3.6 QUANTUM LIMIT IN OPTICAL COMMUNICATION SYSTEMS
3.6.2 Actual Quantum Limit from Poisson Distribution
In the previous section, the results for AWGN were applied to a fictitious example to get an estimate of the quantum limit in terms of parameters that are familiar to a circuit designer. In this section, we will determine the actual quantum limit based on the random arrival statistics of photons. To determine the absolute minimum power needed, we will assume that we have the capability to detect a single photon.
Furthermore, we’ll assume that we have a light source with a 100% extinction ratio (when the light is off, it is really off). The received signal will consist of either a light pulse, or no light pulse. When there is no light pulse, there is absolutely no way we can detect a photon (this assumes that there is no ISI). Therefore the conditional probability density function when no pulse is sent is
P
0(r) =(r); (3.87)
which means that the received signal is identically zero with probability one. We can express the optimal decision rule as
R
0; r= 0
R
1; r6= 0: (3.88)
132 Chapter 3
That is, we chooses1(t)when the received signal is anything but zero. The total error probability for this system is
P
e= 12 Pr[d(r) = 1jH0] + 12 Pr[d(r) = 0jH1]; (3.89)
and since there is no noise when there is no light, there is no chance of making an error when no light is transmitted. The only chance of making an error is when we don’t detect any photons when we were supposed to. In other words, we turn on the light source, but because of the random nature of photon emission, no photons are emitted in a timeT, even when the light source is on. The total error probability is just due to the later situation, and is given by
P
e= 12 Pr[d(r) = 0jH1]: (3.90)
To find the probability of this event, we must consider the arrival statistics of the photons themselves.
Derivation of Error Probability If we look at very short time intervalst, we
will assume that the probability of the arrival of one photon in this time interval is proportional tot,
P
1(t) =at; (3.91)
where the significance of the proportionality constantawill be demonstrated later.
Since the time interval is short, either one photon arrives or it doesn’t, but the time interval is too short to allow more than one arrival. Therefore,
P
0(t) +P1(t) = 1
P
0(t) = 1;at (3.92)
We are interested in finding the probability that no arrivals occurred in a time interval of lengthT. We can consider an interval of lengthT+t, and we further assume that the arrival of a photon in the timetis independent of the arrival of a photon in the previous time intervalT. The probability of no emission in a time period ofT + t
is then given by
P
0(T+ t) =P0(T)P0(t) =P0(T)[1;at]; (3.93)
and writing this as a difference equation we get
P
0(T+ t);P0(T)
t =;aP0(T): (3.94)
In the limit astgoes to zero, the difference equation becomes a differential equation
dP
0(T) =;aP0(T); (3.95)
Optimal Decision Theory 133
with the solution of
P
0(T) =e;aT; (3.96)
where we have made use of the boundary condition
P
0(0) = lim
t!0 P
0(t) = 1: (3.97)
Relationship Between Parameteraand Observable Statistics (3.96) gives the desired result, but is expressed in terms of the parametera. In order to determine a relationship for this parameter in terms of observable statistics it is necessary to continue the derivation to determinePn(T), the probability of obtaining preciselynphotons in a given time intervalT. Following Davenport and Root [6, ch. 7, pp. 115–118], we will consider the probability of observingnphotons in a time interval of lengthT + t.
We can restricttto be so small that no more than one photon can arrive in this time;
therefore are only two possibilities exist: either one photon is emitted, or none are.
Since we have assumed that emissions at any timetare independent of emissions in the past, for smalltthe probability of observingnphotons in an interval of length
T+ tis simply given by
P
n(T+ t) =Pn(T)P0(t) +Pn;1(T)P1(t): (3.98)
Recalling thatP1(t) =at, andP0(t) = 1;at, it follows that
P
n(T+ t);Pn(T)
t +aPn(T) =aPn;1(T): (3.99)
In the limit ast!0, we obtain a differential recursion equation
dP
n(T)
dT
+aPn(T) =aPn;1(T); (3.100)
which has a solution given by3
P
n(T) =ae;aTZ T
0 P
n;1()ead; (3.101)
where we have utilized the boundary conditionPn(0) = 0. For the case ofn= 1we
can make use of the resultP0(T) =e;aTfrom (3.96) to obtain
P
1(T) =ae;aT
Z
T
0 e
;a
e a
d = (aT)e;aT: (3.102)
3Davenport and Root reference Richard Courant, “Differential and Integral Calculus,” I. rev. ed., 1937;
II, 1936, Interscience Publishers, New York.
134 Chapter 3
For the case ofn= 2, we can substitute the previous result to obtain
P
2(T) =ae;aTZ T
0
(a)e;aead = (aT)2e;aT
2 : (3.103)
It is not difficult to see the pattern that emerges from the recursion and therefore determine the probability for any arbitrarynas
P
n(T) = (aT)ne;aT
n! (3.104)
Using (3.104) we can find the expected number of arrivals in timeTas
n
1=X1
n=0 nP
n(T) =X1
n=0
n(aT)ne;aT
n! (3.105)
This sum can be evaluated explicitly as follows. First consider the Taylor series expansion foreaT.
