OPTICAL FIBER Introduction Fiber optics deals with the light propagation through thin glass fibers. Fiber optics plays an important role in the field of communication to transmit voice, television and digital data signals fro one place to another. The transmission of light along the thin cylindrical glass fiber by total internal reflection was first demonstrated by John Tyndall in 1870 and the application of this phenomenon in the field of communication is tried only from 1927. Today the applications of fiber optics are also extended to medical field in the form of endoscopes and to instrumentation engineering in the form of optical sensors. The Basic principle of optical fiber Principle: The basic principle of optical fiber in the transmission of optical signal is total internal reflection. Total internal reflection:- When the light ray travels from denser medium to rarer medium the refracted ray bends away from the normal. When the angle of incidence is greater than the critical angle, the refracted ray again reflects into the same medium. This phenomenon is called total internal reflection. The refracted ray bends towards the normal as the ray travels from rarer medium to denser medium. The refracted ray bends away from the normal as it travels from denser medium to rarer medium. Conditions for Total Internal Reflection (a) the refractive index n1 of the core must always be greater than the refractive index n2 of the cladding. (b) The angle of incidence i must be greater than critical angle C it can be define as when light travels from a more optically dense material [larger index of refraction] to a less dense material the angle of refraction is larger than the incident angle.
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Transcript
OPTICAL FIBER
Introduction Fiber optics deals with the light propagation through thin glass fibers.
Fiber optics plays an important role in the field of communication to transmit voice,
television and digital data signals fro one place to another. The transmission of light
along the thin cylindrical glass fiber by total internal reflection was first demonstrated
by John Tyndall in 1870 and the application of this phenomenon in the field of
communication is tried only from 1927. Today the applications of fiber optics
are also extended to medical field in the form of endoscopes and to instrumentation
engineering in the form of optical sensors. The Basic principle of optical fiber
Principle: The basic principle of optical fiber in the transmission of optical signal is total
internal reflection.
Total internal reflection:-
When the light ray travels from denser medium to rarer medium the refracted
ray bends away from the normal. When the angle of incidence is greater than the
critical angle, the refracted ray again reflects into the same medium. This
phenomenon is called total internal reflection.
The refracted ray bends towards the normal as the ray travels from rarer medium
to denser medium. The refracted ray bends away from the normal as it travels
from denser medium to rarer medium.
Conditions for Total Internal Reflection
(a) the refractive index n1 of the core must always be greater than the refractive index n2
of the cladding.
(b) The angle of incidence i must be greater than critical angle C
it can be define as when light travels from a more optically dense material [larger index of
refraction] to a less dense material the angle of refraction is larger than the incident angle.
Because the refracted angle is always larger than the incident angle, it is possible for the
refracted angle to reach 90° before the incident angle reaches 90°. If the light were to refract out
of the denser medium, it would then run along the surface. Larger angles would then yield
situations which would force the sine function to be larger than 1.00, which is mathematically
impossible.
When the incident angle reaches the condition whereby the refracted ray would bend to an angle
of 90°, it is called the CRITICAL ANGLE. The critical angle obeys the following equation:
This reflected ray changes in intensity as we vary the angle of incidence. At small incident
angles (almost perpendicular to the surface) the reflected ray is weak and the refracted ray is
strong.
Construction of optical fiber:-
The optical fiber mainly consists the following six parts as shown in figure Core:
A typical glass fiber consists of a central core material. Generally core
diameter is 50 . The core is surrounded by cladding. The core medium
refractive is always greater than the cladding refractive index.
Cladding Cladding refractive index is lesser than the cores refractive index. The
over all diameter of cladding is 125 to 200 .
Silicon Coating Silicon coating is provided between buffer jacket and cladding. It
improves the quality of transmission of light.
Buffer Jacket Silicon coating is surrounded by buffer jacket. Buffer jacket is made of
plastic and protects the fiber cable from moisture.
Strength Member Buffer jacket is surrounded by strength member. It provides strength to the
fiber cable.
Outer Jacket Finally the fiber cable is covered by polyurethane outer jacket. Because
of this arrangement fiber cable will not be damaged during pulling,
bending, stretching and rolling through the fiber cable is made up of glasses.
NA & ACCEPTANCE ANGLE DERIVATION
“In optics, the numerical aperture (NA) of an optical system is a dimensionless number
that characterizes the range of angles over which the system can accept or emit light.”
optical fiber will only propagate light that enters the fiber within a certain cone, known as
the acceptance cone of the fiber. The half-angle of this cone is called the acceptance angle,
θmax. on
where n1 is the refractive index of the fiber core, and n2 is the refractive index of the cladding.
When a light ray is incident from a medium of refractive index n to the core of index n1, Snell's
law at medium-core interface gives
From the above figure and using trigonometry, we get :
Where is the critical angle for total internal reflection, since
Substituting for sin θr in Snell's law we get:
By squaring both sides
Thus,
from where the formula given above follows.
θmax =
This has the same form as the numerical aperture in other optical systems, so it has become
common to define the NA of any type of fiber.
Definition:-
Acceptance angle:- Acceptance angle is defined as the maximum angle of incidence at the interface
of air medium and core medium for which the light ray enters into the core and
travels along the interface of core and cladding.
Acceptance Cone:-
There is an imaginary cone of acceptance with an angle .The light that enters the fiber at
angles within the acceptance cone are guided down the fiber core Numerical aperture:- Numerical aperture is defined as the light gathering capacity of an optical fiber and it is
directly proportional to the acceptance angle.
Numerically it is equal to the sin of the acceptance angle
Classification of fibers:-
Based on the refractive index of core medium, optical fibers are classified into
two categories.
i. Step index fiber ii. Graded index fiber
Based on the number of modes of transmission, optical fibers are classified into
two categories
i. Single mode fiber
ii. Multimode fiber
Based on the material used, optical fibers are may broadly classified into four
categories
i. All glass fibers
ii. All plastic fibers iii. Glass core with plastic cladding fibers iv.
Polymer clad silica fibers.
Step index fiber:- In step index fibers the refractive index of the core medium is uniform and
undergoes an abrupt change at the interface of core and cladding as shown in figure.
The diameter of core is about 10micrometers in case of single mode fiber and 50 to 200
micrometers in multi mode fiber. Attenuation is more for step index multi mode fibers but less in step index single
mode fibers
Numerical aperture is more for step index multi mode fibers but it is less in step
index single mode fibers
Graded index fiber:- In graded index fibers, the refractive index of the core medium is varying in the
parabolic manner such that the maximum refractive index is present at the center
of the core.
The diameter of the core is about 50 micro meters.
Attenuation is very less in graded index fibers Numerical aperture is less in graded index
fibers Graded index Figure Two types of fiber: (Top) step index fiber; (Bottom) Graded index fiber
Single mode optical fiber In single mode optical fibers only one mode of propagation is possible.In case of single mode
fiber the diameter of core is about 10micrometers.The difference between the refractive
indices of core and cladding is very small. In single mode fibers there is no dispersion,
so these are more suitable for
Communication. The single mode optical fibers are costly, because the fabrication is
difficult.The process of launching of light into single mode fibers is very difficult.
Multi mode optical fiber
In multi mode optical fibers many mummer of modes of propagation are possible. In case of
in multi mode fiber the diameter of core is 50 to 200 micrometers. The difference between
the refractive indices of core and cladding is also large compared to the single mode
fibers. Due to multi mode transmission, the dispersion is large, so these fibers are not used
for communication purposes. The multi mode optical fibers are cheap than single mode
fibers, because the fabrication is easy. The process of launching of light into single mode
fibers is very easy.
