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: 60

Department: Department of Engineering Physics and Mathematics Pro-fessorship: Tfy-3. Material Physics

Supervisor & Instructor: Matti KrusiusIn the thesis, rotational aspects and vibrational analysis of Cryo I Helsinki

cryostat 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

Master’s Thesis

Kaarle Aleksanteri Kulvik

52664T

24th April 2008

Vibrational Study of Cryo I Helsinki, and Testing of Homogeneity of aSuperconducting Magnet

Supervisor M. Krusius

2

Contents

1 Preface 2

2 Introduction 3

3 Fourier Methods 3

4 Piezoelectric Accelerator 12

5 Low-pass filter and measurement setup 23

6 Vibrations 23

7 Torsional vibrations 40

8 Balancing rotation 49

9 Studies of the Noise of Old Rota I Cryostat 60

10 Superconducting high-homogeneity magnet for NMR mea-

surements 67

11 Superconductivity 68

12 Superconductor quenching 71

13 Nuclear Magnetic Resonance, and Imaging 73

14 Cooling of the Cryostat 75

15 Preparing the NMR measurements 79

16 Conclusions 80

1

1 Preface

Dedicated 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 Churchill

2

2 Introduction

This 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 Methods

Fourier 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 formula

f(t) = a0 +∞

∑

n=1

an cos(nωt) +∞

∑

n=1

bn sin(nωt), (1)

where t is an independent variable which in our case represents time, and Tp

is 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 formula

a0 =1

Tp

∫

Tp2

−Tp2

f(t)dt, (2)

an by

an =2

Tp

∫

Tp2

−Tp2

f(t) cos(nωt)dt, (3)

and bn by

3

bn =2

Tp

∫

Tp2

−Tp2

f(t) sin(nωt)dt. (4)

The frequencies nω are known as the nth harmonics of ω. The series may bewritten in exponential form

f(t) =∞

∑

n=−∞

dneinωt (5)

in which

dn =1

Tp

∫

Tp2

−Tp2

f(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 in

f(t) are finite.The squared modulus of a transform is referred as the energy spectrum.

|F (ω)|2 is 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 following

dn = |dn|eiφn , (7)

where

|dn| = (a2n + b2

n)1/2

(8)

and

φn = − tan(bn/an), (9)

where φn is the phase angle of the nth harmonic component.

4

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) =1

2π

∫

−∞

∞

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 the

Fourier 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 1

2F (s −

ω/2π) + 12F (s + ω/2π). If f(x) and g(x) have FTs F (s) and G(s), then

convolution 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

∫

∞

−∞

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 + 2α

∑

k=1

(

ak cos(2πkt

T) + bk sin(

2πkt

T)

)

(16)

where k ≥ 0 is an integer, and

6

Figure 1: Sampling a continuous function at regular intervals.

ak =1

T

∫ T

0

x(t) cos(2πkt

Tdtbk =

1

T

∫ T

0

x(t) sin(2πkt

Tdt. (17)

The previous can be combined into a single equation:

Xk = ak − ibk (18)

and putting

e−i2πkt/T = cos2πkt

T− i sin

2πkt

T(19)

from we get

Xk =1

T

∫ T

0

x(t)e−i2πkt/T dt. (20)

Knowing only the equally spaced samples of the continuous time seriesx(t) represented by the discrete series xr, r = 0, 1, . . . , (N − 1), where

7

Figure 2: Calculating Fourier coefficients from a discrete series using approximation.

t = r∆, and ∆ = T/N , then the integral may be replaced approzimately bythe summation

Xk =1

T

N−1∑

r=0

xre(−i2πk/T )(r∆)∆. (21)

This is just assuming the total area under the curve in Figure 2. PuttingT = N∆ gives

Xk =1

N

∑

r=0

N − 1xre−i2πkr/N . (22)

This may be regarded as approximation for calculating the Fourier series.The inverse formula for the series xr is

xr =N−1∑

k=0

Xke2iπkr/N . (23)

8

This can be seen

N−1∑

k=0

Xke2iπkr/N =

N−1∑

k=0

(

1

N

N−1∑

s=0

xse−2iπks/N

)

e2iπkr/N =N−1∑

k=0

N−1∑

s=0

1

Nxse

(−2iπk/N)(s−r)(24)

and by interchanging the summation

=

N−1∑

s=0

( N−1∑

k=0

e−2i(πk/N)(s−r)

)

1

Nxs (25)

and the exponentials all sum to zero unless s = r when the summationequals N and hence

N−1∑

s=0

( N−1∑

k=0

e−2i(πk/N)(s−r)

)

1

Nxs = 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 as

Xk =1

N

N−1∑

r=0

xre−2iπkr/N (27)

for k = 0, 1, . . . , N − 1. Calculating values of Xk for k is greater thanN − 1. Letting k = N + l then

XN+l =1

N

N−1∑

r=0

xre−(2iπr/N)(N+l) =

1

N

N−1∑

r=0

xre−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 = X l (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 to

9

ωk =2πk

N∆=

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 so

π

∆> ω0 (30)

or f0 = ω0/2π giving

1

2∆> 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 N2 multiplications. 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 sequencesyr and zr where yr = x2r and zr = x2r+1, r = 0, 1, . . . , (N/2 − 1). TheDFT’s of these are

Yk =1

N/2

N/2−1∑

r=0

yre−i 2πkr

N/2 Zk =1

N/2

N/2−1∑

r=0

zre−i 2πkr

N/2 , k = 0, 1, . . . , N/2 − 1.(32)

10

Firstly we separate the odd and the even terms in xr getting

Xk =1

N

N−1∑

r=0

xre−2iπkr/N =

1

N

( N/2−1∑

r=0

x2re−i 2π(2r)k

N +

N/2−1∑

r=0

x2r+1e−i 2π(2r+1)k

N

)

=1

N

( N/2−1∑

r=0

yre−i 2πrk

N/2 + e

from which

Xk =1

2

(

Yk + 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.

