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. 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 well known, wave number relation between x direction and the y direction and z direction. This must be satisfied. Therefore, in in the z direction, we can see this kind of relation, and x and y has to satisfy this relation. Therefore 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. If it is less than zero, then mathematically it means that the kz can be imaginary. 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. Could op, happen in this area the way tunneling can happen in this area. Okay. But in most case for plane wave case, 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 because of the impedance mismatch exists between this and that, we can expect that some of the wave such as the wave that has. 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. 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, 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. So, in this case, there is the chamber and this is the length. 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. Okay? And the magnitude of transmission loss is controlled by the area ratio between this inlet and the area over 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. 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 wavelengths, what is it? 343 divided by 500. So that is about 60 centimeter. Therefore the, the chamber has to be 2.4 meter, which is, which is fairly large. So, it's not likely possible to put 2.4 meter size the muffler to control the engine noise. It's too long. 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, 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. Okay? So, again, controlling low frequency noise, coming from the, coming from the fan or the engine, is a very challenging subject to control. 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. And, let's move to another topic Suppose we have engine over here and apply, if you have a this kind of resonator or side-branch sometimes we call. Again the role of side branch is to introducing quick impedance mismatch at this pole position. 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. Right? So that's why we call side branch, often quarter wavelengths. Quarter wavelengths or quarter wavelengths, what? Quarter wavelengths side branch, okay, so there are many applications for example if you see the, the, anechoic chamber you, you can see a lot of wedges you can see in anechoic chamber, right. How to determine the size of this wedge? 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, I mean lowest frequency, that we want to have a non-reflection condition. 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. Like this is a still play. But still you have some sort of, the, effective side branch, therefore you could have very, not totally effective, but some 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? For example, if you use the material that has, that has a slowly changing impedance over here, then your effective length would be increased. For example, back to the pipe case, okay? When you have a experiment, you want to have infinite pipe over there, and what you have to do is, you have to put some anechoic chamber. 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. And this has to do with something like a quarter wavelength plus some effective, you know effective length. If you have like if you have this horn. Horn type one. This gradually changing horn, meaning that you have a gradually changing impedance. So, compare this, this one with that one. You have abrupt change at, at, this, this, this, this, this, you know this part. 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. Okay, let me back to the quarter wavelengths issue. 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. 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? So if you put a series of side branch over here. 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. 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 the, those, about to relative wide branches, wider bandwidths. And and basically what we can use the the silencer is the dissipative type as well as reactive type, and reactive type basically use the impedance mismatch. All right, so Webster horn equation as I mentioned and you can see this. All right, chapter summary. In this chapter, we talked about the acoustically large space, as well as acoustically small space. Acoustically large space means that the space that is surrounded by impedance mismatch, basically what impedance? And the sound inside of the acoustically large space follows very surprisingly the the parameter what we call Sabin's, reverberation time, T 60, and T 60 is controlled by mainly size of room, and inversely proportional to the open area window. That is obviously summing up the how much energy is absorbed, so absorption coefficient multiplied by the area, 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 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. 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 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. And depending on whether it expand and contraction, 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 impedance mismatch at space. And that provide us the understanding of the expansion chamber. And then we examined how to design muffler for the for the fan noise, as well as the engine noise. And that's all about we studied in this, in this chapter five. And, and that's all about, I'd like to, I'd like to I'd like to, 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]