. So,
welcome to the NPTEL lecture on Architectural Acoustics the 5th lecture today. The fifth
lecture is all about the Near and Far Field Propagation and Loudness. And this fifth lecture
is the last lecture of our first week course and you will go to the some other topic
in the second week. And the the architectural physics the architectural in the building
physics is going to be end with this lecture. So, if you remember in the last lecture what
we discussed is the sound pressure level, sound intensity level, and how they can be
added and what is the way? . The particular conversion from the level to the the intensity
or from the intensity to the level is going to concern I am going to happen.
So, in this lecture lecture number 5 again I have to learning objective, the first
1 is you have to differentiate the near and far field propagation of the sound by
this lecture can you differentiate I mean after the this lecture, if you are going
through properly you must differentiate between that.
And, then you must correlate between the frequency sound level and this loudness of
the sound . And, that will be the main major focus to this particular lecture .
So, let us go to the sound intensity once more in a different way . What is intensity?
Intensity is nothing, but the power per unit area. So, I can write like the power by area
is equal to the intensity. And, if the power is now further converted to the energy part
time, because the energy per time is your joule per second, which is your watt which
is your power and power per area that is watt per meter square is your intensity .
Now, this area can be rewrite as the length by volume. Now, further if I see this total
things can be rewrite like this volume can be taken into the below this. So, the energy
by volume into length by time and length by time is nothing, but the velocity of the wave
. So, let us now see how this conversion
of the energy and volume can be done. As, we know that the energy is nothing, but the
kinetic energy of the particles, which is under the motion . So, it is half into mass
into V max square, because the velocity of the particle also we will fluctuate and
it is minima and there is a maxima. So, the the maximum velocity of the particle
is A into omegas omega and the this is square of that and the the the volume of mass in
into density. So, if this 2 are added together in this particular
form, then the intensity can be rewrite as half into m A omega square into rho by m because
it is volume now and into cc is the the the wave velocity. So, the intensity can be written
in this form. So, you see this is very interesting formula and you see the intensity of sound
is depend upon the velocity, is depend upon the rho density of the the media, it is depend
upon the amplitude of the motion amplitude of the propagation and also it is depend upon
the frequency or the the angular the velocity omega and this 2 are in square term.
So, if I know the frequency, if I know the amplitude and velocity of the sound and the
density of the media, I can actually find out what is the sound intensity . Now, go
further head. So, this intensity is again rewrite as the
same equation can be rewrite as this rho c A omega square by rho c, because the I have
taken another rho c in the here and the and dividing that by the numerator. And, here
that I can get 2 such product and 2 such parameter separately, and I can write is half into rho
P square P 0 square and Z, what is this 2? The P 0 or the P 0 is the pressure amplitude
of the wave propagation, which is nothing, but your rho c A and omega and this is Z is
called acoustical impedance which is the product of the medium density rho and the wave velocity
. Now, why it is wave this pressure? So, let
us find out from the the the units of them rho is kg per meters cube, and the c is meter
per second velocity amplitude is meter and this is radian per second is the omega. So,
if you just finally, come into that I mean if you just finally, rewrite those things
and rearrange the the units finally, you can come down to the the Mpa, the mega Pascal
the mega Pascal is nothing, but your pressure. So, you can see from this particular equation
that the I the intensity is proportional to square of the pressure and that is why if
you remember in the last lecture lecture number 4, we have discussed that the pressure
square are the the intensity is proportional to pressure square. And, that is why if the
if you convert the pressure level to the decibel level, you have to multiply that by virtue
of 20 not 10. And, if you convert the intensity to the intensity,
level you have to multiply that by 10 and if it, because it is logarithmic .
So, next slide in this slide we will see the the how the propagation of the sound is occur.
The propagation of sound is spherical; it is from a point source and it is going in
a spherical way. So, in a in a air or in a space it is going in a spherical way . So,
from this particular point s source is going in 3 dimension in this spherical way.
So, the intensity is the what output divided by the surface area of the sphere and as in
when you go move further further from the source to the sum further distance, your radius
of the the distance is assume as a radius your total sphere, that is going to increased,
your surface area of the sphere is going to increased and your total intensity is going
to decrease. So, if you are very near to the source the
density is very high, because of sphere the area of sphere is very small, if you go higher
distance larger distance further more from the source or so, your sphere is bigger, your
radius is bigger, your surface area of the sphere is bigger, your intensity is go
going down and you hear less amount of sound in the level point of view .
So, the intensity is can be written as W w is the output watt by 4 pi R square as you
know the 4 pi R square is the surface are of a sphere . Let us see small problem
suppose this is a sound source of sound as and it is giving a sound output as
0.005 very very small wattage . And, you are standing at the receiver is standing over
here which is 7 meter away . And, if it is so, I will find out what is the the surface
area of a sphere, which is having 7 meter of radius.
