Acoustical Impedances: Calculations and Measurements on a Trumpet

by Jonathan Kipp

A Bachelor’s Thesis submitted in

October 2015

to the Faculty of Mathematics, Computer Science and Natural Science Department of Physics at RWTH Aachen University with Prof. Dr. rer. nat. J¨ org Pretz

Declaration of Authorship Hiermit erkl¨ are ich, Jonathan Kipp, an Eides statt, dass ich die vorliegende Bachelorarbeit selbst¨andig verfasst und keine anderen als die angegebenen Quellen und Hilfsmittel benutzt sowie Zitate kenntlich gemacht habe.

Signed:

Date:

ii

Symbols and Constants

k S p u U = Su Z = Upaa z = upaa

Symbols Wave number Cross section Pressure Particle velocity Volume flow Impedance Acoustical Impedance

m−1 m2 N m−2 ms−1 m3 s−1 N sm−5 N sm−3

ω

angular frequency

rads−1

Constants Speed of Sound c = Density of air ρ =

343 ms−1 1.3 kgm−3

iii

Contents Declaration of Authorship

ii

Symbols and Constants

iii

List of Figures

vii

1 Introduction

1

2 Theoretical Background 2.1 Defining a Horn . . . . . . . . . . . . . . . . 2.2 Sound Propagation in Air . . . . . . . . . . 2.2.1 The Infinite Cylindrical Tube . . . . 2.3 Finite Tubes with Flares . . . . . . . . . . . 2.3.1 Impedance of the Cylindrical Tube . 2.3.2 Impedance of the Exponential Tube 2.3.3 Impedance of the Conical Tube . . .

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3 Simulation 3.1 Cylindrical Model . . . . . . . . . . . . . . . . . . . . . 3.2 Conical Model . . . . . . . . . . . . . . . . . . . . . . 3.3 Exponential Model . . . . . . . . . . . . . . . . . . . . 3.3.1 Influence of single Diameters on the Spectrum 4 Measurement 4.1 Diameter Measurement 4.2 Measuring Technique . . 4.3 Measuring Setup . . . . 4.4 Results . . . . . . . . . .

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3 3 3 4 6 9 10 12

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25 25 27 28 30

5 Comparison of Simulation and Measurement

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6 Conclusion

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A Appendix

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v

Contents Bibliography

vi 43

List of Figures 2.1 2.2 2.3 2.4 2.5 2.6

Cross Section and Section of Infinite, Cylindrical Tube First higher Mode . . . . . . . . . . . . . . . . . . . . Sound Propagation in Cylindrical Tube . . . . . . . . Finite Cylindrical Tube . . . . . . . . . . . . . . . . . Finite Exponential Tube . . . . . . . . . . . . . . . . . Finite Conical Tube . . . . . . . . . . . . . . . . . . .

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. 4 . 6 . 7 . 9 . 10 . 12

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10

Diameter along Horn Axis . . . . . . . . . . . . . . Cylindrical Model for infinite ZL . . . . . . . . . . Exponential and Cylindrical Model in Comparison Impedance, Conical Model . . . . . . . . . . . . . . Impedance, Exponential Model . . . . . . . . . . . Horn in Section . . . . . . . . . . . . . . . . . . . . Influence of 24th Diameter . . . . . . . . . . . . . . Simulation with new Value, Exponential Model . . Forth Resonance vs 24th Radius . . . . . . . . . . . Forth Resonance vs 10th Radius . . . . . . . . . . .

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17 18 18 20 21 21 22 22 24 24

4.1 4.2 4.3 4.4 4.5 4.6

Shape of the Trumpet . . . . . . . . . . . . Diameter Measurement . . . . . . . . . . . Measurement Setup . . . . . . . . . . . . . Flange for Trumpet . . . . . . . . . . . . . . Impedance of Trumpet without Mouthpiece Impedance of Trumpet without Mouthpiece,

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26 27 29 30 31 31

5.1 5.2

Impedance of Conical Model and Measurement . . . . . . . . . . . . . . . 35 Impedance of Exponential Model and Measurement . . . . . . . . . . . . 35

A.1 A.2 A.3 A.4

Influence of Friction, Exponential Model . . . . Influence of Friction, Conical Model . . . . . . Comparison of Conical and Exponential Model Influence of 24th Diameter . . . . . . . . . . . .

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All photographs are taken by the author. All figures are compiled by the author.

vii

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40 40 41 41

Chapter 1

Introduction Manufacturing brass instruments is a highly complicated matter. The manufacturers main goal is to build an instrument with good tuning, which means that the instrument’s natural resonances match the frequencies of the corresponding notes. These natural resonances are influenced not only by the instrument’s flare and length, but also by e.g. the exact position and diameter of valves and tuning slides. To cover all these effects would require a great amount of work, which is not compatible with the scope of this work. This thesis aims on predicting and measuring the natural resonances of brass instruments within a simple model. The comparison of prediction and measurement is another goal as well as giving an outlook on which parameters will need more careful treatment in further going simulations of these natural resonances. The focus will be on the instruments properties, not taking into account that the stimulation of sound waves by the players lips is highly complicated: professional players are able to play up to three tones out of center just using tongue and lips1 . The term horn will be of great importance, it will be necessary to define the term horn in the context of this work as well as developing ideas on how this term must be extended to improve the results of e.g. simulations. The natural resonances can be obtained by measuring the acoustical impedance Zin = pa /ua , where pa is the acoustical pressure and ua is the acoustical particle velocity. The natural resonances occur at those frequencies, where |Zin,throat |, the impedance at the throat of the horn, reaches a local maximum. So the task will be to simulate and measure the impedance at the throat of the instrument, because the excitation takes place at this point. The impedance depends on the flare of the instrument from throat to mouth. Modeling the shape of the instrument in different ways will give different results for the simulation. One part of this work is to study the weaknesses and advantages of the two models proposed, the exponential and the conical 1

Playing out of center: the musician bends the note and excites acoustic waves not with the initial frequency, but with another frequency. Hence the musician forces excitation at a frequency which is not intended to be played with the valve combination that is chosen.

