Chapt.3 Basics

7/28/2019 Chapt.3 Basics

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Chapter 3Basics

A Few important definitions and terms

Wavelength

Unlike electrical engineers who use frequency, opticians use wavelength. When nothing

is specified, the wavelength refers to free space wavelength,

=

=

0

0

/cf

f/c

(3.1)

where c is the velocity of light in free space ( s/m103 8 ) and f is the frequency.

Example 3.1:Consider a light emitting diode emitting a wavelength of 800 nm.

The corresponding frequency is = Hz)10800/(10398

375 THz (1 THz = 1210 Hz)

a very high frequency indeed.

Coherence

A wave in time and space has a variation in space and time as)zt(j

e

. If it is a perfectsinusoidal function of time of frequency, rad/s all the time, it is called perfectlycoherent. This is depicted in Fig.3.1 (a). There is of course no perfectly coherent source.

In practice, the source may emit a burst of a pure sine wave over a period of time, t , asshown in Fig,3.1(b). Then this time period is called the coherence time. After this time,the source may emit the same frequency but with an abrupt change in phase over another

length of time t . If we split light from the source into two parts with equal power, letthem travel along different paths and combine them, the intensity of the combined light

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will be between dark and very bright depending on whether the path difference is an odd

or even multiple of the wavelength this the well known interference of light. However,

interference giving dark and bright spots is possible for coherent light. If the source is notperfectly coherent, i.e., it is coherent over time t , or corresponding to a path c t infree space, the interference phenomenon can be obtained (in free space) only if the path

length difference does not exceed this length. The length c t is called the coherencelength and is illustrated in Fig.3.1.b.The time variation of an incoherent light, such as white light is random as shown in

Fig.3.1c.

The coherent characteristic of light can also be seen in the frequency domain (blue RHS

plots in Fig.1.1. The amplitude variation in the frequency domain is called spectrum.

Thus, a perfectly coherent light will have a single line spectrum as shown in Fig.1.1a, apartially coherent light will have a band limited spectrum as shown in Fig.1.1b, and

completely incoherent light will have a flat spectrum as shown in Fig.1.1c. You are

familiar with the last one in electronics as the spectrum of white or Gaussian noise. In

fact, the term white noise comes from optics where it is the spectrum of white light.

TimeP

Q

Field

Amplitude

Time

(a)

Amplitude

= 1/t

Time

(b)

P Q

l = ct

Space

t

(c)

Amplitude

(a) A sine wave is perfectly coherent and contains a well-defined frequencyo. (b) A finite

wave train lasts for a durationt

and has a lengthl

. Its frequency spectrum extends over= 1/t. It has a coherence timetand a coherence lengthl. (c) White light exhibitspractically no coherence.

1999 S.O. Kasap,Optoelectronics(Prentice Hall)

Fig.3.1

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Spectral width of a source

We just saw that no source is perfectly coherent. Hence they are characterized by their

spectral width. If the spectrum is plotted as power against wavelength the difference in

wavelengths of the points at which the power is half the peak power is called the spectralwidth (also called full width at half maximum or FWHM in optics). Again this is familiar

to electrical engineers who call it the 3dB bandwidth.

Fig.3.2 : represents optical power. 0 is the nominal or mean wavelength and 2/1

is the spectral width (FWHM)

In optics, the spectral width is expressed in wavelengths, while in electrical engineering,

we are more familiar with the quantity expressed in Hz.. The relationship between

spectral width, 2/1f in Hz and spectral width, 2/1 in wavelengths can be easily

obtained by differentiating (3.1).

=

=

2/1202/1

2/12

2/1

)/c(f

f)f/c((3.2)

Example 3.2

The light source in example 3.1 has a spectral width of 30 nm. Its spectral width in Hz is

=

Hz1030)10800(

103 929

8

14.06 THz.

Note that the frequency was 375 THz. So the spectral width is 3.75% of the nominal

frequency (its the same in terms of wavelength, i.e., 30/800).

We will see later that this width is not good for a high speed optical communicationsystem.

Refractive index

If pv is the phase velocity of light in a medium, the refractive index, n of the medium is

defined as:

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pv

cn = (3.3)

where, c is the velocity of light in free space.

In an unbounded medium, light as an electromagnetic wave travels as a TEM wave.

Recall from electromagnetics that its velocity is given byr0r0

p1v

= .

rr and are called relative permittivity and relative permittivity respectively. r is also

called the dielectric constant.

In free space, rr 1 == and the velocity is c, while in a dielectric, 1r = and the

velocity is pv .

So00

1c

= and

r00

p

1v

=

Hence

==2

r

r

n

n (3.4)

Example 3.3

We must be careful to note that dielectric constant varies with frequency. For example, atlow frequencies, the dielectric constant of water is about 80. At optical frequencies

(around 800 nm), the refractive index of water is about 1.3 and so its dielectric constant is

1.69 !

