Learners might have learned the basic concepts of the acoustics from the ‘Introduction to Acoustics (Part 1).’ Now it is time to apply to the real situation and develop their own acoustical application. Learners will analyze the radiation, scattering, and diffraction phenomenon with the Kirchhoff –Helmholtz Equation. Then learners will design their own reverberation room or ducts that fulfill the condition they have set up.
Lecture 2-1 Part 3. Kirchhoff-Helmholtz Equation 2
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Из курса от партнера Korea Advanced Institute of Science and Technology
Introduction to Acoustics (Part 2)
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K-H Equation & Baffled Piston Problem
First, you will learn Kirchhoff-Helmholtz equation, then you will adapt into the baffled circular piston problem.
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Yang-Hann KimProfessor
Mechanical Engineering
First,
[SOUND] G.
[SOUND] What is G.
Green's function.
G(r|r0).
[SOUND] Simply, sound
due to the delta function.
Excitation.
Its like a like, [SOUND] this point.
This is r0,
and any point over there would be r.
[SOUND] If this
satisfy, [SOUND]
we can use it.
So we can use monopole, like G.
[SOUND] Select, for example,
G is equal to 1 over R e to the jkR
where R vector is r minus r0.
[SOUND] It's completely okay.
[BLANK_AUDIO]
You can use, for example, in this room
you can use impulse response function at any point.
So you can measure it.
[SOUND]
Okay?
So this G can be
anything.
Very strange.
Very
strange.
But you have to see this P and this P and this P,
are all
the same function.
Therefore, this is integral equation because P, P, P.
All right?
[SOUND] Right?
That's why this integral equation.
This is integral equation.
[SOUND] Because of P
appears in integral as
well as over here.
Right?
Second.
[SOUND] second.
What is the regions of validity to use that integral equation?
This integration, [SOUND] has to be,
the surface that is simply connected.
In other words the surface cannot be like the surface.
Like this kind of surface.
No.
But as long as, as the surface like that, or like that.
[SOUND] It doesn't matter.
We can do it.
Therefore, we can use that solution, I mean
that Kirchhoff, Kirchhoff-Helmholtz integral equation to solve this problem.
[SOUND] Yeah.
That is possible.
Now let's use that, famous well-known
Kirchhoff-Helmholtz integral equation to solve this problem.
Oh, before I go into, I mean, solving the,
baffled piston radiation case.
So let's, let's, let's discuss more about this.
[BLANK_AUDIO]
Because there is a freedom of selecting Green's function.
I can select Green's function that can make, one of these terms 0.
On, the boundary, right?
Because I have a degree of freedom.
So, if I select a Green's function.
That this term goes away, that means, just to measuring pressure on
the boundary and use Green's function propagating information to
the position where I do want to estimate or predict some pressure.
Okay, if this is 0, then,
Case
A, select,
G on
S0 equals 0, then, [SOUND] Kirchhoff-Helmholtz integral equation
says, the pressure at any point would be surface integral
P dG/dn
dG/ds0, ds0.
That means if I measure, [SOUND]
some pressure, for example,
at certain discrete point,
maybe if I use coordinate, x,
y and z, [SOUND] then s0, ds0
turns out to be dx, dy 0 0.
And this is, this is, x0,
and this is x0, and this is y0.
[SOUND] And then measuring pressure at x0 and y0,
and propriating using this Green's function.
And I can obtain the pressure at any z, or at any point over there.
This is the basis of acoustic holography, everything.
Because it's rather easy to build yours on pressure instead of.
Velocity, which is over there dP/dn.
dP/dr0 is the velocity.
So holography use microphone over here, and then predict sound at any
other unmeasured positon and we call this prediction forward prediction, we
call, the prediction close to the source we call that a backwards prediction.
And some student may ask me.
And why then you have so many papers, related with
acoustic holography, acoustic holography theory
is very straight forward and simple.
Yes.
Yes.
[BLANK_AUDIO]
In this case, this surface has to
be infinity, theoretically but we cannot measure
infinite number of pressure so there are
a lot of issues associate with signal processing.
Finite structure.
Finite aperture size effect.
Now how we close to, how, how can we
predict the sound pressure close to the source problem.
So.
So called near-field acoustic holography.
I mean, addresses, that problem.
So on and so
on.
If this surface is arbitrary.
Then this equation, support boundary element method.
Okay?
So, if you use boundary element method, it's just to measure the pressure of the
surface and use this Green's function then you
can say that I can predict any pressure,
that's not measured.
So boundary element essentially use this kind of singular function, dG/dn
and dG/dn, dG/dr requires only to satisfy
interesting inhomogeneous wave equation.
Nothing else.
It does not have to satisfy the boundary condition.
[BLANK_AUDIO]
Of course you can use the Green's function that satisfy the boundary condition too.
It doesn't, doesn't really matter.
Okay. For example, if you measure the Green's function
by exciting at one point by pistol for example.
And you measure the pressure over there.
That' solution satisfy both boundary condition as
to what is in homogeneous wave equation.
Interesting.
And also this is the foundation of many sound manipulation problems.
[LAUGH] Okay.
Suppose you are making sound at this point.