Based on the material:- Three common type of fiber in terms of the material used:
Glass core with glass cladding –all glass or silica fiber Glass core with plastic cladding –plastic cladded/coated silica (PCS) Plastic core with plastic cladding – all plastic or polymer fib
Attenuation:-
Definition: a loss of signal strength in a lightwave, electrical or radio signal usually related to
the distance the signal must travel.
Attenuation is caused by:
Absorption
Scattering
Radiative loss
Losses:-
Losses in optical fiber result from attenuation in the material itself and from scattering,
which causes some light to strike the cladding at less than the critical angle
Bending the optical fiber too sharply can also cause losses by causing some of the light to
meet the cladding at less than the critical angle
Losses vary greatly depending upon the type of fiber
Plastic fiber may have losses of several hundred dB per kilometer
Graded-index multimode glass fiber has a loss of about 2–4 dB
per kilometer
Single-mode fiber has a loss of 0.4 dB/km or less
Macrobending Loss:
The curvature of the bend is much larger than fiber diameter. Lightwave suffers sever loss due
to radiation of the evanescent field in the cladding region. As the radius of the curvature
decreases, the loss increases exponentially until it reaches at a certain critical radius. For any
radius a bit smaller than this point, the losses suddenly becomes extremely large. Higher order
modes radiate away faster than lower order modes.
Microbending Loss:
microscopic bends of the fiber axis that can arise when the fibers are incorporated into cables.
The power is dissipated through the microbended fiber, because of the repetitive coupling of
energy between guided modes & the leaky or radiation modes in the fiber.
Dispersion:-
The phenomenon in an optical fibre whereby light photons arrive at a distant point in different
phase than they entered the fibre. Dispersion causes receive signal distortion that ultimately
limits the bandwidth and usable length of the fiBer cable
The two main causes of dispersion are:
Material (Chromatic) dispersion
Waveguide dispersion
Intermodal delay (in multimode fibres)
Dispersion in fiber optics results from the fact that in multimode propagation, the signal travels
faster in some modes than it would in others.Single-mode fibers are relatively free from
dispersion except for intramodal dispersion .Graded-index fibers reduce dispersion by taking
advantage of higher-order modes.One form of intramodal dispersion is called material
dispersion because it depends upon the material of the core.Another form of dispersion is called
waveguide dispersion .Dispersion increases with the bandwidth of the light source
The advantage of fiber optic cable over metallic cable:-
1. Extremely wide (large) bandwidth.
The bandwidth available with a single glass fibre is more than 100GHZ. With such a large
bandwidth, it is possible to transmit thousands of voice conversations or dozens of video signals
over the same fibre simultaneously. Irrespective of whether the information is voice, data or
video or a combination of these, it can be transmitted easily over the optical fibre. Less no of
independent signals alone can be sent through metallic cables.
2. Immunity to electrostatic interference.
As optical fibres are being made of either glass or plastic external electric noise and lightning
do not affect the energy in a cable. The result is noise free transmission. While this is not true
for metallic cables made up of metals, as they are good conductors of electricity.
3. Elimination of cross Talk.
Fibre systems are immune to cross talk between cables caused by magnetic induction. Whereas
in a metallic cable cross talk results from the electromagnetic coupling between two adjacent
wires.
4. Lighter weight and smaller size.
Fibres are very smaller in size. This size reduction makes fibre the ideal transmission medium
for ships, aircraft and high rise buildings where bulky copper cables occupy to much space.
Reduction in size so reduction in weight also.
5. Lower cost.
The material used in fibres is silica glass or silicon dioxide which is one of the most abundant
materials on earth. So available in lower cost.
6. Security.
Fibre cables are more secure than metallic cables. Due to its immunity to electromagnetic
coupling and radiation, optical fibre can be used in most secure environment. Although it can be
intercepted or tapped, it is very difficult to do so because, at the receiving users end an alarm
would be sounded.
7. Greater safety.
In many wired system the potential hazard of short circuits requires precautionary designs.
Whereas, the dielectric nature of optical fibres eliminates the spark hazard.
8. Corrosion
Fibre cables are more resistive to environmental extremes. They operate over large temperature
variation than their metallic counter parts, and are less affected by corrosive liquids and gases.
9. Longer life span and ease of maintenance.
A longer life span of 20 to 30 years is predicted for the fibre optic cables as compare to 12to 15
years of metallic cables.
Differences between step index fibers and graded index fibers:-
Step index fiber
Graded index fiber
1. In step index fibers the refractive index of the
core medium is uniform through and
undergoes an abrupt change at the interface of
core and cladding.
1. In graded index fibers, the refractive index of
the core medium is varying in the parabolic
manner such that the maximum refractive index
is present at the center of the core.
2. The diameter of core is about
10micrometers in case of single mode fiber and
50 to 200 micrometers in multi mode fiber.
2. The diameter of the core is about 50 micro meters.
3. The transmitted optical signal will cross the
fiber axis during every reflection at the core
cladding boundary.
3. The transmitted optical signal will never cross
the fiber axis at any time.
4. The shape of propagation of the optical
signal is in zigzag manner.
4. The shape of propagation of the optical signal
appears in the helical or spiral manner
5. Attenuation is more for multi mode step
index fibers but Attenuation is less in single
mode step index fibers
5. Attenuation is very less in graded index fibers
6. Numerical aperture is more for multi
mode step index fibers but it is less in single
mode step index fibers
6. Numerical aperture is less in graded index fibers
Differences between single mode fibers and Multy mode fibers:-
Single mode fiber Multimode fiber
Single Mode cable is a single strand (most
applications use 2 fibers) of glass fiber with a
diameter of 8.3 to 10 microns that has one mode of
transmission.
Multi-Mode cable has a little bit bigger diameter,
with a common diameters in the 50-to-100 micron
range for the light carry component
Single Modem fiber is used in many applications
where data is sent at multi-frequency (WDM
Wave-Division-Multiplexing) so only one cable is
needed
Most applications in which Multi-mode fiber is used,
2 fibers are used (WDM is not normally used on
multi-mode fiber).
Example:- step index fiber Example:- multimode step index fiber
The small core and single light-wave virtually
eliminate any distortion that could result from
overlapping light pulses, providing the least signal
attenuation and the highest transmission speeds of
any fiber cable type.
multiple paths of light can cause signal distortion at
the receiving end, resulting in an unclear and
incomplete data transmission
Applications of optical fibers
1. Optical fibers are extensively used in communication system. 2. Optical fibers are in exchange of information between different computers 3. Optical fibers are used for exchange of information in cable televisions,
space vehicles, submarines etc.
4. Optical fibers are used in industry in security alarm systems, process control
and industrial auto machine.
5. Optical fibers are used in pressure sensors in biomedical and engine control. 6. Optical fibers are used in medicine, in the fabrication in endoscopy for
the visualization of internal parts of the human body.
7. Sensing applications of optical fibers are
Displacement sensor
Fluid level detector Liquid
Temperature and pressure sensor
Chemical sensors
8. Medical applications of optical fibers are
Gastroscope
Orthoscope Couldo
EXAMPLE:-
1.
A silica optical fiber has a core of refractive index 1.55 and a cladding of refractive index
1.47. Determine (i) the critical angle at the core-cladding interface (ii) the numerical
aperture for the fiber and (iii) the acceptance angle in the air for the fiber.
Given,
n1=1.55,
n2=1.47
Øin(max)=?
NA=?
Øc=?
Acceptance angle Øin(max)= sin-1
(n12 – n2
2)
1/2
Øin(max)= sin-1
(1.552 –1.47
2)
1/2
= sin-1
(2.41-2.16)1/2
= sin-1
(0.25)1/2
=
sin
-1 (0.316)
Øin(max) =30°00’
Numerical aperture NA= (n12 – n2
2)
1/2
= 1.552 –1.47
2)
1/2
= (2.41-2.16)1/2
= (0.25)1/2
= 0.316
critical angle Øc = sin-1
(n2 / n1)
= sin-1
(1.47 / 1.55)
= sin-1
(0.9483)
= 71°.55’
2.
