rotation. The rotation speeds were 250, 1000, 2000, 3000 mrad/s.The accelerometer used is a Bruel and Kjaer Accelerometer Type 4370 [37],and Bruel and Kjaer preamplifier was used before collecting the data withAgilent 54641 Oscilloscope [38]. The data aquired was done using TCL, andtransferred by GPIB.The results were similar to previously done measurements [36]. The nois-iest direction by far was the tangent of the rotation vector, this means thatthe noise was rather torsional rather than back and forth swaying of thecryostat. The 250 mrad/s speed is a problematic speed as it showed somerisen noise levels. This was propably due that some part of the cryostatexperiences its first harmonic.In Figure 24. at 250 mrad/s some peaks in the fourier transform seemedto be quite visible and large compared to the 1000 mrad/s case. Some ofthe peaks have lowered and also disappeared at 1000 mrad/s, but the overallnoise seems to have increased at all frequencies. This is taken with all theelectronics on, and in the tangential direction. In the normal direction thelargest peaks are half to that of tangential and the overall noise is somewhatlower at all frequencies.As the speed was increased the noise-levels rose, but not all frequenciesseemed to be affected by this. In Figure 25. the noise-levels are additivelyintegrated over region up to 55 Hz at different speeds. In this tangentialnoise-plot one is able to see the unbalanced increases at different frequencies.The removal of electronics had an unexpected effect on the noiselevels.Vibrational noise fell sharply in the measurements after the removal of theelectronics. This was probably due to the fact that load from the air bearingswas lifted, and thus removing some friction. This means that the air bearingsare experiencing their maximum load, and when the electonics are mountedon the cryostat the friction rises.The integration without the electronics reveals that the noise at differentspeeds in the tangential directions does not vary as expected. Rather the250 mrad/s case seems to dominate in this situation as can be seen in Figure28. The moment of inertia has been changed, so the high noise-level at 250mrad/s is not necessarily a first harmonic that the whole cryostat or the beltof the rotating motor is experiencing. It could also be that motor is notfunctioning perfectly at this speed.61
  64. 64. 0 5 10 15 20 25 30 35 40 45 50 5500.511.522.533.5x 104V(10−3)Tangential Noise with Electronics at 250 mrad/s0 5 10 15 20 25 30 35 40 45 50 5501234x 104HzV(10−3)Tangential Noise with Electronics at 1000 mrad/sFigure 24: Two fourier spectra show measured tangential noise at rotating speeds 250and 1000 mrad/s.62
  65. 65. 0 5 10 15 20 25 30 35 40 45 50012345678x 104HzVHz(10−3)Tangential Additive Noise Integral with Electronics250100020003000Figure 25: The integrated additive plot shows the complexity of rising noise level atdifferent speeds in tangential direction.63
  66. 66. 0 5 10 15 20 25 30 35 40 45 5000.511.522.5x 104HzVHz(10−3)Normal Noise Additive Integral Plot with Electronics250100020003000Figure 26: The integrated additive plot shows the complexity of rising noiselevel atdifferent speeds in normal direction.64
  67. 67. 0 5 10 15 20 25 30 35 40 45 50 5502000400060008000V(10−3)Tangential Noise without Electronics at 250 mrad/s0 5 10 15 20 25 30 35 40 45 50 5502000400060008000HzV(10−3)Normal Noise without Electronics at 250 mrad/sFigure 27: This plot shows how the normal and tangential directions differ taken at 250mrad/s without electronics.65
  68. 68. 0 5 10 15 20 25 30 35 40 45 50 5501234x 104V(10−3)Tangential Noise With Electronics at 3000 mrad/s0 5 10 15 20 25 30 35 40 45 50 5501234x 104HzV(10−3)Tangential Noise Without Electronics at 3000 mrad/sFigure 28: This plot shows how the tangential noise varies with and without theelectronics taken at 3000 mrad/s.66