So, that is why I am finding out the intensity as this W by 4 pi R square and this is
like this and finally, it is 8.12 into 10 to the power minus 7 watt per meter square.
And, as and when you got this intensity, which you are actually experiencing here at a distance
7 meter by virtue of a sound output of 0.005 here, it is easy to convert these to the sound
level . In that very old formula the 10 log I by I reference is now your your here and
your listening 59 dB sound . Let us see the this particular problem or
another problem in a different way. Suppose this is a sound source and your with a dB
meter at point a your at c bar act A of the that sound and your receiving eighty dB at
A. The sound pressure sound pressure level at A or is 80 dB or the sound intensity level
is eighty dB at A distance 3 meter. The my question is what will this particular
person get which is who is standing in B? What will be the SPL or the SIL at B? Who
that is B is 12 meter from the sound source. I have I am going assume it is the
spherical propagation it is the point source. So, first find out what is the SPL is 80 dB.
So, first try to find out the what will be the intensity at this particular level by
virtue of this formula . So, by this it is 10 to the power minus 4,
with the intensity at here . If this is 10 to the power minus 4 watt per meter square,
which is on this particular sphere of radius of 3, then what is the wattage of the sound
output at the source? Which I can find out from this equation because W by 4 pi 3 square
is equal to this 10 to the power minus 4, and solving this I can get the watt output
or the power output of this particular source is 0.0113 1. And if this is the wattage or
the output wattage at the source, what could be the intensity at the B, which is at at
A distance of 12 meter 12 meter mix sphere. So, that I can find out this W is now this
and this 4 pi 12 square and this is the intensity level. We can compare the intensity is now
very very low with compare to the intensity as A. So, I am expecting something less than
80 dB over here of definitely and as I got this intensity B I B, it is very easy to find
out this particular level as your this old formula by log this by I reference at 68 dB
. So, this particular problem gives you how
from intensity level to thus taking consideration of the spherical wave front propagation, how
to find out the, what output? And from what output to again a intensity in some different
distance and from that you convert that intensity to a intensity level or the sphere .
So, the sound output from a suppose a source is W is s and it at a distance R 1, which
is nearer is the intensity is this and the SPL is suppose just I 1 by I reference. Now,
you go further further distance R 2, which is higher than R 1 and definitely you
will be experiencing less amount of intensity, because the sphere is larger area is larger
and your SPL 2 is your this. And, then let us see this is going to be happened
because your if your near to the source the intensity is high, your decibel is high SPL
1 will be higher than the SPL 2 . So, what is the difference between them I am going
to find out? So, difference is SPL 1 is this SPL 2 is this and if you just operate this
mathematical operation is 10 log A by B. So, I 1 by I reference by I 2 by I reference.
So, we can get it like this I reference get cancelled. So, finally, I got 10 log I 1 by
I 2 and now you replace the I 1 and I 2 from here and got SPL 1 minus SPL to the difference
between this 2 is R 2 by R 1 squares this 2 goes here.
So, 20 log R 2 by R 1 and suppose if R 2 equal to twice of R 1, if the distance is doubled,
R 2 and this ratio is 2 log 2 is 3 I am sorry log 2 is 0.3 0.3 into 26 dB .
So, in case of a spherical propagation, if the distance is double your decibel level
is going to drop by 6 Db. This is the final outcome of this particular the slide . Now,
spherical propagation and cylindrical propagation. Suppose, this is the unit source and it is
give me a spherical propagation and the examples of the spherical propagations are the some
loudspeakers some human voice, some noise from and small car or something like that,
but if suppose there are so, many point multiple point source in a very congested line or may
be very close less are distance, linear format.
So, then you can superimpose those are the sphericals and it will you will get a the
cylindrical propagation. So, the watt are the the examples of that may be a noise
from a moving rail, maybe a traffic noise, where there are lot of cars moving 1 after
another which is the point source, but very closely separated point sources in a road
or may be a array of loudspeaker in any theatre or any may be any any public address
gathering field. So, those are the the cylindrical propagation
example of the cylindrical propagation. So, we can actually say it is not as point it
is a cylinder. So, in case of a cylindrical propagation we
will see the the propagation is from a cylinder not from a point source so; that means, it
is a line source. So, the length of the line or the length of the source is L and by virtue
of it has in having a radius, because if you go further away from that line L your radius
is increasing your surface area of the cylinder is increasing.
So, there I will take the same principal I will what output that is the the power output,
I have to divide that by the surface area of the cylinder. And surface area of the cylinder
for virtue of this particular geometrical condition is 2 pi R, that is the perimeter
into the L . And, now this will be the intensity. So, suppose a length of the source
is 100 meter output is 0.005 watt what will be the SPL at 7 meter from the line source.