1

Theoretical Background

2

flare. The measurement of the impedance will be helpful to interpret the simulation with respect to these weaknesses or advantages. This work also offers insight into the dependence of the natural resonances on the instruments radius at one particular point. This is done by varying single radii and computing the impedance for an instrument shape with these variated radii. The aspiration of this work is to do a prediction with a simple understanding of the term horn, which is within a halftone range of the actual measurement. If this aspiration is met, all effects neglected will be just minor corrections to the simple understanding. A next step, which is beyond the scope of this work, would be not just to study the natural resonances, but the radiation from the bell and the spectrum of harmonics, too. These aspects are of great interest when studying, if the sound of the instrument fulfills a certain ideal.

Chapter 2

Theoretical Background 2.1

Defining a Horn

In the context of this work, a horn is a straight tube, which has a defined length L, ideally rigid walls and a flare which can be described analytically. It has one open, driven end, where the acoustical excitation takes place. The other end would be ideally closed on the first look, but we will introduce a description to the behaviour of a radiating pipe. In this understanding of the term horn, neither the brass instruments curvature nor the material from which the horn is built is taken into account. This simple understanding will be tested with measurements on an ordinary trumpet to learn if the model respects the important characteristics for the natural tone scale and which neglected facts may be worth considering. The most important characteristics, which are not respected here, are the curvature of wave fronts, vibration of the walls, radiation effects at the bell and the complicated stimulation at the throat.

2.2

Sound Propagation in Air

The phenomenon called sound is the modulation of pressure pabs = patm + pa and local particle velocity vabs = vatm + va in a medium dependent on time and position. A differential equation describing the particle displacement ξ in a tube with a flare from a simple, one dimensional model, will be derived. However, propagation of sound in three dimensions will first be described in a very academical case, to learn why this simple one dimensional model is valid. If we want to describe the propagation of sound waves with a simple time dependence eiωt (which is obviously an assumption) in any medium,

3

Theoretical Background

4

we have to solve Helmholtzs Equation for the acoustical pressure: ∆pa + k 2 pa = 0, [1], where k =

ω c.

(2.1)

Note that ρ

∂u = −∇p. ∂t

(2.2)

The focus is set on the natural resonances of our system, which occur at those frequencies maximizing the reflexion coefficient R (the systems border reflects most of the wave energy back into the system). While the reflection coefficient is also measurable and maybe the plausible property of a sound system, it is common to work with the impedance z. It is defined as z(~x) = pa /ua or Z(~x) = pa /Ua , where ua is the local particle velocity and Ua = S · ua is the volume flow (S is the general cross section here).

2.2.1

The Infinite Cylindrical Tube

Now a solution to Helmholtz’s equation in an infinite, cylindrical pipe is proposed. Taking advantage of the system’s translational symmetry along the pipe axis, the equation will be solved in cylindrical coordinates. Choosing the coordinate system with the z-axis being aligned to the pipes axis, the boundary conditions for this case are as follows: ∂p =0 ∂ρ ρ=a,φ,z

(2.3)

Φ(φ) = Φ(φ + 2πn)

(2.4)

Figure 2.1: Cross section of infinite, cylindrical tube of radius a and infinite, cylindrical tube of radius a in section

Theoretical Background

5

Using the Laplace Operator in cylindrical coordinates1 and assuming the solution p to be separable, p = R(ρ) · Φ(φ) · Z(z), yields ρ2 R00 + ρR0 + (µ2 ρ2 − α2 )R = 0,

(2.5)

where: • Z(z) ∼ eiγz • Φ(φ) ∼ eiαφ • µ=

p

k2 − γ 2

Note, that Φ fulfills condition 2.4. In the following, r = µρ is the new, dimensionless coordinate, which gives: r2 R(r)00 + ρR(r)0 + (r2 − α2 )R(r) = 0.

(2.6)

This is Bessels differential equation, which is solved by the Bessel and Neumann functions. The latter are not interesting because we do not expect the solution to have poles on the z-axis. Here α is an integer and indicates the order of the Bessel function. Hence our general solutions are of the form pα (r, φ, z, t) = const · Jα (r) · ei(γz+αφ+ωt) ,

(2.7)

Jα being the Bessel function of order α. The solution for the pressure in an infinite cylindrical pipe will be of no practical use within the context of this work, since the focus is set on finite pipes with arbitrary flare. However, important facts can be determined from this result, if boundary condition 2.3 is respected, which yields

Jα (µa) = 0 ⇒ µ =

qm,α ∨ µ = 0, a

(2.8)

where qm,α is the mth root of the Bessel function of order α. Solving for γm,α yields

2 γm,α = k2 −

2 qm,α . a2

(2.9)

2 The first mode (0,0) with γm,α = k 2 ( µ = 0, pa is constant over the cross section) will

always propagate, but higher modes (m,α) will only propagate if the condition 1

∆=

1 ∂ ρ ∂ρ

  ∂ ρ ∂ρ +

1 ∂2 ρ2 ∂φ2

+

∂2 ∂z 2

Theoretical Background

6

Figure 2.2: First higher mode (0,1), appearing at approximately 6300 Hz in a tube of radius 21 mm. The figure shows the pressure amplitude over the pipes cross section, markers are parts of the tubes radius.

k>

qm,α a

(2.10)

is fulfilled. Only then k is real and the wave is not damped along the axis. This result is central for this work, because it legitimates a crucial simplification to the three dimensional problem of sound propagation in pipes. The pipes used for trumpets have radii in the cm range, while the excited wavelengths are at least two orders larger, about 1 m. In example, the minimum frequency for propagation of the first higher mode (0,1), appearing at q0,1 = 2.40482 , is approximately f0,1 =

2.4048 · 343.3 1 = 6256.82 Hz, 0.021 · 2π s

(2.11)

for a tube of radius 21 mm (typical radius in the conical section of a trumpet, see figure 3.1, while the played frequencies range from approximately 50 Hz up to 2000 Hz. This constellation allows to neglect all higher modes of (0,0) and therefore a one dimensional model along the z-axis can be used to describe the influence of the flare on the propagation. This model and its results are discussed in section 2.3.1, 2.3.2 and 2.3.3.