An optical fiber has refractive index of core and cladding is 1.514 and 1.48 Respectively.
Calculate the acceptance angle and the fractional index Change
Given
,n1=1.514,
n2=1.48
Øin(max)=? ∆=?
Acceptance angle Øin(max ) = sin-1
(n12 – n2
2)
1/2
Øin(max) = sin-1
(1.5142 –1.48
2)
1/2
= sin-1
(2.29-2.19)1/2
= sin
-1 (0.1)
1/2
=
sin
-1 (0.316)
Øin(max)=18°42’
Numerical aperture NA= (n12 – n2
2)
1/2
=(1.5142 –1.48
2)
1/2
=(2.29-2.19)1/2
=(0.1)1/2
=0.316
NA=n1√2∆
0.316/1.514=√2∆
(0.2087)2=2∆
∆=0.0435/2
∆=0.0217
Dielectrics
Introduction
Dielectrics are the materials having electric dipole moment permanently.
Dipole: A dipole is an entity in which equal positive and negative charges are separated by a small distance..
DIPOLE moment (µEle ):The product of magnitude of either of the charges and separation distance b/w them
is called Dipole moment.
µe = q . x coulmb.m
All dielectrics are electrical insulators and they are mainly used to store electrical energy.
Ex: Mica, glass, plastic, water & polar molecules…
Dielectric const. of medium
The relative permittivity(εr) is often known as dielectric const. of medium it can given by,
εr=ε/ε0
Dielectric constant is ratio of permittivity of medium to permittivity of free space.
The value of capacitance of capacitor is given by,
C0=εrε0A/d
By this eqn we can say that high εr increases capacity of capacitor.
Polar and Nonpolarized Molecules
Non-polar Molecules : The Dielectric material in which there is no permanent dipole existence in absence
of an external field is …..
2 – Compounds made of molecules which are symmetrically shaped
Polar Molecules :The Dielectric material in which there is permanent dipole existence even in absence of
an external field is …..
Polarization of Dielectrics
As shown in fig. when an electric field is applied to dielectric material their negative & positive charges tend
to align in equilibrium position.
Gauss’s Law In Dielectrics
In absence of dielectric In presence of dielectric
0 0
d
0
0
0 0
0 0 0
0
0
E Vk
E V
E qE
k kA
q q 'E
A A
q q q 'So,
kA A A
1then , q ' q (1 )
k
So, E.ds
V Ed
S
q q '
1q q (1 )
k
q
k
k E.ds q
o,
N ow
This relation true is for parallel plate capacitor Which is Gauss’s law for dielectrics.
0
0
0
0
0
E.ds q
qE A
qE
A
0
0 0
0 0
E.ds q q '
q q 'E A
q q 'E
A A
Three Electric vectors
The resultant dielectric field is given by,
Where,
E=Electric field
D=Flux Density or
Displacement vector
P=Polarization
Electric susceptibility:
The polarization vector P is proportional to the total electric flux density and direction of
electric field.
Therefore the polarization vector can be written as:
Relation between εr &
Displacement vector,
Types of polarization
1. Electron polarization
2. Ionic polarization
3. Orientation polarization
4. Space charge polarization
0 0
0 0
0
0
'
',
,
, D
p
q qE
A A
qnow P
A
q PE
A
qE P
A
qnow D
A
So E P
0
0
0
0
( 1)
1
e
e
r
e r
P E
P
E
E
E
0
0
0
r 0 0
0
D E P
N ow ,P=
( - ) E P
(or) ( . - ) E P
( 1) . P
W here,( 1)
r
r
E
E
1. Electronic polarization
When no external field is applied nucleus of atom is like in fig. (a)
When external field is applied, displacement in opposite direction is observed between nucleus &
electrons due to this dipole moment is induced.
This type of polarization is called Electronic polarization.
Ex. Germanium, Silicon, Diamond etc…
2. Ionic polarization
Some materials like ionic crystal does not possess permanent dipole moment.
Fig. (a) shows natural arrangement of ionic crystal. When Ele. Field is applied on this type of
material displacement of ions is observed.
Due to an external electric field a positive & negative ion displaces in the direction opposite to
each other due to which distance between them is reduced & ionic polarization is generated.
Ionic polarization is observed in materials like NaCl, KBr, KCl etc…
Let us consider simple example of NaCl crystal.
As shown in fig. when crystal is placed in an external electric field Na+ ion displaces in one
direction & Cl- ion goes in opposite direction.
3. Orientation polarization
Some molecules like H2O, HCl having permanent dipole moment p0.
In the absence of a field, individual dipoles are arranged in random way, so net average dipole
moment in a unit volume is zero as shown in fig. (b).
A dipole such as HCl placed in a field experiences a torque that tries to rotate it to align p0 with
the field E.
In the presence of an applied field, the dipoles try to rotate to align parallel to each other in
direction of electric field fig (d).
This type of polarization is Orientation polarization.
This type of polarization occurs only in polar substances like H2O, CH3Cl when they are placed
in external field.
4. Space charge polarization (Interfacial polarization)
A crystal with equal number of mobile positive ions and fixed negative ions.
In the absence of a field, there is no net separation between all the positive charges and all the
negative charges.
In the presence of an applied field, the mobile positive ions migrate toward the negative charges
and positive charges in the dielectric.
The dielectric therefore exhibits Space charge or interfacial polarization.
Energy stored in dielectric field
Work done is,
.
?
.
.
dW F dr
F
dW qE dr
dW E dp
p pP
lA V
0
0
0
2
0
2
0
( 1) .
. .( 1) .
. .( 1) .
1( 1) E
2
1( 1) E
2
?
r
r
r
r
r
p PV
dW EVdP
P E
dW E V dE
dW E V dE
W V
W
V
U
Band Theory of Solid
Objectives
• Effective Mass of electron
• Concept of Holes
• Energy Band Structure of Solids:
Conductors, Insulators and Semiconductors
• Semiconductors
Intrinsic and Extrinsic Semiconductors
• Type of diodes
Simple Diode
Zener Diode
Effective Mass of electron
An electron moving in the solid under the influence of the crystal potential is subjected to
an electric field.
We expect an external field to accelerate the electron, increasing E and k and change the
electron’s state.
------ (1)
But, dx/dt = vg ------ (2)
------ (3)
dk
dgv
1
gvdx
dVe
dt
dk
dk
d
dt
dx
dx
dVe
dt
dVe
dt
d
eV
and
----- (4)
------ (5)
------ (6)
------ (7)
----- (8)
Concept of Holes
Consider a semiconductor with a small number of electrons excited from the valence
band into the conduction band.
If an electric field is applied,
• The conduction band electrons will participate in the electrical current
• The valence band electrons can “move into” the empty states, and thus can also
contribute to the current.
If we describe such changes via “movement” of the “empty” states – the picture will be
significantly simplified. This “empty space” is called a Hole.
“Deficiency” of negative charge can be treated as a positive charge.
dx
dVek
dt
d
gvdx
dVe
dt
dkgv
eEkdt
d
dt
dk
dk
d
dk
d
dk
d
dt
d
dt
dva
g
11
k
dt
d
dk
d
dt
dk
dk
d
2
2
22
211
Holes act as charge carriers in the sense that electrons from nearby sites can “move” into
the hole.
Holes are usually heavier than electrons since they depict collective behavior of many
electrons.
To understand hole motion, one requires another view of the holes, which represent them
as electrons with negative effective mass m*.
For example the movement of the hole think of a row of chairs occupied by people with
one chair empty, and to move all people rise all together and move in one direction, so the
empty spot moves in the same direction.
