Optical Fibres Flashcards

Question

How are the magnitudes of a plane wave and electric field related?

Answer 1

𝐸=|𝐻| sqrt(𝜇**/**𝜖) The plane wave will also have a magnetic field orthogonal to both the E-field and z

Answer 2

Wave impedance: 𝑍=sqrt(𝜇/𝜖). The wave impedance of free space is 𝑍₀=sqrt(𝜇₀𝜖₀)=377Ω.

Answer 3

The intensity (irradiance) is: 𝐼=|𝐸₀|²/2𝑍. Where E₀ is the peak electric field.

Answer 4

1. Sample time-varying analogue signal (at baseband sampling rate must be ≥ 2fmax) 2. Quantisethese samples giving rise to quantisation noise 3. Express each sample as a binary number 4. Send these bits followed by the bits for the next sample down your channel 5. Do the reverse of this at the end of the system, but there will be errors due to noise

Answer 5

Channels often don’t transmit “DC”, in other words, a continuous train of “1”s or a continuous train of “0”s, very well.

Answer 6

Coding is the science of replacing the original sequence with another to match the data to the channel, often it includes adding bits for error detection and correction as well.

Answer 7

Information on the timing of pulses Error Detection and Correction

Answer 8

So that it can align its clock and sample the received digital signal at the right points in time. Timing information is obtained by locking a clock oscillator at the receiving end to the transitions on the signal. If there are few transitions, the clock timing information is lost and sampling and reconstructing the signal is impossible. Appropriate coding ensures the probability of too few transitions is vanishingly small.

Answer 9

Typically use OOK or on-off keying From OOK, the system may use the return to zero(RZ) format, or the non-return to zero(NRZ) format.

Answer 10

In RZ, the signal returns to the value representing zero for the second half of the bit time slot. The duty cycle of the signal may not be 0.5, i.e. divided into the second and first half. The division may be more like 1/3 and 2/3.

Answer 11

In NRZ, the signal keeps its value for the whole time slot.

Answer 12

RZ has simpler clock recovery at the receiver as the data stream contains the clock frequency

Answer 13

NRZ uses less bandwidth and gives the detector more photons/bit -important for detectability

Answer 14

A clock is not transmitted with the data –it has to be recovered at the receiver.

Answer 15

Peak at centre, smaller peaks surround it similar to normal function. The envelope of Power Spectrum has a sine shape - Matches Fourier transform of a square pulse Extend to infinity

Answer 16

We don’t attempt to transmit square pulses, we pre-filter to reduce bandwidth and avoid cross-talk between channels in DWDM systems

Answer 17

A carrier might be single-frequency (monochromatic in optics) but a modulated carrier cannot be single frequency.

Answer 18

Dispersion will result in pulse-spreading and thus intersymbol interference, unless compensated

Answer 19

Dispersion results in pulse-spreading because refractive index and hence velocity varies with wavelength/frequency A real data stream must contain a band of frequencies, so the pulses spread.

Answer 20

An eye pattern, also known as an eye diagram, is an oscilloscope display in which a digital signal from a receiver is repetitively sampled and applied to the vertical input, while the data rate is used to trigger the horizontal sweep. Wikipedia

Answer 21

They sho pictorially and quantitatively, the constraints on deciding whether a received bit is 1 or 0 in the presence of attenuation, noise, jitter, pulse-spreading etc.

Answer 22

A properly constructed eye should contain every possible bit sequence from simple alternate 1’s and 0’s to isolated 1’s after long runs of 0’s and all other patterns that may show up weaknesses in the design. The larger the opening the more reliably a design in. And less affected by errors..

