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# Verbale N 1

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Verbale N 1

PROBLEMS CHAPTER 8

Problem 8-1 A weak signal, P=1-nW (or, respectively: 1-pW, 1-?W) is

detected by a coherent homodyne scheme, using a silicon photodiode with =1A/W, I= 0.1 A, terminated on a 1-k load. Find: (i) the minimum ?dark

local oscillator strength that is required; (ii) the signal amplitude that is obtained, as compared to direct detection; (iii) the S/N ratio and its improvement respect to direct detection. Assume =1 and B=1 GHz. ?

Answ: The condition I >> I + I + 2kT/eR yields for the local OLdark

oscillator detected current I: OL

I >> 0.1?A + 1pA +50mV/1k = 50 ?A OL

(note that it is the third term, associated to the load resistance, to dominate). By satisfying the disequality with a multiplicative factor 10 and as we have =1A/W, we get for the required local oscillator power:

P = 500 ?W (and I = 500 ?A). OLOL

For P=1 fW and 1 nW, the local oscillator power remains unchanged. The signal amplitude is:

RP = 1 ?V (resp.: 1 nV, 1mV) (direct detection), RPG = RI (1+2?I /I) (coherent detection) OL..5 = 1?V [1+2?(510)] = 1.4 mV (resp: 44?V, 44 mV).

The S/N ratio is:

2. -9. -19. 93(S/N) = 4I/2eB = 410/3.21010 = 12.5 (resp: 0.0125, 1.25 10)

(coherent detection)

22and (S/N)= I/2eB(0.1?A + 1pA +50mV/1k：？；

-18 -19. 9.-6-4 .-7. = 10/3.2 1010 50 10= 0.6 10(resp: 0.610, 60)

(direct detection)

Note that, with the values assumed in this example, neither the coherent nor the direct detection are in the reach of the 1-pW signal at the given bandwidth.

About the 1-nW signal, coherent detection becomes adequate while direct detection is not yet.

Problem 8-2 A phase-modulated signal, of the type:

E exp i[?t+?+2ks(t)] 0

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is detected by a homodyne coherent scheme. Write the expression of the output signal and guess the likely regime of detection.

Answ: The homodyne detection scheme gives:

I = I + 2?II cos[?-?+2ks(t)] 00Slo

Apart from a dc term I and a constant phase?-?, the coherent detection 0 lo

of a phase-modulated signal supplies as a result the cosine function of the driving term s(t) of the phase-modulation.

This result describes what exactly happens in a conventional interfero-meter: a laser beam is splitted in a beam divider, one portion of it goes to a distant target and comes back with a phase 2ks (optical pathlength by k=2?，)). Upon recombination with the fixed-phase, reference beam, a truly homodyne detection scheme is clearly recognized.

As a consequence, interferometers are most likely to work in the quantum-limit regime of detection (unless the reference beam power is kept unusually weak).

Problem 8-3 How large is the signal associated with a N=10-photon-per-bit

-9rate, the one giving a BER=10? Consider bit rates R= 1Mbit/s, 1Gbit/s.

Assume )=1.24 ?m, and a spectral sensitivity = 0.8 A/W. What is the

quantum noise and the S/N ratio of this signal ?

Answ: The photodetected signal is:

-19. .6 .I = e N R；！；1.6 101010= 1.6 pA @ 1Mbit/s,

and = 1.6 nA @ 1Gbit/s; the corresponding powers, being h=1eV @ )=1.24 ?m are:

P = h，； N R；！；，；，；：；= 2pW and 2 nW, respectively.

Quantum noise is (note that it is B=R/2 from Nyquist theorem):

1/21/2-19.-12 .61/2 ..i = [2eI B]= [2eI R/2]= [1.6101.61010]n = 0.51 pA @

1Mbit/s,

and = 5.1 nW @ 1Gbit/s.

From these numbers, the signal to noise is found back as:

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2 22(S/N)=(I /i)= (1.6/0.51) = 10 (in both cases). n

Problem 8-4 Given the data of Probl.8-1, what is the theoretical minimum

that would be tolerable to still have an advantage in using coherent ?

versus direct detection?

Answ: By comparing the S/N ratio of the two cases, we get the break-point coherence factor ? as:

-4 -2 ?；！；(S/N)/(S/N)= ?(0.6 10/12.5) = 0.22 10(P=1-nW) dircoh .-7-2 and ?0.610/0.0125 = 0.22 10 (P= 1 pW)

3-2 ?60 /1.25 10 = 0.22 10(P=1 ?W)

of course, these are always the same value because the noise is the same in all cases.