Energy Band Structure of Solids Conductor, Semiconductor and Insulator
In isolated atoms the electrons are arranged in energy levels.
Energy Band in Solid
The following are the important energy band in solids:
Valence band
Conduction band
Forbidden energy gap or Forbidden band
Valance band
The band of energy occupied by the valance electrons is called valence band. The
electrons in the outermost orbit of an atom are known as valance electrons. This band may be
completely or partial filled.
Electron can be move from one valance band to the conduction band by the
application of external energy.
Conduction band
The band of energy occupied by the conduction electrons is called conduction
band. This is the uppermost band and all electrons in the conduction band are free electrons.
The conduction band is empty for insulator and partially filled for conductors.
Forbidden Energy Gap or Forbidden band
The gap between the valance band and conduction band on energy level diagram
known as forbidden band or energy gap.
Electrons are never found in the gap. Electrons may jump from back and forth
from the bottom of valance band to the top of the conduction band. But they never come to rest
in the forbidden band.
According to the classical free electron theory, materials are classified in to three types:
Conductors
Semiconductors
Insulators
Conductors
There is no forbidden gap and the conduction band and valence band are
overlapping each other between and hence electrons are free to move about. Examples are Ag,
Cu, Fe, Al, Pb ….
Conductor are highly electrical conductivity
So, in general electrical resistivity of conductor is very low and it is of the order of 10-6
Ω
cm.
Due to the absence of the forbidden gap, there is no structure for holes.
The total current in conductor is simply a flow of electrons.
For conductors, the energy gap is of the order of 0.01 eV.
Semiconductors:
Semiconductors are materials whose electrical resistivity lies between insulator
and conductor. Examples are silicon (Si), germanium (Ge) ….
The resistivity of semiconductors lies between 10-4
Ω cm to 103 Ω cm at room
temperature.
At low temperature, the valence band is all most full and conduction band is almost
empty. The forbidden gap is very small equal to 1 eV.
Semiconductor behaves like an insulator at low temperature. The most commonly used
semiconductor is silicon and its band gap is 1.21 eV and germanium band gap is 0.785
eV.
When a conductor is heated its resistance increases; the atoms vibrate more and
the electrons find it more difficult to move through the conductor but, in a semiconductor
the resistance decreases with an increase in temperature. Electrons can be excited up to the
conduction band and Conductivity increases.
Insulators
Here the valence band is full but the conduction band is totally empty. So, a free electron
from conduction band is not available.
In insulator the energy gap between the valence and conduction band is very large and
it’s approximately equal to 5 eV or more.
Hence electrons cannot jump from valence band to the conduction band. So, a very high
energy is required to push the electrons to the conduction band.
Therefore the electrical conductivity is extremely small.
The resistivity of insulator lie between 103 to 10
17 Ωm, at the room temperature
Examples are plastics, paper …..
Types of semiconductors
Intrinsic Semiconductor
The intrinsic semiconductors are pure semiconductor materials. These
semiconductors possess poor conductivity. The elemental and compound semiconductor can be
intrinsic type. The energy gap in semiconductor is very small. So, even at the room temperature,
some of electrons from valance band can jump to the conduction band by thermal energy.
The jump of electron in conduction band adds one conduction electron in
conduction band and creates a hole in the valence band. The process is called as “generation of
an electron–hole pair”.
In pure semiconductor the no. of electrons in conduction band and holes in holes
in valence bands are equal.
Extrinsic Semiconductor
Extrinsic semiconductor is an impure semiconductor formed from an intrinsic
semiconductor by adding a small quantity of impurity atoms called dopants.
The process of adding impurities to the semiconductor crystal is known as doping.
This added impurity is very small of the order of one atom per million atoms of
pure semiconductor.
Depending upon the type of impurity added the extrinsic semiconductors are
classified as:
• p – type semiconductor
• n – type semiconductor
p – type semiconductor
The addition of trivalent impurities such as boron, aluminum or gallium to
an intrinsic semiconductor creates deficiencies of valence electrons, called "holes". It is typical to
use B2H6 di-borane gas to diffuse boron into the silicon material.
n – type semiconductor
The addition of pentavalent impurities such as antimony, arsenic or phosphorous
contributes free electrons, greatly increasing the conductivity of the intrinsic semiconductor.
Phosphorous may be added by diffusion of phosphine gas (PH3).
Simple Diode
The two terminals are called Anode and Cathode. At the instant the two materials
are “joined”, electrons and holes near the junction cross over and combine with each other.
Holes cross from P-side to N-side and free electrons cross from N-side to P-side.
affecting acoustics of building and their remedies
Introduction :
Sound is always produced by some vibrating body. The vibrating
body generates mechanical waves and these waves spreads in the
surrounding medium. We are aware that these waves propagate in the
form of a series of compressions and rarefactions in air or the surrounding
medium. When reached upto the human ear drum it causes a sensation of
hearing. As far as architectural acoustics are concerned, we are interested
the combined effect of sound waves which creates a sense of sound on
human ear.
Some important characteristics
(1) The propagation of sound requires the presence of an elastic
medium.
(2) Sound can not travel through vacuum
(3) The compression and rarefactions due to a sound modulate the
normal atmospheric pressure with small pressure changes occuring
regularly above and below it.
(4) The velocity of sound depends on the nature and temperature of the
medium.
1-2 Architectural Acoustics
1.1 Classification of Sound :
Based upon frequency of sound waves, it can be classified into its
three main categories.
(a) Audible waves : Sound waves with frequency in the range of
20Hz to 20KHz.
(b) Infrasonic waves : Sound waves below audible range i.e. below
20Hz.
(c) Ultrasonic waves : Sound waves above audible range i.e.
20KHz.
1.1.1 Characteristics of Musical Sound :
Musical sounds & Noise sound:-
Musical sound are distinguished from noises in that they are
composed of regular, uniform vibrations, while noises are irregular
and disordered vibrations. One musical tone is distinguished from
another on the basis of pitch, intensity, or loudness and quality, or
timbre.
Noise sound that Produce Jarring effect on the ear is called Noise
sound.Noice sound make unpleasent to hear .Example are sound
produce by flying aeroplane,road traffic,cracker etc
Pitch describes how high or low a tone is and depends upon the
rapidity with which a sounding body vibrates, i.e. upon the frequency
of vibration. The higher the frequency of vibration, the higher the tone;
the pitch of a siren gets higher and higher as the frequency of vibration
increases. The apparent change in the pitch of a sound as a source
approaches or moves away from an observer is described by the Doppler
effect.
The intensity or loudness of a sound depends upon the extent to
which the sounding body vibrates, i.e. the amplitude of vibration. A
1-3 Architectural Acoustics
sound is louder as the amplitude of vibration is greater, and the intensity
decreases as the distance from the source increases. Loudness is measured
in units called decibels.
Timber is the Quality of the sound which Enable us to
distinguish between two sound having the same loudness & pitch.The
sound waves given off by different vibrating bodies differ in quality, or
timbre. A note from a saxophone, for instance, differs from a note of the
same pitch and intensity produced by a violin or a xylophone; similarly
vibrating reeds, columns of air, and strings all differ. Quality is dependent
on the number and relative intensity of overtones produced by the
vibrating body (see harmonic), and these in turn depend upon the nature
of the vibrating body.
1.2 Important Terms Used :
In the study of sound waves we come across various terms like Pitch
(This law does not hold good near the upper and lower limits of
audiability), Timber which basically deals with the quality of the sound
waves and source. At the same time for technical assessment, we make
use of important parameters like intensity and loudness.
1.2.1 Weber Fechner Law :
This law has it roots hidden in psychology and proved scientifically
according to which : The loudness of sound sensed by ear is directly
proportional to logarithm of its intensity.
