# Mod-01 Lec-33 Impact of Carrier Frequency Offset (CFO) in OFDM

Hello, welcome to another lecture in this
course on 3 G 4 G wireless communication systems. In the last lecture we had talked
about MIMO OFDM that is that is the use of OFDM technology MIMO wireless communications.
We said that MIMO OFDM converts a MIMO frequency selective channel into a set of
parallel MIMO flat fading channels, such that across each sub carrier, the net wireless
communication systems looks like a MIMO flat fading systems that is like vector Y 0 is
now H 0 times X 0, where X 0 is the transmit vector across the 0 th sub carrier, H 0 is
the channel matrix and Y 0 is the received vector that across the 0 th sub carrier. Similarly,
the first carrier so on so on so on till the N minus 1 sub carrier, where Y N
minus 1 is r dimensional received vector which is equal to H, the matrix H N minus 1 across
the N minus 1 sub carrier and X bar N minus 1, where X bar N minus 1 is the transmit vector
across the N minus 1 th sub carrier all right and. We are also seen that for the purposes of
receiving across each sub carrier; similar to MIMO, we can either use 0 minus force thing
or the MIMO MMSE equalizer that is we can use same similarly to what we had seen in
the case of MIMO system, we can use both 0 forcing and MMSE equalizer, and with that
we had also said that this channel matrix, do you obtain channel matrix across each sub
carrier? We said that if you consider the u v th element of all this channel matrices
across each sub carrier; that is there are N such elements from 0 to N minus 1. They are
nothing but then N point FFT of if you take all the channel taps matrices that is H 0,
H 1, H L minus 1; take their u v th element their L such elements then you 0 pad them,
and take the N point N point FFT of these elements then you will get the u v th element
of the channel matrices across the sub carriers.
You can similarly you do that for all the elements and thus construct essentially the channel matrices for the sub carriers all
right. So, that is how we construct these channel matrices in this MIMO OFDM system.
It is fairly straight forward and its extension of the case for the single input
single output OFDM system, but it is much more helpful in MIMO system, because MIMO
equalization that is MIMO frequency selective channel selective channel poses much more
bigger challenge for communication compare single input single output frequency selective
systems all right. So, that is the basic idea behind MIMO OFDM. And then with that, we had moved on the effect
of frequency offset, what is the distorting effect of frequency offset in OFDM
system? We said OFDM divides the frequency band into a set of orthogonal overlapping
yet orthogonal frequency get us. The orthogonality is a very critical aspect actually
of any OFDM system which means once you have an OFDM system as we had depicted in
this figure last time. The orthogonality is lost which means now there will be ICI or
inter carrier interface. The carriers will start interfering with each
other and the larger the frequency offset, the greater is this inter carrier inter carrier
interferences, this is a very how you deal with frequency set is in fact a very
important aspect of any OFDM system and we had also started seeing them how to model the
affect of this inter carrier interference well as its frequency offset. We had set for a frequency offset of epsilon,
normalized with respect to sub carrier bandwidth. The received signal sample y n
is given as summation X k H k e power j 2 pi N k plus epsilon over capital N number of sub
carriers plus noise that is we said the system model for this. And then, we had started looking at what is
the affect when you take FFT of these received samples? Now with the presence carrier
frequency offset what is the affect of this on the FFT that is what we will start
looking at with that let us go on to today is lecture So, we said that y l that is what we said
we said y l is nothing but you consider that is the received symbol across l sub carrier
is nothing but consider the FFT of the received samples as in this fashion, that
is summation received samples y n e power minus j 2 pi n l over N, let me remind you
this is the l th FFT point which is the l th received symbol across the l th sub carrier.
This is the received symbol across l th sub carrier this using the expression that we
had earlier, this can be simplified as follows this is nothing but 1 over N summation n summation
k equals minus N over 2 to N over 2 X k H k e power j 2 pi k minus l plus epsilon
over N, that is I am doing nothing but I am simply substituting the earlier expression
that we had for the N th sample y n in this expression. That is X k H k e power j 2 pi.
The expression that we had earlier I am simply substituting over here plus the noise part which is plus summation over n w n e
power minus j 2 pi n l over N. So, I have expression for, I have the expression for,
for y n that is N th received sample, I am simply substituting that expression in this
