0:00

Okay,

duct is considered to be a space that has a very short

distance, acoustically small space in, in, in, in, in

the in the cross sectional area as you can see over here but

rather a very long distance compared with the cross sectional area.

So in this case we can expects that the the wave is mainly propagated in, in.

0:35

So in this case we can expect that the wave is mainly propagated in the z

direction, therefore we can write the, this kind of exponentialfunction.

And in x and y direction, there will be a some standing wave.

They can be considered as cosine and sine function.

So in terms of wave number

the, in, in, in, x direction should

we anticipate this kind of wave, and the y

direction we anticipate this kind of wave.

So as you can see here, in z

direction, we can expect, due to the the

2:09

the following relation has to be satisfied from this relation we can see that.

If kx and ky exceed some number, for example, free space wave number kz,

then what's going to happen is kz can be, can be, kz square can

be, less than 0 or greater than 0.

2:49

That means in this case, the wave will propagate

jkz, and z, if kz can

be, if kz has some imaginary value then

the wave in z direction will exponentially decay evanescent wave.

Okay if kz is greater than 0, then the wave

is propagating in z direction with a certain wave number.

Okay?

All right.

So observing that kind of situation.

If this is the wave number in x and y, x and z

direction, in this case, in this if the

the pipe up to here is a small, compare

with a that direction then the, as demonstrated

over here, the sum of the wave, which has

the real part, over here but could have a,

the different wave number, in this case the wave can

propagate in this, in this region but can not

propagate in this region because of the wave number relation.

Therefore, wave blocked in time.

The other way, the other way around in this case some of wave which is,

which is, which is, you know, which does

not really propagating well in this section, may propagating

very well in this section, which we call wave tunneling, phenomenon.

Mixing up these two case, would have here, we're blocking.

5:10

Because we have a big impedance mismatch over here,

and another big impedance mismatch over here, so some of

wave will be reflected, and some of wave will be

transmitted, again, here because there is a big impedance mismatch.

Some of the wave will reflected, and some of the wave will be transmitted.

An interesting thing is that, impedance mismatch occurs at

z equals 0, as well as where z equal L.

Therefore, the, some of the wave will

be reflected, and then, and then back reflected

6:11

Like a sine wave like this

or cosine wave if you, if you like, like that or to

have a maximum pressure over here and the maximum pressure over here can not

really, coming out from this, from this, the chamber.

So the this wave and the some of the wave that has maximum and then maximum

over there, this kind of wave cannot, cannot

coming out from this box or from this chamber.

So if you handle this mathematically, assuming

incident wave like that and then, reflected wave over here like that

and, again, another wave, Psi, which is considered to be a incident wave.

To this impedance mismatch, and the reflected wave and another

one transmitted wave like that, and then solve by using

the condition over here, pressure continuity and the velocity continuity

over here and again pressure continuity and velocity continuity over there.

7:32

In this case we have four boundary conditions to apply, therefore we have

one two three four, and five unknowns, therefore what we can

obtain from this condition is the ratio between the, how much wave

can be reflected, compared with the incident wave, and its vice versa.

So we can solve it and the result shows the very interesting result.

Transmission loss again is defined as the like

this and the tau is the ratio between incident and transmitted wave.

Mathematically is, it is written like that and,

8:21

this is the, actually, this is the result.

Okay?

This is the .5 when is the .5, L over lambda is .5.

So that's the case, the lambda is about two times of the length.

8:52

And the lambda is two times of the length therefore, the wave looks like this.

Therefore, there is a maximum over here

and a minimum over here, minimum over here.

Therefore the wave can, would not be affected

by the impedance mismatch over here, and over here.

Therefore the transmission loss would be zero.

Okay, every wave we will be, will be will be easily passed through this chamber.

So transmission loss is 0 and the transmission loss over here is 0, in

other words the frequency corresponding to this value will not

be effectively controlled by this this expansion chamber.

The frequency that can be effectively controlled by

this expansion chamber, would be something related over here.

In other words, wavelengths is a quarter wavelengths, of the chamber.

10:08

there.

Okay.

So that is a very interesting result, therefore

what we can see from this picture is

that when we have this kind of noise as, as I said in the previous lecture.

If you want to control this frequency.

10:27

Okay, this is frequency one, and that frequency would have the wavelengths

lambda one, because there's a relation between frequency and wavelength.

Recall the dispersion relation and the lambda has to be 343 divided

by frequency and, and if you want to control this frequency using

expansion chamber and that is the point you'd like to use.

So therefore the wave chamber length has to be

four times of the, frequency because of this relation.

OKay?

So if the frequency is like, 500 Hz, then

11:48

So what we can do?

So first thing we can attempt to do is the

if there is a engine, we can put the, like a

2.4 meter long pipe over there, and I put

the muffler over here, but 2.4 is still

very long compared with the total size of

a car.

So what we can do, is we can, we can, we can make this kind of wavy pipe,

12:25

or some, some other way to, to, to get a, longer longer pipe.

Okay?

Or [COUGH], you could think that okay, I have a expansion chamber over here.

I have a pipe and I have a pipe over there, therefore the total effective length

could be somehow matched with the 2.4 length, that could be effective.

12:54

Okay?

So, again, controlling low frequency noise, coming from the,

coming from the fan or the engine,

is a very challenging subject to control.