2.3

Finite Tubes with Flares

In the frequency range of interest (50Hz up to 2000Hz) no higher modes will propagate, so pressure and particle velocity will be in good approximation constant over the area of the propagating wavefront (at given time and position along the tube’s axis). A one dimensional model will be introduced now, which describes the influence of the cross sectional flare of a tube on the acoustical impedance neglecting the influence of higher 2

Value taken from http://mathworld.wolfram.com/BesselFunctionZeros.html, from 17th of June, 2015

Theoretical Background

7

Figure 2.3: A tube with cross section S. A wave travels along the tube and displaces the volume V = S · dz by dξ. New borders of V are dashed red.

modes. For the one dimensional model the starting point is the following idea: When a plane wave travels through a cylindrical pipe, a volume element of thickness dx moves from ABCD to abcd and will be displaced by dξ (take a look at figure 2.3). So the change in volume is given by   ∂ξ V + dV = Sdz 1 + , ∂z

(2.12)

because the cross section does not change with z. Respecting that the total pressure is ptot = patm + pa and using the definition of the bulk module dptot = −K dV V , yields: pa = −K

∂ξ ∂z

(2.13)

Respecting that the elements motion must obey Newton’s equations (pressure gradient in z direction must be equal to mass times acceleration) and substituting 2.13 we get: −S

∂pa ∂2ξ ∂2ξ K ∂2ξ dz = ρSdz 2 ⇔ = ∂z ∂t ∂t2 ρ ∂z 2

(2.14)

This differential equation describes the propagation of a plane wave in a cylindrical pipe in one dimension.

Theoretical Background

8

Now for the case with a flare: the change in volume is   1 ∂(Sξ) , V + dV = Sdz 1 + S ∂z

(2.15)

which gives us pa = −

K ∂(Sξ) . S ∂z

(2.16)

Again using Newton’s equation yields −S

∂pa ∂2ξ dz = ρSdz 2 . ∂z ∂t

(2.17)

Differentiating again with respect to z and swapping the differential operators ∂2 ∂t2

∂ ∂z

and

gives: −

∂ ∂pa ∂ 2 ∂ξ S = ρS 2 ∂z ∂z ∂t ∂z

(2.18)

Substituting 2.16 yields Webbsters Equation: ρ ∂ 2 pa 1 ∂ ∂pa S = S ∂z ∂z K ∂t2

(2.19)

The solution to this differential equation is of great value. It describes the propagation of plane sound waves in tubes with arbitrary cross sectional flare, which will allow the computation of the acoustical impedance Z = pa /Ua and prediction of the natural resonances for such a tube. Now solutions to this equation will be presented for a cylindrical tube as well as an exponentially and a conically flaring tube. For convenience, 1

2.19 will be brought into a more convenient form. Substituting P = S 2 ψ and writing S = πa2 (a being the radius of the tube at position z) as well as ρ/K = c yields:   ∂2ψ 1 ∂2a 2 + k − ψ=0 ∂z 2 a ∂z 2

(2.20)

Note, that the wave ψ and hence the real pressure pa is non propagating if k 2 < F , where F is the horn function F =

1 ∂2a a ∂z 2 .

This fact will be discussed for the exponential

tube later on. Please note that the discussion following is made exclusively for plane waves. This is not necessarily correct, since a flare of the pipe forces the wavefronts to be curved, but the diameter of the instruments described here are quite small. Hence the difference between area of the wavefront and the pipes cross section is small enough to neglect the curvature for the goals of this work.

Theoretical Background

9

Figure 2.4: A finite, cylindrical tube with length L and radius a

2.3.1

Impedance of the Cylindrical Tube

The cylindrical pipe of radius a has the easiest of shapes one could think of. The horn function F is zero in this case, so no cutoff is obtained. Equation 2.20 takes the form: ∂2ψ + k2 ψ = 0 ∂z 2

(2.21)

ψ(z) = Aeikz + Be−ikz .

(2.22)

The solution to this equation is

This yields the pressure: pa =

ψ 1

=

S2

 1  ikz · p0 e + p1 e−ikz eiωt a

(2.23)

Using 2.2 gives the volume flow Ua : S Ua = − · ρ

Z 0

t

 ∂pa 0 S  ikz dt = − p0 e − p1 e−ikz eiωt + const ∂z ρc

(2.24)

Theoretical Background

10

If the computation of the input impedance Zin of the cylindrical tube of length L at z = 0 is needed, two equations to eliminate p0 and p1 are required. Simple application of the boundary condition determining the impedance at z = L to be equal to ZL gives the two equations: pa (L) ρc p0 eikL + p1 e−ikL =− · = ZL Ua (L) S p0 eikL − p1 e−ikL

(2.25)

pa (0) ρc p0 + p1 = Zin =− · Ua (0) S p0 − p1

(2.26)

Solving 2.25 for p0 /p1 and substituting into 2.26 results in: Zin,cyl =

2.3.2

ρc ρc ZL cos(kL) + i S sin(kL) · S iZL sin(kL) + ρc S cos(kL)

(2.27)

Impedance of the Exponential Tube

One of the many models for the flare of a tube is the exponential one, a(z) = a0 · emz , where z is measured from the narrow part of the pipe and a0 is the radius at the narrow

Figure 2.5: A finite, exponential tube with length L and radii a and a2

Theoretical Background

11

end of the pipe. For this kind of slope, the equation 2.20 takes the form:
∂2ψ + k 2 − m2 ψ = 0 2 ∂z

(2.28)

This kind of differential equation has the solution (with λ2 = k 2 − m2 ) ψ(z) = Aeiλz + Be−iλz

(2.29)

This gives for the pressure (with the assumed time dependence eiωt and a = a0 · emz ): pa =

ψ 1

S2

   1  iλz =√ Ae + Be−iλz · eiωt = p0 ei(λ+im)z + p1 e−i(λ−im)z · eiωt πa

(2.30)

