This course introduces acoustics by using the concept of impedance. The course starts with vibrations and waves, demonstrating how vibration can be envisaged as a kind of wave, mathematically and physically. They are realized by one-dimensional examples, which provide mathematically simplest but clear enough physical insights. Then the part 1 ends with explaining waves on a flat surface of discontinuity, demonstrating how propagation characteristics of waves change in space where there is a distributed impedance mismatch.
Lecture 4-1 Part 2. Normal Incidence on a Flat Surface of Discontinuity 2
Loading...
From the course by Korea Advanced Institute of Science and Technology(KAIST)
Intro to Acoustics (Part 1)
35 ratings
From the lesson
Waves on a Flat Surface of Discontinuity 1
You will now learn when the waves meet on a flat surface of discontinuity. It will be explained with the mass law of wave phenomena.
Meet the Instructors
Yang-Hann KimProfessor
Mechanical Engineering
Where is the max integer?
Because our objective is to find the reflection, and
transmission coefficient we need two equations.
Very simple mathematical conclusion, and these two equation has to come from here,
this boundary condition or discontinuity condition, okay?
At x=0.
We can see that velocity must to be continuous.
And because there is no mass at this continuity.
Just two fluids separated by simple discontinuity.
Therefore, I can say
pressure must be continues.
If I write this physical requirement
in terms of the mathematical form and I can write this.
Velocity. What is the velocity?
Of medium one?
So one is the velocity induced
by the incident wave,
that is PI divided by Z1,
Because we assume the plane wave, if it is not plane wave we cannot use this.
Okay, and then velocity due to reflected work look like that.
And because we assume the velocity of instant wave in this direction and
the velocity due to reflected wave that direction,
we have to put the minus sign over here.
That is the total velocity in u sub i.
Instant as well as reflected to it.
And that has to be equal to the velocity
of transmitter which is Pt over Z2.
And so we have equation number one, right?
And actually, I re-omitted over
here all exponential minus j omega t.
In other words, this equality has to
be hold before any instance of time, any frequency.
Now, what about pressure?
Pressure at median one is Pi and Pr composed by two pressure.
And the Pi propagating this way.
So, if I say, this is plus, then I have to put the minus,
and that has to be equal to Pt.
Is it correct?
Is it correct?
No, because pressure is a scalar.
And the pressure by definition is acting on the surface normal direction.
Therefore the pressure at medium one is Pi + Pr, okay?
Now, let's say this equation two and
then we have to find out the reflection quotient and
the transmission quotient from these two equations, okay?
And everybody can do it, how?
Maybe one simple way to do it is
maybe I divide this, Z1, Z1 and Z2.
Then add all, then this will go away, and then I, no, no, no.
This will go away, and
I have 2PI over z1 is equal to Pt,
well this has to be Z1,
1 over Z1 plus 1 over Z2.
So I can get Pt over Pi Is
equal to Z1 + Z2, 2Z1.
Actually I remember I didn't try.
So derivation is very straightforward as everyone can do it.
So let me go back to the results.
So from, I say from the equations one and
two, we obtain reflection coefficient
is Z1 + Z2 Z1 minus Z2 and
transmission coefficient is
equal to Z1 plus Z2 and 2Z1.
Which is, as I said before, exactly same as
What we have over here.
The only difference is this Z1 is equal to low LC1 and
this Z2 is equal to low LC2
and this Z1 is equal to rho1 c1 and
this Z2 is equal to rho2 c2.
Why this is the same?
Because both cases are one dimensional cases.
All right, now let's look at more about this.
Okay, special cases.
It's always nice to look at the rather
complicated mathematical form
in more, in simpler cases.
How do we do it?
Generally looking at extreme cases often provides
very intuitive physical understanding.
So what if Z1 is much,
much smaller than Z2?
Okay, Z1 is much, much more smaller than Z2.
Then reflection coefficient,
Z1 is much, much smaller than Z2,
meaning that if Z1 is $1 and Z2 is $1 million.
Then this one is minus $1 million plus $1 and
the denominator is $1 million and $1.
So that approximately what?
Minus one.
What about the transmission coefficient?
Transmission current, Z1 is $1,
so 2 multiplied by $1 is $2 divided
by $1 million $1, that is zero.
What does it mean?
If Z1 is very small compared to Z2, this is the case when we shout
In air, we'll want to hear
what is transmitted in water.
Okay, mathematical derivation shows
that the fish cannot hear our song.
Therefore when you fish, when you enjoy fishing, you are allowed to play music.
Because it does not, the wave is not transmitted.
But this continuity produce reflection coefficient to minus 1, what it means?
Total reflection?
Okay, let's see another case two.
Z1 is very, very large than Z2.
Then or is it R, now Z1 is very large,
then R = 1 and what about the top?
