harmonic stepped waveguides and their application to music ...j.-p. dalmontet j....

Harmonic stepped waveguides and their application to music…

…and more

Jean-Pierre DalmontLaboratoire d’Acoustique de l’Université du Maine

UMR CNRS 6613

Le Mans, France

Harmonic stepped waveguides

I. General properties and examples

A bit theoretical

II. Stepped cones, conical reed instruments and analogies

More applied

III. Networks of tubes of equal length and diameter

More fun!

2

What is it about?

• Cylinders of same length and various cross sections aj connected together having harmonic eigenfrequencies

• Closed input => reed instruments

• Open input => flutes

• NB : lengths have to be equal to ensure harmonicity of all eigenfrequencies

…

open or

closedopen …a1 a2 aj aN-1 aN

3

I General properties and examples

Why is it interesting?

• For wind instruments harmonicity of eigenfrequencies is the guarantee of an easy playability

• Searching for discrete resonators may give insights for continuous resonators design (especially conical reed instruments)

4


Stepped wave guides as a discretizationof continuous horns

Examples• Stepped cone:

• Exponential horn:

�� = �(� + 1)��

��

= cte

sn-1 sn sn+1 sn+2

an-1 an an+1

�� = ��

Continuous cross section

Discrete cross section

a0 = 0 a1 =2s1 a2 = 6s1 = 3a1 …

s0 = 0 s1 2²s1 3²s1

5


Transfer matrix of discrete waveguides(N cylinders)

ia )cos(kLx )tan(kLjt with the ith cross section, and

.

1/j

/j1

)cos(/)sin(j

/)sin(j)cos(

11 cta

actx

kLcakL

akLckL

DC

BA

i

iN

N

i

iN

NN

NN

]2/[

1

2

...1

]2/)1[(

1

12

...1

]2/)1[(

1

12

...1

]2/[

1

2

...1

11 1

1

111 1

1

111 1

1

11 1

1

...

...1/

...

...

...

...

...

...1

N

i

i

Njiji jj

iiN

i

i

Njiij ii

jj

N

i

i

Njiij jj

iiN

i

i

Njiji ii

jj

N

taa

aact

aa

aa

ctaa

aat

aa

aa

xDC

BA

ii i

i

ii i

i

ii i

i

ii i

i

It can be shown that:

J.-P. Dalmont et G Le Vey "Discrete acoustical resonators with harmonic eigenfrequencies", Acta Acustica united with Acustica, 103 (1), 94-105 (2017).

6

…a1 a2 aN-1 aN


Transfer matrices of discrete waveguides(examples for 1 to 4 cylinders)

2

3

2

3

1

2

12

2

31321

2

31

2

321

2

2

3

1

3

1

2

3

33

33

1/

1111

ta

a

a

a

a

act

a

aaaaat

ctaa

a

aaatt

a

a

a

a

a

a

xDC

BA

4

42

312

4

3

4

2

4

1

3

2

3

1

2

12

3

42

3

41

2

41

2

314321

2

42

3

41

3

41

2

31

2

4321

4

31

422

3

4

2

4

1

4

2

3

1

3

1

2

4

44

44

1/

11111

taa

aat

a

a

a

a

a

a

a

a

a

a

a

act

a

aa

a

aa

a

aa

a

aaaaaat

ctaa

a

aa

a

aa

a

aa

a

aaaatt

aa

aat

a

a

a

a

a

a

a

a

a

a

a

a

xDC

BA

1/

1

1

111

11

cta

ca

t

xDC

BA

2

2

121

21

2

1

2

2

22

22

1/

111

ta

acaat

caa

tta

a

xDC

BA

ia )cos(kLx )tan(kLjt with the ith cross section, and

7


How to find the eigenfrequencies?(lossless case, boundary conditions p = 0 or v = 0)

The transfer matrix (Pressure – Volume velocity) being given

finding the eigenfrequencies leads to solve:

A=0 if open-closed, considering:

B=0 if open-open, considering:

C=0 if closed-closed, considering:

D=0 if closed-open, considering:

or else, considering Z=B/D, eigenfrequencies are given by:

Z = 0 if open-openZ = ∞ if closed-open

NB : dispersion and radiation not considered

DC

BA

s

s

e

e

Q

P

DC

BA

Q

P

0 ;0

0

0

0 ;0

se

se

se

se

PQ

QQ

PP

QP

8


General properties of the input impedance(lossless case, boundary conditions p = 0 or v = 0)

• Spectrum is periodic: period is • is always a minimum of Z (eigenfrequency of open-open guide)• It is symmetrical with respect to the Shannon frequency• is a maximum of Z if N is odd and a minimum of Z if N is even• If the [N / 2] natural frequencies below Shannon are fixed arbitrarily all the other

modes arise by symmetry and periodicity • Hence, there is an infinity of possible geometries for the same natural frequencies