Answer 23

* Pulses are not square because that would need a lot of unnecessary bandwidth * They are broad in the vertical (detector current) direction because of noise and the electronic response to different numbers of 1’s or 0’s in succession * They are broad in the width (time) direction because of timing jitter

Answer 24

* Attenuation, noise, more jitter, and pulse-spreading have further closed the eye * In presence of noise, the decision between 0 and 1 is more prone to an error (poor S/N) * The timing of the decision is more critical * Ultimately for a long channel or a high bit rate the eye closes and the system doesn’t work

Answer 25

Jitter occurs when riding or falling edges occur at times that differ from the ideal time. Some edges occur early, some occur late. In a digital circuit, all signals are transmitted in reference to clock signals. The deviation of the digital signals as a result of reflections, intersymbol interference, crosstalk, PVT (process-voltage-temperature) variations, and other factors amounts to jitter. Some jitter is simply random.

Answer 26

“Jitter” causes errors in pulse timing with the clock, yet again increasing the probability of an error

Answer 27

Attenuation (loss) reduces the distinction between the level of a “1” and a “0” so that receiver noise has an increased probability of causing an error

Answer 28

These three phenomena need remediating in a repeater after some length of fibre–amplification, pulse reshaping and pulse retiming

Answer 29

A range of phenomena associated with the superposition of waves

Answer 30

intensity will be ## Footnote **I = sqrt(ε₀/μ) \*(4A² /2)cos(φ/2)**

Answer 31

Total E-field is : E = Ae^{-j(k₁z -ω₁t)} + Ae^{-j(k₂z - ω₂t)} Which may be rearranged to: E = 2Ae^-j(kz-ωt)cos(Δkz - Δωt) where k =( k₁+k₂)/2 and Δk = (k₁-k₂)/2 same for ω and Δω Thus becomes a plane wave with velocity ω/k modulated by a cos wave of frequency Δω. Cos wave travels with a velocity of Δω/Δk In turn, the group velocity is V_g = dω/dk.

Answer 32

v_g = d𝜔/dk, 1/v_g= d/d𝜔[𝜔/c \* **n(𝜔)**] as k = 2π/𝜆 n and 𝜆 = (2πc)/𝜔 so k = (𝜔/c)\*n Differentiating with respect to 𝜔 1/v_g= 1/c[**n(𝜔) +** 𝜔dn/d𝜔] We recognise the first term as being equal to 1/v_p and rearrange the 2nd term in terms of λ: 𝜔dn/d𝜔 = 2πc/𝜆[dn/d𝜆 \* d𝜆/d𝜔] = -2πc/𝜆\* 2πc/𝜔²\* dn/d𝜆 = -[2πc/𝜔]² \* 1/𝜆 \*dn/d𝜆 = -𝜆(dn/d𝜆) Such that: 1/v_g = 1/c[n(𝜆) - 𝜆dn/d𝜆] = 1/v_p -(𝜆/c)dn/d𝜆

Answer 33

n = sqrt(ε_r)

Answer 34

In dispersive materials (where velocity changes with wavelength) the dependence of the refractive index of the material leads to vg ≠ vp. This means the modulation envelope does not travel at the same speed as the carrier

Answer 35

Pulses travelling at a different velocity to v_p doesn’t mean they spread, just that the delay in transmission (latency) is different.

Answer 36

Not by itself, however, if the group velocity is in a dispersive medium, i.e. frequency dependent, then pulse spreading will occur. This is because each frequency component of a pulse travels at a different group velocity and so they disperse and arrive at different times at the end of the fibre.

Answer 37

Latency is approximated using the equation (D = T \* V) where D is the distance the fibre travels, T is the time it takes to travel, and V is the velocity of light.

Answer 38

Material dispersion and waveguide dispersion is used to do so.

Answer 39

Pulse-broadening leads to inter-symbol interference, increasing the error rate It occurs as modulated carriers have a non-zero bandwidth.

Answer 40

n₁ is the refractive index of the medium from which the wave is incident n₂ is the refractive index on the other side of the dielectric interface α is the angle of incidence

Answer 41

n₁ is the refractive index of the medium from which the wave is incident n₂ is the refractive index on the other side of the dielectric interface α is the angle of incidence

Answer 42

All states of polarisation can be expressed by superposing E_//andE_⊥ in different amplitudes and phases.