Because G1 is very, very large, the taller equal to.
That means, if [INAUDIBLE] has
a party underwater and we can hear.
Because tau is two.
When fish say let's have fun, then what we will hear is let's have fun.
Okay, hearing has to do with not
only pressure but also with velocity.
Hearing means that my sensation has to be activated and
that means we needed some power not only pressure.
All right, for example in electrical engineering,
if you have an electric shock.
I mean not only high voltage is necessary,
but also high current that provide enough power.
So we have to see the power.
So what about the power reflection and the power transmission coefficient has to be?
That we have to find out, how to get the power transmission coefficient.
Power transmission coefficient is
the ratio between instant velocity and
instant, ratio between instant power and
transmitted power or reflective power.
That is simply multiplied by this pressure
reflection coefficient with the velocity reflection coefficient.
And the velocity reflection coefficient is simply this one
This is reflected way, so this one divided by Z1 and this one divided by Z1.
Why?
Because we are handling plane wave, the velocity is pressure divided by Z1.
Therefore, what we obtain from this calculation, we obtain that.
Okay, this is the velocity due to instant wave and
this is the velocity due to reflected wave, right?
So reflection coefficient
due to the velocity,
I would say velocity has to be
equal to the velocity at media one
Velocity at vd1 due to reflection,
divided by velocity at vd1 due to instant.
That is correct.
The velocity due to reflection I can't
write that is that pr divide by Z1 and
the velocity two two incident that's a pi divide by z1, okay?
That isn't the reflection coefficient of velocity.
So I have to multiply this one to here,
not just to divide those terms by Z1.
Okay?
So if I multiply this one and this one or to there,
what I will obtain is let me move my blackboard over here.
So the power transmission coefficient
has to be Z1- Z2 over Z1 + Z2.
So this is velocity transmission coefficient and
compare with the velocity, the power.
Velocity transmission coefficient and
I have to multiply this one to there, that will produce
Z1- Z2 divided by Z1 + Z2.
Does it produce a Z1- Z2 squared divided by Z1 + Z2 squared.
Power Transmitted power
has to be pressure multiplied
by velocity conjugate.
Okay, as we studied yesterday.
As we studied it yesterday.
So, what we obtain would be Z1 + Z2,
Z1- Z2 conjugate multiplied by Z1+Z2, Z1- Z2.
That's the reflected power.
And the transmitted power will be
the transmitted velocity,
that would be Pt divided by Z2.
That give me the transmission coefficient
would be 2Z2 divided by Z1+Z2.
Therefore, what I would get for the transmission coefficient,
power transmission coefficient,
which is P multiplied by new conjugate would be this
one multiplied by the conjugate 2Z2, Z1+Z2.
That gives me the picture you can see in the,
In our book.
So as I said, we have a pressure continuity at x=0.
That looks like that.
And the velocity continuity that looks like that.
The reason why we have a minus over here is because
the direction of velocity due to reflection
has to be opposite sign than compared with the incident wave.
Okay?
And then we assume three different waves.
As I said, because those two are right going waves,
we put the minus sign over there.
And, as I explained, the omega has to be same.
But, because we have two different mediums, the wave number can be different.
In other words, the number of waves per
unit lens would be different, even though we have the same frequency.
And the boundary condition for below still boundary condition,
we can rewrite this as we explained before.
Therefore, we've got reflection coefficient.
And this is transmission coefficient.
And we discussed this with a special case, but
velocity reflection coefficient looks like that.
And the power reflection coefficient turns out to be like that.
Therefore, even if the Z0 is very,
very smaller than Z1,
which I said the fish has a party and
we or we are trying to listen.
There are noise, that is the case,
Z0 is much, much smaller than Z1,
then the transmission coefficient over here,
since the Z1 is much, much smaller than Z0,
We cannot hear fishy party noise.
And also if Z0 is very, very large compared to the Z1,
still fish cannot hear our voice, right?
But, if you have rather different experience.
If you go to the swimming pool,
sometimes you can hear the sound outside of your water.
I mean, when you jump into the water.
And often you can hear the sound coming from
some areas beyond the wall of water, why?
It's because the sound excite,
somehow, in very small amount the wall of a swimming pool.
Therefore, the wall will excite the water, therefore you can hear it.
But this model assumes that we have infinite water and infinite air.
Therefore, the predicted wall is somehow different
what we normally can experience in swimming pool.
Okay, maybe you swimming pool with your friend and
ask your friend to make applaud over the surface of the water.
And then, you go down, and
then you cannot here the sound that's coming from above the water.
Okay, this means highlight of this
derivation in forced reflection and
transmission is dominated by
the characteristic impedance of median.
Explore our Catalog
Join for free and get personalized recommendations, updates and offers.
Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.
© 2019 Coursera Inc. All rights reserved.