Note that a cutoff is obtained here (as mentioned in the last section). For k < m waves are not propagating. This happens especially for small frequencies, if the tube is quite short and the flare then is to rapid, which results in a big flaring parameter m. This is a central problem for the simulation, because small pipe segments are used, the impedance being computed for these segments separately. This method requires very short pipe segments, because we want to describe the flare as exact as possible. Choosing the segments to be satisfyingly infinitesimal will produce difficulties measuring the segments length and the radii, but will solve the problem of imaginary λ at frequencies above 50Hz. Using 2.2 gives the volume flow Ua : Z t S ∂pa 0 · dt ρ 0 ∂z  S  =− p0 (λ + im)ei(λ+im)z + p1 (−λ + im)e−i(λ−im)z · eiωt + const ωρa0

Ua = −

Again, two equations are needed to eliminate p0 and p1 . Simple application of the boundary condition determining the impedance at z = L to be equal to ZL (as in section 2.3.1) gives the two equations: pa (L) ρω p0 eiλL + p1 e−iλL = ZL =− · Ua (L) S(L) p0 (λ + im)eiλL + p1 (−λ + im)e−iλL

(2.31)

pa (0) ρω p0 + p1 =− · = Zin Ua (0) S(0) p0 (λ + im) + p1 (−λ + im)

(2.32)

Now solving 2.32 for

p0 p1

(λ − im)ZL − p0 = e−2iλL · p1 (λ + im)ZL +

ρω S(L) ρω S(L)

(2.33)

Theoretical Background

12

and substituting into 2.31 (note that λ2 + m2 = k 2 , see 2.22):   ρω −2iλL (λ − im)Z − ρω e L S(L) + (λ + im)ZL + S(L) ρω   Zin = · S(0) e−2iλL (λ − im)Z − ρω (λ + im) + (λ + im)Z + ρω (−λ + im) L L S(L) S(L)   ρω ρω e−iλL (λ − im)ZL − S(L) + ((λ + im)ZL + S(L) )     = ρω ρω k 2 ZL − S(L) (λ + im) e−iλL − k 2 ZL − S(L) (λ − im) eiλL Writing λ + im and λ − im in their polar forms, which are(Θ = arctan( m λ )):

λ + im = k · eiΘ

(2.34)

λ − im = k · e−iΘ

(2.35)

will simplify this equation. Substituting these forms into Zin gives (using k = Zin,exp =

2.3.3

ρc ρc ZL cos(bL + Θ) + i S2 sin bL · . S1 iZL sin bL + Sρc2 cos(bL − Θ)

ω c)

(2.36)

Impedance of the Conical Tube

A more simple flare is the conical flare, a(z) = a0 · z, where the distance z is measured from the conical apex, see 2.6. 2.20 now takes this form: ∂2ψ + k 2 ψ = 0. ∂z 2

Figure 2.6: A finite, conical tube with length L and radii a and a2

(2.37)

Simulation

13

The solution to this differential equation is well known to be ψ(z) = Aeikz + Be−ikz .

(2.38)

Using equation 2.2, the volume flow Ua is given by: S Ua = − · ρ

Z 0

t

∂pa 0 S dt = − ∂z ρca

  1 1 ikz −ikz p0 e (1 − ) − p1 e (1 + ) eiωt ikz ikz

(2.39)

Analogous to the section before, the boundary conditions pa (z1 ) ρc =− · Ua (z1 ) S(z1 ) p0 (1 −

p0 eikz1 + p1 e−ikz1 = Zin 1 ikz1 + p (1 + 1 )e−ikz1 1 ikz1 )e ikz1

(2.40)

pa (z2 ) ρc =− · Ua (z2 ) S(z2 ) p0 (1 −

p0 eikz2 + p1 e−ikz2 = ZL 1 ikz2 + p (1 + 1 )e−ikz2 1 ikz2 )e ikz2

(2.41)

are used. Again solving 2.41 for

Zin

p0 p1

and substituting into 2.40 yields:

    −ik(z1 −z2 ) (1 − 1 )Z + ρc ik(z1 −z2 ) (1 + 1 )Z − ρc + e e L L ikz2 ikz2 S(z2 ) S(z2 ) ρc      =− 1 S(z1 ) eik(z1 −z2 ) (1 + 1 )Z − ρc −ik(z1 −z2 ) (1 − 1 ) + ρc 1 − 1+ − e L ikz2 ikz1 ikz2 S(z2 ) S(z2 ) (2.42)

We now introduce 1 ±

1 ikzj

1 ikz1



in their polar form:

1 1 1± = ikzj i

  1 1 i± = q kzj i 1+

1 k2 zj2

e±iΘj ,

(2.43)

where Θj = arctan(kzj ). Accordingly: 1 q 1+

1 tan2 (Θj )

tan(Θj ) e±iΘj = p e±iΘj = sin(Θj )e±iΘj 1 + tan2 (Θj )

(2.44)

Substituting and using L = z2 − z1 yields

Zin

 i −ikL −Z iΘ2 − e L sin(Θ2 ) e ρc  =− i S(z1 ) i e−i(kL+Θ1 ) −Z iΘ L sin(Θ2 ) e 2 − sin(Θ1 )

  i −iΘ2 + ρc + eikL ZL sin(Θ e S(z2 ) 2)   ρc i i i(kL+Θ1 ) Z −iΘ2 + L sin(Θ2 ) e S(z2 ) − sin(Θ1 ) e (2.45) ρc S(z2 )



Now all terms with the same exponent are collected and Euler’s formula is used, yielding: Zin,con = −

ρc i ρc ZL sin(Θ2 ) sin(kL − Θ2 ) + S(x2 ) sin(kL) S(z1 ) ZL sin(kL+Θ1 −Θ2 ) − i ρc sin(kL+Θ1 ) sin(Θ1 ) sin(Θ2 )

S(z2 )

(2.46)

sin(Θ1 )

This formula describes the impedance at one end of a conical horn, while the other end

ρc S(z2 )

.

Simulation

14

is determined by ZL . It is, like the other formulas for the cylindrical and exponential tube, starting point for the simulation of the impedance (see chapter 3).

