Transcript Introduction to Loudness

What should we be reading??

Johnston
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Interlude - 2 piano
Interlude - 6 percussion
Chapter 7 – hearing, the ear, loudness
Appendix II – Logarithms, etc,
Initial Handout – Logarithms and Scientific Notation
Roederer
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2.3
the Ear
3.1, 3.2 material covered in class only
3.4 loudness (Friday)
Upcoming Topics
 Psychophysics
– Sound perception
– Tricks of the
musician
– Tricks of the mind
 Room
Acoustics
October 14,2005
The Process
At the Eardrum
Pressure wave arrives at the eardrum
 It exerts a force
 The drum moves so that WORK IS DONE
 The Sound Wave delivers ENERGY to the
EARDRUM at a measurable RATE.
 We call the RATE of Energy delivery a
new quantity: POWER

POWER
energy
Joule
Power 

 Watt
second second
Example: How much energy does a 60 watt light bulb consume
in 1 minute?
joules
60 watt  60
second
joules
60
 60 seconds  3600 J
second
We PAY for Kilowatt Hours
energy
KWH 
 time  energy
time
We PAY for ENERGY!!
More Stuff on Power
10 Watt
INTENSITY = power/unit area
Intensity
P
I
A
sphere :
P
I
2
4r
So….
ENERGY



Same energy (and power) goes through surface
(1) as through surface (2)
Sphere area increases with r2 (A=4r2)
Power level DECREASES with distance from the
source of the sound.
 Goes
as (1/r2)
To the ear ….
Area of Sphere =r2
=3.14 x 50 x 50
= 7850 m2
50m
30 watt
Ear Area = 0.000025 m2
Continuing
30watt
2
Power / Unit  Area 
 0.004w / m
2
7850m
Power to ear 
watt
.004 2  0.000025m 2
m
At Ear
power  .000000095 watts
Scientific Notation = 9.5 x 10-8 watts
Huh??
Move the decimal point
over by 8 places.
Scientific Notation = 9.5 x 10-8
Another example: 6,326,865=6.3 x 106
Move decimal point
to the RIGHT by 6 places.
REFERENCE: See the Appendix in the Johnston Test
Scientific Notation
Appendix 2 in Johnston
0.000000095 watts = 9.5 x 10-8 watts
Decibels - dB
 The
decibel (dB) is used to measure
sound level, but it is also widely used
in electronics, signals and
communication.
 It is a very important topic for
audiophiles.
Decibel (dB)
Suppose we have two loudspeakers, the first playing a
sound with power P1, and another playing a louder
version of the same sound with power P2, but
everything else (how far away, frequency) kept the
same.
The difference in decibels between the two is defined
to be
?
10 log (P /P ) dB
2
where the log is to base 10.
1
What the **#& is a logarithm?
 Bindell’s definition:
 Take a big number … like 23094800394





Round it to one digit: 20000000000
Count the number of zeros … 10
The log of this number is about equal to the number
of zeros … 10.
Actual answer is 10.3
Good enough for us!
Back to the definition of dB:
10 log (P2/P1)
The dB is proportional to the LOG10 of a
ratio of intensities.
 Let’s take P1=Threshold Level of Hearing
which is 10-12 watts/m2
 Take P2=P=The power level we are
interested in.

An example:

The threshold of pain is 1 w/m2
dB rating for the threshold of PAIN :
 1 
12
10 log  -12   10  log( 10 )  10  12  120
 10 
Another Example
Example :
1
1
2
 2  10  .01
100 10
Look at the dB Column
DAMAGE TO EAR
Continuous dB
85 dB
88 dB
91 dB
94 dB
97 dB
100 dB
103 dB
106 dB
109 dB
112 dB
115 dB
Permissible Exposure Time
8 hours
4 hours
2 hours
1 hour
30 minutes
15 minutes
7.5 minutes
3.75 min (< 4min)
1.875 min (< 2min)
.9375 min (~1 min)
.46875 min (~30 sec)
Can you Hear Me???
Frequency Dependence
Why all of this stuff???
We do NOT hear loudness in a linear
fashion …. we hear logarithmically
 Think about one person singing.

 Add
a second person and it gets a louder.
 Add a third and the addition is not so much.
 Again ….
Let’s look at an example.
 This
is Joe the
Jackhammerer.
 He makes a lot
of noise.
 Assume that he
makes a noise
of 100 dB.
At night he goes to a party with his
Jackhammering friends.
All Ten of them!
Start at the beginning

Remember those logarithms?
Take the number 1000000=106
The log of this number is the number of zeros or
is equal to “6”.
Let’s multiply the number by 1000=103
New number = 106 x 103=109
The exponent of these numbers is the log.

The log of {A (106)xB(103)}=log A + log B

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

9
6
3
Remember the definition
P
dB  10 log
P0
P0  10
100  10 log( P / 10
12
12
watt / m
2

)  10 log( P)  log( 10 )
100  10 log( P)  10 log( 1012 )
100  10 log( P)  120
10 log( P)  20
log( P)  2
2
P  10 Watt
12

Continuing On


The power level for a single jackhammer is 10-2
watt.
The POWER for 10 of them is
 10
x 10-2 = 10-1 watts.
 101 
dB  10 log  12   10 log( 1011 )  110
 10 
A 10% increase in dB!