that gives me this, now what I am going to do is from this term over here, in this first
term over here. I am going to isolate the term corresponding to k equals to 1; this
is similar to what we had done earlier. Because I want to look at this as 2 terms;
one is the desired part which is what I am suppose to receive on the l th sub carrier
that corresponds to k equals to l and then the interference part that corresponds to
all k not equals to l. So, I am simply going to write this as summation
of 2, 2 terms. Now, when k equals l, this goes to 0; this simply reduces to X l
H l e power j 2 pi epsilon over 2 pi epsilon N, it should be N here; there is a N missing
2 pi epsilon N over. So, I am going to write this as summation 1 over N n X l X l
H l e power j 2 pi N epsilon over l. Now, remember if there was no carrier frequency
offset then this epsilon would have been 0; this would have been 1; this would have been
summation 1 which would have been N over N; so this is this would have yielded or X l
H l similarly to what we had earlier. Now, because of this carrier frequency offset,
this is not exactly 1 all right that is the problem here, plus summation 1 over N, now
k equals minus N over 2 to N over 2 however not equals l, because we have isolated the
term k equals l, summation over N this is X k H k e power j 2 pi N over N into k minus l
plus epsilon plus some noise which is w tilde of l plus.
So, what we are doing is, I am taking the expression for the received N sample, I am substituting the FFT expression FFT at the
FFT expression that is the l th FFT point and thus I receive I am formulating or I am deriving
the expression for y l which is nothing but the expression for received sub carrier
received symbol across the l th sub carrier and that is given by this expression and that
can be further simplified before I simplify that further. I want to illustrate a result the result.
I want to write here is the result that I will use in simplification that is as follows summation
n equals 0 to N minus 1 e power j theta n equals, you can verify this and it
is a simple problem of geometric progression and this is sin N theta over 2 divided by
sin theta over 2 into e power j phi tilde, where this phi tilde e power j phi tilde is
factor, because it does not affect the power what I want to illustrate is what is
the impact on the signal to noise power ratio.
So, magnitude e power j phi is 1, so it does not affect the phase however it does not affect the power, however this is the sin
N theta by 2 sin theta by 2. This is the critical factor for us all right, so this
for us this is do not care as it does not affect power that is why I am not particular
about this, this is do not care as it does not affect and it does not affect the net
received signal power. So, this now can be writ10 as y l, this can
be writ10 as y l equals H l X l sin pi epsilon divided by sin pi epsilon divided
by n into 1 over N into e power j some phi I tilde all right. Now, again you can verify
that if epsilon equals to 0, limit epsilon tends to 0; this tends to N, N into 1by N;
this tends to 1; this is H l X l, this is the same thing we had before plus what we have
here is summation k equals minus N by 2 to N by 2 k not equals l H k X k sin pi epsilon
divided by N sin pi l minus k plus epsilon divided by N into e power j phi k l tilde.
This is the interference term I l, this is the interference term; this is the net interference now which is not equals to 0,
because of loss of orthogonality, because the presence of this frequency offset, because
you can see if this epsilon is 0; this whole interference this goes to 0. Now, because
of the presence of a carrier frequency offset, you end up having this inter carrier
interference and this is nothing but the interference from all the other subcarriers
other than the l th sub carrier plus w tilde all right.
This is the expression that we have. So, this I would like to say so this is the signal part for this, this here, this is the signal
part that is the desired signal part, this part here this is the inter carrier interference
part and this of course, this is our Gaussian noise. This is our thermal noise
or Gaussian noise all right so let me just call it as Gaussian noise. So, there is three
parts; one is the desired signal, other is the inter carrier interference and then again
you have the Gaussian noise part all right. Now, similar to CDMA, now we said now,
now we do not not only have signal noise but we also have signal interference noise,
hence in such scenario we use the signal not signal to noise ratio but signal to interference
plus noise power ratio. So, now we want to characterize the SINR as
SINR equals signal power
divided by interference
plus noise power equals signal power
divided by expected interference power plus the noise signal power divided by interference
power is nothing but expected I l square plus sigma N squared which is the noise
power. Now, let us start characterizing each term that is first I want to characterize
signal power then the interference power then the noise power. So, let us start with the signal power aspect