Calculating

Xk =1

2

(

Yk + e−2iπk/NZk

)

, k = 0, 1, . . . , N/2 − 1Xk =1

2

(

Yk−N/2 + e−2iπk/NZk−N/2

)

, k = N/2, N/2 + 1

or

Xk =1

2

(

Yk + e−2iπk/NZk

)

Xk+N/2 =1

2

(

Yk + e−2iπ(k+N/2)/NZk

)

=1

2

(

Yk − e−2iπk/NZk

)

, k = 0, 1, . . . , N/

These are the formulas occuring in most FFT programs, and definingW = e−2iπ/N we 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 =220 samples calculated directly with DFT compared to FFT has the ratio

11

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 Accelerator

Piezoelectric 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 compression

12

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 = SEijσj + dmiEm (37)

13

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

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 = §Dijσj + 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

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 eσ

11 = eσ22, and eσ

33. One can write thesein matrix form to give:

ε1

ε2

ε3

ε4

ε5

ε6

=

S11 S12 S13 0 0 0S12 S11 S13 0 0 0S13 S13 S33 0 0 00 0 0 S44

0 0 0 0 S44 00 0 0 0 0 2(S11 − S12)

σ1

σ2

σ3

σ4

σ5

σ6

+

0 0 d31

0 0 d31

0 0 d33

0 d15 0d15 0 00 0 0

E1

E2

E3

(41)

and

D1

D2

D3

=

0 0 0 0 d15 00 0 0 d15 0 0

d31 d31 d33 0 0 0

σ1

σ2

σ3

σ4

σ5

σ6

. (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 =V

t. (43)

This strains the transducer, e.g.

ε1 =∆l

l(44)

and

∆l =d31V l

t. (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 short

16

circuit charge per unit area flowing between connected electrodes perpendic-ular ~j to the stress applied along ~i. A force F~k is applied to the transducergenerates the stress

σ3 =F

lw(46)

resulting in the elecric charge

q = d33F (47)

flowing through the short circuit. The constant gij denotes the electric

field along ~i when the material is stressed along ~j. Force F applied in thepositive ~i, resulting in the voltage:

V =g31F

w. (48)

The other way to intepret gij is to take the ratio of strain along ~j to the

charge per unit area deposited on electrodes perpendicular to ~i. Placing anelectric charge Q on the surface electrodes (plates are on top and bottom

perpendicular to ~k) changes the thickness by:

∆l =g31Q

w. (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 SE

ij measured with theelectrodes short-circuited, and similarly D denotes that the measurementswere done with electrods left open-circuited. SE

ij 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 eσ

11 is the permittivity for a field applied along x-axis, whenthe material is not restrained.

17

Ability of a piezoceramic to transform electrical energy to mechanicalenergy and vice verca is denoted by kij . The stress or strain is ~j-oriented

and 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∆z

2. (50)

Electric charges accumulate on the electrodes due to piezoelectric effectamounting to the elecrical energy in the piezoelectric capacitor:

WE =Q2

2CP

. (51)

From this we get

k33 =

√

WE

WM=

Q√

F∆zCp

. (52)

The coupling can be written otherwise as

k2ij =

d2ij

Sij

E

eσij = 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:

D1

D2

D3

=

0 0 0 0 d1500 0 0 d15 0 0

d31 d31 d33 0 0 0

σ1

σ2

σ3

σ4

σ5

σ6

(54)

18

The charge generated can be determined

q =

∫ ∫

[

D1 D2 D3

]

dA1

dA2

dA3

, (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 charge

VP =q

CP

, (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 x1

and 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 m1

and m2 under the dynamic forces are:

m1x1 + ka(x1 − x2) = 0 (57)

m2x2 + ka(x2 − x1) + ks(x2 − x3) = 0. (58)

19

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 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 = −A1ω2 cos 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 − m2ω

2) − k2a = 0. (64)

This can be written as quadratic in ω2,

ω4 − [ka(1

m1+

1

m2) + ks

1

m2]ω2 +

kaks

m1m2= 0 (65)

Now we simplify the calculations by taking new constants a = m2/m1,r = ks/ka, and ω2

0 = ka/m1. Substituting these into the above equation andcalculating the frequency:

ω = ω20

√

√

√

√

[1 + 1a(1 + r)] −

√

[1 + 1a(1 + r)]2 − 4r

a

2. (66)

21

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 get

ω = 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

5 Low-pass filter and measurement setup

Taking 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 definedas

fc =1

2πR2C(68)

or equivalently in radians per second

ωc =1

R2C. (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 Vibrations

The types of vibrations in our case can be divided into two main categories:the unbalanced rotation of the cryostat creates harmonic oscillations and

23

Figure 8: An active low-pass filter.

Figure 9: Butterworth low-pass filter in a circuit used to obtain vibration spectra [19].

24

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 +k

mx = 0 (70)

giving

x = 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 gives

x = x0 cos

√

k

mt. (72)

25

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 =∑

i

F/ki (73)

so that

1/ke =∑

i

1/ki. (74)

Similarly parallel springs give

ke =∑

i

ki. (75)

Let us consider a beam with m as the mass unit length, and y is theamplitude of the deflection curve (see Figure 10) then

Tmax =1

2

∫

y2maxdm =

1

2ω2

∫

y2dm, (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 is

V =1

2

∫

Mdθ. (77)

Assuming the deflection of beams small

Rdθ = dx, (78)

thus

1

R=

dθ

dx=

d2y

dx2. (79)

26

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 =1

2

∫

M

Rdx =

1

2

∫

EI(d2y

dx2

)2

dx. (80)

Now Tmax = Vmax;

ω2 =

∫

EI(

d2ydx2

)2

dx∫

y2dm. (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 of

27

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:

mx + cx + kx = 0. (82)

Assuming solution of the form x = Xest 6= 0, and substituting for roots:

s1,2 = − c

2m±

√

(c2 − 4mk)

2m, (83)

hence

x = X1es1t + X2e

s2t, (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 by

ζ = c/cc, (85)

and

s1,2 = (−ζ ±√

(ζ2 − 1))ω. (86)

When damping is less critical ζ < 1

s1,2 = −ζω ± i√

(1 − ζ2)ω (87)

so

x = Xe−ζωt sin (√

(1 − ζ2)ωt + φ). (88)