Chapter 3

Simulation The impedance at one end of a pipe of length L is determined by the pipe’s shape and length, but also by its impedance at the opposite end, ZL . The simulation of the input impedance at the throat of a horn will be done by separating the horn into satisfyingly short segments, so that the actual flare is described in good approximation by the model we apply on that segment. The previous segments input impedance is used as ZL . At the bell, were a small part of the vibrational energy is radiated to the surrounding air, the impedance takes a more complicated form. The impedance in general has the form Z = R + iX.

(3.1)

Here, R is the reflection coefficient and X is the transmission coefficient (determining, which amount of vibrational energy is transmitted through the border plane). In this case, the mouth of the horn is not ideally closed, where R = ∞ and X = 0, but it can be understood as a pipe being flanged to an infinite plane. The impedance then takes the form

  H1 (2ka) J1 (2ka) +i . Z = ρcS 1 − ka ka

(3.2)

Here H1 is the Struve Function of first order and J1 is the Bessel Function of first order. This expression will determine the impedance at the bell, which we assume to be a ’baffle in an infinite plane’. This model is accurate in our context, because the space in front of the bell is quite large compared to the bell diameter of 139 mm. The infinite plane represents nothing more than the boundary condition of the acoustic pressure being fixed to atmospheric pressure. From another point of view the volume flow is fixed to the atmospheric value, which is zero or at least very low in comparison to the acoustic volume flow, because of the enormous cross sectional jump from the bell to the outside area. This model is based on the assumption, that the pressure and volume flow are constant over the (bell’s) cross sectional area, which is a problematic simplification 15

Simulation

16

especially at the rapid flaring bell. Since this work is entirely based on ideas neglecting higher modes, they are not considered here either. One more interesting point is friction. Since the walls will never be ideally rigid and smooth, this effect is never to be eradicated. Because dealing with vibrating walls in the context of this work is not possible, there will be at least a simple attempt on dealing with friction. The frictional losses at the walls are described with a simple correction to the phase velocity and the wavenumber k, which is extended by an imaginary part. So for the case with friction:   1.65 · 10−3 √ vph = c 1 − fa √ 3 · 10−5 f α= a ω k = + iα c

(3.3)

(3.4) (3.5)

Here a is the tubes radius and f is the frequency. Please note that the information about friction and radiation which is given here is minimal. For further reading take a look at [1], from where the method and numerical values are taken. Now the input impedance for the horn is computed with two different models (conical, exponential), using the diameters from figure 3.1. For the corresponding measurement technique see section 4.1. Frictional losses as well as radiation effects are respected as described above.

Simulation

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20.45

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Figure 3.1: Diameter along the horn axis in mm. The horn is shown in section.

Simulation

3.1

18

Cylindrical Model

There are two boundary cases which can be considered to test, weather the simulation works correctly. The simulation for the cylindrical tube will return evenly spaced resonances, which are uneven multiples of the fundamental frequency f0 = c/ (4L), if the impedance at the bell is ZL = ∞. Figure 3.2 shows the expected behavior, reproducing the required boundary case. The second case is the exponential prediction for equal diameters. If the simulation with the exponential model is done for segments which all

Figure 3.2: Cylindrical model for infinite ZL . All resonances are uneven multiples of f0 = 50 Hz. c = 200 in this example for simple values.

Figure 3.3: 30 segments with same radius after exponential model and cylindrical tube of length equal to sum of segment’s lengths vs frequency. Exponential prediction was multiplied by 10 to optimize overview. Both predictions are equal over all frequencies.

Simulation

19

have the same diameter, the result must be the same as the one of one cylindrical segment with same radius and the length equal to the sum of the segment’s lengths. Figure 3.3 displays exactly this behavior, which allows to accept the simulation and compute the impedance with the measured diameters.

3.2

Conical Model

The impedance at the throat of a conical segment of length L is given by (see chapter 2.3.3): Zin,con

iZL sin(bL−Θ2 ) + Sρc2 sin bL ρc sin Θ2 · = S1 ZL sin(bL+Θ1 −Θ2 ) − i ρc sin(bL+Θ1 ) sin Θ1 sin Θ2 S2 sin Θ1

(3.6)

Where • Sj is the pipes cross section at zj • ZL is the input impedance at the mouth of the horn • Θj = arg( ωc + izj ) Because the horn function equals zero for the conical case, we obtain no cutoff frequency in this model. Below cutoff, the wavenumber k is imaginary, the wave does not propagate (see Webbsters equation, Chapter 2.3) and the impedance is zero. Since waves are propagating for each wavenumber k in a conical segment, the expression for Zf l (developed above) can be used without modification. Figure 3.4 now shows the absolute value of Zin,con at the throat of the horn plotted against frequency f on a log-scale. The resonances are marked with black dashed lines, the tolerances are quarter tones. The decrease in amplitude is caused by the real part of Zf l , which is anti proportional to ω and therefore decreases with growing frequency. Recall, that the resonances are uneven multiples of a ground frequency f0 = c/(4L) = 64 Hz for a cylindrical segment of length L = 1.34m (typical length of a trumpet) with ideally closed end. The first resonance of the conical model however is located at a much higher frequency. Whereas all higher resonances are approximately multiples of the second resonance frequency f2 divided by two (3/2f2 , 2f2 , 5/2f2 , 3f2 . . . ), the first resonance breaks this pattern. One would expect it to be at 1/2f2 , which equals one octave.1 If compared to the second resonance, it can be seen that the first resonance is not one octave, but approximately one octave plus one quart (17 half tones) lower than the second. This surprising fact needs careful comparison with actual measurements because this behavior is not expected. 1 One octave is divided into halftones of 100ct,where the tonal difference I in cent between two frequencies f1 and f2 is I = log2 (f1 /f2 ). So one octave (1200ct) equals a frequency ratio of 2/1.