of this so there are three aspects, so first is so as we saw there are three aspects
that is signal power, interference power, and the noise power. I am starting by characterizing
the signal power. The signal power is nothing but we can see this is expected
magnitude H l square into expected magnitude X l square into sin pi epsilon divided by
divided by N sin pi epsilon divided by N whole square all right, so this is the signal part
that is expected magnitude H l square that is the gain of the channel across the l sub
carrier expected X l square which is nothing but p l which is the power transmitted across
the l subcarrier or if you are using uniform power then this is simply p times
sin pi epsilon divided by N sin pi epsilon by there are large number of sub carriers that is I am going to simplify these expression
further to get an idea of how this expressions look like when the number of sub
carriers is typically the number of sub carriers is very large; it is the order of
N equals 512 or N equals 1024. So, how does this expression look when N tends to infinity
so for large N
that is number of sub carriers, we have limit
N tending to infinity sin pi epsilon over
N
as N tends to infinity pi epsilon over N approximately tends
to 0 that is limit sin theta becomes very small and the limit theta becoming very small
sin theta approximately tends to theta. Hence this is approximately pi epsilon over
N for very large N implies N sin pi epsilon over N tends to N into pi epsilon over N equals
pi epsilon, hence for a large number of sub carrier number of sub carrier N sin pi
epsilon over N approximately looks like pi epsilon. Hence this expression for signal
power can be simplified as follows. Hence, the signal power for a large number
of sub carriers for N is given as expected mod H square power sin pi epsilon divided
by pi epsilon square, where p remember this p is the power in the signal, this is p is nothing
but expected X l square which is the power in the symbol transmitted across the
l th sub carrier. This is power this is nothing but the power the signal power or
the symbol or the symbol power. Hence this nothing but and I am going to assume that
expected H l square is the same across all sub carriers as this is p magnitude H square,
where magnitude H square is nothing but expected magnitude H l square that is the
average gain across the sub carrier times sin pi epsilon divided by pi epsilon whole square.
So, this is nothing but the
this is nothing but the
this is nothing but the signal power. So, we have derived simplified the
expression for the signal signal power in presence of large number of sub carrier and
the signal power in the presence of a carrier frequency offset of epsilon approximately
looks as p, where p is the transmitted power magnitude H square is the average gain
across the sub carrier times sin pi epsilon divided by pi epsilon whole square this is
the key factor. Now, let us characterize slightly difficult
part which is the interference for the more elaborate part which is the interference.
So, how does the interference power look like, how does the interference power look like?
If you look at the interference power we said interference power is nothing but expected
magnitude I l square that is equal to expected
magnitude X square expected magnitude H l
square into summation k equals minus N by 2 k not equals l k equals N by 2 times
sin pi epsilon N sin pi l minus k plus epsilon divided by N this whole square. Remember
when this simplification sin pi epsilon by sin pi l minus k plus epsilon, this is
arriving from this property that we looked at here all right. Remember we have introduced
a special property over here and this simplification is simply arising from this
property. Now, I will have to do, now I will do a series
of manipulation which are slightly, but I asked you to but fairly simple also I ask
you to just pay attention, because there are more mathematical manipulation rather than
anything else but the idea is essentially simplify this expression. So, what I am going
to do here as a first step to simplify this, I am going to set k minus l equals u,
just to a simple substitution k minus l equals to u and also assume and assume n tends
to infinity that is the number of sub carriers N is very large and hence this expression
can be simplifies as follows. Expected magnitude I l square equals p magnitude
H square summation u equals minus infinity. Remember I have now substituted
k minus l equals u and at the same time k not equals l which means u not equals 0, sin pi
epsilon whole square divided by N sin pi u divided by N whole square, where in the denominator
I have used the approximation, u plus epsilon approximately equal to u, because
remember that epsilon is a very small quantity and u is varying from minus infinity
to infinity. So, for large u u plus epsilon is approximately the same as u u is
small fraction. So, this is the net expression that we that
we get and hence this can be further simplifies as follows that is the interference
power which is p times magnitude H square also sin pi epsilon come pi epsilon square.