28

When ζ = 1 the damping is critical and equation x = (A + Bt)e−ωt isvalid. Finally damping greater than critical ζ > 1 gives two negative realvalues of s so that x = X1e

s1t + X2es2t. Substituting for damping constant

a constant friction force Fd that represents dry friction (Coulomb damping)applicaple in many mechanisms:

mx + kx = Fd. (89)

Getting a solution

x = A sin ωt + B cos ωt +Fd

k. (90)

Hence

x = (x0 +Fd

kcos ωt +

Fd

k, (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 :

mx + cx + 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 get

mXν2 sin(νt − φ + π) + cXν sin(νt − φ + π/2) + kX sin(νt − φ) = F sin(νt).(93)

29

From this

F 2 = (kX − mXν2)2 + (cXν)2, (94)

or

X = F/√

((k − mν2)2 + (cν)2), (95)

and

tan(φ) = cXν/(kX − mXν2). (96)

The steady state solution

x =F

√

((k − mν2)2 + (cν)2)sin(νt − φ), (97)

where

φ = tan−1( cν

k − mν2

)

. (98)

The complete solution includes the transient motion given by the com-plementary function:

x = Ae−ζωt sin(ω√

(1 − ζ2)t + α), (99)

where ω =√

k/m and Xs = F/k so that

X

Xs

=1

√

(1 − (ν/ω)2)2 + (2ζν/ω)2, (100)

and

φ = tan−1( 2ζ(ν/ω)

1 − (ν/ω)2

)

. (101)

30

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

The force transmitted to the foundation is the sum of the spring forceand the damper force. Thus the transmitted force is given by

FT =√

(kX)2 + (cνX)2. (104)

The transmissibility is given by

TR =FT

F=

X√

k2 + (cν)2

F(105)

since

X =F/k

√

(1 −(

νω

)2)2 +

(

2ζ νω)2

, (106)

TR =

√

(

1 + (2ζ νω)2

)

√

(

1 − ( νω)2

)

+(

2ζ νω

)2. (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 ω

is 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 frequencies

32

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√2πσ

e−12

(

x−xσ

)2

, (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 probability

Prob−λσ ≤ x(t) ≤ λσ =

∫ λσ

−λσ

1√2πσ

e(− 12

x2

σ2 )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 change

33

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:

MX = −KX − k(X − x) + F sin νt (110)

and for the vibration absorber

mx = k(X − x), (111)

where X = X0 sin νt and x = x0 sin νt.

34

Figure 12: System with undamped vibration absorber.

It can be easily seen that

X0 =F (k − mν2)

∆, (112)

and

x0 =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 vibration

35

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

ω

)4

− (2 + µ)

(

ν

ω

)2

+ 1 = 0, (114)

hence

Ω1,2

ω=

[(

1 +µ

2

)

±√

(

µ +µ2

4

)]1/2

. (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:

(

Ω1

ω

)2

=

(

1 +µ

2

)

−√

µ +µ2

4(116)

and

(

Ω2

ω

)2

=

(

1 +µ

2

)

+

√

µ +µ2

4(117)

then multiplying gives

Ω1Ω2 = ω2 (118)

and

(

Ω1

ω

)2

+

(

Ω2

ω

)2

= 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 ωp

rev/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

Figure 13: Model of a pump.

mpx1 + kpx1 = F sin νt, (120)

gettin

x1 = X1F

k1 − 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:

m2x2 = −k2(x2 − x1) (122)

and

mpx1 = k2(x2 − x1) − k1x1 + F sin νt. (123)

37

Figure 14: Adding the absorber on the pump.

Taking x1,2 = X1,2 sin νt giving

X1(k1 + k2 − m1ν2) − X2k2 = F (124)

and

−X1k2 + X2(k2 − m2ν2) = 0. (125)

From this

X1 =F (k2 − m2ν

2

(k2 + k1 − mpν2)(k2 − mmν2) − k22

. (126)

If ν2 = k2/m2, X1 = 0 then the frequency equation is (k1+k2−mpν2)(k2−

m2ν2) − k2

2 = 0. Putting µ = m2/mp = k2/k1 and Ω =√

k2/m2 =√

k1/mp,giving

(

ν

Ω

)2

=2 + µ

2±

√

µ2 + 4µ

4. (127)

38

From this and smallest absorber mass ν1/Ω = 0.8 as then ν2/Ω = 1.25,which is acceptable. Thus µ = 0.2 and hence

m2 = µmp = 0.2mp, (128)

and

k2 = 2πωp. (129)

One good system is a ciscous damped absorber such as in Figure 15.Equations of motion are:

MX = F sin νt − KX − k(X − x) − c(X − x) (130)

and

mx = 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 then

X0 =F

K − 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

Figure 15: System with damped vibration absorber.

X0 =F

K − Mν2, (135)

and when c is very large,

X0 =F

K − (M + m)ν2. (136)

7 Torsional vibrations

Historically 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 determine

40

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 is

J =πd4lγ

32g, (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 disk

41

Figure 16: Circular shaft subject to torsional loading.

Figure 17: Free-body diagram of the differential element of shaft at arbitrary instant.

42

(

∑

M

)

ext

=

(

∑

M

)

eff

(139)

gives

T (x, t)dx − Tr(x, t) + Tr(x, t) +δTr(x, t)

δxdx = ρJdx

δ2θ(x, t)

δt2(140)

or

T (x, t) +δTr(x, t)

δx= ρJ

δ2θ

δt2. (141)

From mechanics of materials,

Tr(x, t) = JGδθ(x, t)

δx(142)

which leads to

T (x, t) + JGδ2θ

δx2= ρJ

δ2δ

δt2. (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 these

(

L2Tm

JG

)

T (x, t) +δ2θ

δx2=

δ2θ

δt2, (144)

where the ∗ has been dropped from nondimensional variables. The prob-lem formulation is completed by specifying appropriate initial conditions ofthe form

θ(x, 0) = g1(x) (145)

and

δθ(x, 0)

δt= g2(x). (146)

43

Consider

δ2θ

δx2=

δ2θ

δt2. (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 gives

1

X(x)

d2X

dx2=

1

T (t)

d2T

dt2. (148)

leading to

d2T

dt2+ λT = 0, (149)

and

d2X

dx2+ λX = 0, (150)

where λ is the separation constant. The solution is

T (t) = A cos√

λt + B sin√

λt, (151)

where A and B are arbitrary constants of integration. Similarly

T (t) = C cos√

λx + D sin√

λx (152)