13:13

So, what I, what I, what I show you here, in

principle, the expansion, I mean the muffler

uses the, for the reactive type of muffler uses

the the impedance mismatch.

And the, for the, for other this kind of wave, which is above, like a 700 hertz.

Then we can, we can use the absorptive treatment, type of matterial over there.

And then that can be achieved.

14:24

Having side branch again this side branch will the, its own natural frequency, okay.

If we can estimate very roughly the, the natural

frequency of side branch here, by seeing that, this is rigid wall,

therefore there will be a maximum pressure has to be, has to be applied.

And, over here is sort of, free space, therefore

minimum pressure will be acting at this region or point.

Therefore, the wave which will satisfy this kind of boundary

condition would be, pressure maximum and pressure minimum over here.

Therefore, this has to be quarter wavelengths.

15:23

Quarter wavelengths or quarter wavelengths, what?

Quarter wavelengths side branch, okay, so there are

many applications for example if you see the, the,

15:55

Again, the purpose of anechoic chamber is to make a wall that cannot reflect.

The, the frequency that we have interest.

So low frequency, the minimum frequency, or,

16:24

Okay, say if it is 100 hertz, then the wavelengths is

how much, 3.4 meter right, approximately.

What is the quarter wavelengths of this case?

Quarter wavelengths is simply dividing by, this one by 4.

And that is about a, 85 centimeter, so theoretically,

if you have the side branch, if your wedge that has

about 85 centimeter long, no absorbing material.

17:27

of the effect that you can have less, reflection from the

wall, but that is not enough, therefore you use some sound absorbing material.

Sound absorbing material actually increased the oh

sorry, decreased the quarter wave effect, okay?

18:23

Anechoic termination over there.

How you make anechoic termination?

The size of this wedge has to do with some

lowest of frequency you want to have, I mean lowest

frequency that will not reflect from the, from the end of a pipe.

19:16

So, compare this, this one with that one.

You have abrupt change at, at, this, this, this, this, this, you know this part.

19:26

So if you have a gradually changing horn, then because you have a gradually changing

impedance at the end, you have less reflection from the end, okay?

So if you look at my book, you will see the Webster-Horn equation.

Then based on Webster horn equation, you can

design the Webster horn to have the anechoic termination.

If you put some absorbing material over there, the

performance of the Webster, Webster horn termination will be increased.

20:06

So by using quarter wavelength side branch, you can somehow control the,

transmission loss coming from, I mean, transmission

loss desired transmission loss over there.

20:22

What is the principle?

The principle is very simple.

The corresponding quarter wavelength branch, side branch, and the corresponding

frequency of noise would see the very big impedance mismatch over here.

Therefore, the wave will be reflected, okay?

20:49

Then you could have a, the the wave reflected, wave reflected, wave reflected

back to the engine based on the size of this side branch.

That would be another interesting topic for, for, for the world.

21:14

So in duct acoustics, as I, as I summarized

over here in general, dissipative silencer is known to be more

effective for controlling high frequency noise, and the absorbing

23:19

and interesting number we learned is 0.161, so

Reverberation time is 0.161 multiplied

by the volume of room and open area window.

And the we had a very unique experience by calculating the

23:44

reverberation period of this room very, very rough calculation.

And we came out with the number .7 with actual measurement show that about .9.

That's what I remember, is it, was it, is it correct?

Yeah, so the lesson we got from that kind of

demonstration is that the as an engineer, we can, we can

very roughly estimate an open area window as well

as volume, provide us a very rough but good guideline to have the

sort of, sort of the number that will

guide us to, to evaluate this room without

using expensive instrument, but anyway, so

that the reverberation period is the measure

or primary parameter measure, in other words to, to

characterize the acoustic, acoustically large room.

And that is major understanding coming from the acoustically large room.

In acoustically large room, we learned about reverberation, what is

reverberation, and then what is a diffused field, and then what is a free field.

And that is very important concept to understand the

reverberation, those kind of characteristics, we invite the

the radius of reverberation, that is regarded to very useful.

And then we move on to the acoustically small space.

Which means that the size of the space is much, much smaller than wavelengths.

25:49

Okay, in this case the first very interesting example was

Helmholtz's resonator, and the Helmholtz resonator is the device

that produce very abrupt impedance mismatch.

For the, Helmholtz, at the Helmholtz frequency, and the Helmholtz frequency

can be estimated very simply by the formula we just examined,

and in that case in the last lecture we attempted to.

Calculate the frequency of a famous, the

26:32

the of a drinking bottle, and we failed by some reason, but

this time, we brought up that question again, and

With the help of our assistant Mr. Jung, we came up with the the

Helmholtz frequency of our soldier battery is about 175.

Which is close to 180.

And the we move on to another type of acoustically small space case.

Which is duct, duct, as a small, of course, a small space in cross direction,

but let very large direction in, in, in, in, in the other way around.

27:26

We saw that the wave tunneling or other

types of the interesting phenomenon could occur, but for the plane wave

region we found that the the the

wave propagating in a duct would follow very simply, the, as if there are two

28:28

I will plan to teach you.

You guys, during this term.

So I mean, this is the end of our lecture, so we are ready to celebrate ourselves.

[BLANK_AUDIO]