This comes out of the summation summation what do we have here u equals minus infinity
to infinity, u not equals 0, 1 over sin pi u divided by N whole square. Now, sin theta
now I want to use some approximation here, the first approximation I am going to use
is that sin theta is greater than equal to 2 theta over pi. So, sin theta is greater than
or equal to 2 theta over pi. I want to use this for this term pi u by n
which means I can write this as sin pi u divided by N is greater than or equal to 2
pi that is 2 theta, where theta is pi u divided by N, 2 pi pi u divided by 2 theta
divided by pi equals 2 u divided by N implies N sin pi u divided by N greater than or equal
to 2 u, and hence now I can simplify this expression here. by simply approximating bounding
this expression as follows. I can bound this expression as expected I l square
as since N sin pi u by N greater than equal to 2 u, expected I l square is in fact less
than or equal to because this term is in the denominator. This is p magnitude H square
sin pi epsilon square summation u equals minus infinity to infinity, u not equals 0 1 over
2 u whole square all right. So, that is the net approximation that we
have over here and this can be further simplified as follows, this is equal to I
will do a series of manipulations p magnitude H square sin pi epsilon square times 2. I
will write the integral from u equals minus infinity to infinity as 2 times the integral
u equals 1 to infinity 1 over 2 u square. This is nothing but half p magnitude H square
sin pi epsilon square summation u equals 1 to infinity 1 over u square all right. And
this is simply the sum 1 over u square, where I use the set of all positive integers and
this has a standard this. We know the result of this summation; this
is simply pi square over 6. Hence this is nothing but pi square over 12 into p into
magnitude H square sin square pi E which is essentially nothing but 0.822 times p magnitude
H square sin square pi into epsilon all right. So, this is the net expression for
the interference power. Hence now, I can write the SINR, hence what we have done is we have,
we have again derive the power the entire expression for the interference power is 0.822
times p times norm H square sin square pi epsilon, and also you can observe as a if
epsilon equal to 0, this is 0 that there is no inter carrier interference, and hence interference
point is 0 which is what we expect and hence this is nothing but the SINR or
hence the net SINR. Hence, SINR in the presence of carrier frequency
offset
of epsilon is is given as, SINR SINR equals p magnitude H squares in pi epsilon
divided by pi epsilon whole square divided by 0.822 p magnitude H square sin
pi epsilon whole square plus sigma N square, where sigma N square is noise power and this
is the expression for the SINR in the presence of carrier frequency offset. Let
me remind you again, this is the signal power; this is the interference power from inter
carrier interference; this is the noise power; this is the noise power.
Now, we have derived the signal to interference noise ratio in the presence of carrier frequency offset of epsilon all right which
is p magnitude H square, where magnitude H square; remember is the average gain across
each sub carrier that is expected H l square sin pi epsilon by pi epsilon whole square
divided by 0.822 p magnitude H square plus sin pi epsilon square plus sigma N square, which
is the thermal noise. So, this is the net expression and you should verify that in absence
of carrier frequency offset that is the epsilon equals to 0.
This reduces to simply p magnitude H square divided by sigma N square which is essentially the expression for the original
OFDM signal to power ratio without any inter carrier interference all right. You should
verify that in fact it is straight forward. If you look at this its directly visible,
if you look at this expression; this tends to 1; the numerator sin pi epsilon by pi epsilon
tends to 1; if epsilon is 0 and this denominator the interference tends to 0. Hence,
this is the expression all right, and also remember this expression is an approximation.
It is valid under reasonably good approximation in which the number of sub carrier
N is tending to infinity. The approximation one of the approximations is that is large number of
that is large number of sub carriers. So, that is what we have here, now let us
do a small example. So, let us illustrate small example to illustrate, illustrate the
performance of the SINR under the presence of carrier frequency offset. Consider a system
with average sub carrier power gain as unity. Let us consider consider an OFDM system
such that magnitude H square equals unity that is average gain of, and we also consider
the scenario with signal power or transmitted symbol power p equals 10 d B and
sigma n square equals 0 d B. That is the signal the symbol to noise power
ratio p over sigma N square equals 10 d B. Now, in absence of a carrier frequency offset,
if carrier frequency offset; if epsilon equals 0 then we have SNR receive is nothing
