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In the last lecture, we investigated the characteristics of a photo detector. Now, we see if the photo
detector is used in optical receivers, what are the performance parameters of an optical
receiver and in this case, essentially we are going to discuss the digital data that
means, if the information is sent in the form of bits, how do we measure the performance
of an optical receiver? So, in this lecture, essentially we are going to discuss the performance
of the optical receivers.
We have seen in the previous lecture that in a photo detector or in general, in an optical
receiver, there are various types of noises which are present. So, we have a shot noise,
which is because of the statistical nature of the photons and the statistical nature
of the interaction of the photon with the matters. This we call also as the quantum
noise. We saw that this noise has a poisson distribution, and this noise is multiplicative
in nature. We also saw that there is a noise now, which is what is called the dark current
noise that in the presence of signal, we have this noise. But when the signal is not present,
then there is some ambient light falling on the photo detector, and that gives some fluctuations
in the photo current, which we call as the dark current noise.
So, dark current noise is divided into two categories, what is called the bulk dark current
noise which is essentially inside the device; whereas, we have the current which is flowing
almost on the surface of the device, which we call as the surface current. And then we
have fluctuation in that current, which we call as the surface dark current noise. And
then we saw that because of the resistances used in the electronic circuitry, we have
the so called thermal noise, and this thermal noise is additive in nature, and also it is
distribution is Gaussian. So, in any optical signal, essentially we have combination of
all three noises present and depending upon the optical intensity, either this noise will
dominate or this noise will dominate and so on. So, last time we wrote down the expressions
for these noises.
So, we have seen that the quantum noise or the shot noise has the variance or mean square
value, which is proportional to the mean photo current and it is proportional to the bandwidth.
Also if we use the photo detector which is avalanche detector, then there is internal
gain which is given by this factor M and then we have a noise figure; because again the
avalanche process also is a statistical process. Similarly, we had the expression for the dark
current noise. So, this is the expression for the bulk dark current noise. We see the
amplification because of the avalanche process and the surface current does not see amplification
because of the avalanche process. And then we have the thermal noise, which has a variance
equal to 4 times the Boltzmann constant and the thermal temperature of the receiver, the
bandwidth of the receiver and the load resistance and then we said that assuming that these
noises are independent of each other.
So, total variance will be sum of their variances. So, we essentially got the noise power which
is the quantum noise plus the dark current bulk noise plus the dark current surface noise
plus the thermal noise. And then we got a signal power, which is proportional to i p
square and if you are using the avalanche photo detector, then this quantity will be
having some value M. If you are using simple p i n detector, then M is equal to 1 and then
there is no internal amplification in this process. So, essentially now we are going
to make use of this quantity to find out the system performance, which is what is called
the bit error ratio. I mentioned last time that if you are considering
the analog communication system, then signal to noise ratio is the parameter which is of
importance; whereas, if you are using the data which is digital in nature, then more
useful quantity is what is called the bit error ratio; that means is the ratio of number
of bits wrongly detected to the total number of bit transmitted or received by the receiver.
So, today essentially we are going to do the analysis of the bit error ratio. So, we want
to now calculate what is called the bit error ratio and for that, we can make certain justifiable
assumptions about the data.
So, firstly we are going to do the analysis of this for the data, which is binary in nature;
that means we are sending the information in the form of only two levels, 0 and 1. And
let us say, 0 means no optical power and 1 means some optical power. While discussing
the optical fibers, we have discussed that if you consider the normal semiconductor lasers,
the spectral width of this laser is very large. And because of that, sophisticated communication
modulation techniques cannot be employed; because the spectrum of the signal is completely
washed out. So, for normal optical communication system, essentially we use the amplitude modulation;
because that is the modulation, which can be recovered in time domain.
It does not require information about the spectral domain and that modulation if we
convert in to corresponding digital form that is the one which we call as the amplitude
shift keying; that means now the data 0 and 1 is transmitted in the form of optical pulses.
So, if 0 bit is transmitted, there is no optical pulse; if 1 bit is transmitted, there is an
optical pulse. So, we are assuming now the data is binary. So, data is send only either
presence of a optical pulse or absence of a optical pulse. Also we are assuming that
the data is unbiased; that means there is no preference given to 1 level or 0 level.