(from N = 3).• In order to have a unique solution, it is necessary to set the amplitudes

2/c

4/c

…a1 a2 aN-1 aN

0 10 20 30

f/f1

-40

-20

0

20

40

20lo

g1

0(|

Z/Z

c1|)

2/c

4/cN=5

2/c

4/c

9


Duality

• The dual of a resonator is that which reduced admittance is equal to the reduced

impedance of the resonator: ��

��=

�

��

• For a stepped wave guide this leads to ��

��=

��

��

• The open-open eigenfrequencies of a guide are that of his dual closed-closed and conversely

• The closed-open eigenfrequencies of a guide are that of his dual open-closed and conversely

• So a closed-open guide and its dual reversed have the same eigenfrequencies

• An exponential horn is its own dual reversed

a1 a2 aj aN-1 aNa1

��

��

��

��

��

��aN

Dual reversed

10

Same input/output ratio

Same eigenfrequencies


• Eigenfrequencies being given for given boundary conditions, what are the possible geometries?

• Harmonic N-natural series: �� = �� but i non multiple of N+1– 1-natural series: 1 3 5 7 9 11…

– 2-natural series: 1 2 4 5 7 8 10 11…

– 3-natural series: 1 2 3 5 6 7 9 10 11…

– …

– ∞-natural series: 1 2 3 4 5 6 7 8 9 10 11… (only for open-open case)

• Solve� = 0 or � = 0 for given �� ⟹ �� =?

11

Finding harmonic resonator


Closed-open => for kℓ=np/4 and n € (1 2 3 5 6 7 …)

This implies: =>

1 2 3 4 5 61.55

6

7.5

6.66

a2/a

1

a3/a

1

[1 2 6]Dual of the stepped cone

[3 12 20]

[2 3 15]

[3 5 20]

[2 10 15]

0)(tan1)(cos 2

3

2

3

1

2

133

k

a

a

a

a

a

akD

[1 3 6]Stepped cone

[1 ∞ ∞ ][1 1 ∞ ]

13

2

3

1

2

1

a

a

a

a

a

a

)1/(

)1/(//

12

121213

aa

aaaaaa

[1 1+√2 (1+√2)²]Stepped exponential

Finding harmonic resonatorexample : stepped saxophone made with 3 cylinders

12


0 1 2 3 4 5 6

-80

-60

-40

-20

0

20

40

60

80

100

120

f/f1

20lo

g1

0(|

Z/Z

c1|)

[1 2 6] dual of the stepped cone

[1 3 6] stepped cone

[1 1.1 23.1]

[1 20 22.1]

Tends to the cylinderof length 2ℓ

1 2 3 4 5 4

c

Finding harmonic resonatorexample : stepped saxophone made with 3 cylinders

Tends to the cylinder of length ℓ

13


0.5 1 2 30.8

1

1.2

1.5

a2/a

1

a3/a

1

Closed-open => for kℓ=np/6 and n € (1 3 5 7 …)

This implies: =>

Finding harmonic resonator example : stepped clarinet made with 3 cylinders

[5 10 6]

[2 1 3]

[5 3 6]

[2 6 3]

0)(tan1)(cos 2

3

2

3

1

2

133

k

a

a

a

a

a

akD

33

2

3

1

2

1

a

a

a

a

a

a

)1/3(

)1/(//

12

121213

aa

aaaaaa

[1 1 1]Cylindre of length 3ℓ

14


Open-open => for kℓ=np/3 and n € N

This implies =>

[3 6 10]

[3 1]

[1 3 6]Stepped cone

[1 1 1]Cylindre of length 3ℓ

Fermé-ouvert

Ouvert-ouvert

[10 15 21]

1/3

0)(tan111

)tan()(cos 2

31

2

321

33

k

aa

a

aaakkcjB

Finding harmonic resonator example : stepped flute made with 3 cylinders

Tru

nca

ted

step

ped

con

es

31

2

321

3111

aa

a

aaa

)1/(

)1/3(//

12

121213

aa

aaaaaa

Saxophone curve

0 1 2 3 4 5 6

a2/a

1

0

1

2

3

4

5

6

7

8

9

10

a3/a

1

Clarinet curve

15


Time domain illustrations of harmonicity

• 3 examples

– Inharmonic stepped waveguide

– Harmonic stepped waveguide

– Stepped cone

16


An impulse propagating in a non harmonic discrete resonator (7 cylinders, closed-open)

17


An impulse propagating in a harmonic discrete resonator (6 cylinders, closed-open)

18


An impulse propagating ina stepped cone (7 cylinders, closed-open)

19


Stepped cones

• Closed-open: with n non multiple of N+1

a0 = 0 a1 3a1 6a1 10a1 ………. N(N+1)a1/2

�� = � + 1 ℓ

� = �ℓ

� =2��

� + 1 ��sin �� = 0 ⇒ �� =

��

2�

� =1

� + 1

sin ��′

sin(�ℓ)= 0 ⇒ �� =

��

2�′

J.-P. Dalmont et J. Kergomard "Lattices of sound tubes with harmonically related eigenfrequencies", Acta Acustica, 2, 421-430 (1994).