According to Weber-Fechner law :
Suppose the loudness is S for intensity I and S0 for intensity I0,
S = K log10 I
S0 = K log10 I0
The intensity level L is the difference in loudness.
L = S – S0
1-4 Architectural Acoustics
= K log10 I – K log10 I0 = K log10 I
I0
take, K = 1
L = log10 I
I0 …(1.1)
Intensity and loudness are the two words which are similar but with
slight difference.
Table 1.1
Sr.
No.
Intensity Loudness
1. Defined as the quantity of energy
propagating through a unit area per
unit time, in the direction of
propagation being perpendicular to
the area (unit : watt/m2).
It is just an aural
sensation and it a
physiological
phenomenon rather than a
physical one.
2. It refers to the external or the
objective measurement.
It refers to an internal or
subjective aspect.
3. It is a physical quantity. Merely a degree of
sensation.
Loudness ‘S’ increases with intensity ‘I’ as per the following
relation*
or S log I …(1.2)
dS
dI =
K
I …(1.3)
Where K is proportionality constant
Here ds
dI is called the sensitiveness of the ear.
In practice, it is the relative intensity that is important and not the
absolute value. Hence the intensity of sound is often measured as the
ratio to a standard intensity I0. The intensity level is I / I0.
1-5 Architectural Acoustics
The standard intensity taken is I0 = 10–12
watts / m2. (It is an
arbitrarily selected value. It is an intensity that can just be heard
at frequency 1 kHz)
1.2.2 Bel :
As discussed in art 1.2.1, whenever the intensity of sound increases
by a factor of 10, the increase in the intensity is said to be 1 bel (A unit named after Alexander Graham Bell, the inventor of telephone)
Therefore dynamic range of audibility of the human ear is 12 bels or
120 dB. When the intensity increases by a factor of 100.1
, the
increase in intensity is 0.1 bel or 1dB.
From Equation 1.1
L = log10 I
I0
in decibel
L = 10 log10 I
I0
For the intensity level change = 1 dB
1 = 10 log10 I
I0
I
I0 = 1.26 …(1.4)
If I = I0,
L = 10 log 1 = 0
This represents the threshold of audibility.
It means that intensity level alters by 1dB when intensity of sound
changes by 26%
1-6 Architectural Acoustics
Table 1.2 : Intensity levels of different sounds
Sr. No. Sound Intensity level (in db)
(1) Threshold of hearing 0
(2) Rustle of leaves 10
(3) Whisper 15 – 20
(4) Normal conversation 60 – 65
(5) Heavy traffic 70 – 80
(6) Thunder 100 – 110
(7) Painful sound 130 and above
1.2.3 Phon :
The intensity levels given in the above Table 1.2 refer to the
loudness in decibels with the assumption that the threshold of
audibility is the same irrespective of the pitch (Pitch is a subjective
sensation perceived when a tone of a given frequency is sounded. It
enables us to classify a note as high or low and to distinguish a shrill
sound from a flat sound of the same intensity on the same instrument.) of the sound.
However, the sensitivity of the ear and the threshold audibility vary
over wide ranges of frequency and intensity.
Hence the intensity level will be different at different frequencies
even for the same value of I0.
For measuring the intensity level a different unit called phon is used.
The measure of loudness in phons of any sound is equal to the
intensity level in decibels of an equally loud pure tone of frequency
1000 Hz.
Hence Phon scale and decibel scale agree for a frequency of 1000 Hz
but the two values differ at other frequencies.
Suppose the intensity level of a note of frequency 480 Hz is to be
determined. A standard source of frequency 1000 Hz is sounded and
the intensity of the standard source is adjusted so that it is equal to
1-7 Architectural Acoustics
the loudness of the given note of frequency
480 Hz.
The intensity level of the standard source in decibels is numerically
equal to the loudness of the given source in phons.
Ex. 1.1 : Calculate the change in intensity level when the intensity of
sound increases 100 times its original intensity.
Soln. :
Given :
Initial intensity = I0
Final intensity = I
I
I0 = 100
Increase in intensity level = L
L = 10 log10 I
I0 (in dB)
L = 10 log10 100 = 20 dB …Ans.
Ex. 1.2 : Find the intensity level in phons if 3000 Hz with intensity level
of 70 dB produces the same loudness as a standard source of
frequency 1000 Hz at a intensity level 67 dB.
Soln. :
As the 3000 Hz source has the same loudness of standard source of
1000 Hz with 67 db, the intensity level of the note of frequency 3000 Hz
is 67 phons. …Ans.
1.3 Architectural Acoustics :
Lets try to understand what exactly acoustics of a hall means.
Consider the following cases :
(a) Imagine a hall, it is easy for any one to understand that sound
produced at a point will reach the other point directly as well as after
reflections from walls, roof etc. The intensity of the sound depends
1-8 Architectural Acoustics
on the distance covered by sound on different paths. These sounds
are generally out of phase and due to interference the distribution of
intensity in the room is not uniform.
(b) It is also important to consider a possibility that the different
frequency sounds of a musical instrument may interfere differently at
some point and quality of music may become unpleasent.
(c) It is known that sound persists for some time due to multiple
reflections, even when the original sound has ceased. During this
time if any other syllable is received, superimposition of these two
will affect audiability as both will remain indistinct. If this takes
place during a speech, a confusion will be created.
(d) Concentration of sound taking place at any part of the hall.
The above mentioned points are very common but needs a special
scientific attention. Prof. W.C. Sabine was the first person who took it
seriously.
1.4 Reverberation Time :
Reverberation means the prolonged reflection of sound from walls,
floor or roof of a hall. In simple language it is nothing but persistence of
sound even after the sources of the sound has stopped.
Reverberation time :
The time gap between the initial direct note and the reflected
note upto a minimum audibility level is called reverberation time.
More precisely, the interval of time taken by a sustained or
continuous sound to fall to an intensity level equal to one millionth
of its original value. (i.e. fall by 60 db in loudness) is called
reverberation time.
In a good auditorium it is necessary to keep the reverberation time as
small as possible. The intensity of the sound as received by listener
is shown graphically in Fig. 1.1.
1-9 Architectural Acoustics
Fig. 1.1
When a source emits sound, the waves spread out and the listener is
aware of the commencement of sound when the direct waves reach
his ears. Subsequently the listener receives sound energy due to
reflected waves also. If the note is continuously sounded, the
intensity of sound at the listener’s ears gradually increases. After
sometime, a balance is reached between the energy emitted per
second by the source and energy lost or dissipated by walls or other
materials.
The resultant energy attains an average steady value and to the
listener the intensity of sound appears to be steady and constant.
This is represented by a portion BC of the curve ABCD.
If at C, the source stops emitting sound, the intensity of sound falls
exponentially as shown by the curve CD.
Fig. 1.2
When intensity of sound falls below the minimum audibility level,
the listener will not get the sound.
When a series of notes are produced in an auditorium each note will
give rise to its own intensity curve with respect to time. The curve
for these notes are shown in Fig. 1.2.
1-10 Architectural Acoustics
In order to maintain distinctness in speech it is necessary that :
(a) Each separate note should give sufficient intensity of sound in
every part of the auditorium.
(b) Each note should die down rapidly before the maximum
average intensity due to the next note is heared by the listener.
1.5 Absorption :
When a sound wave strikes a surface there are three possibilities.
(a) Part of energy is absorbed
(b) Part of it is transmitted
(c) Remaining energy is reflected
The effectiveness of surface in absorbing sound energy is expressed
by absorption coefficient denoted by a.
a = Sound energy absorbed by the surface
Total sound energy incident on the surface …(1.5)
For the comparison of relative efficiencies of different absorbing
material, it is necessary to select a standard or reference.