There is equal possibility of getting 0 bit or 1 bit. Also what you are assuming here
is that the noise is additive in nature and it is Gaussian in nature.
So, recall when we talk about the shot noise, we said that actually the shot noise is poisson
in nature; also it is multiplicative in nature. However to do the analysis simpler, we take
the appropriate variance value for the noise in two levels 0 and 1. But beyond that point,
we assume that this noise is distributed with Gaussian distribution and then we can do the
analysis for the bit error ratio. So, to account for the shot noise appropriately into the
analysis, essentially what we are doing now is we are saying that the noise is different
for the 0 level and for the 1 level. So, appropriately we can provide these variances for the two
levels for the data. If we do that, then essentially we have now the bits which are going in two
levels.
So, we have a 0 level here and we have 1 level here and the noise is superimposed on this.
So, these are the mean levels which are denoted level 1 and level 0 and this is the photo
current. And photo current is fluctuating around this mean value corresponding to 0
level and around this mean value, which corresponds to 1 level. And in general, as we said we
are assuming that these quantities are variances of these two are not equal and that is what
is shown here, if you look at this distribution here. This distribution looks narrower compared
to this distribution. And there is for the simple fact that in this case, we will have
the shot noise present; whereas, in this case we will have only noise which is dark current
noise or the thermal noise. So, what is plotted here is the probability
of the current lying around this mean position which is 0. So, the current will fluctuate
around this value which is this and there is the density function for that current for
the 0. So, that is the density function which we say is p y given 0 corresponding to 0 level.
Similarly, we can get here p y corresponding to level 1, where y is the quantity which
is representing current or corresponding voltage, if you are measuring the voltage across the
load resistance. So, then we can write here the probability of the current for level 1
and level 0.
So, we say that the density function; so p of let us say photocurrent I given 0. So,
that is the density function for the 0 level; that is equal to 1 upon square root 2 pi into
sigma 0 e to the power minus I 1 or I minus I 0 whole square divided by 2 sigma 0 square;
where sigma 0 is the standard deviation of this Gaussian noise or sigma 0 square is the
variance of this Gaussian noise corresponding to 0 level. And I 0 is the mean value of the
current which can be 0. If we assume that, no light was transmitted in the corresponding
to 0 bit. But due to some factors, there may be certain low value of light which will be
transmitted even during the 0 bit. So, in general let us leave it that this level
is not 0; but this level is given by some I 0. Similarly, we have a probability of the
current given 1 level and that will be equal to 1 upon square root 2 pi sigma 1 e to the
power minus I minus I 1 square divided by 2 sigma 1 square, where sigma 1 is the standard
deviation of the noise for 1 level. Say you seen here if the shot noise is present, then
you will have a value of the standard deviation larger for this compared to the 0 level. So,
sigma 1 square and sigma 0 square are the total variances of noise corresponding to
1 and 0 level. So, now you can say that if the one 0 level is transmitted and seen the
data is binary. Essentially, what we do? We have a threshold
level for decision. If the signal lies above the threshold value, then we detect the bit
as 1. If the signal lies below the threshold value, then we detect the level at 0. So,
we have this level here which we call as some I threshold and this compared at the instant,
when the bit is expected. If the current is greater than this value which is I threshold,
then we assign the bit as 1. If the current is less than I threshold, then we assign the
bit as 0. Now, with this density function which is given here; that means, we can now
calculate what is the probability that 0 level was transmitted.
But the signal would have excited this value, which is I threshold and that probability
essentially is given by this area under this. Similarly, if 1 was transmitted and if the
signal happens to lie below the threshold value, then will take 1 as 0 and then there
will a bit error. So, for 0 level transmission, if the signal exceeds I threshold, there is
bit error; whereas for 1 level transmission, if signal lies below I threshold, then there
is a bit error. Now, we can say that the total bit error probability is the probability of
bit error for 1 level plus the probability of bit error for 0 level and ofcourse 0 then
1, the appearances also will have their own probabilities.