  69. 69. 0 5 10 15 20 25 30 35 40 45 5002000400060008000100001200014000HzVHz(10−3)Tangential Additive Noise Integration without Electronics250100020003000Figure 28: Additive integration is taken without electronics in the tangential direction.10 Superconducting high-homogeneity mag-net for NMR measurementsChemical analysis, and imaging of biological samples are commonly probedusing nuclear magnetic resenance methods. The samples in a magnetic fieldare studied using continuous wave NMR causing the samples’s magnetic mo-ments of the atomic nuclei to arrange in a fashion minimizing the magneticpotential energy. The NMR frequency of the sample is changed according tothe dependent magnitude, thus the polarization field strenght is sweeped [39]and the resulting spectrum measured gives information about the structureGetting a good NMR signal requires the magnetic moments to be as uniformas possible. A very homogenous field is thus a must, which requires thatthe polarizing magnet is constructed with great care. The homogeneity isdefined as the relative deviation from the center point value B0:|∆BB0| = |B − B0B0|. (202)One must also take into account the space where the measurements areto be made. They are in volume very constricted, and the sample as in our67
  70. 70. case is helium is to be measured in a cylindrical cell. Solenoid magnet isthus a very good choice for this experiment, and the field can be representedin closed form ( [40], [41]). Increasing the coil length and optimizing thediameter of the solenoid improves the homogeneity, which in general meansdecreasing the diameter. The homogeneity is reduces when one moves axiallyaway from the center, and trying to reduce this short compensation coils atboth ends are placed. The most common of the compensation methods isthe sixth order end compensation [42], in which the correct choice of magnetdimensions cancels the first five derivative terms in the expansion series ofthe field. The homogeneity can also be improved using the Meissner effect,which occurs as superconducting material is placed in a magnetic field. Asin the case of perfect diamgnet the superconductor sets up surface currentscancelling the field within the material by opposing magnetization. Thesolenoid is surrounded by superconducting material in order to force thefield lines to concentrate in center of the solenoid center [43] and also acts asa shield for external unwanted fields. The magnet used for the measurementswas already provided [44]. The wire consist of quite fragile Super Con Inc. 18filament NbTi inside CuNi alloy matrix [45]. As the measurements are donenotably under the critical 4.2 K temperature, there should be no problem ofproper thermalization. The dimensions of the magnet can be seen in Figure29. The magnet was previously tested merged in liquid helium with maximuncurrent load of 12.5 A without quenching. The absolute value of the magneticfield strength for the magnet was not accurately measured as the Hall probeused by Vesa Lammela was designated to work at temperature range of −10to 125oC. The homogeneity of the magnet was therefore to be tested withactual NMR-measurements.11 SuperconductivityThe phenomenon of superconductivity occurs in specific materials at ex-tremely low temperatures. The characterization includes exactly zero elec-trical resistance, and the Meissner effect [46]. The resistivity of the metallicconductor gradually decreases as temperature is lowered, and drops abruptlyto zero when the material is cooled below the critical temperature. It is aquantum mechanical phenomenon. The physical properties vary from mate-rial to material, especially the heat capacity and the critical temperatures.Regular conductors have a fluid of electrons moving across a heavy ioniclattice, where the electrons constantly collide with the ions, thus dissipat-ing phonon energy to the lattice converting into heat. The superconductorsituation differs in that the electronic fluid cannot be distinguished into in-68
  71. 71. Figure 29: Number of turns on each layer, and the dimensions of the magnet.dividual electrons, but it consists of Cooper pairs [47] caused by electronsexchanging phonons. The Cooper pair fluid energy spectrum possesses anenergy gap amounting to a minimum energy ∆E that is needed to excite thefluid. When ∆E is larger than the lattice thermal energy kT, there will beno scattering by the lattice as the Cooper pari fluid is a superfluid withoutphonon dissipation. The critical temperature Tc varies with the material,and convetionally are in the range of 20K to less than 1K. The transition tosuperconductivity is accompanied by changes in various physical properties.In the normal regime the heat capacity is proportional to the temperature,but at the transition it suffers a discontinous jump, and linearity is impaired.At the low temperature range it varies as e−α/Twhere α is some constantvarying with the material. The transition as indicated by experimental datais of second-order, but the recent theretical improvements (which are ongo-ing) show that within the type II regime transition is of second order andwithin the type I regime first order, and the two regions are separated by atricritical point [48]. Considering the Gibbs free energy per unit volume g,which is related to internal energy per unit volume u and entropy s:g = u − Ts (203)69
