28 January 1978

Dear Mr. Manger,

In order to soothe my bad conscience a little, I examined your loudspeakerstheoretically last weekend. I was surprised to find that the radiation principlechosen by you gives a radiation time history that corresponds to the timehistory of the current (at least in the idealisation I examined); i.e. transientoscillations and other disturbing effects do not occur. If I interpret theequations correctly, it seems feasible that a plate (membrane) with a stiffnessdecreasing towards the edge (thickness variation), leads to a slightlyimproved radiation efficiency.

With many thanks for your Christmas surprise,Yours,(M. Heckl)

Radiation from a very pliable large plate with circular excitation

1. The problemWe consider a very large plate which is excited along a ring. The excitationforce is uniform along the ring, so that the problem can be regarded asradially symmetric.

The aim is to determine the sound radiation of such a configuration, with thetime history of the sound pressure being of particular interest.

2. Mathematical modelIn order to keep the mathematical effort low, the following assumptions aremade in the calculations:(a) The plate is large enough and strongly enough damped that the bendingwaves excited by the ring do not get reflected by any edge that may bepresent. (Because of the high damping of the material, this condition shouldbe satisfied at least above a few hundred Hertz.)(b) The motion of the plate can be described by the bending wave equation;i.e. membrane tensions (which would lead to nonlinearities) must be absent.(In the present configuration, this condition is probably satisfied.)(c) The bending wavelength of the plate should be at least 25 kHz below thesound wavelength in air. (Since the material is very pliable, one can assumethat the bending wavelength is below the sound wavelength by as much asabout 50 kHz.)(d) In the frequency range of interest, i.e. above about 300 Hz, the radiationloading by the surrounding air can be neglected. (Since the weight of theoscillating foil is high compared with that of loudspeaker membranes, thiscondition, too, should be satisfied.)

3. Calculation of the sound pressure3.1. Basic equationsThe problem is solved if solutions can be found to the sound wave equationand the bending wave equation with ring-shaped excitation, under theassumptions given above. It is known from literature [e.g. Heckl, Acustica 9(1959), p. 371] that the sound pressure radiated by a very large plane plate isgiven for the radially symmetric case by the formula

p r z t e p k J k r e k dk dj tr r

j k k zr r

r( , , ) ( , ) ( ) .= − −∞

−∞

∞

∫∫12 0

0

02 2

πω ωω (1)

2

r

z2

F

The following notation has been used:p r z t( , , ) = sound pressure at position r z, at time tr = perpendicular distance from the symmetry axis which passes

through the centre of the ringz = distance from the platet = timeω = integration variable (angular frequency)J0 (..) = Bessel function of order zero

j = − 1kr = integration variable (wave number)

k0 = ωc0

c0 = propagation speed of sound waves in airp(..) = wave number spectrum of the pressure

If the plate oscillates with the radially symmetric velocity distribution v r t( , ) ,then

p k v kk k

r ro

r

( , ) ( , ) ,ω ωρ ω

=−0

2 2(2)

with

v k v r t J k r e rdrdtr rj t( , ) ( , ) ( ) .ω ω= −

∞

−∞

∞

∫∫ 00

(3)

ρ0 is the density of the air. If the plate is excited by a radial pressuredistribution of the form p r tA ( , ) , then

p kj p k

B k k k kr

A r

r B r

( , )( , )

( ),ω

ω ωπ

ωρ=

− −2 4 40

02 2

(4)

with

p k p r t J k r e rdrdtA r A rj t( , ) ( , ) ( ) .ω ω= −

∞

−∞

∞

∫∫ 00

(5)

The following notation has been used:B = bending stiffness of the platem' ' = mass per unit area of the plate

kB = ω 24 m B' '/ = wave number of free bending waves

In the present case of a ring-shaped excitation, pA is given by

p r tF t

ar aA ( , )

( )( ),= −0

2πδ (6)

hence

p kJ k a

F t e dtJ k a

FA rr j t r( , )

( )( )

( )( ).ω

π πωω= =−

−∞

∞

∫00

002 2

(7)

If equation (7) is inserted into (4) and the resulting equation inserted into (1),one obtains

p r z tj

B

J k a F

k k k kJ k r e e k dk dr

r B r

rj k k z j t

r rr( , , )

( ) ( )

( )( ) .=

− −

∞

−∞

∞− −∫∫

ρπ

ω ωωω0

2

20 0

4 402 2 0

0402 2

(8)

Because of the assumption (b), we always have k kr B<< in the far field, i.e.for k kr0 > ; therefore

p r z tj

m

F

k kJ k a J k r e e k dk d

r

r rj k k z j t

r rr( , , )

' '( )

( ) ( ) .=−

− −∞

−∞

∞

∫∫ρ

πω

ωω02

0

02 2 0 0

0402 2

(9)

3.2. Evaluation of equation (9)The integral over kr can be obtained from a comparison with the calculationof the radiation from a small piston membrane, and one can show that

J k a J k r

k ke k dk

j

r zer r

r

j k k zr r

jk r zr0 0

02 2 2 2

0

02 2

02 2

2

( ) ( ).

−≈

−+

− − − +∞

∫ (10)

Strictly speaking, this formula applies only if a is less than half a soundwavelength; along the symmetry axis, equation (10) also holds for largervalues of a . Substitution of (10) into (9) gives

p r z tm r z

F e e dm r z

F tr z

cj t

jc

r z

( , , )' '

( )' '

( )=+

=+

−+− +

−∞

∞

∫ρ

πω ω

ρπ

ωω

0

2 2 2 00

2 2 0

2 2

08 40

2 2

(11)

This surprisingly simple result says that the time history of the sound pressurenear the symmetry axis corresponds exactly to the time history of the forceacting on the plate and hence to the time history of the current in the movingcoil. The time-lag due to sound wave propagation in the air is expressed by

the argument tr z

c−

+2 2

0

.

4. ConclusionIt can be shown theoretically, with the above simplifying assumptions, that thesound pressure radiated by a large, very pliable, damped plate along thesymmetry axis, has a time history that corresponds to that of the excitationforce in the moving coil (which has to be very light) and hence to the timehistory of the coil current. Therefore, time histories of currents with suddenchanges (square wave) are reproduced correctly in the radiated soundpressure.

The time history of the sound pressure at positions off the symmetry axis canbe obtained from the above equations only after lengthy numericalevaluations. Presumably, the gradual decay of the function J k ar0 ( ) forincreasing arguments causes a "rounding-off of the corners in the timehistory" for small distances from the symmetry axis, and a completelydifferent time history for points further away.