So, we can write what is called the bit error ratio, which is nothing but the probability
of any bit detected wrongly. So, we can write BER is equal to the error probability of detecting
0, when 1 was transmitted. So, that is probability of getting signal 0, when 1 was transmitted
multiplied by the probability of transmission of 1. So, which is probability of 1 plus the
probability of detecting 1, when 0 is transmitted multiplied by the probability of transmission
of 0. Now, note here we were assume that the data is unbiased; that means, there is a probability
of 1 level and 0 level transmission equal. So, we have here p of 1 is equal to p of 0
is equal to 1 by 2; that means in a data stream which we are receiving 0 and 1 levels are
equi probable. Therefore, we can write here the BER that
is equal to 1 by 2 into this probability. So, that means probability of detecting 0
when actually 1 was transmitted plus probability of detecting 1 when actually 0 was transmitted.
And as we have seen probability of this, when 1 was transmitted detecting 0 is essentially
given by this area. So, if I take this density function corresponding to 1 and if I integrate
for the currents manage infinity up to this point I threshold; that is the area which
gives me the probability that the current will lie below threshold value and therefore,
1 will be detected as 0. So, we can write that then that this probability
of detecting 0, when actually 1 is transmitted; that is 1 upon square root 2 pi sigma 1 minus
infinity to I threshold e to the power minus I minus I 1 whole square divided by 2 sigma
1 square in to d I. Similarly, the probability of this which is probability of detecting
1, when actually 0 was transmitted that will be given by this area. So, if I take a density
function corresponding to 0 level and if I integrate this from the I threshold up to
infinity, this area is the error probability which will be p 1 0.
So, we can write p 1 0 that is equal to 1 upon square root 2 pi sigma 0 integral I threshold
to infinity e to the power minus I minus I 0 whole square divided by 2 sigma 0 square d
I. So, once I get these two probabilities, there I can substitute into this expression
and then we got the bit error probability or bit error ratio. Now, these integrals cannot
be solved in the close form. So, essentially they are represented by functions what are
called the error functions or complementary error functions. So, the complementary error
function is defined as follows.
erfc as a function of x that is defined as 2 upon root pi integral x to infinity e to
the power minus y square d y. So, taking this definition for the complementary error function,
these integrals now can be written as error functions. So, we get the probability of getting
0, when actually 1 is transmitted; that is nothing but half half error function I 1 minus
I threshold divided by sigma 1 root 2. Similarly, this quantity here p 1 0, we can write as
half complementary error function of I threshold minus I zero divided by sigma 0 root 2. So,
infact there are standard complementary error function tables are available. So, if we know
this quantity the threshold current and the level I 1 and I 0, then we can find out this
quantity here. Then we can go to the tables for the complementary
error function and from there, then we can find out this probability of getting zero
or getting 1. And then we can find out the total probability of error in the data, which
we call as the bit error rate. So, combining these two and substituting into this expression
for the BER, we get now the BER for the data which is equal to 1 upon 4 erfc of I 1 minus
I threshold upon sigma 1 root 2 plus erfc I threshold minus I 0 upon sigma 0 root 2.
So, this is the expression for the bit error ratio. Now, as one can see here the bit error
ratio now is going to depend upon the threshold current and which make sense. Because if you
look at these plots here, if the threshold value is brought down, then there will be
more a errors in detection of 0. Because now there will be more probability
of signal crossing the threshold level. So, more 0’s will be detected wrongly there
will be lesser 1’s detected wrongly; so more 1’s will be detected correctly. Similarly,
if I bring the threshold value higher, then more 1’s will go wrong and more 0 will be
detected correctly. So, there has to be some optimum value for which essentially we get
the BER minimized. So, now we have questioned that if we are having the bit error rate given
by this quantity which is the function of I threshold, then what is this value of optimum
value for I threshold? For which, this bit error ratio would go minimum. One can show
that this bit error ratio was minimum, when we have this condition satisfied.