The initial conditions give C = 0, B = 0 and the only reasonable solution

λk = [(2k − 1)π

2]2k = 1, 2, . . . (153)

44

Now infinity of solutions arise corresponding to

Xk(x) = Dk sin (2k − 1)π

2x (154)

for any Dk. The modes are orthogonal giving

(Xk(x), Xj(x)) =

∫ 1

0

DjDk sin (2k − 1)π

2x sin (2j − 1)

π

2xdx = 0 (155)

for j 6= k, but when k = j we get

1 = (Xk, Xk) =D2

k

2(156)

leading to

θ(x, t) =

∞∑

k=1

√2 sin (2k − 1)

π

2x[Ak cos (2k − 1)

π

2t]. (157)

From the initial conditions we get the last

Ak =4γ

√2(−1)k+1

π2(2k − 1)2(158)

yielding in total

θ(x, t) =8γ

π2

∞∑

k=1

(−1)k+1 1

(2k − 1)2sin((2k − 1)

π

2x) cos((2k − 1)

π

2t). (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= −β

δ2θ(1, t)

δt2(160)

where β = I/(ρJL). Separation of variables give:

45

dX(1)

dx= βλX(1). (161)

The solution is

X(x) = D sin√

λx (162)

giving

tan√

λ =1

β√

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):

d2Xi

dx2+ λiXi = 0 (164)

and

d2Xj

dx2+ λjXj = 0. (165)

Multplying the first of these by Xj(x) and integrating from 0 to 1 gives

∫ 1

0

d2Xi

dx2Xjdx + λi

∫ 1

0

XiXjdx = 0 (166)

and integrating by parts leads to:

Xj(1)dXi

dx(1) − Xj(0)

dXi

dx(0) −

∫ 1

0

dXi

dx

dXj

dxdx + λi

∫ 1

0

XiXjdx = 0. (167)

so that

βλiXi(1)Xj(1) −∫ 1

0

dXi

dx

dXj

dxdx + λi

∫ 1

0

XjXidx = 0. (168)

46

Integrating the second equation, after the multiplication by Xi(x), from0 to 1:

βλjXj(1)Xi(1) −∫ 1

0

dXi

dx

dXj

dxdx + λj

∫ 1

0

XiXjdx = 0 (169)

and subtracting the last two equations leads to

(λi − λj)

(

βXi(1)Xj(1) +

∫ 1

0

XiXjdx

)

= 0. (170)

This implies since λi 6= λj

βXi(1)Xj(1) +

∫ 1

0

XiXjdx = 0 (171)

and defining scalar product of g and f by

(f, g) =

∫ 1

0

f(x)g(x)dx + βf(1)g(1) (172)

then (Xj, Xk) = 0.The mode shape is normalized

1 = (Xk, Xk) =

∫ 1

0

D2k sin2(

√

λkx)dx + D2kβ sin2

√

λk = D2k

[∫ 1

0

1

2(1 − cos(2

√

λkx))dx + β sin2 λk

]

=

Little manipulation produces

Dk =√

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 problems

with 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

The torsional oscillations, in terms of nondimensional variables, withθ(0, t) = 0 are

δθ

δx(1, t) = −β

δ2θ

δt2(1, t) +

T0L

JGsin ωt (176)

where ω = L√

ρ/Gω. Since the external excitation is harmonic, thesteady state response is assumed as

θ(x, t) = u(x) sin ωt. (177)

This leads to

d2u

dx2sin ωt = −ω2u sin ωt (178)

or

d2

dx2+ ω2u = 0. (179)

From the boundary conditions given in the previous example and u(0) = 0leads to

du

dx(1) − βω2u(1) =

T0L

JG. (180)

The solution is

u(x) =T0L

(ω cos ω − βω2 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

8 Balancing rotation

The 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 the

49

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, cosΩt 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

Figure 18: Rotor lateral motion measured by two displacement proximity transducers inorthogonal orientation and the Keyphasor phase reference transducer.

51

Figure 19: Time-base waveforms of rotor response to unbalance inertia force. Note thatresponse lags the force by the phase difference.

52

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 response

RF = Bei(Ωt+β). (184)

The corresponding vectors are obtained when the periodic function oftime, eiΩt is eliminated:

−→F = Feiδ = mrΩ2ejδ (185)

53

Figure 20: Unbalance force vector and rotor fundamental response vector in polarcoordinate format. Note conventional direction of response angle, β, lagging the force

vector 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 are

54

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,

Beiβ and 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 − MΩ2 and 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 Dynamic

Stiffness,−→k . The inverse of the complex dynamic dtiffness is also known

receptance. 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 = Feiδ and

−→B = Beiβ , for the

unbalance force vector and response respectively, as well as,−→H = 1/

−→k for

the rotor transfer function vector. The original unbalance response at aconstant speed, Ω is:

55

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 the

angles are most often omitted and replaced by lag.

56

Figure 22: Balancing in one-plane.

−→F−→H =

−→B . (189)

In this relationship, there are two unknown vectors,−→H and

−→F . The

response vector,−→B is measured thus its amplitude and phase are known. In

order 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 additional

input force vector,−→F = mτrτΩ

2eiδτ . This run is called a calibration run.