but SNR at receiver equals p magnitude H square divided by sigma N square which is
equal to 10 d B. So, we have a 10 d B SNR in the absence of any carrier frequency offset
that is the receiver end transmitter that is carrier oscillator is perfectly synchronized
to the incoming carrier wave then we have SNR of 10 d B. There is no inter carrier interference
however if the carrier frequency offset. Consider now a slight distortion. Consider a carrier frequency offset epsilon
approximately equal to 5 percent. Consider a carrier frequency offset epsilon equals 5
percent equals 0.05. Consider a carrier frequency offset epsilon equals 5 percent
equal 0.05, then we have the SINR equals remember SINR equals p that is 10 into magnitude
H square into sin pi epsilon divided by pi epsilon whole square; that is sin pi into
0.05 divided by pi into 0.05 whole square divided by 0.822 into p that is 10 into sin
pi epsilon which is 0.05 sin pi of 0.05 whole square plus sigma n square, which is
the noise power; which is one and this can be shown to be you can verify this by computing
this is 8.25. Previously in the absence of carrier frequency
offset the SINR was 10 d B which is 10. Now, as carrier frequency offset has increased
to that is we have a carrier frequency offset of 5 percent, the SINR has dropped
to 8.25. Hence, the reduction in SINR equals 10 minus 8.25, which is 1.75 which is essentially
17 percent, which is equal to 17 percent, because out of 10 the degree is 1.75
which is essentially corresponds to 17.5 percent reduction in SINR. So, because of
the carrier frequency offset of 5 percent, epsilon equals 5 percent then there is a 17.5
reduction SINR. Let us look at what it means to have a carrier
frequency offset of 5 percent. Now, we said 5 percent is the net is normalized to
the bandwidth of the sub carrier. Let us go back to our WiMAX example WiMAX; we said that
is fixed profile WiMAX. It has roughly, it has let me remind you. It has 256 sub carriers
all right each of bandwidth; the sub carrier bandwidth B s equals sub carrier bandwidth
equals 15.625 kilo hertz. We said the bandwidth per sub carrier is 15.625
kilo hertz. So, 5 percent of 15.625 kilo hertz that is the frequency offset. So, the
frequency offset net absolute terms is equals 5 by 100 into 15.625 kilo hertz that
is equal
to 0.78125 kilo hertz that is approximately 0.78 kilo hertz all right. So,
the net frequency the absolute frequency offset of 5 percent in this WiMAX system it
means that it is about 0.78 kilo hertz, but WiMAX we said operates at 2.4 Giga hertz byte.
So, if you look at this absolute frequency offset as if percentage of net carrier
frequency. So, what is the percentage of this frequency offset at 2.4 Giga hertz? So, the percentage at 2.4 Giga hertz carrier
equal 0.78 divided by 0.78 kilo hertz divided by 2.4 Giga hertz which is essentially
2.4 into 10 to the power of, 10 to the power of 9, and if you look at this is approximately
this is equal to essentially 0.78 divided by 2.4 into 10 power 6. So, if you
can look, if you look at if you look at this, this is an extremely small fraction; if you
look at this, this is 0.78 kilo hertz and the 2.4 Giga hertz. This is roughly the fraction,
so if you look at this this is an extremely small fraction. In fact this is
approximately equal to one-third into 10 power minus 6 that is it is of the order of 10 power
minus 6. So, what this means is at the Giga hertz frequency
when your carrier at 4 G wireless system is at Giga hertz, even if you have
a carrier frequency offset that is roughly 10 power minus 6 fraction of that carrier frequency
or roughly 10 power minus 4 percent that is 0.0001 percent of that carrier frequency
that can result in huge s N in fact 17.5 or 20 percent decrease in SINR the receiver
end; which means that carrier synchronization at the receiver end. This
4 G OFDM system is a very critical aspect. Because if the number of sub if the number
of sub carriers is of course, typically large that is 256 and as you progress to higher
and higher bandwidth. The number of sub carriers is going to increase that is 256,
512, 1024 which means that you synchronization also has to improve. Because,
even if you have a small carrier frequency offset at the receiver that means inter carrier
interference which means interference from a large number of sub carriers and the
tremendous decrease in the SINR at the receiver that is what this seems to imply.
So, carrier synchronization is a very, very critical aspect of any OFDM system. Because, if the carriers at transmitter and receiver
are not synchronized then we have serious interference inter carrier interference which
results in a a tremendous decrease in the SINR at the receiver. So, with this we finish
the topic of the impact of distorting effects on OFDM; that is the impact of carrier
frequency offset again to go through this this material again to enhance your understanding
of the various expression that we have derived for the impact of the SINR affects
the signal power, the interference power and what is the single to interference noise ratio