Sabine selected a unit area of open window, as standard. For any
open window the sound falling on it completely passes out no
reflection, and more importantly no absorption.
Hence open window is an ideal absorber of the sound. The
absorption coefficient is measured in open window unit.
(OWU) or Sabine :
The absorption coefficient of a material is defined as the reciprocal
of its area which absorbs the same sound energy as absorbed by unit
area of open window.
Effective absorbing area A of the surface having total area S and
absorption coefficient ‘a’ is given by
A = a S …(1.6)
1-11 Architectural Acoustics
If the a1, a2, a3, …. , an are the absorption coefficients for each
reflecting surface and S1, S2, S3, …. Sn are the corresponding areas,
then the average value of absorption co-efficient is
a + 89 = a1 S1 + a2 S2 + a3 S3 + …… + an Sn
S1 + S2 + S3 + …. + Sn
=
n
i = 1 ai Si
S …(1.7)
Where S is total surface area.
1.6 Sabine’s Formula :
Prof. W.C. Sabine observed the concept of reverberation time for
varieties of conditions like empty room, furnished room, small room,
auditorium etc.
He concluded the following,
(a) Reverberation time depends upon reflectivity of sound form
various surfaces available in side the hall. If the reflection is
good, reverberation time of the hall will be longer as sound take
more time to die out.
(b) Reverberation time depends upon volume of the hall.
i.e. T V
(c) Reverberation time depends upon coefficient of absorption of
various surfaces present in the hall. For shorter reverberation,
absorption should be more.
(d) As absorption coefficient is found to be increased with increase
in frequency, reverberation time decreases with frequency.
Reverberation time T V
A
where, V = Volume of hall
A = Absorption
1-12 Architectural Acoustics
or T = K V
A
where, K = Proportionality constant
It has been further observed that is all the parameters are taken in SI
then, proportionality constant is found to be 0.161.
T = 0.161 V
A …(1.8)
Equation (1.8) is Sabine’s formula.
Absorption A given in Equation (1.8) represents overall absorption
which is given as
A =
n
i = 1 a S = a1 S1 + a2 S2 + …… + an Sn
Ex. 1.3 : For an empty assembly hall of size 20 15 10 cubic meter
with absorption coefficient 0.106 . Calculate reverberation
time.
Soln. :
Given :
(i) Size of the room = 20 15 10
= 3000 cubic meter
(ii) a = 0.106
Formula T = 0.161 V
A
= 0.161 V
aS
Here S = Total surface area of the hall is given by
2 (20 15 + 15 10 + 20 10)
= 1300 sqm
Reverberation time T = 0.161 3000
0.106 1300
1-13 Architectural Acoustics
Reverberation time = 3.5 sec ...Ans.
1.7 Determination of Absorption Coefficient :*(only for reference)
Step 1 : Using a source of sound inside the hall, reverberation time is
measured with the help of chronograph without inserting any test
material (whose co-efficient of absorption is to be calculated). Let
the reverberation time be T1,
T1 = 0.161 V
A
= 0.161 V
aS
1
T1 =
aS
0.161 V …(1.9)
Step 2 : Now consider a material like curtain or stage screen whose
co-efficient of absorption is to be found out suspended inside the
room and reverberation time T2 is obtained. Since the material is
suspended in hall, surface area from both the side are to be
considered.
1
T2 =
0.161 V
aS + 2a2 S2
where a2 = Co-efficient of absorption of the material under
investigation
S2 = Surface of the material (since both the sides are used,
it is multiplied by 2)
1
T2 =
aS + 2a2 S2
0.161 V …(1.10)
From Equation (1.9) and (1.10)
1
T2 –
1
T1 =
1
0.161
2a2 S2
V
2a2 S2 = 0.161 V 1
T2 –
1
T1
1-14 Architectural Acoustics
a2 = 0.161 V
2 S2
1
T2 –
1
T1 …(1.11)
All the quantities on RHS are known, co-efficient of absorption of an
absorbing material which is suspended in hall with both the surfaces open
can be calculated.
Table 1.3 : Absorption coefficients of some materials
Material Absorption coefficient per m2 at 500 Hz
Open window 1.0
Stage curtain 0.2
Common plaster 0.3
Carpet 0.4
Heavy curtain 0.5
Perforated cellulose fiber tiles 0.85
1.8 Conditions for Good Acoustic :
As already introduced in art 1.3, a lecture hall or auditorium should
satisfy the following conditions in order to be acoustically good.
(a) The initial sound from the source should be of adequate intensity.
(b) The sound should spread evenly with proper loudness every where is
the hall
(c) The sound of speech or music should be clear and words of or
musical notes must be distinctly audible to all.
(d) All undesired or extraneous noise must be reduced to the extent that
it will not interfere with normal hearing of speech or hearing.
(e) Any distortion due to shape and size must be absent.
1.9 Methods of Design for Good Acoustics :
In order to make acoustically correct hall following points may be
considered. These are merely the guidelines, depending upon specific
requirement a justified step be taken.
1-15 Architectural Acoustics
(a) Selection of proper site :
Avoid noisy places like railway tack, roads with heavy traffic,
airports, industrial vicinity for auditorium.
(b) Volume :
Size of the hall/ auditorium should be such that it remains
optimum.
Small halls leads to irregular distribution of sound because of
formation of standing waves.
Too big halls may also create a weaker intensity and larger
reverberation time which is a very serious issue.
(c) Shape :
It is one of the most important parameter to be considered for
acoustically correct hall.
As the reflections are created by roof and side walls, they
should be designed in such a way that echos are not allowed to
generate.
In place of parallel walls, splayed side walls are preferred.
Curved surface on walls, ceilings or floor produce concentration
of sound into particular region and absence of sound in other
regions.
Hence curved surface must be designed with proper care.
(d) Use of absorbents :
Once the construction of hall is completed certain errors are
found or the hall requires further correction as far as acoustics
are concerned. For this use of absorbents is very common.
As the reflections from rear wall are of no use. It must be
covered with absorbents, so as the ceiling.
1-16 Architectural Acoustics
False ceiling provided in large halls solves this problem
effectively. The floor needs to be covered with carpet so as
unwanted reflections and the noise created by audience is
suppressed.
1-17 Architectural Acoustics
(e) Reverberation :
Reverberation time must be maintained in such a that it does not
remain too short or too large i.e. nearly 0.5 seconds for lecture
hall, around 1.2 for concerts hall and around 2 for cinema halls.
Proper use of absorbing materials, sufficient people as audience,
presence of open windows presence of furniture etc are the
major components which can decide the reverberation time.
Calculated use of such components will be helpful to either
increase or decrease the reverberation time.
(f) Echelon effect :
Fig. 1.3 : Echelon effect
A set of railings or staircase or any regular spacing of reflected
surfaces may produce a musical note due to regular succession
of echoes of the original sound to listener.
This makes original sound to appear confused. Either one
should avoid use of such surfaces or keep them covered with
thick carpet.
1.10 Solved Problems :
Ex. 1.4 : Calculate the change in intensity level when intensity level
increases by 106 times its original intensity.
Soln. :
1-18 Architectural Acoustics
Given :
Initial intensity = I0
Final intensity = I
I
I0 = 10
6
Increase in intensity level in dB
L = 10 log10 I
I0 = 10 log10 (10
6)
L = 60 dB …Ans.
Ex. 1.5 : A room has dimensions 6 4 5 meters calculate :
(a) the mean free path of the sound waves in the room
(b) the number of reflections made per second by the sound
wave with the
walls of the room
Given : Velocity of sound in air = 350 m/sec
Soln. :
(a) The mean free path of sound waves is defined as the average
distance travelled by sound wave through air between any two
consecutive encounters with the walls of the room. Jaeger had
calculated as
l = 4V
S =
4 (Volume of the room)
Total surface area
Here V = 6 4 5 = 120 m3
S = 2 [6 4 + 4 5 + 5 6] = 148 m2
l = 4 120
148 = 3.243 m …Ans.