1
X
n=0
(aT)n
n! =eaT (3.106a)
Taking the derivative with respect to both sides gives
1
X
n=0
n(aT)n;1
n! =eaT; (3.106b)
and multiplying by(aT)gives a series expansion for(aT)eaT:
1
X
n=0
n(aT)n
n! = (aT)eaT (3.106c)
Finally, multiplying bye;aT puts this in the form of (3.105).
n
1=X1
n=0
n(aT)ne;aT
n! =aT (3.106d)
Now we can see the significance of the parameteraand substituteaT =n1into (3.96) to obtain
P
0(T) =e;n1; (3.107)
which is the probability of not getting any photons in a timeT, when on average we getn1.
Quantum Limit in Terms of Number of Photons per Bit Using the above results, the desired probabilityP0(T)can now be expressed in terms ofn1, which is an observable
Optimal Decision Theory 135
P
e n
1
10;3 6:2 3n:1 10;6 13:1 6:6 10;9 20:0 10:0 10;12 26:9 13:5 10;15 33:8 16:9
Table 3.2 Quantum limit in terms of average photons per one symboln1and average photons per bitnincident on the photo-detector in one bit periodTto insure a given error probabilityPe.
quantity. The total error probability for a fiber-optic receiver operating at the quantum limit is given by
P
e= 12P0(T) = 12e;n1: (3.108)
We have now arrived at the desired result that, due to quantum noise, we require on averagen1photons per one symbol to insure an error probability ofPe, where
n
1=;ln(2Pe) (3.109)
Since there are no photons transmitted for a zero symbol, the average number of photons per bitnis given by
n= 12 (n0+n1) = 12n1; (3.110)
therefore
n=;ln(2Pe)
2 (3.111)
The quantum limits n1andnare given in table 3.2, where we see that, approximately, an additional 7 photons per one symbol are required on average to reducePeby 3 decades.
Plots ofPn(T)are given in Figs. 3.13(a) and (b) forn1= 10and20respectively. It can also be shown using the same method as outlined in (3.106) that the variance of the Poisson distribution is also equal ton1. Therefore the standard deviation is equal to the square-root of the average number of photons, and the average SNR is equal to
1=pn1, which is a familiar result for independent random variables.
136 Chapter 3
0
Number of Photons Observed in Time T
Probability
5 10 15 20 25 30 35 40
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
0
Number of Photons Observed in Time T
Probability
5 10 15 20 25 30 35 40
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
(a) (b)
Figure 3.13 Possion distribution forn1equal: (a) 10, (b) 20.
Quantum Limit in Terms of Optical Power We can relate the quantum limit to optical power for a given bit intervalT. Continuing with our example of a 10 Gb/s system, the bit-intervalTis equal to 100 ps. The energy in a photon is given by
eph= hc
; (3.112)
wherehis Planck’s constant, andcis the speed of light. Substituting these constants the photon energy is
eph= 198:610;12(nJm)
: (3.113)
The photon power is the energy divided by the intervalT;
pph= eph
T
= 1:986(nWm)
for T = 100ps: (3.114)
Since the light pulse is only on half of the time, the average optical power is
pav= 12(0)+ 1
2(n1pph) =npph; (3.115)
or substituting (3.109), we can write the average power at the quantum limit in terms of the desired error probability, for a bit-rate of 10-Gb/s such that,
pav=;log(2Pe)
2:29(nWm)
: (3.116)
The following equation expresses the result for an arbitrary bit-rateBT,
pav= ;ln(2Pe) 2
hc
B
T
; (3.117)
Optimal Decision Theory 137
P
e n
1
pav(= 0:80m) pav(= 1:30m) pav(= 1:55m)
dBm dBm dBm
10;3 6:2 ;51:1 ;53:2 ;54:0 10;6 13:1 ;47:9 ;50:0 ;50:8 10;9 20:0 ;46:0 ;48:1 ;48:9 10;12 26:9 ;44:7 ;46:9 ;47:6 10;15 33:8 ;43:8 ;45:9 ;46:6
Table 3.3 Quantum limit for various wavelengths of light in terms of average optical power incident on the photo-detector for a 10 Gb/s optical receiver (T=100 ps) to insure a given error probability.
and substituting for the numerical constants we obtain the general expression for the quantum limit in an optical system using on-off modulation and no coding.
pav=;log(2Pe)
0:229(nWm)
B
1-Gb/sT
(3.118) Table 3.3 gives the quantum limit in terms of the minimum average optical power required to achieve various error probabilities for different wavelengths of light at a data rate of 10-Gb/s. We can see that, at best, we need an optical power of about -48 dBm forPe= 10;9. This analysis gives us a theoretical limit on the minimum received power. However practical implementation problems will limit the sensitivity of the receiver such that many more photons above the quantum limit will be needed for accurate communication.