The measured rotor response vector is now−→B 1 = B1e

iβ1 , which is differentfrom the response vector. For this second run, the following input/outputrelationship holds true:

(−→F +

−→F τ )

−→H =

−→B 1. (190)

In the above the unknown vectors are−→H and

−→F , and the others are

known. The last two equations are suffiecient to solve the one-place balancing

57

problem and calculate the unknown parameters. The unkonwn vector −−→F

and the corrective mass, mc are calculated, therefore, as follows:

−−→F = mcrcΩ

2eiδc =

−→F τ

−→B

−→B −−→

B 1

(191)

or

− F

Ω2= mcrce

iδc =

−→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 = mτe

iδτ

−→B

−→B −−→

B 1

. (193)

Finally, note that the balancing procedure does not require calculation of

the second unknown parameter, the rotor transfer function vector−→H provides

this 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 stiffness

vector is often used in the form−→k /Ω2 and 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 = mτrτe

iδτBeiβ

Beiβ − B1eiβ1= mτrτe

iδτB

B − B1ei(β1−β)(195)

58

from where

mcrc(cos δc + i sin δc) = mτrτ (cos δτ + i sin δτ )B

B − B1(cos (β1 − β) + i sin (β1 − β))= mτrτ (cos δτ + i

Now the real and imaginary parts are:

mcrc cos δc = mτrτB(B − B1 cos (β − β1)) cos δτ − B1 sin (β − β1) sin δτ

B2 + B21 − 2BB1 cos (β − β1)

(197)

and

mcrc sin δc = mτrτB(B − B1 cos (β − β1)) sin δτ + B1 sin (β − β1) cos δτ

B2 + B21 − BB1 cos (β − β1)

.(198)

From these we get:

mcrc = mτrτB

√

B2 + B21 − 2BB1 cos (β − β1)

(199)

and

δc = 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 the

rotor and measure its original synchronous response vector,−→B = Beiβ, at the

rotational speed Ω. After this stop the rotor and choose a radial and angular

scale for plotting vectors. Draw the vector−→B = Beiβ in 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 = Fτe

iβτ = mτrτΩ2eiδτ . Run the rotor at the same speed Ω. Measure

the new rotor synchronous response vector−→B 1 = B1e

iβ1 and draw it in thepolar plot using the same scale, and then stop the rotor. Subtract vectorially−→B 1 from

−→B in the plot; draw a vector

−→B −−→

B 1. Find the corrective weight

angular position as δc = δτ + θ. The angle θ is between the vectors−→B and

59

Figure 23: One plane balancing of a rotor using polar plot.

−→B −−→

B 1. Since the response is proportional to the input force, the triangles

‖−→B ‖, ‖−→B 1‖, ‖−→B 1 −

−→B‖, and ‖−→F ‖, ‖−→F +

−→F τ‖, ‖

−→F τ‖ are similar; they have

the same angles. Now measure on the plot the length of the vector ‖−→B −−→B 1‖

using the assigned scale. Next calculate the corrective mass, mc, applyingthe formula:

mc = mτrτB

rc‖−→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 Cryostat

As Rota I Cryostat was to be modified, and also partly rebuilt, rotationalcharacteristics were measured focusing on noise. The measurements were

60

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

0 5 10 15 20 25 30 35 40 45 50 550

0.5

1

1.5

2

2.5

3

3.5x 10

4

V(1

0−3 )

Tangential Noise with Electronics at 250 mrad/s

0 5 10 15 20 25 30 35 40 45 50 550

1

2

3

4x 10

4

Hz

V(1

0−3 )

Tangential Noise with Electronics at 1000 mrad/s

Figure 24: Two fourier spectra show measured tangential noise at rotating speeds 250and 1000 mrad/s.

62

0 5 10 15 20 25 30 35 40 45 500

1

2

3

4

5

6

7

8x 10

4

Hz

V H

z(10

−3 )

Tangential Additive Noise Integral with Electronics

250100020003000

Figure 25: The integrated additive plot shows the complexity of rising noise level atdifferent speeds in tangential direction.

63

0 5 10 15 20 25 30 35 40 45 500

0.5

1

1.5

2

2.5x 10

4

Hz

V H

z(10

−3 )

Normal Noise Additive Integral Plot with Electronics

250100020003000

Figure 26: The integrated additive plot shows the complexity of rising noiselevel atdifferent speeds in normal direction.

64

0 5 10 15 20 25 30 35 40 45 50 550

2000

4000

6000

8000

V(1

0−3 )

Tangential Noise without Electronics at 250 mrad/s

0 5 10 15 20 25 30 35 40 45 50 550

2000

4000

6000

8000

Hz

V(1

0−3 )

Normal Noise without Electronics at 250 mrad/s

Figure 27: This plot shows how the normal and tangential directions differ taken at 250mrad/s without electronics.

65

0 5 10 15 20 25 30 35 40 45 50 550

1

2

3

4x 10

4

V(1

0−3 )

Tangential Noise With Electronics at 3000 mrad/s

0 5 10 15 20 25 30 35 40 45 50 550

1

2

3

4x 10

4

Hz

V(1

0−3 )

Tangential Noise Without Electronics at 3000 mrad/s

Figure 28: This plot shows how the tangential noise varies with and without theelectronics taken at 3000 mrad/s.

66

0 5 10 15 20 25 30 35 40 45 500

2000

4000

6000

8000

10000

12000

14000

Hz

V H

z(10

−3 )

Tangential Additive Noise Integration without Electronics

250100020003000

Figure 28: Additive integration is taken without electronics in the tangential direction.

10 Superconducting high-homogeneity mag-

net for NMR measurements

Chemical 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:

|∆B

B0| = |B − B0

B0|. (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 our

67

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 Superconductivity

The 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

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/T where 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

where the volume term has been neglected. The associated magneticenergy in the presence of an applied magnetic field Ba is −M ·dBa, where Mis the magnetic dipole moment per unit volume. A change in the free energyis given by

dg = −M · dBa − sdT. (204)

Integrating yields:

g(Ba, T ) = g(0, T )−∫ Ba

0

M · dBa. (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 ) +

∫ Ba

0

BadBa

µ0= gS(0, T ) +

B2a

2µ0. (206)

We see B2a/2µ0 as the extra magnetic energy stored in the field as resulting

from 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 =T

µ0

[

Bcd2Bc

dT 2+

(dBc

dT

)2], (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 than

70

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 quenching

We 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. The

71

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)2

Acu(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 coil

V (t) = I(t)R(t) + L(I)dI(t)

dt−

∑

i

MidI

dt− UPC , (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)dI

dt, (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

13 Nuclear Magnetic Resonance, and Imag-

ing

Quantum 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 by

µ = 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 is

µz = γIz = mγ~. (216)

I2 has 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:

dE

dt= 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 be

73

zero for a net absorption. Now in a similar fashion using a alternating fieldwe get for the absorption energy [57]

dE

dt= n~ωW = n0~ω

W

1 + 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 is

dµ

dt= µ × γ[H0 + H1(t)]. (223)