Number of reflections made per second
n = Velocity of sound
Mean free path
1-19 Architectural Acoustics
n = 350
3.243 = 107.9 …Ans.
Ex. 1.6 : The sound from a drill gives a noise level 90 dB at a point
short distance from it. What is the noise level at this point if
four such drills are working simultaneously at the same
distance from the point ?
Soln. : Acoustic intensity level is given by
L = 10 log10 I
I0 dB …(1)
Reference to I0 in watts / m2
Let I1 be the intensity level due to one drill and I2 be the intensity
level due to four such drills.
I2
I1 = 4 …(2)
Consider one drill on
L1 = 10 log I1
I0 dB …(3)
In second case with four drills on
L2 = 10 log I2
I0 dB …(4)
Increase in noise level (in dB)
L2 – L1 = 10 log I2
I0 – log
I1
I0
= 10 log I2
I1
but I2
I1 = 4
L2 – L1 = 10 log 4 = 6.021 dB
Final intensity level
1-20 Architectural Acoustics
= L1 + 6.021 = 90 + 6.021
Final intensity level = 96.021 dB …Ans.
Ex. 1.7 : Calculate the increase in the acoustic intensity level in dB.
When the sound is doubled.
Soln. :
Intensity level in dB is
L = 10 log I
I0
Let the intensity level in case 1 be I1 and the in case 2 be I2
For case – 1
L1 = 10 log I1
I0 dB
For case – 2
L2 = 10 log I2
I0 dB
Change in intensity level in dB
L2 – L1 = 10 log I2
I0 – log
I1
I0
= 10 log I2
I1
but I2
I1 = 2 (given)
L2 – L1 = 10 log 2
= 10 (0.3010)
L2 – L1 = 3.01 dB …Ans.
Ex. 1.8 : An air conditioner unit operates at a sound intensity level of 70
dB. If it is operated in room with an existing sound intensity
level of 80 dB, what will be the resultant intensity level.(4 Marks)
1-21 Architectural Acoustics
Soln. :
Here for case – 1
Intensity level is 70 dB
70 = 10 log L1 = 10 log I1
I0
I1
I0 = Antilog 7.0
or I1 = 107 I0 watts/m
2 …(1)
Similarly for Case – 2, intensity level is 80 dB.
80 = 10 log L2 = 10 log I2
I0
I2
I0 = Antilog 8.0
I2 = 1 108 I0 watts/m
2 …(2)
Resultant intensity
I = I1 + I2
= 107 I0 + 10
8 I0
= I0 (1.1 108)
Resultant intensity level in dB
L = 10 log I
I0
= 10 log 1.1 10
8 I0
I0 = 10 log (1.1 10
8)
= 80.41 dB
Resultant intensity level (in dB) is 80.41 …Ans.
Ex. 1.9 : The noise form an aeroplane engine 100 m from an observer is
40 dB in intensity. What will be the intensity when the
aeroplane flies overhead at an altitude of 2 km ?
1-22 Architectural Acoustics
Soln. : Intensity of sound is given by formula
I = P
4 R2
Where P = Acoustic pressure level
R = Radial distance
Here, for case – 1
I1 = P
4 R2
1
And for case – 2
I2 = P
4 R2
2
I2
I1 =
R2
1
R2
2
Now R1 = 100 m, R2 = 2000 m (given)
I2
I1 =
1002
20002 =
1
400
or I1
I2 = 400 …(1)
For the case – 1, intensity level in dB is given by
L1 = 10 log I1
I0
…(2)
and for case – 2
L2 = 10 log I1
I0
…(3)
as intensity level is suppose to decrease, we will take L1 – L2
L1 – L2 = 10 log I1
I0 – log
I2
I0
1-23 Architectural Acoustics
= 10 log I1
I2
= 10 log 400 = 26.021 dB
as L1 = 40 dB given
L2 = L1 – (L1 – L2)
L2 = 40 – 26.021 = 13.97 dB …Ans.
Ex. 1.10 : A hall of volume 5500 m3 is found to have a reverberation
time of 2.3 sec. The sound absorbing surface of the hall has an
area of 750 m2. Calculate the average absorption coefficient.
Soln. :
Given : V = 5500 m3
T = 2.3 sec
S = 750 m2
Let absorption coefficient be a
Using Sabine’s formula
T = 00.161 V
aS
a = 0.161 V
ST
= 0.161 5500
750 2.3
a = 0.513 …Ans.
Ex. 1.11 : For an empty hall of size 20 12 12 cubic meter, the
reverberation time is
2.5 sec. Calculate the average absorption co-efficient of the
hall. What area of the floor should be covered by carpet so as
to reduce the reverberation time to 2.0 sec. Given that
absorption co-efficient of carpet is 0.5.
Soln. :
1-24 Architectural Acoustics
(a) Reverberation time
T1 = 0.161 V
aS …(1)
aS = 0.161 V
T1
= 0.161 (20 12 12)
2.5
= 185.47
Now total surface area of the hall,
S = 2 (20 12 + 12 12 + 20 12)
= 1248 m2
a = 185.47
1248 = 0.1486 …Ans.
(b) By using the carpet of surface area S1 whose absorption coefficient is
0.5, the reverberation time is reduced to 2.0 sec.
Let T2 = 2.0 sec
Carpet surface = S1
Co-efficient of absorption of carpet ac = 0.5
Writing Sabine’s formula
T2 = 0.161 V
aS + aC S1– aS1 …(2)
(Here Total surface area = S, now if carpet is used of area S1, the
area covered by the material with co-efficient of absorption a is a (S – S1)
= aS – aS1)
From Equation (1)
1
T1 =
aS
0.161 V …(3)
1-25 Architectural Acoustics
From Equation (2)
1
T1 =
aS + aC S1– aS1
0.161 V …(4)
1
T2 –
1
T1 =
1
0.161 V [aC S1 – aS1]
= S1 (aC – a)
0.161 V
S1 = 0.161 V
aC – a
1
T2 –
1
T1
Substituting various value
S1 = 0.161 (20 12 12)
0.5 – 0.1486
1
2 –
1
2.5
= 131.95 m2
Carpet area required to reduce reverberation time up to 2.0 sec is
131.95 m2 …Ans.
Ex. 1.12 : Calculate the reverberation time for the seminar hall with
(a) No one inside.
(b) 50 persons inside
(c) Full capacity of audience.
Given that
Sr.
No.
Surface Area Absorption
co-efficient
1. Carpet covering entire
floor (10 12)
sqm
0.06
2. False ceiling (10 12)
sqm
0.03
3. Cushioned seats 100 Nos 1.00
4. Walls covered with
absorbent
346 sqm 0.2
1-26 Architectural Acoustics
Sr.
No.
Surface Area Absorption
co-efficient
5. Audience occupying
seats
– 0.46 /
person
6. Wooden door (3 2) sqm 0.2
Soln. :
Let us calculate total absorption in the hall in case – 1 i.e. for empty hall
(1) Absorption due to carpet 120 0.06 = 7.2
(2) Absorption due to false ceiling 120 0.03 = 3.6
(3) Absorption due to seats 100 1 = 100
(4) Walls covered with absorbent 346 0.2 = 69.2
(5) Wooden door 6 0.2 = 1.2
aS = 181.2 …(1)
Now Area of floor = Area of ceiling = (l b)
= 120 sq.m
Area of wall + Area of door = 346 + 6 = 352
= 2 [(b h) + (l h)]
as l b = 120 m2
let us take l = 12 m, b = 10 m
352 = 2 [(10 h) + (12 h)]
h = 8 m …(2)
hence volume V = 12 10 8 = 960 m3 …(3)
Case 1 :
For empty hall
Reverberation time T1 = 0.161 V
aS
= 0.161 960
181.2
1-27 Architectural Acoustics
T1 = 0.85 sec …Ans.