  72. 72. where the volume term has been neglected. The associated magneticenergy in the presence of an applied magnetic field ¯Ba is − ¯M ·d ¯Ba, where ¯Mis the magnetic dipole moment per unit volume. A change in the free energyis given bydg = − ¯M · d ¯Ba − sdT. (204)Integrating yields:g(Ba, T) = g(0, T) −Ba0¯M · d ¯Ba. (205)For type-I-superconductor ¯M = − ¯H [50] due to the Meisness effect, andthus the previous can be written as:gS(Ba, T) = gS(0, T) +Ba0BadBaµ0= gS(0, T) +B2a2µ0. (206)We see B2a/2µ0 as the extra magnetic energy stored in the field as resultingfrom the exclusion from the superconductor. At the transition between thesuperconductivity and normal state we have gS(Bc, T) = gN (0, T), whereBc is the critical magnetic field and the magnetization of the normal statehas been neglected. The entropy difference ∆s < 0 between normal andsuperconducting states can be obtained from this using s = −(δg)/δT. Nowthe heat capacity per unit volume c = Tδs/δT is:∆c =Tµ0Bcd2BcdT2+dBcdT2, (207)which shows that there is a discontinuous jump in ∆c even for Bc = 0.As our superconducting filament behaves as convetional superconductiv-ity the pairing can be explained by the microscopic BCS theory. The assump-tion of the BSC theory [49] is from the assupmtion that electrons have someattration between them overcoming the Coulomb repulsion. This attractionin most materials is brought indirectly by the interaction between the elec-trons and the vibrating crystal lattice as the opposite spins becoming paired.As electron moves through a conductor nearby a positive lattice point causesanother electron with opposite spin to move into the region of higher positivecharge density held together by binding energy Eb. When Eb is higher than70
  73. 73. phonon energy from oscillating atoms in the lattice, then the electron pairwill stick together, thus not experiencing resistance and describing an s-wavesuperconducting state. Let us look more closely to the electron-phonon in-teraction. Electron in a crystal with wavevector ¯k1 scatters to ¯k′1 emitting aphonon ¯q. Then this phonon is absorbed by a second electron ¯k2 to ¯k′2 hencethe conservation of crystal momentum:k1 + k2 = k′1 + k′2 = k0. (208)States in the ¯k-space can interact, but are restricted by the Pauli exclusionprinciple corresponding to electron energies between EF and EF + ωD, whereEF is the Fermi energy and ωD is the Debye frequency. The number of allowedstates occur when ¯k1 = −¯k2, and they are called the Cooper pairs [51]. Thetransition temperature Tc is given by BSC model [52]kBTc = 1.14 ωDe−1/V0g(EF ), (209)where V0 is the phonon-electron interaction strenght. In the BCS groundstate (T = 0), there is a binding energy 2∆ to the first allowed one-electronstate, in which the Cooper pairing is broken. The gap energy in conventionalsuperconductors are in the range of 0.2-3 meV, thus very much smaller thanEF . Next we will consider quenching mechanisms for the breaking up of thesuperconducting state to normal state. It is exactly this that the Cooperpairing is broken as some disturbance overcomes this binding energy.12 Superconductor quenchingWe will deal with the matter of quenching here, as it will be quite a impor-tant factor affecting the work. Quenching occurs when a superconductivefilament goes to normal resistive state. There are three critical parametersnamely temperature, current density, and magnetic field affecting this behav-ior. When one of these parameters’ critical value 1is exceeded by some phys-ical process, superconductor becomes normal-conducting. When the coolingpower for the filament is not sufficient the zone of the normal conductor ex-pands. Usually in a quench case the entire energy stored in the magnet isdissipated as heat, even burning the filament. There are two main distur-bances namaley transient and continuous, which can be again divided intotwo more causes point and distributed. In the case of continuous disturbancethe steady power becomes a problem due to e.g. bad joint, or soldering. The71