So, we get BER minimum, when we have I 1 minus I threshold divided by sigma 1 is equal to
I threshold minus I 0 divided by sigma 0. So, essentially what we are saying is that
when these two areas this area which gives me the error probability of 0 bit and this
area which gives me the error probability of 1 bit. When these two areas become equal;
that means, when the bit error probability for 0 and 1 become equal, that is a time we
will get the minimum bit error rate. And that is what essentially given by this condition
here that when these two become equal, the two areas become equal and that is what gives
you the minimum bit error rate. So, this is the optimum value of the threshold
current. So, we can now say that then the optimum threshold current I th that by solving
this, we get essentially as sigma 0 I 1 plus sigma 1 I 0 divided by sigma 0 plus sigma
1. One can verify that if we have a situation that the noise is equal on the 0 bit and the
1 bit; that means when sigma 0 is equal to sigma 1 and that is what will happen if we
consider the noise which is only thermal noise, then the sigma 0 equal to sigma 1. So, I threshold
essentially will become I 1 plus I 0 divided by 2. So, for a binary data for which the
noise is equal on both the levels, that time the threshold level has to be half way between
these two levels. So, it is I 1 plus I 0 divided by 2; whereas,
when the sigma 1 and sigma 0 are not equal, that time we have to adjust the threshold
appropriately to minimize the bit error ratio. And in the optical communication, it is more
appropriate to treat the sigma 0’s and sigma 1 different. Because there may be a possibility
that in this level, you may have a shot noise which may be substantially larger compared
to the noise which we you get here because of the essentially thermal origin. So, in
general essentially then we are having the optimum value of threshold current, which
essentially is given by this. Now, one can consider the case when sigma 1 is much much
greater than sigma 0, there could be one extreme condition or other condition could be sigma
0 could be equal to sigma 1, which we call as the thermal noise condition.
So, firstly for optimum value of this thing if I substitute this I th into this now, we
can get this quantity which we call as the Q parameter of the data. So, we define a quantity
Q which is this quantity essentially. Say if I substitute this optimum value in this,
these two will become equal right and that Q quantity will be given by I 1 minus I 0
divided by sigma 1 plus sigma 0. So, now we are saying this is the quantity Q, which is
the parameter which is now deciding the bit error rate on the data. Now, if you look at
this thing, what it it is telling you physically? If I look at the levels here, we have a 0
level some fluctuations are there; we have 1 level, on this some fluctuations are there.
So, now if I say this is my 0 level, and this is my 1 level, and there is some noise which
is present on this and there is some noise which is present on this. This quantity, which
is the level difference between 0 and 1 that quantity is I 1 minus I 0; I 1 this level
is I 0. So, the difference between the mean value of this 1 and mean value of 0 level
that is the swing, which we have in the two levels of the data and sigma 1 and sigma 0
are essentially deviations from this mean level. So, if you consider a Gaussian distribution,
typically this would be of the order of about 2.5 sigma sigma 0 from the mean value. Similarly,
from this mean value this deviation peak deviation would be something like 2.5 in to sigma 1.
So, now we are having encroachment in this swing, which is from I 0 to I 1 between the
two levels and the encroachment is now from both sides. This side it is 2.5 time sigma
1 and this is 2.5 time sigma 2 or this encroachment is proportional to sigma 0 plus sigma 1. So,
we have this quantity which is the swing in the two levels, which is I 1 minus I 0 and
then we are having encroachment because of the noise in this swing; that is proportional
to sigma 0 plus sigma 1. So, essentially this range which is available now clean. One can
call this as noise margin; that is the quantity which essentially is this parameter Q.
So, we are essentially talking this quantity Q which is saying that we have a total swing,
which is available between two levels I 1 minus I 0 and then you are having this is
the encroachment parameter. So, this is a quantity which is in some sense is a measure
of what is the noise margin present in to your system and that is the quantity, which
is now going to decide the bit error rate of the receiver. So, now if I substitute this
Q seen these two quantities have become equal now for the minimum bit error ratio, the BER
expression now can be simplified further. So, for this optimum condition these two terms
have become equal and these quantity here I 1 minus I threshold divided by sigma 1;
that quantity is nothing but Q.
So, we can write down now the bit error rate which is BER; that is now equal to half erfc
Q divided by root 2. This can be approximated to an exponential function for large values
of Q. So, we can say this is approximately e to the power minus Q square by 2 divided
by Q root of 2 pi and if Q is greater than about 3, then the approximation is quite accurate.
So, within few percent, we get this quantity equal to the complementary error function.
So, now essentially what we are saying is we have now derived the expression for the
bit error ratio. And we have defined this parameter Q, which essentially the measure
of the current swing between the two level 0 and 1 and the noise present at the two levels.
So, for a data quality if you can measure this quantity Q than under the assumption
that the noise is of Gaussian nature, we can calculate the bit error rate for the data.
Now, as we can see here this quantity is a very rapidly decreasing function of Q. It
is going is e to the power minus Q square.