Problem 8-5 A coherent detection uses a non-ideal local oscillator, that has: a circular polarization instead of the linear one of the signal, a mode size (Gaussian) twice as wide as that of the signal, and a phase rms deviation = 0.5 rad. (

Calculate the total coherence factor and the S/N penalty. ?

Answ: We have:

?= [1,0];;? [1/?2, 1/?2] = 1/?2= 0.707; pol

?= ? gauss (r,w) gauss (r,2w) 2?rdr = 1/?5= 0.48sp 00

2?= 1 - /2 = 0.75 ( (

and collecting the terms

?= ??；?= 0.707 0.48 0.75 = 0.24 ( sppol

S/N penalty is ?= 0.24 = 6.2 dB

Problem 8-6 Evaluate the number of photons per bit to achieve a

-9 BER=10in a SOPSK (state-of-polarization-shift-keying) transmission with homodyne detection, where the ‘1’ is transmitted with a given state of

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polarization (e.g., a linear one) polarization (for example, same as that of the local oscillator) and the ‘0’ with the polarization state orthogonal to that of the ‘1’.

Answ: The ‘1’ signal is N=2?NN , the ‘0’ signal is zero, so that the 0S

optimum threshold is S=N/2. Noise on both levels is ?N (see also N0

p.258).

Thus we have:

BER = erfc S/ = erfc ?N NS

and we obtain N =36 photons per bit. S

Problem 8-7 Again consider a SOPSK coherent detection, and assume using a balanced detector with an input Glan-beamsplitter dividing the incoming polarization. What is the number of photons per bit to achieve a

-9 BER=10?

Answ: In this case, each photodiode will detect a symbol, the ‘0’ and the ‘1’.

The difference signal at the output of the balance detector swings from N

to +N and is twice as large the individual ones, N=4?NN, while noise 0S

on the –N to +N levels is always ?N. 0

Optimum threshold is S= N/2.

Thus we have:

BER = erfc S/ = erfc 2?NNS

and accordingly we get N =9 photons per bit. S

Problem 8-8 Should optical receivers for a 100 Gbit/s transmission rate become developed and available for system use, which would be their expected sensitivity (in ?W or dBm) ?

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Answ: Looking the practical results reported in the diagram of Fig.8-4 and by extrapolating the bit-rate to 100 Gbit/s, we may expect that the sensitivity of actual receivers should be:

3 - 5 ?W (or -25 to 23 dBm) for direct detection

0.30.5 ?W (or -35 to 33 dBm) for coherent detection

?；5 ?W (or 23 dBm) for optical preamplified detection

3Problem 8-9 An optical amplifier with a gain G=10 is intended for use as

a preamplifier at )=1500 nm. Calculate: the output ASE, the input-equivalent ASE, the NEP (noise-equivalent- input) for the full bandwidth available (40 nm) and for a B=10 GHz electrical bandwidth. What is changed if we filter the amplifier output with a narrowband filter with ，)=0.5 nm ?

Answ: The ASE for a single polarization state is, from Eq.(8.25): ASE = n h，；(G-1) ，， outsp

where n,the inversion factor, that can be taken n =0.9 as a typical value. spsp

For both polarizations, or, when no polarization selection is performed, we have ASE=2ASE and accordingly out/tot out. .-19.3. 12ASE = 2 0.9 1.6 10 (1.24/1.55) 10(40/1500) 200 10 out/tot

(in the above expression, we compute h as 1eV by the ratio of

wavelength ) to 1.24 ?m, and ，， as the fractional wavelength ，) /) by the

optical frequency 200 THz that corresponds to ) =1500nm)

So we have:

.-19.3. 12ASE= 1.8 1.28 10105.33 10= 1.23 mW out/tot

This is the dc optical power at the output of the amplifier. Referred to the input, we have

ASE= ASE/G = 1.23 ?W in/totout/tot

and the associated noise for a bandwidth B=10GHz is (Eq.8.26):

2NEP = 2 h，； ASEB ASEin/tot -19. -6 . 10 -15 = 1.28 101.23 1010= 1.57 10

or, NEP = 39.6 nW. ASE

To get the total noise, we shall add the excess term (Fig.8-11), F?2 in a

good amplifier and for a not too small signal P>ASE. Sin/tot

Thus, we may conclude that our optical amplifier has, for P>1.23 ?W S

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either: (i) a quantum-noise performance with excess factor F, or: (ii) a

. noise 239.6 nW = 80 nW (or equivalently 41 dBm) whichever is

larger.