So, you are making sound you are generating P over there
using that Green's function then you design sound field P r.
So that's the theory of sound manipulation.
That's the fundamental.
You know, foundation that support the theory of sound manipulation.
So this is, really exciting, [SOUND] expression.
What is ds0?
Actually this has to be vector, in strict sense.
That means I have to integrate the sound normal to the surface.
Of course normal to the surface because we do not allow viscosity.
[BLANK_AUDIO]
Okay, so let's use this equation.
Or having sound from baffle to piston.
Using the beauty of,
linearity.
The beauty of linearity is, it allows us to, summing up.
[BLANK_AUDIO]
OKay?
[SOUND] In nature there are a lot of things are nonlinear.
[SOUND] But because we are handling
linear acoustics, that is valid.
Up to at least 120 decibel, so we have many applications still.
So for, baffled piston, I have piston.
And I have wall over here.
And this is oscillating with the velocity profile right that.
[SOUND] This is the magnitude of the velocity
profile and these oscillating with frequency of omega.
With respect to time, okay?
Using that notation, we could say,
[SOUND] this is r0, and this is r.
And this is s0, [SOUND] surface.
Okay.
Now, our job is first to select, G, Green's function.
We have to select Green's function.
And we already did.
This Green's function
has to be, [SOUND] has to
be 0 on s0, okay?
Then we can use this.
Okay.
And this physical situation requires a boundary
condition that over here the velocity has to be 0.
Therefore dG/dn has to be 0 on this surface.
Right?
Because this is rigid wall boundary condition.
All right.
And one clever, student would think that this is
equivalent, to have, piston in open air not baffle,
but as a velocity profile, like that.
[SOUND] Okay then, of
course this is r0.
And this is r.
Okay?
Then the Green's func, I mean
then the pressure will be 0 on
this surface because of the symmetry.
Right?
Due to the symmetry, pressure is 0, or velocity is 0.
Velocity has to be 0 to satisfy this boundary condition.
[SOUND] Right.
Therefore, the Green's function we
Select would be the something that
satisfy this boundary condition, okay?
And lets say we have a surface S minus and S plus, okay.
And the, the Green's function we
would like to select that
satisfy what I just described
would be, G dP/ds0, ds0.
In other words, in other words, I want to,
the velocity on this surface has to be
0, therefore dG/dn dG/ds0, on
this boundary has to be 0 therefore instead
of using this we would like to use that form.
Okay?
This is more feasible than this formation.
Now, you have to integrate this.
This ds0 has two part.
One is s0 plus and one is s0 minus as I indicate over there.
All right?
Performing, many details mathematics, you, we can
predict sound radiation due to baffled piston.
Again,
it will be the function of, k multiply this size, and it
will be the function of the near field and the far field.
And we will see very interesting directivity pattern, due to
baffled piston, what we can envisage at least, at this moment.
If ka, a is the size of this, is very large, in other words,
the wavelengths is small than, the size of a radiator.
It will radiate very well.
Okay.
And, if wavelength is small compared with
the that's the case which I just solved, which I just-
if the wavelength is very large compared with
that, then it would radiate like a monopole.
So it may look like, monopole, right.
So when ka is small, look like a monopole, and k is getting large and
large, then there will be very interesting things
happen because if you look at the fluid particle.
Over here.
Wavelength is large, this fluid particle and that fluid particle would be happily
affected by the fluid particle over there
because wavelength is very large, so therefore
looks like monopole radiation where ka is getting larger
and larger meaning then wavelength is small compared with a, then wavelength
is small therefore the fluid particle over here would not
be effected by the motion of fluid particle over there, therefore.
It has some directivity because no less radiation this
part and that part, large radiation over that part.
Things like that.
So when ka is getting larger and larger, which means
that high frequency component, radiator, then we have a directivity.
Therefore, therefore for high frequency we have a directivity pattern,
therefore for high frequency component in audio system, we do need two separate,
speakers so that we can feel about the
Stereo sound.
Okay?
So stereo sound mainly has to do with when ka is large.
For low frequency component, then ka is small and
the wavelength is very large compared with that size.
The sound is radiating as if it, as if it is monopole.
Therefore, for low frequency noise, sound, we do not actually need two speakers.
We may need one speaker because it does not have a signficant directivity.
So in old days, some, some people who do not have enough money, like myself.
When I was a graduate student, I bought one, subwoofer and
two mid-range speaker and I was happy about that, okay?
So even if you, if you, if you know
very well about acoustics, you can save some money.
[LAUGH]
Okay, I think I will stop today's lecture over here.
Because what I delivered today is, is very complicated it's not easy to understand
at once.
So take time and look at again and again, and then you will gradually understand.
To me it took, and I still confuse sometimes,
but I thought about this formulation for 30, 30 years.
In the beginning when I learned this, when I, when
I was at graduate school, I was, I was really confused.
That's why I developed to derived this, equation based on one dimensional case.
Okay, one dimensional approach is, is very rare to, to see in other literature.
Personally, when I succeeded to derive all of these equation in one dimensional, one
dimensional case, I was really happy because,
certainly I can understand [LAUGH] one dimensional case.
[BLANK_AUDIO]
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