Case 2 :
With occupancy of 50 persons.
Absorption = aS + 50 (0.46)
Reverberation time T2 = 0.161 V
aS + 50 (0.46)
= 0.161 960
181.2 + 23
T2 = 0.757 sec …Ans.
Case 3 :
With full occupancy. i.e. 100 persons here, the absorption is = aS +
100 (0.46)
Reverberation time T3 = 0.161 V
aS + 100 (0.46)
T3 = 0.68 sec …Ans.
1.11 Solved Examples :
Ex. 1.11.1 :The volume of room is 600 m3. The wall area of the room is 220 m
2, the floor area is 120 m
2 and the ceiling area is 120 m
2. The average sound absorption coefficient, (a) for the walls is 0.03 (b) for the ceiling is 0.8 (c) the floor it is 0.06. Calculate the average sound
absorption coefficient and the reverberation time.
Soln. :
Given :
Let S1 = 220 m2 a1 = 0.03
S2 = 120 m2
a2 = 0.8
S3 = 120 m2 a3 = 0.06
The average sound absorption coefficient is
a = a1 S1 + a2 S2 + a3 S3
S1 + S2 +S3
1-28 Architectural Acoustics
= 220 0.03 + 120 0.8 + 0.06 120
220 + 120 + 120 =
0.238
a = 0.24 …Ans.
Total sound absorption of the room = aS
= 0.24 460
= 110.4 Sabine
Reverberation time, using Sabine’s formula
T = 0.161 V
aS =
0.161 600
110.4
T = 0.875 sec. …Ans.
Ex. 1.11.2 :What is the resultant sound level when a 70 dB sound is added to a 80 dB
sound ? (4 Marks)
Soln. : Increase in intensity level = L = 70 dB
Say, resultant intensity increased by x times the original intensity
Hence, L = 10 log10 x Io
Io dB
70 = 10 log10 (x )
7 = log10 x
or x = 107
So, Resultant sound level is increased 107 times the original
intensity.
1-29 Architectural Acoustics
example :
Q. 1 A class room has dimensions 20 15 5 m3. The reverberation
time is 3.5 sec. Calculate the total absorption of its surface and
average absorption co-efficient.
Ans. :(0.07)
Q. 2 The reverberation time is found to be 1.5 sec for an empty hall and
itis found to be
1 sec when a curtain of 20 m2 is suspended at the center of the hall.
If the dimensions of the hall are 10 8 6 m3, calculate co-
efficient of absorption of curtain.
Ans. :(0.64)
Q. 3 For an empty assembly hall of size 20 15 10 m3, the
reverberation time is 3.5 sec. Calculate the average absorption co-
efficient of the hall. What area of the wall should be covered by the
curtain so as to reduce the reverberation time to 2.5 sec. Given the
absorption co-efficient of the curtain cloth is 0.5.
Ans. :(0.106, 140.12 m2)
ULTRASONIC WAVE
Introduction :
The term ultrasonics applies to sound waves that vibrate at a frequency higher than
the frequency that can be heard by the human ear (or higher than about 20,000
hertz).
Sound is transmitted from one place to another by means of waves. The
character of any wave can be described by identifying two related properties:
its wavelength (lambda, λ) or its frequency (f). The unit used to measure the
frequency of any wave is hertz. One hertz is defined as the passage of a single
wave per second.
Ultrasonics, then, deals with sound waves that pass a given point at least 20,000
times per second. Since ultrasonic waves vibrate very rapidly, additional units also
are used to indicate their frequency. The kilohertz (kHz), for example, can be
used to measure sound waves vibrating at the rate of 1,000 times per second, and
the unit megahertz (MHz) stands for a million vibrations per second. Some
ultrasonic
devices have been constructed that produce waves with frequencies of more than
a billion hertz.
PROPERTIES OF ULTRASONIC WAVES
(1) They have a high energy content.
(2) Just like ordinary sound waves, ultrasonic waves get reflected, refracted and
absorbed.
(3) They can be transmitted over large distances with no appreciable loss of
energy.
(4) If an arrangement is made to form stationary waves of ultrasonics in a liquid,
it serves as a diffraction grating. It is called an acoustic grating.
(5) They produce intense heating effect when passed through a substance.
Ultrasonic Production :
There are three methods for producing Ultrasonic waves. They are:
(i) Mechanical generator or Galton’s whistle.
(ii) Magnetostriction generator.
(iii) Piezo-electric generator.
Magnetostriction method:
Principle:
“When a magnetic field is applied parallel to the length of a ferromagnetic rod
made of material such as iron or nickel, a small elongation or contraction
occurs in its length. This is known as magnetostriction. The change in length
depends on the intensity of the applied magnetic field and nature of the
ferromagnetic material. The change in length is independent of the direction
of the field. “
The change in length (increase or decrease) produced in the rod depends upon the
strength of the magnetic field, the nature of the materials and is independent of the
direction of the magnetic field applied.
Construction:-
The experimental arrangement is shown in Figure
XY is a rod of ferromagnetic materials like iron or nickel. The rod is
clamped in the middle.
The alternating magnetic field is generated by electronic oscillator.
The coil L1 wound on the right hand portion of the rod along with a variable
capacitor C.
This forms the resonant circuit of the collector tuned oscillator. The
frequency of oscillator is controlled by the variable capacitor.
The coil L2 wound on the left hand portion of the rod is connected to the
base circuit. The coil L2 acts as feed –back loop.
Working:-
When High Tension (H.T) battery is switched on, the collector circuit
oscillates with a frequency,
f =
This alternating current flowing through the coil L1 produces an alternating
magnetic field along the length of the rod. The result is that the rod starts
vibrating due to magnetostrictive effect.
The frequency of vibration of the rod is given by
n =
where l = length of the rod
Y = Young’s modulus of the rod material and
=density of rod material
• The capacitor C is adjusted so that the frequency of the oscillatory circuit is
equal to natural frequency of the rod and thus resonance takes plate.
• Now the rod vibrates longitudinally with maximum amplitude and generates
ultrasonic waves of high frequency from its ends.
Condition for resonance:
Frequency of the oscillatory circuit = Frequency of the vibrating rod
Merits: 1. Magnetostrictive materials are easily available and inexpensive.
2. Oscillatory circuit is simple to construct.
3. Large output power can be generated.
Limitations 1. It can produce frequencies upto 3 MHz only.
2. It is not possible to get a constant single frequency, because rod depends on
temperature and the degree of magnetization.
3. As the frequency is inversely proportional to the length of the vibrating rod, to
increase the frequency, the length of the rod should be decreased which is
practically impossible.
piezo electric ossilator
Introduction
Can all the crystals exhibit piezoelectric effect? What is special about the
piezoelectric crystal?
Is the piezoelectric effect direction dependent?
Learning Objectives On completion of this chapter you will be able to:
1. define piezoelectric effect
2. define inverse piezoelectric effect
3. know what type of crystals will exhibit piezoelectric effect
4. Understand the working of piezoelectric generator
Piezoelectric effect:
When crystals like quartz or tourmaline are stressed along any pair of opposite
faces, electric charges of opposite polarity are induced in the opposite faces
perpendicular to the stress. This is known as Piezoelectric effect.
Inverse piezoelectric effect:
When an alternating e.m.f is applied to the opposite faces of a quartz or tourmaline
crystal it undergoes contraction and expansion alternatively in the perpendicular
direction. This is known as inverse piezoelectric effect. This is made use of in the