If an optical filter with ，)=0.5 nm is used, the ASE decreases by a factor 0.5/40=1/80 [so that ASE=15 nW], and the NEP by 1/?80?1/9 in/tot

(reaching 50 dBm).

This figures (compare with data in Fig.8-4) are very interesting for applications, albeit limited to a specific wavelength of operation.

Problem 8-10 A signal from a narrow-line DBR laser around 1500 nmn is fed in another similar laser, detuned of f=20 GHz from the first..The

-1-1laser gain is =c 200 cm, loss is =c10 cm, cavity length is L=200 ?m,

mirror facets reflectivity is R=0.3, power emitted is P= 1mW. Calculate out

the heterodyne injection gain and the signal amplitude for a detected power P=1pW. S

Answ: The power in the laser cavity is P= P/(1-R)=1.4mW; mirror 00out

field transmittance is T=?(1-R)=0.84.

Using Eq.8.39, we can compute the injection gain as:

G = 2?[P/P] [Tc/2L(；？；)] 00S. -3-12 1/2 . . 4. = 2 [1.410/10]0.84/(2 0.02 190) = 7.48100.11= 8200

If a photodiode is place on the rear mirror of the laser where 1mW is emitted as well as from the front mirror, the photodetected current is (assuming =1A/W and after Eq.8.38):

I = I + G I cos (-)t 000S00S

= 1mA+ 8200 1pA [dc] [20 GHz]

= dc term + 8.2 nA at signal frequency.

As a verification, we shall check that the signal frequency f is within the

bandwidth of injection response. In the laser diode, the gain linewidth is much larger than f (usually THz’s), while the cavity linewidth is given by f=f/Q, Q being the quality factor of the cavity. line

Calculating Q = 4?L/)(1-R)

.. = 12.560.02/0.000150.7= 2300

we get:

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；，f= 200 THz/2300= 870 GHz, line

a value much larger than f.

Problem 8-11 In a heterodyne or homodyne injection detection, what is the minimum signal that can be detected by a 1-mW semiconductor laser with the parameters of Probl.8-10 ? Evaluate it both for a small detuned signal f injected in the laser and for a weak echo at the same frequency of the laser.

Answ: As injection detection, no matter if heterodyne or homodyne, is a coherent detection albeit with a penalty factor not much less than unity, the detectable signal is the same as for coherent detection. Typical values reported in experiments (see References on page 293-94), both with He-Ne and GaAlAs and InGaAsP diodes are, e.g.:

for heterodyne injection:

- pW’s for small/moderate bandwidth (<1MHz)

- nW’s for high bandwidth (?1GHz)

for homodyne injection:

- -90 dB of attenuation (respect to P) at small bandwidth (?1kHz) 00

- -50 dB of attenuation, high bandwidth (?1GHz)

Problem 8-12 Could the QND scheme of detection be implemented with a pump at the same frequency of the signal? How shall the setup of Fig.8-16 be modified?

Answ: Entering in the setup of Fig.8-16 with orthogonal pump and signal polarizations, and using a Glan polarization s in place of the IF filters, we can indeed superpose pump and signal with virtually no loss. However, we then need an index of refraction nonlinearity n associated with orthogonal 2(3)axes, that is a material with non-vanishing nonlinear permettivity ~2221

along the signal (1) and pump(2) polarization axes.

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Problem 8-13 What is the squeezing factor we can obtain from a diode laser converting the injected electrons into photons with an efficiency ?=0.95 ? What is the squeezing factor when this laser signal is launched in a fiber?

Answ: Ideally, if all of the electrons were converted into photons, we would have F=0. Whatever the nature, a loss ?=1-?；is accounted for by

Eq.8-56:

F’ = 1- ?；(1- F) = 1- (1-?)(1- F)

and, introducing in this equation F=0, ?=0.05, we get F’= 0.05.

If we attempt to launch in a fiber such a squeezed signal, as the launching efficiency laser-to-fiber is seldom ?>80%, we would immediately get launch

a degradation, again from the above equation, to a squeezing factor F: launch

F= 1- ?(1- F’) = 1 – 0.8 0.95 = 0.24. launch launch

The same result would supply the observed squeezing factor if we attempt to detect the laser emission with a photodiode having a quantum efficiency ?=80%.

Problem 8-14 Why the squeezed radiation actually improves the

interferometer readout (pages 286-287) while it does not help improving in data transmission and detection ?

Answ: Because these are physically different situations.