So, this we just convert with that I got this particle intensity level and again I will
take the particular equation intensity or the intensity reference, I will get 60.5 dB
as the the decibel level . Now, from the line source If, I go R 1 distance
from R 1 distance while get the intensity as the the cylinder of the radius R 1 of length
L, if I go to R 2 I get this value of I 2 and let us find find out the what is the difference
between the L the level at the at a distance R 1 and R 2 . And I know this R 1
will be higher the level sound level at this is higher, because it is nearer to the source.
So, again I will take the operation of that and very similarly I can find out the what
will be the values of I 1 and I 2 from here and what is the difference between the SPL
a modern SPL 2 and this gives me a formula like SPL 1 and SPL 2 is 10 log or 2 by R 1
. And, if the distance is doubled suppose R 2 is twice of R 1 then this is 10 log 2,
which is 3 dB . So, if it is a cylindrical propagation, if
the distance is doubled, the dB level drop by 3 dB, but if you remember in the in the
in the the spherical propagation it was 6 dB . So, next let us go to the near and far
field propagation. Suppose it is a it is a the the BC road
sections and that is gives me a cylindrical propagation.
But, why virtue of this propagation if you now go further in this away from the this
line away from the road . This way front is again going to get a spherical of suppose,
if you are very standing very far distance from the road since road noise will be act
as a point source 1 ones again. So, some point source gives you a line source
and if you are very near to that particular line source this is a cylindrical propagation.
And, now if you go further that particular line will act as a point and then it is trans
form to a or to a thus spherical propagation. So, there is a near field zone and there is
a transition zone and there are the far field. So, what happened in the far field zone? In
the far field zone sound is spherical propagation sound the the sound propagation is spherical
. What happened in the near field zone the propagation is cylindrical?
So, if in this near field zone I want to calculate something I have to assume that this source
is a cylinder linear source propagation is cylindrical . Suppose I am here, if I want
to find out the same effect of this particular road noise, then I have to assume this as
a point source. And I have to assume this is as a spherical propagation.
So, if the spherical and the cylindrical propagation or the near field and the th far field has
to be taken into account. So, in the near field zone it is drop by 3 dB by if the developed
the distance and in the far field which is spherical is 6 GB as per the distance is doubled.
So, by virtue of this there are 2 type of sources in case of any kind of the analysis
or any kind of the the measurement, if you take for any kind of sound at in the
the environmental sound. Then you have to know that we were in a far field or your
in a near field and you have to take the specific formula, and you to take the specific condition
and then analyzed . Now, next go to the the loudness, loudness
is a combination of the frequency and the SPL . Now, what happened if in the X axis
I have marked the all the octave band frequencies, it is the starting from the 31.5 to the 16,000
. And the SPL is actually written in the Y axis that is in the dB.
Now, the human being perceive sound differently suppose I plot some points. The first point
let us see this is the when 1000 hertz and 60 dB, it is gives me some kind of sensation,
some kind of sensation. And the sensation is very similar, when the the frequency of
the sound is 125 dP and more or less 65 65 dB and 125 hertz.
Sorry 125 hertz sound with 65 dB for 63 hertz, the same sensation what I am experiencing
here? I am getting the same sensation for 63 hertz at around 80 dB . So, if I just plot
all those point, which gives me a same similar type of sensation, with respect to 1000 hertz
and some amount of dB and I can actually create a contour line or a kind of a line.
And, this is a kind of a loudness it is a similar loudness or the same loudness . So,
loudness of the sound is vary from with this 2 fundamental parameter of frequency and the
SPL. So, this is the loudness and corresponding to this 1000 and 60, the loudness is this
particular loudness graph is called 60 Phon. So, by definition 1 Phon is equal to 1 dB
at 1000 hertz . Similarly, can plot lot of such plots of in
the frequency and the SPL level so, it is starts from 10, which is threshold of audibility
and it is goes up to 100 may be 120 also, which is the threshold pen and those blue
numbers are your Phons and this red lines are of your the equal or loudness path or
the equal loudness contour . So, you see in this I can say that this
particular band, which is 2000 to 8000 hertz is very sensitive this frequency is very sensitive
to our here, because you see this loudness contour curve dipping in dipping over here.