and what is for some reasonable carrier frequency offsets in fact.
We have looked at the examples, what is the impact in a practical scenario? What is the impact of having a carrier frequency offset
of 5 percent at the receiver and so on all right, and next I want to address one more
practical issue. Another equally important not more significant practical issue in OFDM
which is related to the PAPR that is peak to average power ratios. So, I want next issue I want to address in
this context is the PAPR issue in OFDM. This is the PAPR issue in an OFDM system. What
does PAPR stand for? PAPR stands for PAPR equals peak; it requires peak to average.
PAPR stands for simply stands for peak to average power ratio and simply put this is
a critical this is a critical factor in OFDM. This is critical, this is critical in an OFDM
system. What does this PAPR issue for instance? Consider a typical non OFDM transmission
system. Let us go back for a moment to understand. Let us go back to a single
carrier system. Consider a non OFDM or single carrier system
OFDM is a multi carrier system. So, I am saying just lets go back and a consider our
single, let us go back and consider our single carrier system for instance. Consider
a typical BPSK with BPSK modulated symbols. Consider a single carrier system with BPSK
modulated systems. We have let us consider its transmitted symbols sequence of x 0 x
1 x 2. Now, each of this is plus or minus. us say the amplitude level is a and the information
depending on the phase is plus a or minus a that is for a plus one your transmitted
plus a; for minus you transmit minus a. So, we have plus a let us say the next symbol
is minus a, plus a and so on. Now, if you look at this sequence, what is
the power in each symbol? The power in each symbol is each symbol is either plus or minus a in both
cases the power is simply a square that is expected a square which is
simply a square. So, the power in each symbol is a square which is also the peak power.
Because, if you look at power in each symbol is the same; so the peak power is a square
as well as the average power is also a square. So, this is also the peak power, how
equals expected magnitude x k square; the average power is
also a square. Hence, both peak power and average power in
this single carrier system is a square when a is the transmitted amplitude. Hence, both
peak and average power equal to a square. Hence the ratio of peak peak power to average
power equals 1. In practice it is not exactly 1. It is close to 1, because of pulse
shipping and so on in the base band; but for all practical purposes, I can say the
peak to average power in a single carrier system is close to 1, which is essentially
you can write this also in d B terms as 0 d B that is there is no.
So, if you look at the mean power level that is there is no significant fluctuation of the power over the mean, that is both the
mean and the peak power that is; if you look at the power instantaneously, there is it
its it is closely tight to the mean that is there is no significant deviation about that
mean. Sometimes, you can have a signal which has some mean but there can be significant
positive and negative variation about the mean so that the net average is the same
but the deviation is high. What we are saying here is that both the mean
in same and peak is also same which means the deviation about this mean are also controlled
essentially that is what this mean that is what this says. Hence, there is no
significant deviation from the mean power level. Hence, this ratio close to 1 also means,
hence there is no significant deviation from the mean power. Hence there is no significant
deviation from the mean power level. So, we looked at the PAPR the peak to average
power ratio, in the traditional conventional single carrier system and we
said that it is close to 1 that is the mean power is a square. The peak power is also
a square. Hence, there is no significant fluctuation of the instantaneous power over
the average. So, because of lack of time, we are going to stop here and we will start from
this point and continue this discussion in the next lecture when will look at what is
how does the PAPR look like in an OFDM system.
Thank you very much.

### 6 thoughts on “Mod-01 Lec-33 Impact of Carrier Frequency Offset (CFO) in OFDM”

1. Yi Ren says:

May I ask, at 13:27, why the numerator of the ICI term is sin(pi*e) instead of sin[pi*(k+e-l)]? Thank you!!

2. benzarti mejdi says:

mpact of Carrier Frequency Offset (CFO) in *****MIMO-OFDM***** ?????????????????

3. Sylar Lao says:

at 5:32, the DFT shouldn't be divided by N.
Also can anyone recommend some synchronization introductory lectures? Thx.

4. Ameha Tsegaye says:

Wow! I never thought CFO has this much severe effect in OFDM system.

5. MOHAMMED BAHAAELDEN says:

this is stupid really

6. rajendra bareto says:

I am not able to get it for small N=12 without approximation. signal power is okay, but interference power does not work. have taken u=-N to N. Is this fine summing over u = -12 to -1 and 1 to 12.