In transmission, information is carried by the signal, and a squeezed local oscillator can’t help [Eq.(8.57)] because of the cross-multiplication of

signal amplitude and squeezing factor. As the information limit due to photon statistics is already bounded at the transmission side, it is quire expectable that we cannot overcome the quantum limit of detection by squeezed states. Only a squeezed source can improve, but with the severe limitation of propagation attenuation.

In an interferometer, opposedly, information is internal to the optical path and thus a cleaner, squeezed signal can indeed help to read it with less noise.

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PROBLEMS CHAPTER 9

Problem 9-1 With a SbSvidicon tube and a F=50 mm, D= 22 mm 23

objective lens, we pickup a scene illuminated at E=500 lux. What is the sc

target current ? What is the S/N ratio of the image ? Assume an average scene diffusivity；?=0.3, a photoconductive gain G=10, a target load R=10 k and an electron beam current i=1?A. tpe

Answ: The F-number of the lens is F/=22/50=0.44; the typical luminous sensitivity of the SbS= 20 nA/lux (see Fig.9-9; this value target is 23 l

includes the G=10 gain).

From Eq.(9.5), written in terms of luminous unit-area sensitivity , we l

have:

2 i ? E / 4(F/)= t lsc

2 ...= 20 (nA/lux) 0.3 500 (lux) / 4(2.2)= 150 nA

Assuming the typical value of dark current I=10nA, and (=3,the noise 0

(Eq.9.4) is given by:

22 = 2e [1+2(M-1)] (i +I)B + 2e( iB +4kTB/Rt0pet it 2 = 2e{[1+2(M-1)] (i +I)+；( i +50mV/R }B t0pet

-19. 2.-9 .-6.6 = 3.2 10 {[1+29] 16010+ 310 +50mV/10k}5 10

-19 .-6.-6.-6.6 = 3.2 10 {1620.15 10+310+ 510}5 10

(incidentally, comparing the three terms reveals that shot noise is the dominant term). Going on the calculation,

-19 .-6.-6.-6.62 = 3.2 10 {1620.15 10+310+ 510}5 10 it-19 .-6 .6-18 = 3.2 10 {3410}510= 54 10

whence = 7.4 nA; the S/N is finally given by: it

S/N = 150 nA / 7.4 nA = 20 (or, 26 dB, a fair value for an image pickup).

Problem 9-2 Consider a 3-(；Si-CCD with 5 ?m cell size, with an acceptor

14 -3doping level N=10cm, and a dark current per pixel I=0.01 pA. What A0

is the clock driving voltage required? What is the saturation level and the dynamic range? What is the signal for a 100 lux illumination on the CCD?

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5.Answ: Taking a field V/w V/cm, a value close to the maximum =510Gox

set by oxide dielectric strength, Eq.(9.9) gives the stored charge at

saturation as:

-145 . ...Q= ? (V/w)= 3.98.8610 (F/cm)510V/cm max oxGox

.-7 -2 = 1.710Ccm,

or, in terms of the number of electrons,

-7 -2-19 12-2...N= Q/ e= 1.710Ccm/ 1.610C = 1.110 cm. max max

2-7 2 For the elemental cell, of area A=5x5 (?m)= 2.5 10cm,the saturation

charge becomes:

..5NA = 2.75 10 electrons max

Compared to the dark electrons collected in a T=20ms frame,

-12 -3 -19 .N= IT/e = 0.01 10A 20 10s /1.610C dark 0

3. = 1.6 10 electrons

this amounts to a dynamic range of the video signal

5 3 ..N/N= 2.75 10/ 1.6 10= 170 max dark

The required amplitude of clock voltages is (Eq.9.9):

V=Qw/?,Gmaxoxox

that, for a typical value w=100 nm yields: ox

-14-7 -2 -5..V = 210Ccm10 cm / 3.9 8.8610 F/cm = 5.8 V G

Correspondingly, from Eq.(9.7) the well extension is found as:

1/2 W = [ 2 ?(/ e N]s s A

...-14.-19.14 -31/2= [ 2 11.9 5.8 8.86 10 F/cm /1.610 C10cm]

= 8.7 ?m.

Last, the signal for a 10-lux illumination on the CCD is: i = ?；： A E outlsc

where ?；=2/3 is the fill-factor, and can be estimated (Fig.9-15) l

assuming a 0.3A/W average radiant sensitivity and a spectral content of

100 lm/W. Thus,

-11 2 ..-12 .i = 0.66 0.3A/W 2.5 10m10 lux /100 lm/W = 0.510A out

This current yields, in a frame period,

-12 ..-3 -19 4...N= iT/e =0.5102010s /1.610C = 610 electrons out out

a value well within the dynamic range.

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