Now moving to a coordinate system such that the system rotates aboutthe z-direction at frequency ωz then:

δµ

δt= 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 −ω

γ) + H1i. (225)

74

The effective field exactly when the resonance condition is fullfilled theeffective field is iH1, and a magnetic moment parallel to the static field willthen precess in the y − z plane. Turning H1 on for a wave train of durationtw so that moment precesses through an agle θ = γH1tw = π so inverting themoment. Now if θ = π/2 the magnetic moment is turned from z-direction tothe y-direction. Turning H1 off then would make the moment remain at restin the rotating frame. Simple method for observing magnetic resonance canbe done with these remarks. Putting a sample in a coil, the axis of whichis perpendicular to H0. Alternating field applied to the coil produces analternating magnetic field. Adjusting tw and H1 we can apply a π/2 pulse,after which the excess magnetization will be perpendicular to H0 precessingat angular frequency γH0. Now the moments make alternating flux throughthe coil inducing emf that may be observed. Now variations of the similarprinciple can be used to make measurements using NMR. The method inour experiment is using continuous wave method. The receiver coils are setperpendicular to the magnetic field from the magnet. The current is sweepedin the magnet, and the resonance signal should pick-up in our coil. Themagnitude of nmr resonace signals is proportional to the molar concentrationof the sample [58].

In our experiment the magnet is placed within a vacuum, and in themiddle of the magnet one places the sample to be measured with the pick-upcoils in perpendicular direction to the magnetic field (Figure 30)

The resonance circuit in Figure 30 is tuned for both excitation and de-tection. At first there was no cooled preamplifier in the setup. The lock-inamplifier measured the signal from the pick-up coils, which was locked to thefrequency of the oscillator. The oscillator was used to drive the excitationfrequency of the circuit. The magnet was supposed to be swept across toobserve NMR spectrum from the 3He sample inside the magnet. The test-ing was supposed to be done for 25 turns per each pick-up coil, and then50 turns. The pick-up coils were very carefully coiled on glass piece that fitdirectly on the sample tube right in the middle. The sample was place inthe center of the magnet, and thus the magnetic field from the magnet andthe pick-up coils were perpendicular. The gyromagnetic ratio of 3He is 32.43MHz/T [?]

14 Cooling of the Cryostat

The main purpose in our measurement is to measure the homogeneity ofthe superconducting magnet, by doing NMR measurements of normal liquid3He, which requires temperatures below 3.19 K. The experiment is done

75

Figure 30: The setup of the NMR-measurement in our experiment.

in an insulated dewar, see Figure 30. The precooling is done using liquidnitrogen, which boils at 77 K [59], and liquid N2 has about 60 times thelatent of heat evaporation to that of cryo-helium [60]. Liquid nitrogen is alsoan order cheaper than cryo-helium. One very important property of liquidnitrogen is that it freezes at 63 K, and this has to be taken into account,when the next cooling phase is done with liquid 4He. After all the equipmentis placed within the cryostat, and all the necessary preparations have beendone, liquid nitrogen is pumped into the cryostat using a metal tube. Thisis done gradually as the liquid evaporates it creates pressure, and there is avalve to let excess pressure out, but one does not want the pressure to getmuch greater than 1.4 Pa in the cryostat. This is because the cool gas willfreeze the insulating rubber in the valve, thus yielding the insulation useless.After the nitrogen cooling is over the rest of the nitrogen is pumped outusing a narrow nylon tube reaching the bottom of the cryostat. This is solelyfor the purpose that the nitrogen freezes otherwise, and it has a rather largespecific heat. The cryostat setup is seen in Figure 31. The vacuum spaceat this point is filled with 3He exchange gas to hasten the cooling of sampleand the magnet. The heat of evaporation for cryo-helium is about 2.6 kJ/lat its boiling point is 4.2 K. The 4He used as a cryo-liquid for cooling has

76

Figure 31: The setup of the cryostat.

some precautive aspects to take into concideration. The 4He expands veryrapidly if comes too fast into contact with the surroundings even at 77 K.The helium should be pumped into the cryostat in a very slow paste usingthe evaporating gas to cool the upper parts of the cryostat. Again if thepressure gets too high the cool gas will escape throuhg the input line and theemergency valve, thus the rubber insulators will freeze useless. Helium willalso freeze all the pipes and tubings in the cryostat, and one has to be carefulnot to break frozen pipes, and not to touch them with bare hands. Now astemperature of 4.2 K is reached the vacuum-space is then pumped to a closevacuum state. It takes some time to pump, as the exchange gas as it so coldat this point. Getting down into temperature of below boiling point of 3.19for 3He sample to become a liquid we need to pump the 4He. The samplehas been pressurised to 18 Pa in order to get a good NMR signal.

The system for the continuous evaporation cryostat is given in Figure 32.Small fraction of the 4He liquid is sucked by a pump through a suitable

flow impedance into a small pot in the vacuum can. The liquid is expandedthrough the impedance, and will arrive at a lower temperature. The fillsuntil level of liquid is balances the heat transferred from the helium andthe experiment with the cooling power of the refrigirator. The following

77

Figure 32: Continuously operated 4He refrigerator.

balancing equation is obtained:

QHe(≃1

2nL) = Qtube(h) + Qext, (226)

where L is the latent heat of evaporation, and h is the height of liquidin the pot. When the external load increases the level in the pumping tubewill drop, thus there’s a self regulation. The temperature for the refrigiratorremains quite constant, and in our case it was 1.42 K on average. Theproblem is that in order to keep the vacuum can at the wanted frequencythe 4He evaporates at a rate, that the dewar had to be filled every day. Therequirement for the impedance is:

Z =∆P

V ν, (227)

where V is the volume flow rate of the medium with viscosity ν, and ∆Pis the pressure drop. In this experiment, a pre built pot was used that hadbeen in use, and was working properly. The design of the cryostat was suchthat even when pumping on the helium pot, the helium level could be refilled.

78

For the helium level an acoustic level detection was utilized using a thin steeltube with a thin rubber on top. The oscillation differences when submergedinto the dewar can tell the when the tube is in the cold gas compared to theliquid. This method could only be utilized 10 cm above the vacuum can. Soresistive level detection was utilized for the lower parts of the cryostat. Inthis way it was ensured that the level of helium in the dewar never droppedbelow the level of the pot when temperatures below 4.2 K were needed.