So, if you plot this function, the function essentially looks like this. So, what is plotted
here is quantity Q and on the vertical axis, you have plotted this quantity BER. So, here
we have BER. So, that one can see this vertical scale is logarithmic and here every division
is 10 to the power minus 3. So, this is 10 to the power 0 which is 1; this is 10 to the
power minus 3; 10 to the power 6; 10 to the power minus 9; 10 to the power minus 12; minus
15 and so on. So, we can note here that as the Q increases, very rapidly this functions
drops; that means the bit error ratio or bit error rate drop very rapidly as a function
of Q. And if we take some standard number now what bit error ratio will be acceptable
value, then we can find out what is the corresponding value of k.
So, one can do one can say that I have a standard bit error rate acceptable for the data. And
in optical communication, the acceptable bit error rate is 10 to the power minus 9 without
any error codes, error corrections. So, on the raw data you must get a bit error rate
of 10 to the power minus 9. So, if you go to this this quantity 10 to the power minus
9, you will get this value typically about Q equal to 6, where BBR will be 10 to the
power minus 9. So, we get Q equal to approximately 6; for BER 10 to the power minus 9 and then
for every one unit increase in Q, the BER practically dropped by three orders of magnitude.
So, if we increase Q from 6 to 7, the BER would have dropped almost to 10 to the power
minus 12. If we go from 6 to 8, the BER would drop to 10 to the power minus 15.
So, we see that just by increasing the this Q factor for the data, the BER can be very
rapidly reduced. So, by even small increment in this quantity Q, we get substantially small
value of the the quality fact. If we consider now a situation that the noise is let us say
shot noise dominated, then the sigma 0 could be negligibly small and one can make certain
approximation to this data. But in general, as we have seen essentially if then sigmas
0 then sigma 1’s are not equal, then essentially we have to deal in general case and then we
can define this parameter Q, which will now decide the bit error rate of the data. So,
let us see what we have done up till now. We have first said we are having various noises
present in to the receiver. We find out the total variance of the noise,
which is sum of the variances of various noises. From that variance, then we say that we have
the noise various corresponding to 0 level; have the noise variance corresponding to 1
level. And without worrying about what is the actual distribution of this fluctuation,
we say that this fluctuation is of Gaussian nature. And under this assumption, then we
say that we are having two distributions. One corresponding to 1 level; other corresponding
to 0 level and these two distributions have different variances. So, 0 level, we have
the variance which is sigma 0 square and for level 1, we have variance which is sigma 1
square. And then we general we derived the expression for the bit error ratio.
We find the optimum level for which the bit error rate is minimum and then for that, we
get a parameter what is called the Q parameter of the data and then we get the bit error
rate as a function of that Q parameter. As I mentioned for a standard data, the bit error
rate should be less than 10 to the power minus 9. So, that means for a data, we require Q
to be of the order of 6 or more. With this understanding of BER, now one can go back
to our signal to noise ratio expressions. And as I mentioned, we can make certain approximations
two domains; that means one high power domain; other one is low power domain. So, essentially
the operation we can divide into two categories.
One is what we call as the thermal noise dominated and other one, we call as the shot noise dominated.
In thermal noise dominated situation, essentially we are saying that the sigma T is much much
greater compared to sigma S, the shot noise and this essentially happens for the low optical
powers. So, when low optical powers when the shot noise is small, that time essentially
system is limited by the thermal noise. And in that situation, we have sigma T much greater
compared to sigma S. And this application probably more suited for the 0 level of the
data; whereas, if we go to the 1 level of the data or if the data amplitude is large,
then we have what is called shot noise dominated and in that case, we have exactly opposite.
So, we have sigma S which is much much greater than sigma T. Note also in an optical communication
system, if the receiver is very close to transmitter where the power is not attenuated significantly
in the fiber, then the situation probably would be the shot noise dominated; because
optical power will be high; whereas, if you go to a distance significantly far away from
the transmitter, the optical signal would have attenuated significantly. And in that
case, essentially the shot noise contribution will be less and then the system will become
thermal noise dominated. So, essentially we have depending upon the location of the transmitter
and receiver and the distance between them, we may get a situation which is rather thermal
noise dominated or shot noise dominated. So, let us consider now the two limiting cases;
thermal noise and short noise dominated. So, what we do? Now, we consider this expression
which we have got for the signal to noise ratio from here. And then we say that if we
have a situation which is the thermal noise dominated, which is this. Then this is the
term which is present and all these quantities are negligibly small; whereas, when we are
having a shot noise dominated situation, that time we have only this noise which is present.