Simulation

20

Figure 3.4: Absolute value of input impedance after conical model vs. frequency. Resonances are marked with black, dashed lines. Tolerances are quarter tones

3.3

Exponential Model

The impedance at the throat of a exponential segment of length L is given by: Zin,exp =

ρc ρc ZL cos(λL + Φ) + i S2 sin λL · S1 iZL sin λL + Sρc2 cos(λL − Φ)

(3.7)

• Sj is the pipes cross section at zj (where (0,1) is (0,L)) • λ=

√

k 2 − m2

• Φ = arg(b + im) • m is the flare constant given by

√

F (see Webbster equation, Chapter 2.3)

In the exponential model the horn function F is not zero. For an exponential flare, the √ solution to Webbster’s equation is proportional to eibz with b = k 2 − m2 . So a wave will not propagate unless ω/c > m, which means that the acoustical impedance Zin,exp = 0 for ω < mc. In the simulation, ZL = 0 if this condition is true. It is indeed true for a quite large frequency range if the difference between the radii of the considered segment is very large, so the flare is very rapid, which it is especially at the bell. This results in Zin,exp to be not continuous (note the edges). Figure 3.5 shows the absolute value of Zin,exp at the throat of the horn plotted against frequency f on a log-scale. The second resonance immediately catches attention. It is positioned very much out of the pattern. This mismatched position has to be a result of an error in the diameter measurement, because the influence of the segments length and radii on the resonances dominate all

Simulation

21

Figure 3.5: Absolute value of input impedance after exponential model vs. frequency. Resonances are marked with black, dashed lines. Tolerances are quarter tones

Figure 3.6: Horn is shown in section. The 24th radius of 16.65mm is quite large in comparison with its two neighbors.

other effects. Since the lengths are all measured with high precision (take a look at section 4.1), the diameters are more likely responsible for this mismatched position. By looking at the shape displayed against the horn axis in figure 3.6, the diameter most likely being measured wrong can be identified: the 24th diameter of 16.65 mm seems to be too large. Figure 3.7 now shows the exponential prediction with the old value compared to the exponential prediction done with the 24th diameter substituted with the mean of the two neighbors, which is (17.45 + 14.95) /2 mm = 16.2 mm. The simulation with the corrected value is much more consistent with the equally spaced pattern we obtained for the conical prediction, so the obviously wrong diameter is substituted by the new value

Measurement

22

Figure 3.7: Influence of 24th diameter. Exponential prediction with old and new value vs frequency. The new value almost only influences the second resonance, the two curves are nearly identical over the whole frequency range. Measured resonances are in black, dashed lines.

Figure 3.8: Simulation after exponential model with new value for d24 vs frequency. Resonances after this model are marked with black, dashed lines.

of 16.2 mm. The new simulation is shown in figure 3.8 and will be used for comparison with the measurement in chapter 5. As for the conical model the number of resonances is correct, whereas the positions are different especially at high frequencies (see ticks or comparison in Appendix A.4). The exponential model also predicts the first harmonic to be approximately 17 halftones below the second resonance2 . 2

One octave is divided into halftones of 100ct,where the tonal difference I in cent between two frequencies f1 and f2 is I = log2 (f1 /f2 ). So one octave (1200ct) equals a frequency ratio of 2/1.

Measurement

3.3.1

23

Influence of single Diameters on the Spectrum

Simulating the input impedance of a horn, allows to compute the natural resonances, but denies insight on the effect of one particular part of the instrument on this spectrum. It cannot be determined, how the variation of one radius procreates over the following steps of the simulation. To obtain insight on this topic, selected radii ai will be variated in steps of 0.5 percent of the actual value. The corresponding values for the resonances will then be displayed against the radius. From the figures it is clear, that the resonances are strongly influenced by the radii at the almost conical part of the instrument, whereas the bell has not a great influence on the resonances. Figure 3.9 shows, that the forth resonance, which should be highly stable because of its importance for the instrument (note in the mostly played frequency area), decreases in frequency when the 24th radius is variated. The horizontal scale is in steps of 0.005 · r24 . As can be seen in figure 3.1, that equals steps of 0.005 · 20.45 mm ≈ 0.1 mm. The frequency steps are in the order of 5 Hz for a change of about 15 percent, which would be 3 mm. In section 4.1 the measuring uncertainty for the diameters is approximated as 0.54 mm. If 3 mm equal an uncertainty of 5 Hz in the frequency range, then the measuring uncertainty of 0.54 mm equals an uncertainty of roughly 1Hz. This, on the one hand results in a very good spectral resolution, on the other hand the ideas presented here can be seen as not more than rough estimations. One more point can be made in this section. If the correlation of the resonance and one of the first diameters (see figure 3.10) is compared to the correlation shown in figure 3.9, it is clear that the shape of the bell has only little influence on the resonances. Hence measuring the shape at the bell with higher precision than the mostly conical part of the instrument is not recommended. Accordingly the characteristics mostly influencing the resonance spectrum are the total length of the instrument as well as the almost conical part from radius 20-25 (see figure3.1). This however contrasts the aspiration of obtaining an almost continuous impedance for the exponential model by choosing almost infinite segments, as explained above.

Measurement

Figure 3.9: Forth resonance vs 24th radius ± units of 0.5 percent. Left scale in Hz, right scale in ct. The correlation of radius increase/decrease and resonance position is one issue which could be covered in following works.

Figure 3.10: Forth resonance vs 10th radius ± units of 0.5 percent. Left scale in Hz, right scale in ct. The correlation of radius increase/decrease and resonance position is not as dramatic as in figure 3.9.

24

Chapter 4

Measurement 4.1

Diameter Measurement

Figure 4.1 displays the diameter of the instrument along the instruments axis. The measurement is done with a caliper. To carry out the measurements as precise as possible, the trumpet is lying on a sheet of paper. The positions for measurements are first drawn onto this sheet of paper and then transferred onto the trumpet. Figure 4.2 shows the setup for this technique in detail. The straight segments length is determined with very good precision (only the systematic uncertainty of the caliper used, see below), whereas the diameters are, especially at the bell, tainted with a great measuring uncertainty. This uncertainty is caused by the fact, that the caliper may not always be perfectly perpendicular to the pipe axis. In the curvature of the instrument the straight distance between measuring points were chosen as L because in our model we assume the segments to be straight. The uncertainty on the measured diameters must be combined from multiple uncertainties. Obviously, there is the raw measuring uncertainty of the caliper, which is approximately the smallest unit as an upper limit (0.1 mm). While this uncertainty is quite small, the dominating effect is the caliper not being perpendicular to the instruments axis. If the caliper is off center by the angle α, then the measured diameter m would differ from the real diameter r by d = m − r = m (1 − cos α) .