15 Preparing the NMR measurements

The cryostat was meticulously built for the NMR measurements. Tests weredone for the cooling, and there were no problems for cooling. The averagetemperature of 1.42 K was achieved when the pot was pumped, which wasenough for the 3He to become a liquid under the pressure of 18 Pa. Thispressurising was done to get a good NMR-peak. The vacuum was achievedafter a few trials, and the can had to tightened a few times before therewas no leak. Also the pressurising of the sample was done accordingly, andworked. The tesing of the magnet turned out to be quite problematic. Themagnet gave a magnetic field strength per current 8.35 mT/A. The magnetworked properly when the vacuum can was filled with an exchange gas, andthe maximum current that the power supply gave was 9.10 A even with veryfast sweep rates of 0.3 A/s. Next the magnet was tested in vacuum con-ditions for quite a many times, and the magnet quenched at around 2.3 Aevery time, and this was invariant for the slower sweep rates. For higher therates the magnet queched at even lower currents than previously mentioned.The quenching usually drove the whole superconducting magnet unusable fora period from hour up to three hours, if no exchange gas was let in. Thisalso raised the temperature of the sample in the vacuum by some 1 K. Verydetailed study of the cryostat was done, and no problems were found in it.Everything except the magnet seemed to be working properly without anycomplications. The magnet was studied under a microscope or the visibleparts of the magnet. All the visible parts seemed to be in order, and no no-ticeable defects were found. There must have been a defectious portion in thewire like a twist or broken filament that was not broken completely, but bigenough to cause the magnet to go into normal state when in vacuum. Lam-mela tested the magnet after building it only submerged in liquid 4He, whichcooled the small place where the heating occurs driving the magnet normal.This conclusion was supported as well by the fact that when exchange gaswas let into the vacuum, the magnet returned into the superconducting stateliterally within seconds. As the broken part of the magnet’s filament was

79

not visible, the repair of the magnet was not possible, and the NMR-studiescould not be completed without the magnet working in the vacuum.

16 Conclusions

The first part of the work consisted of the rotational aspects of Cryo I cryo-stat. There was a need for very precise rotation [62], and the heat noise wasto be minimized. It was apparent that some of the electronics, were givingout vibrations at quite precise frequencies. These could be dimished withthe use of dissipating materials and methods suggested, without even thatmuch work. Also some of the electronics could have their fans relaplaced withmetal spikes as is done in computers, and also let the cooling be done by airas the cryostat is rotating. This could be a little problematic for some of thedevices. Also some of the pumps gave out vibrations that could diminishedby balacing the pumps with the suggested method. Also the Fourier analysiswas done with data that was not filtered properly, and thus some aliasingaccumulated in the transformed data. This could be avoided by building agood solid highpass filter according to the Nyquist frequency. The rotation ofthe cryostat is done sometimes at quite high speeds, so the balancing of thecryostat becomes quite important. The electronics could be placed accordingto the scheme presented previously, using the one-plane method. Anotherway if the electronics cannot be moved around is to place small balancingweights on the plane carrying the electronics. The balancing method wouldrequire two similar accelerometers, as in this case of this work only one wasavailable. Also a keyphasor would be needed. In the latter part of the worka magnet was to be tested for its homogeneity. The problem was apparentafter the meticulous building of the cryostat. The magnet only worked whenit was submerged in 4He-liquid or 3He exchange gas. The magnet quechedevery time when used in vacuum, and the experiment was thus abandoned,with the result that the magnet is useless.

References

[1] Emmanuel C. Ifeachor and Barrie W. Jervis, ”Digital Signal Processing:A Practical Approach”, Addison-Wesley, p. 53 (1993)

[2] Ronald N. Bracewell, The Fourier Transform and Its Applications,McGraw-Hill, 2000, p. 130.

80

[3] Bendat, J.S. and Piersol, A.G., Measurement and Analysis of RandomData, John Wiley, New York, 1966.

[4] Blackman, R. B. and Tukey, J. W., The Measurements of Power Spectra,Dover, New York, 1956.

[5] Jenkins, G.M. and Watts, D. G., Spectral Analysis and its Applications,Holden-Day, San Francisco, 1968.

[6] Newland, D. E., An introduction to Random vibrations, spectral andwavelet analysis, Longman Scientific and Technical, Essex, 1993.

[7] Cooley, J. W., and Tukey, J. W., An Algorithm for the Machine Cal-culation of Complex Fourier Series, Mathematics of Computation, Vol.19, 1965, 297-301.

[8] Cooley, J. W., Lewis, P. A. W., and Welch, P. D., Historical Notes onthe Fast Fourier Transform, IEEE Trans. Audio and Electroacoustics,Vol. AU-15, 1967, No. 2, 76-9

[9] http://www.fftw.org (February, 13, 2008)

[10] Oppenheim, A. V. and R. W. Schafer, Discrete-Time Signal Processing,Prentice-Hall, 1989, p. 619.

[11] Rader, C. M., ”Discrete Fourier Transforms when the Number of DataSamples Is Prime,” Proceedings of the IEEE, Vol. 56, June 1968, pp.1107-1108.

[12] Institute of Electrical and Electronics Engineers Inc. IEEE standard onpiezoelectricity. ANSI/IEEE Std. 176-1987, 1988.

[13] C. R. Fuller, S. J. Elliot, and P. A. Nelson. Active Control of Vibration.Academic Press, 1996.

[14] Nathan Newman, Considerations in Computing the Useful FrequencyRange of Piezoelectric Accelerometers, NBS Technical Note 487, Na-tional Bureau of Standards, 1969.

[15] Dranetz, A. I., and Orlacchio, A. W., Chapter 16-Piezoelectric andPiezoresistive Pickups, Book, Shock and Vibration Handbook, McGraw-Hill, New York, 1961

[16] Carol F. Vezzetti and Paul S. Lederer, An Experimental Technique forthe Evaluation of Thermal Transient Effects on Piezoelectric Accelerom-eters, NSB Technical Note 855, National Bureau of Standards, 1975.