Answer 43

Total internal reflection occurs (|ρ|=1) if α\>α_c

Answer 44

A typical optical fibre has a core index of 1.4579 and a cladding index of 1.4469 at λ=1560nm

Answer 45

R is the power reflection coefficient (R=|ρ|²) and Rp≡R_//and R_s≡ R_⊥

Answer 46

We have reflection and transmission

Answer 47

we get total reflection (for which n1 \> n2)

Answer 48

The angle of transmission is complex

Answer 49

Then α\>α_cor sinα \> n₂/n₁ This means cosβ is an imaginary angle. Which suggests exponentially decaying or possibly exponentially growing waves in the z-direction in the low-index medium.

Answer 50

Exponentially growing or decaying waves in the low index medium.

Answer 51

φ_t = k₂(xsinβ + zcosβ)

Answer 52

* Phase as a function of time: tω * Phase in the x-direction along the interface: kn₁ xsinα * Complex phase in the z-direction, representing a decaying wave if we choose –ve The complex phase is the evanescent field.

Answer 53

The evanescent field is the complex phase in the z-direction of the electric field.

Answer 54

Brewster's angle is an angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection.

Answer 55

As can be observed from the figure, the reflection coefficient for the R_s polarisation(perpendicular) increases exponentially as the angle of incidence increases. Total internal reflection only occurs when the angle of incidence is 90 degrees.

Answer 56

Is v_p = 1/sqrt(εμ) where ε = ε₀ ε_r similar rule for μ

Answer 57

Rayleigh scattering is due to microscopic fluctuations in the density of the glass (in the case of glass fibre), frozen in place when the glass cools. These random fluctuations in the refractive index scatter light and it is found that the loss is related to a Rayleigh scattering coefficient by: 𝛼_𝑅 = c/𝜆⁴ where C is found to be between 0.7 and 0.9 dB.km^-1.μm⁴ depending upon the fibre composition.

Answer 58

Rayleigh Scattering, Impurities such as OH group, and IR absorption

Answer 59

Material absorption refers to a process where photons are absorbed prematurely preventing light transmission. Absorption can be described by Beer’s Law, P_out/P_in = 10^-εL where ε is the extinction coefficient in units such as ppm^-1 km^-1 (per ppm per km) c is the concentration of the impurity in ppm L is the pathlength of the absorbing material in km

Answer 60

Material absorption is due to impurities in the glass, and the fundamental absorption of the pure silica glass network.

Answer 61

We take an expression for a wave propagating in a medium of refractive index n in the +z direction E(t, z) = E₀ exp{jtωt - jk₀nz} Now we take n to be complex i.e. n' + jn'' E(t,z) = E₀exp{jωt - jk₀n'z}exp{k₀n''z}

Answer 62

The wave decays according to 𝐸(𝐿)=𝐸0exp{𝑘0𝑛"𝐿} 𝑜r 𝑃(𝐿)=𝑃(0)exp{2𝑘₀𝑛"𝐿} Loss in dB is given by 𝛼 =−10log₁₀**|**P(0)**/**P(L)**|** **so that** **𝛼= -**10log₁₀(e)2k₀n"L

Answer 63

V is a form of normalised frequency The equation is below:

Answer 64

b must be between 0 and 1 because, 𝑛₂²eff²1² β normally indicates propagation coefficient where 𝛽=𝑁_𝑒ff𝑘₀

Answer 65

When b = 0 the velocity is that of a wave in the cladding and the mode becomes cut-off. It doesn’t ever quite get there for the fundamental mode, so there is no cut-off for the fundamental mode.

Answer 66

When b = 1 the mode travels with the velocity of a wave in the core material. There is always some evanescent field in the cladding, so b never quite gets to unity.