Reflection and transmission coefficients

In electromagnetics, we are familiar with voltage reflection coefficients

Voltage reflection coefficient, = Reflected wave voltage/Incident wave voltage

Voltage transmission coefficient, = Transmitted wave voltage/Incident wave voltage.From the continuity of voltage,

Incident wave voltage + Reflected wave voltage = Transmitted wave voltage

Or, =+1 (3.5)

In optics, the reflection and transmission coefficients refer to power unless statedotherwise. This can be a source of confusion while reading literature in optics.

coefficient

Power reflection coefficient, R= Reflected wave power/Incident wave power

Power transmission coefficient, T= Reflected wave power/Incident wave powerFrom the conservation of power,

Incident wave power - Reflected wave power = Transmitted wave power

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Or, TR1 = (3.6)

Relationships between the coefficients

Suppose that the wave is traveling from medium 1 to medium 2. Let the intrinsic (or

characteristic) impedance of the two media be 21 and respectively.

R= Reflected wave power/Incident wave power=1

2

1

2

/)voltagewaveincident(

/)voltagewavereflected(

= 2

2

)voltagewaveincident(

)voltagewavereflected(

Or, R =2

(3.7)

T= Transmitted wave power/Incident wave power=2

2

1

2

/)voltagewaveincident(

/)voltagewavedtransmitte(

=1

2

2

2

)voltagewaveincident(

)voltagewavereflected(

Or,1

22T

= (3.8)

Decibels

Communication engineers measure loss and gain in decibels.

Gain in dB=inputPower

outputPowerlog10 10

This will be a positive number for an amplifier because the power output is greater than

the power input. But when the device has loss, this number will be negative, because

power output is less than the power input. Now nobody likes negative numbers (like

negative bank balance!). So we express attenuation in dBs as

Attenuation, dBin = outputPower

inputPowerlog10 10 (3.9)

which will be a positive number.

Thus for a lossy device,

power output = 10/dBinin10 :power input (3.10)

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Nepers

Although decibel is widely used, it does not come naturally in the formulae of physics.This is just like degree which is widely used as measure of angle but the angle measure

that comes in formulae is radians.

In a lossy medium, fields (voltages and currents in transmission lines) decay

exponentially as e per unit length, where is called the attenuation constant inNepers per unit length and z is the direction of propagation.

As power is proportional to square of the field, it decays as 2e per unit length.

Then from the definition of attenuation in decibels,

dBin =Nepersin2

10elog10

= (20 log10 e) Nepersin

i.e., dBin = 8.6859 Nepersin (3.11)

Thus there are more decibels to a Neper(~8.7 decibels for a Neper) just as there are moredegrees for a radian(~57.3 degrees for a radian) or just as there are more sens for a

Ringgit (how many for a Ringgit?).

dBm, dBW

These are units of power often used in fiber optic communications.

1 dBm= 10 log10(Power/1 mW) (3.12)

Thus dBm is the number of dBs by which the power exceeds 1 mW.

I find it more convenient to remember1 dBm= 10 log10(Power in mW) (3.13)

Hence if the power is given as x dBm,

Power in mW= 10x/10 mW (3.14)Similarly, dBW is defined relative to powers of 1W .

1 dBW = 10 log10(Power/1 W) = 10 log10(Power in W) (3.15)

if the power is given as x dB W,Power in W = 10x/10 W (3.16)

Power in dBm or dBW is very convenient when gain or loss has to be taken intoaccount.

Example 3.4

(a) What is the power output of an amplifier with a gain of 10 dB if the power input is2dBm?

(b) If the power output and input of a device are -8 dBm and -2 dBm respectively, what

is the attenuation of the device?

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(a) Power output in dBm=Power input in dBm+ Gain in dB=2 +10=12dBm ( =1012/10

mW= 15.8 mW)

(b) Attenuation in dB= Power input in dBm Power output in dBm = -2-(-8) = 6 dB

Note powers in dBm can not be added to get total power.

Example 3.5

Two devices have power outputs of 20 dBm each. What is the total power output?

(The devices could be loudspeakers each with audio power output of 20 dBm)

The power supplied by each device is 1020/10 mW or 100 mW. Hence the total power

supplied by two devices = 200 mW = 10 log10 (200 mW) = 23 dBm

If we add powers in dBm, we get the wrong answer 40 dBm.

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Optical fiber basics

How does a fiber work?

Fig.3.2

An optical fiber used in fiber optic communication consists of two different types of

highly pure, solid glass (mainly silicon dioxide), composed to form the core and cladding.