We can even assume that the dark noise also is negligibly small and thermal noise is also
negligible. So, what we can do is in these two regimes, we can now get the signal to
noise ratio.
So, if I go for the thermal noise dominated
in this situation, the signal to noise ratio would be equal to R L responsivity square
divided by 4 KTB into p of the data; so we call an p in. So, note here this quantity
we had the photo current square for the signal. Now, if the responsivity of the detector is
known which is R, then the photo current will be R into p in optical power. So, that is
your signal to noise ratio. So, for given receiver all these quantities are constant.
We have responsivity constant, bandwidth constant, temperature constant. So, essentially we can
say that this quantity is proportional to R L and p in square.
So, in thermal noise dominated regime, the signal to noise ratio improves very rapidly
as the optical power. It goes at the square of the optical power. Also the signal to noise
ratio improves with the load resistance and that is the reason as you have seen in the
detector that we use a resistance, which is normally of a large value. Because that essentially
helps you in giving you last signal to noise ratio. Then for a given receiver even if R
L is fixed, then we can define a quantity what is called the noise equivalent power,
which is now a parameter of your receiver in the thermal noise dominated case. Let us
say this is NEP; that is p in divided by square root of the bandwidth.
So, this quantity from this expression, essentially we can get 4 KTB KT into divided by R L R
square responsivity square to the power half. So, for a given load resistance, we have this
noise equivalent power and then this is given as watts per square root hertz and typical
receiver as this quantity ranging between 1 to 10 picowatts per hertz power 1 by 2.
So, the important in the note here is the when we are having a thermal noise dominated
regime, that time the signal to noise ratio can be improved rapidly by increasing the
optical power; because it is proportional to the square of the optical power. If I consider
on the other hand the thermal noise not the short noise dominated regime, then the situation
is different.
So, let us say if I consider now the situation, which is shot noise dominated; that means
sigma S is much much greater than sigma T and in that case, essentially we get SNR,
which will be equal to responsivity times p i n divided by 2 q into B. We can see from
here these quantities negligible; this is negligible. So, only we have this quantity
here and this one is proportional to the photo current. So, one photo current will cancel
with this. So, that is how you got this quantity, which is R into p i n. So, in this case, now
the signal to noise ratio is proportional to the p i n. So, what is important to note
here is that when I go to the low optical powers, at that time the system is dominated
by thermal noise. And then by small increase in the optical
power, improve the signal to noise ratio significantly; whereas, as the power increases in the optical
signal, then slowly the shot noise starts coming into picture. And for high optical
power, the thermal noise becomes negligibly small and then the signal to noise ratio does
not improve that rapidly as it was happening at the low optical powers. So, earlier the
signal to noise ratio was going a square of the power, now it is starts going only linearly
as the optical power. This transition from the dominants of the thermal noise to the
shot noise, somewhere it takes between minus 20 to 30 dbm of power.
So, if you have the optical power in the data, which is more than about minus 20 dbm, then
system essentially tries to go towards the shot noise limited regime; whereas, if you
are having optical powers lying in the range of about minus 40 minus 50 dbm, then the system
essentially is thermal noise dominated regime. So, essentially what we have done in this
lecture? We have analyzed the bit error rate performance of an optical receiver in the
presence of various noises. We assume that all noises can be put together and there will
be equivalent variance we can define for the total noise, which will be different for 0
and 1 level and then assuming that the noise is Gaussian.
We can calculate the bit error ratio for the optimum value of the threshold and we saw
that the bit error rate or bit error probability drops very rapidly as a function of this parameter
what is called q. So, a small change in this quality factor or data quality factor q that
can improve the BER performance significantly. And then we saw that the system can be of
two limiting cases, thermal noise dominated and shot noise dominated. And in thermal noise
dominated case that means a low optical powers, signal to noise ratio improves as the square
of the signal; whereas for the shot noise limited, the signal to noise ratio scales
as linearly. So, it does not increase that rapidly.