81

[17] Williams, Arthur B. and Taylors, Fred J. Electronic Filter Design Hand-book, New York: McGraw-Hill, 1988.

[18] Thomas Kugelstadt, Op Amps for Everyone, Chapter 16: Active FilterDesign Techniques, Texas Instruments, 2008

[19] Paul Michael Walmsley, Textures, Vortices and Persistent Currents in aSlab of 3He-A,The University of Manchester, 2003

[20] C. F. Beards, Engineering Vibration Analysis with Application to Con-trol Systems, Edward Arnold, 1995.

[21] C. F. Beards, Structural Vibration Analysis and Damping, EdwardArnold, New York, 1996.

[22] B. J. Lazan, Damping of Materials and Members in Structural Mechan-ics, Pergamon, 1968.

[23] Cyril M. Harris, Harris’ Vibration Handbook, Chapters 30-33, NewYork: McGraw-Hill, 2002

[24] Paul G. Hoel, Introduction to Mathematical Statistcs, Wiley Publica-tion, 1966

[25] D. J. Cousins, S. E. May, J. H. Naish, P. M. Wamsley, A. I. Golov,Design of a Rotating Cryostat to Minimise Vibrational Heating, Dept.of Physics and Astronomy The Univ. of Manchester, 2003

[26] B. J. Pergamon, Damping of Materials and Members in Structural Mech-anisms, Pergamon, 1984.

[27] A. Muszynska, Rotordynamics, Chapter 2, Taylor and Frnacis, 2005,Minden.

[28] S. G. Kelly, Fundamentals of Mechanical Vibrations, Table 9.1, McGrawHill, Boston, 2000.

[29] N.F. Rieger, Vibration of Rotating Machinery, Vibration Institute,Clarendon Hills, IL, 1982

[30] Agnieszka Muszynska, Rotodynamics, Chapter 6. Mechanical Engineer-ing, The Ohio State University, 2005

[31] http://www.gepower.com/prod_serv/products/oc/en/bently_nevada/acc_vel.htm

8.3.2008

82

[32] http://www.gepower.com/prod_serv/products/oc/en/downloads/141587.pdf

8.3.2008

[33] http://www.imd.us/pdfs/dynamic-stiffness.pdf

13.3.2008

[34] http://www.bpb-co.com/articles/printableFiles/00understand_stiff_print.pdf

15.3.2008

[35] Ripka J.: Theory of the open end rotor spinning system. In:Rohlena V.et al: Open-end spinning. Elsevier, Amsterdam 1975.

[36] Aleksanteri Kulvik, ”Vibrational Analysis of Rota I Cryostat”, 2003

[37] http://www.bksv.com/pdf/Bp0196.pdf

July, 31, 2003

[38] http://we.home.agilent.com/cgi-bin/bvpub/agilent/

July, 28, 2003

[39] Ruutu, Ville: NMR experiments on topological defects in 3He superflu-ids, Ph.D Thesis, 1996

[40] Lindell, Sihvola: Staattinen kenttateoria. Otatieto, 1995.

[41] J. Vanderlinde, Classical Electromagnetic Theory, Wiley and Sons, NewYork, 1993.

[42] Terrell J., Montgomery D.B,: Some Useful Information for the Designof Air-Core Solenoids, National Magnet Laboratory MIT, 1961

[43] Brehm, J.J. Mullin: Introduction to the Structure of Matter: A Coursein Modern Physics, Wiley, 1989

[44] Vesa Lammela, Superconducting high-homogeneity magnet for low-fieldNMR measurements, TKK Materiaalifysiikka, 2001

[45] SW-18 wire by Supercon Inc. 830 Turnpike, Boston Ma. USA

[46] M. Tinkham, Introduction to Superconductivity, 2nd Ed., Dover Bookson Physics (2004)

[47] L.N. Cooper: Bound Electron Pairs in a Degenerate Fermi Gas, Phys.Rev 104(4), (1189), 1956.

83

[48] http://en.wikipedia.org/wiki/Superconductivity

13.4.2008

[49] J. Baardeen, L.N. Cooper, and J.R. Schrieffer, Microscopic Theory ofSuperconductivity, Phys. Rec. 106, 162-164, 1957

[50] Stephen Elliott, The Physics and Chemistry of Solids, Wiley, 2000, p.466

[51] Ibach and Luth, Solid State Physics, Springer-Verlag GmbH and Co.KG, 1995, p. 230-231

[52] [51], p. 245

[53] Surendran Puthiyedath, Quenching and Protection of SuperconductingMagnets for LHC, CERN-INDIA Collaborator, BARC, Mumbai, India,2005

[54] Martin N. Wilson, Superconducting Magnets; Oxford Science Publica-tions, 1987

[55] Rabi, I. I., Zacharias, J. R., Millman, S. and Kusch, P. (1938). A NewMethod of Measuring Nuclear Magnetic Moment. Physical Review 53,318-318

[56] C.P. Slichter, Principles of Magnetic Resonance, Springer, 1980, p. 279

[57] [56], p. 9

[58] http://www.cem.msu.edu/~reusch/VirtTxtJml/Spectrpy/nmr/nmr1.htm

13.4.2008

[59] http://en.wikipedia.org/wiki/Liquid_nitrogen

14.4.2008

[60] F. Pobell, Matter and Methods at Low Temperatures, Springer-Verlag1992. p. 82

[61] E . Ward , L . Hedlund , W . Kurylo , C . Wheeler , G . Cofer , M .Dewhirst , L . Marks , Z . Vujaskovic; Proton and hyperpolarized heliummagnetic resonance imaging of radiation-induced lung injury in rats1 .International Journal of Radiation OncologyBiologyPhysics , Volume 58, Issue 5 , Pages 1562 - 1569

84

[62] U. Parts, V. M. Ruutu, J. H. Koivuniemi, Yu. M. Bunkov, V. V.Dmitriev, M. Fogelstrom, M. Huebner, Y. Kondo, N. B. Kopnin, J.S. Korhonen, M. Krusius, O. V. Lounasmaa, P. I. Soininen and G. E.Volovik; Single-Vortex Nucleation in Rotating Superfluid 3He−B, LTLHUT 1995, Espoo

85