• Open-open:

Analogous to a truncated cone of length L, input section a1/2, output section (N+1)²a1/2

Analogous to a complete cone of length L’, input section 0, output section (N+1)²a1/2

20

II Stepped cones, conical reed instruments and analogies

Stepped cone as a saxophone

0 200 400 600 800 1000 12000

5

10

15

20

25

30

35

40

45

50

Distance to the apex (mm)

dia

mete

r (m

m)

Stepped cone

Bamboo Sax

Ángel Sampedro del Río http://www.unmundodebambu.com.ar

21


• When drilling side holes the harmonicity is theoretically no more ensured

• In practice what happens if the last cylinder is shortened?

• Harmonicity remains very good even when the last cylinder is shortened

0 1 2 3 4 5 6

f/f1

-40

-20

0

20

40

20

log

10(|

Z/Z

c1|)

Stepped saxophone with side holes(dedicated to Ph. Guillemain)

Inharmonicity: <10cents <25cents

�� =�/2

5ℓ�� =

�/2

4ℓ

22


kLkk

acjZ e

ecot3//1

/

3/

1)3/)(1(

1

3/)(1

11

3/)(

1

tan

1cot 2

23

k

kk

kkkkkkk

Truncated cone + mouthpiece

• Closed truncated cone has non harmonic eigenfrequencies• Adding a volume equal to that of the truncation make it closer to the stepped cone• The results can be improved by increasing the taper at the input of the cone (which is done on oboes)

Ll

Truncation

L

l

kLk

acjZe

cotcot

/2 1

Stepped cone vs conical saxophone

Mouthpiece

ℓ L

� = ℓ��/3

ℓL

��

23J.-P. Dalmont, B. Gazengel, J. Gilbert et J. Kergomard "Some aspects of tuning and clean intonation in woodwinds", Applied Acoustics 46, 19-60 (1995).


Patented and realized by Yamaha

kLk

acjZe

cotcot

/2 1

Ll

kLj

a

ckj

a

ctan

2//tan

2

11

Stepped cone vs cylindrical saxophone

L

lℓL

��

ℓ L

2��

24


Ll

L

l

cone vs bowed string

Llℓ L

≈⇔

ℓ

ℓ

L

Llℓ L

L

25


violin – saxophone analogy

Violin Saxophone

String Air column

Bow velocity Mouth pressure

String velocity under the bow Pressure in the mouthpiece

Bow force Input volume velocity

26


Reed instrumentsEpisodes of opening and closing

String velocity under the bow

(from Schelleng)

Sticking

Slipping

Pressure in the mouthpiece

Closing

Opening

Helmholtz motion in bowed strings and conical reed wind instruments

Bowed stringEpisodes of sticking and slipping

27


Influence of the length on Helmholtz motion

t

p(t)

P

-NP

Open

closed

mouth

mouth

N

2L/c 2l/c

L l

t

p(t)

P

-NP

Open

closed

mouth

mouth

N

2L/c 2l/c

L l

Opening time depend on the note, not closing time

28


0 0.002 0.004 0.006 0.008 0.01 0.012 0.014 0.016 0.018-2

-1.5

-1

-0.5

0

0.5

1

Baritone saxophone: first register

Time (s)

Mo

uth

pie

ce p

ress

ure

(a

rbit

rary

sc

ale

)

Invariable

Opening time depend on the note, not closing time

Influence of the length on Helmholtz motion

29


Inverted Helmholtz motion

Inverted Helmholtz motion

p(t)

t

p

-p/N

Ouvert

Fermé

Standard Helmholtz motion

t

p(t)

p

p-N

Ouvert

Fermé Less power fullNever observed on strings

Open N time stepClosed 1 time step

Stepped cone N=2

Closed N time stepOpen 1 time step

30


When does inverted Helmholtz motion occur?

