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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
I General properties and examples
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
I General properties and examples
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
I General properties and examples
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
I General properties and examples
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
PP
QP
8
I General properties and examples
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
I General properties and examples
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
I General properties and examples
• 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
I General properties and examples
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
I General properties and examples
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
I General properties and examples
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
I General properties and examples
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
I General properties and examples
Time domain illustrations of harmonicity
• 3 examples
– Inharmonic stepped waveguide
– Harmonic stepped waveguide
– Stepped cone
16
I General properties and examples
An impulse propagating in a non harmonic discrete resonator (7 cylinders, closed-open)
17
I General properties and examples
An impulse propagating in a harmonic discrete resonator (6 cylinders, closed-open)
18
I General properties and examples
An impulse propagating ina stepped cone (7 cylinders, closed-open)
19
I General properties and examples
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
II Stepped cones, conical reed instruments and analogies
• 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
II Stepped cones, conical reed instruments and analogies
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).
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
Ll
L
l
cone vs bowed string
Llℓ L
≈⇔
ℓ
ℓ
L
Llℓ L
L
25
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
• 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 open episode (in hPa)
Pressure during the closed episode (in hPa)
HM
Inverted HM
Mouthpressure (in hPa)
When does inverted Helmholtz motion occur?
Stepped cone N=2
32
II Stepped cones, conical reed instruments and analogies
Possible scenario?
• When the mouth pressure is increased the amplitude of the HM reaches a maximum and oscillation bifurcates to the inverted HM
Pressure during the open episode (in hPa)
Pressure during the closed episode (in hPa)
HM
Almost inverted HM
Mouthpressure (in hPa)
HMAlmostinvertedHM
Transition
When does inverted Helmholtz motion occur?Experiments on a stepped cone N=2
33
II Stepped cones, conical reed instruments and analogies
What happens for the real cone?
• Raman model no more applicable
• Bifurcations to the inverted HM can however be observed
Pressure during the open episode (in hPa)
Pressure during the closed episode (in hPa)
Mouthpressure (in hPa)
HMInvertedHM
Almost square34
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
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
II Stepped cones, conical reed instruments and analogies
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
III Networks of tubes of equal length and diameter
• 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
III Networks of tubes of equal length and diameter
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
III Networks of tubes of equal length and diameter
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
III Networks of tubes of equal length and diameter
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
III Networks of tubes of equal length and diameter
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
III Networks of tubes of equal length and diameter
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
III Networks of tubes of equal length and diameter
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
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