(4.1)

If the experimentalist is able to keep α under 5◦ , then the maximal measuring uncertainty on diameters is the one for measuring the bell of 139 mm: d = 139 · (1 − cos 5◦ ) mm ≈ 0.53 mm

25

(4.2)

Measurement

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Figure 4.1: Trumpet, diameter in mm measured with caliper. Lengths of the segments are also given in mm.

Measurement

27

Figure 4.2: Measurement of trumpet diameters on the sheet

Quadratic addition gives the total uncertainty σtotal : σtotal =

p 0.532 + 0.12 mm ≈ 0.54 mm.

(4.3)

So the measuring uncertainty for the diameters, which are at least about 12 mm, can be approximated as maximally 0.54 mm. This is an acceptable uncertainty for using these diameters.

4.2

Measuring Technique

Consider a cylindrical tube expanding from z = 0 to z = L. The impedance Zin at z = 0 is given by (for pa and ua have a look at section 2.3.1): Zin =

pa (0) A+B 1+R = ρc = ρc , ua (0) A−B 1−R

(4.4)

where R = B/A is the reflection coefficient. R can be obtained by measuring the pressure amplitude at different positions zj of the tube and computing the transfer

Measurement

28

functions, defined as H1,2 =

pa (z2 ) 1 + Re−2ikz2 ik(z2 −z1 ) ·e . = pa (z1 ) 1 − Rei2ikz1

(4.5)

Solving for R yields R=

H1,2 − eiks , e−iks − H1,2

(4.6)

where s = z2 − z1 is the distance between the two measuring positions. Zin accordingly is Zin


e−iks − H1,2 + H1,2 − eiks e2ikz1 . = −iks e − H1,2 − (H1,2 − eiks ) e2ikz1

(4.7)

This calculation is automatically done by the ITA1 software, the output used for this work already is the complex number Zin . The impedance at a point is anti proportional to the cross section at this particular point. So if the impedance is measured at the point z = 0 of the Kundts tube with radius b and the impedance of the horn of radius a at this point is of interest, the measured impedance has to be multiplied by the ratio r=

Sb Sa :

Zin,horn = r · Zin,measured

4.3

(4.8)

Measuring Setup

We use a Kundt’s tube for our measurements. At one end of this tube the source generates the signal (exponential sweep, sinus signal), the other end holds the test object. The tube with diameter 2 inches, which equals 5.08cm2 , is build from aluminium and has a satisfying wall thickness to prevent vibrations of the walls. It is closed, hence no vibrational energy is lost to the air surrounding the tube. The tube has four holes for microphones, which are closed when the microphone position is not used. For the measurements at the four different positions the same microphone is used, so there are no systematic errors caused by different microphone sensitivities or other inequalities between microphones. A large distance between the microphones is desirable, because the distance between two microphones must exceed 0.05λmax , where λmax is the maximal wavelength for which results are acceptable[2].Therefore a minimal frequency is obtained for each microphone pair, for which measurements are acceptable. The distances of the second, third and forth position are given in the table, with the estimated lower frequency limit for measurements. Each pair of microphone positions (1,2),(1,3),(1,4) allows measurements in different frequency ranges, based on their distance. With these 1

’Institut fuer technische Akustik’ at RWTH 1 inch equals 2.54 cm, value taken from http://www.din-formate.de/kalkulator-berechnung-laengemasse-groesse-einheiten-umrechnung-inch-zoll-in.html from 10.7.2015 2

Measurement

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Table 4.1: Microphone position distances and corresponding lower frequency limit (roughly).

Microphone pair (1,2) (1,3) (1,4)

Distance in mm 17 110 514.05

Lower frequency limit in Hz (roughly) 1009 156 33

frequency ranges combined, we are able to measure the impedance from approximately 40 Hz up to 9000 Hz. The software handling the raw data output of the microphones was developed at ITA3 . As can be seen in figure 4.3, the instrument is connected to the tube by one of two special flange (shown in figure 4.4). The flanges assure that no energy losses occur at the junction from tube to instrument. These flanges, one for measurements with mouthpiece and one for measurements without mouthpiece, were crafted in the workshop of the physics department. When measuring without mouthpiece, the throat must be extended by a distance d = 2 cm, otherwise the flange cannot be connected to the instrument. The measurement is done as follows: The source generates a sinus signal, in

Figure 4.3: Trumpet, flanged to the Kundt’s tube. The instrument is supported by a rock wool cuboid to prevent strains. The signal generator is on the right, red parts are plugs for the microphone position. Note that the tubes diameter is constant, regardless of the outer shape. 3

’ITA Toolbox’, providing a GUI for measurements and data handling with Matlab.

Comparison

30

Figure 4.4: Special flange used to connect trumpet and Kundts tube.

this case with a sampling rate of 44100, from 0−22050 Hz. The sweep consists of roughly 65000 entries, so the step length is approximately 0.3 Hz. First a test measurement is performed, where the nonlinear parts of the signal are controlled. If the amplitude of these nonlinear parts (mainly caused by the source) is too high, then the amplitude of the signal must be reduced. Generally, the intensity of the nonlinear parts should not be greater than one tenth of the linear signal. On the other hand, one must be able to distinct the signal from noise, which requires the amplitude of the signal to be about ten times greater than the noise. For all of the measurements done, the quality of the setup is controlled and rated as being acceptable. Now the sound pressure is measured at the different microphone positions while scanning trough the sweep interval each time. The software then computes the transfer functions H1,2 , H1,3 , H1,4 and returns the impedance.

4.4

Results

This section shows the results of the measurements. The focus is set on the absolute value of the impedance, so in the graphics |Zin,measured | is shown. These measured

Comparison

31

Figure 4.5: |Zin,measured | of the trumpet without mouthpiece vs. frequency. Resonances are marked with black, dashed lines.