• On the basis of Raman model and for a given non linear characteristics (the function which describes the embouchure), the amplitude of the oscillations can be calculated as a function of the mouth pressure

0 10 20 30 40 50 60 70 80 90 100-100

-80

-60

-40

-20

0

20

40

60

80

100

Mouth pressure (in hPa)

Pressure during the open episode (in hPa)

Pressure during the closing episode (in hPa)

Two solutions at the fundamental frequency:the HM andthe inverted HM

HM

Inverted HM

Stepped cone N=2

31


• When mouth pressure is increased the amplitude of the HM reaches a maximum and oscillation bifurcates to the inverted HM

0 10 20 30 40 50 60 70 80 90 100-100

-80

-60

-40

-20

0

20

40

60

80

100

pm

(hPa

p3 (

hP

a)

p1 (

hP

a)


Pressure during the closed episode (in hPa)

HM

Inverted HM

Mouthpressure (in hPa)

When does inverted Helmholtz motion occur?

Stepped cone N=2

32


Possible scenario?

• When the mouth pressure is increased the amplitude of the HM reaches a maximum and oscillation bifurcates to the inverted HM



HM

Almost inverted HM


HMAlmostinvertedHM

Transition

When does inverted Helmholtz motion occur?Experiments on a stepped cone N=2

33


What happens for the real cone?

• Raman model no more applicable

• Bifurcations to the inverted HM can however be observed




HMInvertedHM

Almost square34


Experiments made with the body of a

soprano saxophone

Anomalous regime: double Helmholtz motionSingle Helmholtz motion► 1 slipping episode per period

Double Helmholtz motion► 2 slipping episode per period

35


Soft and Hard D on the Irish Uilleann pipehttp://blog.robertrueger.de/?p=157

Soft D Hard D

Ou

tpu

tP

ress

ure

Re

ed

dis

pla

cem

en

t

Soft D corresponds toStandard Helmholtz motion

hard D corresponds todouble Helmholtz motion

Ou

tpu

tP

ress

ure

Re

ed

dis

pla

cem

en

t

LASERREEDChanter (conical pipe)

Same note, same frequency, same fingering

36


J.-P. Dalmont and G. Le Vey “The irish Uillean pipe: a story of lore, hell and hard D”, International Symposium of Musical Acoustics, Le Mans (July 2014).

From stepped guides to networks

a1 a2 = 3a1

a1 a2 = 3a1 a3 = 6a1

If an=m an-1 an equivalent network can be build

37

III Networks of tubes of equal length and diameter

From stepped guides to networks

No equivalent network

1 1 3 15

1 3 6 10

10

6∉ ℕ

1 3 3 9 18

38


• Replication: when a waveguide is harmonic for both boundary conditions open-open and closed-open it can be mirrored into an new harmonic resonator for open-open conditions

Replication of harmonic resonators

Symmetrical modes

Antisymmetrical modes

39


Variations on a theme [1 3]

• This is the first stepped cone

• Open-open Closed-open n non multiple of 3

• Closed-open n= 1; 5; 7; 11 Closed-closed

• It can be mirrored

1 3

�� =��

4ℓ�� =

��

6ℓ

�� =��

12ℓ�� =

��

4ℓ

⇔

or

or3 1 1 3

3 1 2 6

40


Variations on a theme [1 3]

• This is the first stepped cone

• Open-open Closed-open n non multiple of 3

• Closed-open n= 1; 5; 7; 11 Closed-closed

• It can be mirrored

1 3

�� =��

4ℓ�� =

��

6ℓ

�� =��

12ℓ�� =

��

4ℓ

⇔

or

or3 1 1 3

3 1 2 6

41


Variations on a theme [1 3]• [3 1 2 6] has harmonic resonances for both open-open and closed open

conditions20lo

g1

0(|

Z/Z

c1|)

closed

open

• [3 1 2 6] can be mirrored into [6 2 1 3 3 1 2 6 ]

42


Variations on a theme [1 3]• [6 2 1 3 3 1 2 6 ] has harmonic resonances for both open-open and closed open

conditions

closed

open

• [6 2 1 3 3 1 2 6 ] can be mirrored into [6 2 1 3 3 1 2 6 6 2 1 3 3 1 2 6 ]

43


Variations on a theme [1 3]• [6 2 1 3 3 1 2 6 6 2 1 3 3 1 2 6] has harmonic resonances only for open-open

conditions

Closed-open: non harmonic eigenfrequencies

• [6 2 1 3 3 1 2 6 6 2 1 3 3 1 2 6 ] can not be mirrored

open-open: harmonic eigenfrequencies

46 cylindres44


Thanks to all scientific contributors

• Jean Kergomard• Joël Gilbert• Sébastien Ollivier• Bruno Gazengel• Georges Le Vey• Philippa Dupire• Cyril Frappé• Frédéric Ablitzer• Véronique Dalmont

45

harmonic stepped waveguides and their application to music ...j.-p. dalmontet j....

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