  74. 74. distributed disturbances are usually caused by heat leak to the cryogenicenvironment. Now transient disturbances are sudden, and can be for exam-ple caused by a breaking of a turn in a magnet due to excess Lorenz forcemoving the turn by δ. In this case the work done by the magnetic field isBJδ, where J is the current density, and this energy will heat the magnet tonormal state. Our filament is embedded in a copper matrix, which is a goodabsorber and distributor of heat, thus if the copper gets good enough contactwhere to dissipate the energy, the magnet can stay in superconducting stateeven thought some problems persist. For example cooling of the filament byhelium is given by the adiabatic heat balance equation [53]:ρcu(T(x, t))I(t)2Acu(x)= c(T(x, t))A(x)dT(x, t)dt, (210)where ρcu is the resistivity of copper, Acu is the copper cross-section ofthe composite, I is the current in the filament, T is the temperature of thefilament, and c is the heat capacity. Voltage V (t) is a function of time in themagnet coilV (t) = I(t)R(t) + L(I)dI(t)dt−iMidIdt− UP C, (211)where I(t) is the current, R(t) is the resistance, L(I) is the self inductance,Mi is the mutual induction of a neighboring turn or coil, and Upc is the voltageof the power converter. Now with a good power converter, the quenchingvoltage can be given:VQt) = I(t)R(t) + LQ(t)dIdt, (212)where LQ is the partial inductance and R(t) is the resistance of thequenching zone. Taking mutual inductance zero, we get [54]:VQ(t) = I(t)R(t)(1 − LQ(t)/L). (213)The quench zone thus expands as resistance and partial inductance grow.A good power supply can detect this, and switches itself off.72
  75. 75. 13 Nuclear Magnetic Resonance, and Imag-ingQuantum mechanical magnetic proterties of atom’s nucleus can be studiedwith the nuclear magnetic resonance (NMR). The pehomenon was first dis-covered by Isidor Rabi in 1938 [?]. Neutrons and protons have spin, and theoverall spin is determined by the spin quantum number I, and non-zero spinassociates with a non-zero magnetic moment µ byµ = γI, (214)where, γ, is the gyromagnetic ratio. Angular momentum quantization,and orientation is also quantized. The associated quantum number is knownas the magnetic quantum number m, and can only have values from integralsteps of I to −I thus there are 2I + 1 angular momentum states. Taking thez component Iz:Iz = m . (215)The z-component of the magnetic moment isµz = γIz = mγ . (216)¯I2has eigenvalues I that are either integer or half integer. This can bemeaning(Im µx′ Im′) = γ (Im Ix′ Im′), (217)where µx′ and Ix′ are components of the operators ¯µ and ¯I along thearbitrary x′-direction. This is based on the Wigner-Eckart equation [56]. Forsimplicity we consider system of two m states +1/2 and −1/2 by numbers N+and N−, where the total number N of spins is constant. With the propabilityfor transition W the absorption of energy is given:dEdt= N+W ω − NW ω = ωWn, (218)where ω is the angular frequency of the time dependent interaction (orthe frequency of an alternating field driving the transitions), and n has to be73
  76. 76. zero for a net absorption. Now in a similar fashion using a alternating fieldwe get for the absorption energy [57]dEdt= n ωW = n0 ωW1 + 2WT1(219)where T1 is the spin-lattice relaxation time, and as long as 2WT1 ≪1 we can increase the power absorbed by increasing the amplitude of thealternating field. Taking the alternating magnetic field as Hx(t) = Hx0 cos ωt,and it can be broken into two rotating components with amplitude H1 inopposite directions. They are denoted by:¯Hr = H1(¯i cos ωt + ¯j sin tωt) (220)and¯HL = H1(¯i cos ωt − ¯j sin ωt). (221)We consider only ¯HR as ¯HL is just same with negative ω. Taking ωz ascomponent of ω along z-axis:¯H1 = H1(¯i cos ωzt + ¯j sin ωzt). (222)Now the equation of motion with the static field ¯H0 = ¯kH0 isd¯µdt= ¯µ × γ[ ¯H0 + ¯H1(t)]. (223)Now moving to a coordinate system such that the system rotates aboutthe z-direction at frequency ωz then:δ¯µδt= ¯µ × [¯k(ωz + γH0) +¯iγH1]. (224)Now near resonance ωz + γH0 ≃ and by setting ωz = −ω states that inthe rotating frame moment acts as though it experiences a static magneticfield:¯Heff = ¯k(H0 −ωγ) + H1¯i. (225)74

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