If new medium is same, then z1 = z0, then I've got RTL transmission loss that is 10 log 10. And I have (omega m) square + 4Z0 squared divided by 4Z0 squared. Then I have this is equal to 110 log 10. I have 1 plus omega N 2Z0 square. That's what we have. Z0. Okay. >> [INAUDIBLE] >> This is an interesting formula. If this one is very large compared with one. And this is this case. For example, M for you know, plywood per unit, mass, 1 meter and 1 meter, that will be, how much? Guess. Maybe 2 kilo or 3 kilo. And omega, that is 2 pi f. If you are considering 1 kilohertz, that will be 6,000. Multiply by two or three, hat is a huge and what is Z0 for there. That is about 415. So that is about 1000, right? So this will be normally much much larger than 1. Therefore, I can neglect this therefore I can approximate this as 20 log 10 2 pie fm over 2Z0 okay that is interesting formula. And this is 2 and 2 and pi z0 z0. Fourier is about 415 and pi is about 6. So I can come up with some interesting formula which has very practical value. Okay, and also we can easily find when this value, pi f m / z0 =1. Which is the case when this one has general transmission laws, okay? So that I call, block the pressure, block the frequency fb and block the frequency is simply because this is one that is block frequency then fb is equal to Zo/Pi m. Okay this block frequency concept is very useful. Because, Because I found the blocked frequency. And then if I increase the frequency twice, according to this formula. According to this formula if I increase the frequency twice I will get the transmission log 6 dB increase right. Everybody understand? If I increase f to 2 f 2 f then this will has to be multiplied by two and to the two is six dB so when I increase the frequency twice. I will increase transmission loss six dB. Then if I increase more than twice it's 6 dB more. So what I can do is I adjust the count, for example if blocked frequency by using some plywood, air if I find blacked frequency is 50 hertz. And I increase the frequency or mass per unit area twice. Then I increase 6 dB. So what did I say, black frequency of 50 Hz, right? Increase the frequency twice, then I go to 100. Increase it twice, the frequency R will go to 200. Increase it twice, the frequency R will go to 400. Increase the frequency twice, R will go to 800. What if I have a noise and 800, 100 dB. Then using the material I use I can reduce the noise by six multiply five that is 30 dB. So you can design the wall. By just calculating this frequency and using your thump you can talk to your supervisor using the material available and reduce the noise for example 30 dB.. That make you, Look like a expert in noise control. Okay, let's see some more details using the PowerPoint. Okay, before I show you showing the details I will show you whether this is going to be true or not by demonstrating an example. [SOUND] This is about [SOUND]. That is 72 and I close and this go to around 40 or 50 db and then I will [SOUND]. This is 2t, so I increase the mass twice, then I should get down 60v. That has to be about 45 or 50, so it has to be decreased, [SOUND]. This is about 40 db [SOUND]. So it follow mass law okay our increase [SOUND] 40 that has to go below 40 db and you will see 34, 34. So 34, 40, 46. So that's good, that's good, and the thickness of this wall is very small compared with the wavelength lambda, right. So this theory works. And the best way to reduce noise is to turn of source, of course. And you have a perfect design [LAUGH]. All right so this is what we have. It's a good demonstration. So you have to believe what we obtained over here. So practical variable. So practical variable. This is called mass law. This is a famous mass law in acoustics. What it essentially says, by increasing frequency, or mass twice, transmission laws will increase by 6 dB. That is mass law. And most of the noise control engineering for reducing sound using wall or whatever, they normally use mass law. This is the major concept. So if you understand mass law you can behave as if you are very well trained. Noise control engineer. And now, I will give you some interesting number, so that you can really behave as if you are very well trained noise control engineer. Okay, omega is 2 pi f and 2 pi divided by 2Z 0. Okay and as I said if you want if you know the plot frequency which makes the frequency that is equal to one. Then okay like over there I can use ten over here. And this is block frequency. And this is about 132 divided by mass per unit area. So 132 that's easy to remember. If you don't want to remember this strange number you can just remember 131. Because that looks like a war right, war. Okay war. So 131 is okay. Now, if you remember this blocked frequency, what you can do is this. This is the blocked frequency, and this is the line that expresses the exact RTL mass load, that is 10 log 10, and you have 1 + (omega m 2 Z0) square. And this is the curve. But in the high frequency range In high frequency range over this range, the transmission loss follows the mass law. Okay therefore if I know the blocked frequency, that is simply 123 divided by m. Okay and if the now Gypsum, if you use Gypsum for blocking noise in a factory for example or suppose you are. You want to block the noise coming from your neighborhood in your dormitory. Okay, that will be more sensible. Then if you find some good gypsum and suppose the mass of gypsum is for convenience, is 1 kilo per unit skill. Then your blocked frequency is 123. Then this curve says, At 246 of hers you will gather 6 dB down. 6 dB noise down coming from your neighborhood. And twice the frequency you will have one more 6dv down. So you will have 12 dB down. If you double your Gypsum then you will have additional 6db down, right? So that's very easy way to design your, your noise control or, that will use noise coming from your neighborhood. So Mass Law is the highlight of today's lecture. 6 dB down by doubling frequency or mass okay. That expresses everything you learned today okay. And we will talk more about more complicated to the case, but it's just a correction of mass law. So a central concept, what is associated with a flat surface of this continuity, is mass law. Thank you very much. If you have further question, please send me email or you can ask me after this lecture.