So, even a smallest amount of smallest amount of loudness can here in this particular frequency
zone . So, suppose if you want to here is 6 of 30 dB sound of this 4000, you required
this much of the sound level 22, but same with 63 you required almost about 60 dB you
cannot here, you cannot here . So, 30 loudness sound this is the equal loudness
contour also. Now, how this loudness can be measured. So,
suppose I say that I have a sound from a source which gives me 40 Phon. So, this is the curve
and another source gives me 60 Phon. So, this is the curve. So, can I say if I though both
the sound source if it is on switch on, the final sound which is combination of 60 and
40 the Phon is equal to 100 no it cannot be, it cannot be added like that be, because this
phon is a is a not it it is not a physical parameter. It is a kind of a psychological
parameter, what a human being is receive? Suppose this particular graph which we have
shown you, this is equal loudness contour is for a young gentleman. Suppose, if you
do the same experiment with a the old person of around age of 75 or so, this graph will
be shrink a bit he or she will perceive a different way different type of sum
graph some some some contour and maybe omitted. He or she may not here that particular phon.
So, it is fully psychologically or how the brain is going to map my the sound,
that way it particular it is devised or it is actually analyzed and derived. And that
is why this the unit of loudness this is phon cannot be added and it is not be justified
to add like that. So, we need to add because there are 2 different loudness and why and
this 2 different loudness has to be added to give a resultant count if the sound loudness.
So, we have to take another scale and that scale should be little bit of a linear scale
and that scale is called Sone scale. And in loudness 10 phon increase gives a increment
of the twice of multiplication factor of the sone. So, if I see this is the loudness loudness
plot that is the from 10 to 100 phon, the corresponding and this cannot be added this
phon cannot be added. So, this corresponding phon can be translated
to a linear format of sone, which is showing here in a blue numbers. So, the forty phon
is converted to 1 Sone and correspondingly, if you go to the next higher phon of 10 phon
higher 50, it is multiplied by 2 . If go to another 10 higher 60 this 2 is multiplied
by 2 4 and then 8 16 like that. So, the 100 phon is equivalent to 64 Sone
and now if you go beyond b 1 in below 40 by 10 phon recruitment we have to divide this
1 by 2.5, another 10 decrease 20 from 30.5 divided by 2.25 like that . So, it can be,
now mathematically express like Sone is 2 to the power phon minus 40 by 10 just put
suppose 60, if you put 60 over here 60 minus 40 is 20, 20 by 10 is to 2 to the power 2
is 4. Something like that you can with this particular
formula we can find out . And in the reciprocal way a phon can be find out by just
virtue of this particular equation . So, you just if you want to find out what is 16
Sone equivalent phon put 16 here log 16 multiplied by 33.2 2 and add 40 into it .
So, now we can actually go into this particular measurement. Now forty gives me 1, because
I put 40 over here. So, this is 1 very similarly I convert this 60 phon to it is corresponding
Sone level which is 4. Now, I cannot add phon 60 plus 40 cannot be
100, but I can add this sone. So, 4 plus 1 is 5 total sone, because of this 2 is 5. And
as I got this 5 Sone is the resultant Sone I can convert that Sone to the phon by that
equation I put this Sone as 5 log of 5 into 33.2 2 and I add with that with 40,
it is 63 . So, I can say it is not 100 phon, if you
add 40 and 60 it is 63 . So, that is ends this particular chapter this particular
lecture number 5, which is about the near and far field propagation and also about the
loudness and this fifth lecture also include the first week of our architectural acoustics
course, online course and this total 5 lectures comprises of some historical development of
the sound from the different era, which Doctor Suman Gupta has discussed in the
very first day, and are very first lecture, and then the 4 consecutive lecture on the
pure acoustical physics are the building physics. So, at the end of this lecture let us have
once again some homework and take away to your home let us try this.
The try this particular the formula the sorry this particular problem because this
will give you a kind of a understanding between the near field and far field, suppose a the
sound intensity level S I L and is 10 meter from the sound source SIL at at a distance
10 meter from the sound source, you have to find out what will be the the SIL at the 10
meter from the sound source. If the SIL at 3 meter from the sound source is 75 GB and
first of all you assume it is the near field propagation, where the sound propagate in
a cylindrical form and in second case you assume that has a far field propagation.
So, that is a spherical wave. So, in this particular videos or in this particular PPT
is is sometimes you may sit is a S I L I have written sometimes it is SPL, a sound pressure
level or sound intensity level, both are actually going to give in dB scale and these are related.
So, do not be confused with that thus it is name suggest, it is sound intensity level
and sometimes it is sound pressure level, but finally, by computation end result is
dB is the same . Take a second problem also for your homework,
that is can you find out the loudness of 2 different sound parameter, sound
source a gives you a frequency of 250 hertz and SPL of 60 Db, definitely this 2 combinations
gives you some kind of a phon and a sound level a sound source be gives you a of a frequency
of 4 kilo hertz and SPL of 70 dB, that also converted to a particular the phon
in loudness contour and then you add this 2 together both the the sound source A and
sound source B. The effective or the resultant the loudness
you have to find out . So, those are the references and that is the end of the lecture
number 5, hope you understand the lecture well and if you not go through the lecture
once more and you can also contact us and. Thank you very much .