A protective acrylate coating then surrounds the cladding. This coating protects the glassfrom dust and scratches that can affect fiber strength.

The operation of an optical fiber is based on the principle of total internal reflection.

Light reflects (bounces back) or refracts (alters its direction while penetrating a differentmedium), depending on the angle at which it strikes a surface. That reflection is usuallycaused by creating a higher refractive index in the core of the glass than in the

surrounding cladding glass, creating a waveguide. Usually, The refractive index of the

core is increased by slightly modifying the composition of the core glass generally byadding small amounts of a dopant, while the cladding is pure glass. Alternatively, the

waveguide can be created by reducing the refractive index of the cladding using different

dopants.

How do dopants change the refractive index of glass?

Fig.3.3 shows the refractive index against the percentage of dopants added to silicondioxide (glass).

From the diagram, it is seen that the refractive index can be increased or decreased by

very small amounts by the addition of the dopant. Noting that the core must have higherrefractive index, some of the possibilities are for example:

GeO2- SiO2 core and SiO2 cladding

P2O5- SiO2 core and SiO2 cladding

SiO2 core and B2O3- SiO2 cladding.

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GeO2 -B2O3- SiO2 core and B2O3- SiO2 cladding

Fig.3.3(Fig.4.6 Senior) Effect of various dopants on the refractive index of glass

Total Internal Reflection (TIR): Basic physics

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Fig.3.4(Fig.2.2 Senior): Light rays incident on a high to low refractive index interface (a)

refraction (b) limiting case of refraction showing the critical angle c (c) total internal

reflection where > c

Consider a ray of light traveling from a medium of higher refractive index 1n to amedium of lower refractive index, 2n .

Snells law: 2211 sinnsinn = (3.17)

As the angle of incidence, 1 increases, the refracted ray moves away from the normal

till as in Fig.3.4(b), it is along the interface, i.e.,0

290= . The corresponding angle of

incidence is called the critical angle, c . From (3.17),

1

2

cn

nsin = (3.18)

If the angle of incidence exceeds this value, the ray is turned back. No light is transmitted

in the medium of lower refractive index and light is described to have suffered a totalinternal reflection..

Example 3.6

For the glass-air interface, n1 = 1.5, n2 = 1.0, and the critical angle is given by

01

c 8.415.1

1sin ==

On the other hand, for the glass-water interface, n1 = 1.5, n2 = 1.33, and

01

c5.62

5.1

33.1sin ==

If light can be guided by a material surrounded by a material of lower refractive index,

why not just use glass and air? What is the purpose of the cladding in the fiber?

For transmission of light from one place to another, the fiber must be supported.

Supporting structures, however, may considerably distort the fiber, thereby affecting theguidance of the light wave. A sufficiently thick cladding. protects the total-reflection

surface from contamination. Further, in a fiber bundle, in the absence of the cladding,

light can leak through from one fiber to another. The idea of adding a second layer ofglass (namely, the cladding) came (in 1951) from OBrien (USA) and Van Heel

(Holland).

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The Numerical Aperture (NA) or Acceptance Angle of a fiber

(a)

(b)

Figure 3.5(a) A glass fiber consists of a cylindrical central core clad by a material of slightly lowerrefractive index. (b) Light rays impinging on the core-cladding interface at an angle greater thanthe critical angle are trapped inside the core of the fiber.

For a ray entering the fiber core at its end, if the angle of incidence at the internal core-

cladding interface is greater than the critical angle c [= sin-1 (n2/n1)], the ray will undergo

TIR at that interface. Further, because of the cylindrical symmetry in the fiber structure,

this ray will suffer TIR at the lower interface also and therefore be guided through thecore by repeated total internal reflections. Even for a bent fiber, light guidance can occur

through multiple total internal reflections

Consider a ray that is incident on the entrance face of the fiber core, making an angle i

with the fiber axis. Let the refracted ray make an angle with the same axis. Assuming

the outside medium to have a refractive index n0 (which for most practical cases is unity),

we get from Snells law (3.17)isin)n/n(sin 10= (3.19)

The critical angle is given by (3.10) as 12c n/nsin = and hence2

12c )n/n(1cos =

In order for the light to travel by TIR, c> . But since =090 ,

c

090 > or c

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Hence from (3.11)2

1210 )n/n(1isin)n/n( < ,

Or, 22210 nnisinn ?

For c1 > , 12c1 n/nsinsin =>

So 1sin)n/n( 122

21 > and 0sin)n/n(1 122

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The consideration given here is simplistic. A more rigorous (and hence more

mathematical) analysis of uniform plane waves incident at the boundary between two

dielectrics can be found in books. It was also covered in your electromagnetics course.For the present course, the simple explanation is enough.

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Chapt.3 Basics