PROBLEMS CHAPTER 8
Problem 8-1 A weak signal, P=1-nW (or, respectively: 1-pW, 1-mW) is detected by a coherent homodyne scheme, using a silicon photodiode with s=1A/W, Idark= 0.1 mA, terminated on a 1-kW load. Find: (i) the minimum 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 m=1 and B=1 GHz.
Answ: The condition IOL > Idark + I + 2kT/eR yields for the local oscillator detected current IOL:
IOL > 0.1mA + 1pA +50mV/1kW = 50 mA
(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 s=1A/W, we get for the required local oscillator power:
POL = 500 mW (and IOL = 500 mA).
For P=1 fW and 1 nW, the local oscillator power remains unchanged.
The signal amplitude is:
RsP = 1 mV (resp.: 1 nV, 1mV) (direct detection),
RsPG = RI (1+2√IOL /I) (coherent detection)
= 1mV. [1+2√(5.105)] = 1.4 mV (resp: 44mV, 44 mV).
The S/N ratio is:
(S/N)2 = 4I/2eB = 4. 10-9/3.2. 10-19. 109 = 12.5 (resp: 0.0125, 1.25 103)
(coherent detection)
and (S/N)2= I2/2eB(0.1mA + 1pA +50mV/1kW)
= 10-18/3.2 . 10-19. 109. 50 10-6= 0.6 10-4 (resp: 0.6.10-7, 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:
E0 exp i[wt+f+2ks(t)]
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 = I0 + 2√I0IS cos[f-flo+2ks(t)]
Apart from a dc term I0 and a constant phase f-flo, the coherent detection 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=2p/l). 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
rate, the one giving a BER=10-9? Consider bit rates R= 1Mbit/s, 1Gbit/s. Assume l=1.24 mm, and a spectral sensitivity s= 0.8 A/W. What is the quantum noise and the S/N ratio of this signal ?
Answ: The photodetected signal is:
I = e N R = 1.6 . 10-19. 10 .106 = 1.6 pA @ 1Mbit/s,
and = 1.6 nA @ 1Gbit/s;
the corresponding powers, being hn=1eV @ l=1.24 mm are:
P = hn N R = I / s = 2pW and 2 nW, respectively.
Quantum noise is (note that it is B=R/2 from Nyquist theorem):
in = [2eI B]1/2= [2eI R/2]1/2= [1.6.10-19.1.6.10-12 .106]1/2
= 0.51 pA @ 1Mbit/s,
and = 5.1 nW @ 1Gbit/s.
From these numbers, the signal to noise is found back as:
(S/N)2 =(I /in)2= (1.6/0.51)2 = 10 (in both cases).
Problem 8-4 Given the data of Probl.8-1, what is the theoretical minimum m 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 m as:
m = (S/N)dir /(S/N)coh = √(0.6 10-4 /12.5) = 0.22 10-2 (P=1-nW)
and √0.6.10-7/0.0125 = 0.22 10-2 (P= 1 pW)
√60 /1.25 103 = 0.22 10-2 (P=1 mW)
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 sF= 0.5 rad.
Calculate the total coherence factor m and the S/N penalty.
Answ: We have:
mpol = [1,0] ´ [1/√2, 1/√2] = 1/√2= 0.707;
msp = ∫ gauss (r,w0) gauss (r,2w0) 2prdr = 1/√5= 0.48
mF = 1 - sF2/2 = 0.75
and collecting the terms
m = mF msp mpol = 0.707 0.48 0.75 = 0.24
S/N penalty is m = 0.24 = 6.2 dB
Problem 8-6 Evaluate the number of photons per bit to achieve a BER=10-9 in a SOPSK (state-of-polarization-shift-keying) transmission with homodyne detection, where the ‘1’ is transmitted with a given state of 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√N0NS , the ‘0’ signal is zero, so that the optimum threshold is S=N/2. Noise sN on both levels is √N0 (see also p.258).
Thus we have:
BER = erfc S/sN = erfc √NS
and we obtain NS =36 photons per bit.
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 BER=10-9 ?
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, DN=4√N0NS, while noise on the –N to +N levels is always √N0.
Optimum threshold is S= DN/2.
Thus we have:
BER = erfc S/sN = erfc 2√NS
and accordingly we get NS =9 photons per bit.
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 mW or dBm) ?
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 mW (or -25 to –23 dBm) for direct detection
0.3–0.5 mW (or -35 to –33 dBm) for coherent detection
» 5 mW (or –23 dBm) for optical preamplified detection
Problem 8-9 An optical amplifier with a gain G=103 is intended for use as a preamplifier at l=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 Dl=0.5 nm ?