Figure 4.6: |Zin,measured | of the trumpet without mouthpiece vs. frequency, second measurement. Resonances are marked with black, dashed lines.

impedances will be compared to the simulation in the following chapter. Figures 4.5 and 4.6 show the absolute value of Zin,measured for a trumpet of length L = 1.34 m without mouthpiece. Mark the pattern below the first resonance and the double peak at the second resonance, which are two obvious differences to the simulation. The pattern may be due to the fact that the values are taken very near or even below the measuring border at approximately 40 Hz. The double peak however is not explained so easily, because it is very surprising to have a resonance so consequently out of any pattern. The amplitude of the impedance decreases towards higher frequencies. The most evident characteristic is the amplitude of the forth resonance. It is approximately

Comparison

32

one order greater than all other resonances and quite sharp. This resonance turns out to be the keynote in the middle playing octave and it is very much useful, that this resonance is the clearest one (this note may be the most important note on a brass instrument).

Chapter 5

Comparison of Simulation and Measurement Now the measured impedance is compared with the results of our simulation. The figures 5.1 and 5.2 comparing measurement and simulation show, that the impedances shape is not exact, but the resonances are within a half tone range of the prediction. It is especially evident, that the measurement confirms the position of the first resonance, which is not, as already mentioned in chapter 3, at the half of the second resonance frequency, but nearer to a frequency which is approximately 17 halftones lower than the second resonance. However these values have the greatest difference from measurement to model, the measured resonance being at an even smaller frequency than the model’s prediction. For higher frequencies, measurement as well as prediction decrease in absolute value and the simulation tends to predict the resonances at higher frequencies as well. Also mark, that the measurement confirms the second resonance of the exponential model Table 5.1: Positions of the resonances in Hz, conical as well as exponential prediction and measurements

Resonance 1 2 3 4 5 6 7 8 9 10

Prediction in Hz Conical Exponential 82.44 94.06 226.48 233.32 345.21 352.21 473.17 482.24 598.26 604.63 727.66 735.93 856.25 864.69 986.44 994.88 1115.04 1127.61 1237.11 1245.86 33

Measurements in Hz 1st 2nd 75.37 76.37 243.25 241.91 358.66 354.96 483.48 484.15 605.61 605.28 729.43 728.75 856.60 854.59 977.05 974.70 1099.52 1097.50 1224.68 1220.98

Comparison

34

after correcting the 24th radius. Table 5.1 shows the measured and predicted values for the resonances, contrasting all values. Comparing the measurements with both models, it can be said that both models are equally successful (within the aspirations of this work). Both conical and the exponential model are close to the actual measurement. They also have weaknesses though: the exponential model brings discontinuities with it on the one hand, this is due to the cutoff frequencies for the short segments being not small enough to fall out of the frequency range of interest. On the other hand, the exponential model is more sensitive to variations of the radii along the instruments axis (see appendix figure A.4). It can be seen in this figure, that the exponential model reacts more extremely to changes of the shape than the conical model. In contrast with the exponential model, the conical one has no cutoff and therefore is continuous over all frequencies. It also causes fewer problems with the simulation, but is not as sensitive to changes of radius as the exponential model. The differences between simulation and measurement are acceptable within the frame of this work. Considering additional effects like vibration of the walls and curvature of the wavefronts as well as providing better understanding of the effects considered will much likely improve the result. The understanding of friction and radiation is quite poor in this model used and improving this understanding would surely contribute to further insights.

Conclusion

Figure 5.1: |Zin,measured | of trumpet without mouthpiece and conical prediction vs. frequency. Resonances are marked with black, dashed lines

Figure 5.2: |Zin,measured | of trumpet without mouthpiece and exponential prediction vs. frequency. Resonances are marked with black, dashed lines

35

Chapter 6

Conclusion The goal of this work is to obtain a prediction of the natural resonances of a trumpet, which is within a half tone range of the actual resonance. These actual resonances are extracted from the measured acoustical impedance. Chapter 5, presenting the comparison of prediction and measurement, shows that this aspiration is met. Especially the position of the first resonance is quite dissimilar in the prediction than it was measured though, but definitely not at the half frequency of the second resonance. The shape of |Zmeasured | is in some regions not reproduced with great success also. However, the quite simple model already allows predictions of the resonances with, for the scope of this work, satisfying accuracy. It has been made clear that the shape of the instruments bell (see section 3.3.1) has no great influence on the resonances, whereas the mostly conical section after the bell is of great importance for the spectrum. Measuring the diameters in this region with higher precision would certainly improve the simulations results, but a higher precision is not realistic with the setup chosen for this work (consider measuring uncertainties presented in section 4.1). Using an optical measurement technique, especially to minimize the experimentalists influence on the measured values, would be necessary here. The technique for the impedance measurements however needs no immediate improvement, the results are satisfyingly exact. The only weakness of this setup is that the lower frequency limit for acceptable measurements is in the same order as our first resonance. An additional microphone with greater distance to the first microphone would solve this problem. Overall the aspirations of this work are met and the simple, one dimensional model for sound propagation in tubes turns out to be very successful in predicting the instruments resonances.

37

Appendix A

Appendix

39

Conclusion

Figure A.1: Influence of friction, exponential model with and without friction vs frequency. Influence increases towards greater frequencies but has almost no influence on the resonances positions. The influence on the amplitude however is obvious. Resonances after exponential model with friction are in black, dashed lines.

Figure A.2: Influence of friction, conical model with and without friction vs frequency. Influence increases towards greater frequencies but has almost no influence on the resonances positions. The influence on the amplitude however is obvious, the frictional losses result in smaller amplitude. Resonances after conical model with friction are in black, dashed lines.

40

Conclusion

Figure A.3: Comparison of conical and exponential model. Both models give the same number of resonances, but differ in position and shape. Especially the edges in the exponential prediciton are an obvious difference to the conical prediction.

Figure A.4: Influence of 24th diameter. Exponential prediction with old and new value vs frequency. The new value almost only influences the second resonance, the two curves are nearly identical over the whole frequency range. Measured resonances are in black, dashed lines.

41

Bibliography [1] Neville H. Fletcher and Thomas D. Rossing. The Physics of Musical Instruments. Springer. [2] Komitee ISO/TC 43/SC 2 Bauakustik. EN ISO 10534-2. Springer.

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