Answ: The ASE for a single polarization state is, from Eq.(8.25):
ASEout = nsp hn (G-1) Dn
where nsp,the inversion factor, that can be taken nsp =0.9 as a typical value.
For both polarizations, or, when no polarization selection is performed, we have ASEout/tot=2 ASEout and accordingly
ASEout/tot = 2 . 0.9 .1.6 10-19. (1.24/1.55) 103. (40/1500) 200 1012
(in the above expression, we compute hn as 1eV by the ratio of wavelength l to 1.24 mm, and Dn as the fractional wavelength Dl /l by the optical frequency 200 THz that corresponds to l =1500nm)
So we have:
ASEout/tot= 1.8 .1.28 10-19.103. 5.33 1012= 1.23 mW
This is the dc optical power at the output of the amplifier. Referred to the input, we have
ASEin/tot= ASEout/tot/G = 1.23 mW
and the associated noise for a bandwidth B=10GHz is (Eq.8.26):
NEPASE2 = 2 hn ASEin/tot B
= 1.28 10-19. 1.23 10-6 . 1010 = 1.57 10-15
or, NEPASE = 39.6 nW.
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 PS>ASEin/tot.
Thus, we may conclude that our optical amplifier has, for PS>1.23 mW
either: (i) a quantum-noise performance with excess factor F, or: (ii) a noise 2. 39.6 nW = 80 nW (or equivalently –41 dBm) – whichever is larger.
If an optical filter with Dl=0.5 nm is used, the ASE decreases by a factor 0.5/40=1/80 [so that ASEin/tot=15 nW], and the NEP by 1/√80»1/9 (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 Df=20 GHz from the first..The laser gain is a=c 200 cm-1, loss is G=c10 cm-1, cavity length is L=200 mm, mirror facets reflectivity is R=0.3, power emitted is Pout= 1mW. Calculate the heterodyne injection gain and the signal amplitude for a detected power PS=1pW.
Answ: The power in the laser cavity is P00= Pout/(1-R)=1.4mW; mirror field transmittance is T=√(1-R)=0.84.
Using Eq.8.39, we can compute the injection gain as:
G = 2√[P00/PS] [Tc/2L(a-G)]
= 2 [1.4. 10-3/10-12] 1/2 0.84/(2 0.02 . 190) = 7.48. 104. 0.11= 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 s=1A/W and after Eq.8.38):
I0 = I00 + G IS cos (W00-WS)t
= 1mA[dc] + 8200 1pA[20 GHz]
= dc term + 8.2 nA at signal frequency.
As a verification, we shall check that the signal frequency Df is within the bandwidth of injection response. In the laser diode, the gain linewidth is much larger than Df (usually THz’s), while the cavity linewidth is given by Dfline=f/Q, Q being the quality factor of the cavity.
Calculating Q = 4pL/l(1-R)
= 12.56.0.02/0.00015.0.7= 2300
we get:
Dfline= 200 THz/2300= 870 GHz,
a value much larger than Df.
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 Df 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 P00) at small bandwidth (»1kHz)
- -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 n2 associated with orthogonal axes, that is a material with non-vanishing nonlinear permettivity c(3)2221 along the signal (1) and pump(2) polarization axes.
Problem 8-13 What is the squeezing factor we can obtain from a diode laser converting the injected electrons into photons with an efficiency h=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 e=1-h is accounted for by Eq.8-56:
F’ = 1- e (1- F) = 1- (1-h) (1- F)
and, introducing in this equation F=0, h=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 hlaunch>80%, we would immediately get a degradation, again from the above equation, to a squeezing factor Flaunch:
Flaunch = 1- hlaunch (1- F’) = 1 – 0.8 0.95 = 0.24.
The same result would supply the observed squeezing factor if we attempt to detect the laser emission with a photodiode having a quantum efficiency h=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.
PROBLEMS CHAPTER 9
Problem 9-1 With a Sb2S3 vidicon tube and a F=50 mm, D= 22 mm objective lens, we pickup a scene illuminated at Esc=500 lux. What is the target current ? What is the S/N ratio of the image ? Assume an average scene diffusivity d=0.3, a photoconductive gain G=10, a target load Rt=10 kW and an electron beam current ipe=1mA.
Answ: The F-number of the lens is F/=22/50=0.44; the typical luminous sensitivity of the Sb2S3 target is sl = 20 nA/lux (see Fig.9-9; this value includes the G=10 gain).
From Eq.(9.5), written in terms of luminous unit-area sensitivity sl, we have: