Unformatted text preview: 1 All About Modulation – Part II Intuitive Guide to Principles of Communications
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All About Modulation  Part II
The main Figure of Merit for measuring the quality of digital signals is called the Bit Error Rate (BER) vs.
Eb/N0. Bit error rate is the ratio of number of bits received in error vs. total number of bits sent.
When we talk about bit errors, we need to distinguish between two types of signals.
1. A coded signal  This is a signal that has coding applied to it at the source so that only some of the bits
in the stream are important. We can tolerate a higher gross error rate in this channel, as long we can
decode the information bits with much better integrity. The BER of the information bits is much less
than that of the coded signal.
BERinfo bits << BERcoded bits
If Ebc/N0 is defined as the bit energy of this signal, then Eb/N0 of the information bits is approximately
Eb/N0 = Ebc/N0  Code rate (in dBs) A coded bit stream may have a bit error rate of 101 to 103, but when coding is applied, the rate
experienced by the information bits is quite small, on the order of 104 to 1011.
2. An uncoded signal  There is no coding so all bits are equally important and the stated BER is the BER
of the information bits, and Eb/N0 is equal to Ebc/N0. Since most signals are coded, we need to keep in mind which BER we are talking about. The BER of
the received coded signal or the BER of the information bits? Normally even that distinction is not there
except in simulation and analysis because signals are softdecision demodulated and this softdecision is
handed to the coding decoder for the final bit decoding.
For example, we have two QPSK links, of bit rate Rb, one with coding of rate onehalf and the other
without. Both operate at an Eb/N0 of 7 dB. Both have a gross BER of 103. Bit the one that is coded will
ultimately give a BER of 1010 on the information bits. The only difference is that with link 1 we get full
Rb bits of information whereas with the coded link the price of purity is that only half of the bits are
information bits.
Bit error statistics, although a discrete phenomena, are generally described by continuous functions. In
absence of unusual sources of interference, rf signals are subject primarily to Gaussian random errors. The
Gaussian noise environment is coming from the hardware and from the air/medium in which the signal is
traveling and these errors are generally random, coming in one's, and two's. A large number of consecutive
errors in a Gaussian environment are less likely. A lot of errors at once, called burst errors, are usually
characterized as coming from fading, shadowing and Doppler effects.
Errors don’t actually occur, but are made by the receiver. Prior to the receiver, the signal is in an analog
waveform and in that context bit errors have no meaning. How errors are made in the receiver is a function
of modulation. Modulation determines how tightly various symbols are packed in a signal space and how
sensitive the receiver must be in making a decision.
Let's begin by looking at the transmission of a polar NRZ signal through a communications channel. The
signal has amplitude of ±1 volt, where 1v represents a 1 bit and a 1 v represents a 0 bit. Copyright 2002 Charan Langton www.complextoreal.com 2 All About Modulation – Part II Figure 1  Transmitted binary signal
The signal picks up noise along the way and at the receiver we see instead of ±1 volts, amplitudes that
include noise, sometimes higher than transmitted value, sometimes lower such that
Received signal amplitude = 1 + n, for 1 bit = s1(t)
Received signal amplitude = 1 + n, for 0 bit = s0(t)
where n represents the noise amplitude and is changing from sample to sample. (We are sampling each bit
more than once.) Figure 2  The binary signal has been acted upon by random noise. The received amplitude now
varies at each sample.
Making decision based on the fluctuating values of Figure 2 is like trying to read a tossed coin in a room
where with each toss, the light level goes up or down randomly. Sometimes, the light is bright enough and
you can clearly read the coin and other times, it's too dark to read correctly. Each toss will be either a head
or a tail, and equal number of heads and tails are expected. The fact that an equal number of heads and tails
are expected is called the base probability of the process. But how we read the coin (or decode) depends on
how bright the light was at the moment of the toss.
The fact that there are just two choices to pick between (H or T) is akin to symbol choices of a modulation.
A modulation with fewer choices will lead to smaller errors, this is obvious.
The value of noise, n varies with time and so is different from sample to sample. It is assumed to be
normally distributed with a zero mean and variance of quantity σ 2 , where σ is the standard deviation of
the noise process. Copyright 2002 Charan Langton www.complextoreal.com 3 All About Modulation – Part II Figure 3 – Effect of noise on the symbol
In Figure 3, we see two symbols, a 0 and 1 symbol with the transmitted shape and result of the addition of
random noise at the receiver. Figure 4 – Symbols arrives in various states of distortion at the receiver. By sampling the amplitude
at the decision instant, the receiver makes a decision about what was sent. This decision is made
strictly based on the received amplitude.
In Figure 4, we superimpose several received symbols for bit 1. We can see that the received amplitudes at
the sampling instants vary. For most samples, the receiver would make a correct decision, because the
added noise does not cause signal voltage to change polarity, but there is one instance where noise is large
enough that a positive transmitted voltage became a negative voltage in the eyes of the receiver. The
receiver will make an error for this bit.
The receiver makes decisions by the following rules;
If observed voltage at the receiver is > 0, then transmitted bit was 1 bit.
If observed voltage at the receiver is < 0, then transmitted bit was 0 bit.
We call this process “Threshold Comparison.” The selection of 0 volts as a threshold is called the decision
region boundary or Threshold, T. A threshold is like a net or a line between the players in a game. The
threshold is optimal when it is at the mean of the two values we are trying to distinguish between. However
this is true only if the apriori probability of the two outcomes is equal. In a normal communication signal,
the number of 0's sent (represented by a 1v signal level) is approximately equal to number of 1's sent, so
the optimum threshold is at 0 volts (average of 1v and +1v).
If at the time of decision, the value of noise n exceeds V volts, the transmitted level, then the observed
signal will flip from the correct decision region and a wrong decision will be made. We can write this
probability of error as Copyright 2002 Charan Langton www.complextoreal.com 4 All About Modulation – Part II P (e  1) = P ( n < − V ) 1 Read this as: the probability of making a decision error when a 1 is sent is the probability that noise at this
moment is less than V. It is obvious that the noise present must be negative in order to pull the positive
voltage value into the negative region. The converse holds for the 0 bit.
P (e  0) = P ( n > V ) 2 By Bayes Rule of probability, which states that the overall rate of making a bad decision is
P (e) = P(e  0) P (0) + P (e  1) P(1) 3 This is called the conditional probability. The probability of making a decision error comes from making
errors in interpreting both 0 and 1 bit. If equal number of 1s and 0s are sent (0s and 1s are equiprobable)
then we can rewrite this equation as
P (e ) = 1
1
P (e  0) + P (e  1)
2
2 4 To evaluate this expression, we need to calculate error probabilities P (e  0), P (e  1) which maybe assumed
to be equal when the threshold is exactly in the middle.
Gaussian distribution and density
We know intuitively that probability of an event is defined to be between 0 and 1. The mathematical
definition of a Probability Density Function, f(x), is that it satisfies the following properties.
1. The probability that x, which can be any real quantity, positive or negative, is between two points a and b
is
b P(a ≤ x ≤ b) = ∫ f ( x ) dx 5 a 2. This probability is a positive number for all real x.
3. The area under the Probability Density Function is 1.
+∞ ∫ f ( x) dx =1 6 −∞ The variable x can be discrete or continuous. Height of a person is continuous, dice throws are discrete, as
well as decoding of a 1 or 0 is considered discrete but the received voltage may be thought of as a
continuous variable.
The Probability Density Function (PDF), if you remember from your statistics class, is a histogram. The
y axis is the ration of desired events in the universe of events that take on the x value. For a Gaussian
process, the probability density function of Figure 5 is given by the expression P( x) = 2
2
1
e − ( x −V ) / 2σ
σ 2π 7 where V is the average value and σ2 is the variance of the noise process.
Two different distributions are shown below, 1. for height of women and 2. height of men in inches. Since
the variance is different for them, the height distribution for men spreads out further. Copyright 2002 Charan Langton www.complextoreal.com 5 All About Modulation – Part II 0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
50 55 60 65 70 Height of men 75 80 85 Height of women Figure 5  Gaussian Probability Density Function for variable x, the height of men and women in
inches.
If we integrate the PDF, we get its area. This integrated graph is called, the Cumulative Probability
Function, CDF is a very useful graph, in fact far more than the PDF. This is the graph we use to answer all
kinds of useful questions such as about BER of signals. 1.2
1
0.8
0.6
0.4
0.2
0
55 60 65 70 75 Figure 6  Cumulative Probability Function  Integral of the probability density function
Question: We ask, what is the probability that a woman is between 4’ 11”' and 5'5" tall? By using the CDF
in Figure 5 we can answer this question with ease. This is the difference between the two yaxis values, one
for 5'5" and the other 4'11". P(4 '11" ≤ x ≤ 5'5") = F (5'5") − F (4'11") ≈ .65 or (65%)
The Cumulative probability distribution of Figure 6 is given by
x F ( x) = ∫ f ( x ) dx 8 −∞ Integrating the PDF is difficult. The erf function which is approximated by Eq. 10 can be used to develop
the CDF. Here µ is the mean value of the function, and σ the standard deviation. Copyright 2002 Charan Langton www.complextoreal.com 6 All About Modulation – Part II F ( x) = 1 x − µ 1
µ−x
1 + erf = 2 erfc 2 2σ 2σ 9 The Error function is given by (an approximation) erf ( x ) = 2 π x ∫e − n2 dn 10 0 The complementary error function, erfc function is equal to erfc ( x ) = 1 − erf ( x) 11 The cumulative probability function is always positive since probability is always positive. This is intuitive.
So are the erf and the erfc functions shown below. We can however shift them around and normalize them
to any mean value. The F(x) function shown below is for mean value = 0.
1
erf x
erf cx
F(x)
0.5 0
0 1 2 3 Figure 7 – Plotting erf(x), erfc(x) and the Cumulative probability distribution F(x)
Bit Error Measurement
To apply these ideas to determination of Bit Error rate, we start with the simplest of all modulations, BPSK.
It has just two symbols, one for 0 and another for 1 bit. A simple link consists of a binary signal generator,
a PSK modulator which multiplies the binary signal with a carrier which then picks up noise. The signals is
then down converted to lowpass but the noise remains. The signal is filtered by a filter called “Integrate
and Dump” to give us the waveforms shown below. Carrier Signal 1 M odulator Carrier
Channel
Noise
Signal 2 Copyright 2002 Charan Langton Signal 3 Down
Convert Signal 4 Ideal
Receiver Integrate
& Dump Signal 5 Threshold
Comparator Signal 1
with errors www.complextoreal.com 7 All About Modulation – Part II Figure 8 – A BPSK link (a) Signal 1  Baseband binary signal and its spectrum , 1 0 0 0 0 1 1 1 0 1 0 0 0 0 (b) Signal 2  Modulated signal and its spectrum. Note that spectrum has shifted to carrier frequency. ( c) Signal 3  Modulated signal with noise. The signal shape is the same as Signal 2 and the spectrum
is noisy. (d) Signal 4  Converted signal at the front end of receiver, same as Signal 1 but has noise. (e) Signal 5  Output of Integrate and Dump filter. Bit decision is made at then edge of the symbol
period based on accumulated value.
Figure 9 – Signal transformation through the link Copyright 2002 Charan Langton www.complextoreal.com 8 All About Modulation – Part II 1 bits 0 bits Figure 10  Threshold Comparator output = 1 0 0 0 0 0 0 1 0 0 0 0 1 1 1 0 0 0 , Underlined bits
received in error.
The Integrate and Dump filter works by adding up the individual voltage values at each sample during the
bit. In that it is an integrator. At the end of the bit period it resets itself to start anew for the next bit. The
sum of all the values give a trianglelike shape and bit decision is made based on the last value.
This basic link assumes that this system has infinite bandwidth and has no other distortions such as intersymbol interference, nonlinearities, etc. And most importantly the noise is purely Gaussian, or AWGN,
Additive White Gaussian Noise, and this is a very common assumption in communications.
Below we draw two events; one of transmitting a signal of 1 volt peak amplitude and the other a signal of The figures shows three cases. In each case the nominal voltages are +1 and 1 volts but the variance (the
amount of noise) is different. In the first case it is 1, the second case it is .5 and third case .1 v. The pictures
show the distribution of received voltages for each of the two signals. Probability of mistaking a 1 for a 0 1 Probability of mistaking a 0 for a 1 1 Figure 11  The received voltage of two signals of 1 and 1 volts are distributed around the mean
value with variance equal to the noise variance of the channel.
Even though all three cases have the same mean values, they overlap by a different amount. The amount of
overlap is a function of how much noise is present. The larger the noise variance, the larger the overlapping
and intuitively that means a larger probability of error.
Computing the Bit Error Probability
How do we figure out the probability of being in this overlapped area?
Let’s take a binary signal, where a 0 is sent by signal of amplitude V0 and a 1 with amplitude of V1. Copyright 2002 Charan Langton www.complextoreal.com 9 All About Modulation – Part II Received amp litudes
in this region mean
a 1 was sent. Received amplitudes
in this region mean
a 0 was sent. V0 V1 T Probability of error
of misdecoding a 1 bit Probability of error
of misdecoding a 0 bit Figure 12 – Two signals with an arbitrary threshold
s1 is the signal received.(It is function of time t, but I am leaving out the index to make it easier to read.) It
includes noise, n (also function of time, t.) so that it is really equal to the original amplitude V1 plus noise. s1 (t ) = V1 + n(t )
12
and the noise is given by n = s1 − V1
Similarly for a 0 bit, the received signal is a sum of the original amplitude of V0 plus noise. s0 = V0 + n 13 n = s0 − V0 We set the decision boundary at an arbitrary point between the two signals, at amplitude T.
A decision error is made when noise at the sampling instant exceeds onehalf the difference between the
two discriminating amplitudes. It is also obvious we want this difference to be as large as possible to
minimize errors. The BER, we will see shortly is a function of this “distance” between the signals and we
want this distance large. Signals like countries want as few neighbors and as far away as possible! This
noise level is given by n≥ V1 − V0
2 14 If V1 = 1 v and V0 = 1 v, then the noise level that will cause error is greater than 1. Two different errors
can happen, one for mistaking a 1 for a 0 and vice versa. The probability of each is the area under the curve
starting at the threshold, T and going to infinity in the opposite direction as we can see in Figure 10.
Let’s write the expression for this error. We recognize that the received voltages are randomly distributed
with mean equal to the transmit voltage and variance equal to noise power. We use the probability density
function (Eq. 7) of a Gaussian variable as our starting point with mean of the process equal to V0 and V1
and s0 an s1 being the received random variables. The variance of the process is σ. P(e 1) = T 1 −∞ σ 2π ∫ e −( s1 − V1 )2 / 2σ 2 ds1 15 This is equal to the shaded area shown on the left in Figure 12. Similarly, the probability of error when a 0
is sent is, shown in right shaded in Figure 12. Copyright 2002 Charan Langton www.complextoreal.com 10 All About Modulation – Part II ∫ P(e 0) = 1 ∞ T σ 2π e −( s0 − V0 )2 / 2σ 2 ds0 16 The total probability of both events, using Eq.7 is P (e ) = 1T
1
1T
1
−( s − V )2 / 2σ 2
−( s − V )2 / 2σ 2
e11
ds1 + ∫
e00
ds0
2 ∫−∞ σ 2π
2 −∞ σ 2π 17
Without making it any more complicated, let me say that when we take the differential of the above
equations and set it to zero, we obtain the following value for the optimum threshold, T. Threshold = V1 + V0
2 18 Makes sense! This is the average value, assuming that both 0s and 1s are equiprobable as they usually are.
Now substitute the value of the threshold T, into the Eq. 17, and using Eq. 9 as a guide, we can write Eq. 17
in at least shorter if not totally friendly form as P (e ) = 1 V −V erfc 1 0 2 2 2σ 19 This can also be expressed in the alternate Q function format, which is equal to the erfc function by erfc ( x ) = 2Q ( 2 x) 20 Q function is often used in books but since erf and erfc tabulations are more readily available in Excel and
Matlab, Q function is used less often. Here is the above BER equation in Q format. V −V P (e ) = Q 1 0 2σ 21 We can further manipulate this equation by noting that we can write this equation as P (e ) = 1
A
erfc 2 2 2σ 22 where A = V1 − V0
By Parsevals’ Theorem, we determine the power of this difference signal as Copyright 2002 Charan Langton www.complextoreal.com 11 All About Modulation – Part II A2 ∞ ∫ (V − V ) dt = 2 1 0 −∞ ∞
∞
∞ 2
2
= ∫ (V1 ) dt + ∫ (V0 ) dt − 2 ∫ (V1V0 ) dt −∞ −∞
−∞ 23 = E1 + E0 − 2 E1 E0 σ 10
We note that bit energy is defined as the product of bit time with the square of the instantaneous amplitude
squared.
∞ ∫ (V ) dt = E0 2 24 0 −∞ To deal with the last term in Eq, 23, we define correlation factor ρ, as ρ10 = 1
E0 E1 ∞ ∫ VV 10 dt 25 −∞ Now rewrite the Eq. 23 as P (e ) = E + E −2 E E ρ
1
0
1 0 10
erfc 1
2
2N0 where
N0 = 26 σ2
2 The average bit energy of a signal is the average energy of all its individual bit energies. In this binary case,
we have Eb ( avg ) =
Now set E1 + E0
2 27 ρ10 = −1 because this will give us the conditions for the lowest BER. P (e ) = Eb (1 − R
1
erfc 2
2 N0 ) 28 where R= 2 E1 E0
E1 + E0 ρ10 The value of R is lowest when E1 = E0 and Copyright 2002 Charan Langton 29 ρ10 = −1 , all these conditions are true for BPSK. www.complextoreal.com 12 All About Modulation – Part II So now we get the easy to use (Ha!) BER equation for a binary BPSK signal with unshaped pulses
(unlimited bandwidth), no distortions and Gaussian noise environment using an “optimum receiver”. We
know that BPSK satisfies the important condition that its ρ = 1, since cos wt and sin wt, the two pulses
used are exactly antipodal. P (e ) = Eb 1
erfc N 2
0 30 Since erfc is a monotonically decreasing function, we conclude
1. The bit error rate experienced by the link is proportional to the “distance” or dissimilarity
between the symbol.
2. The most dissimilar signals are BPSK because we use a sine and a cosine to represent the
symbols. Hence BPSK BER forms the limit of lowest BER for a signal.
3. All other modulations have a ρ which is larger than 1 and so their BERs must also be
larger than BPSK.
4. The bit error rate is proportional to the variance of the noise. Large variance means larger
BER.
Bit error of a QPSK signal
A QPSK signal can be seen as two independent BPSK signal, on I and Q channels. If PeI is the probability
of error of the I channel (computed by Eq. 31 ) and PeQ is the probability of errors of the Q channels, then
we can write the probability of QPSK symbol error (that is either the I or Q or both channels make an
incorrect decision, causing the QPSK symbol to be misdecoded.) P(e) = PeI + PeQ − PeI PeQ 31 Assuming that these probabilities are small, and both channels have same error rate, we can ignore the last
term. P(e)QPSK ≈ 2 PeI = 2 PeQ
This says that the QPSK symbol error rate is twice that of BPSK. Now write the error probability of the
QPSK. If P (e) sym − BPSK = 1
Es
erfc
2
N0 then P(e) sym−QPSK = erfc Es
N0 32 Note that although I have been kind of loose with my terms, mentioning bits and symbols in the same
breath, I need now to explicitly state that the error rate we computed for BPSK case was the error rate of its
symbol and not bit. But since for BPSK, a symbol is equal to just 1 bit, it causes no errors. That’s no longer
true for QPSK which has 2 bits per symbol. An error in a symbol can mean 1 or 2 errors. So the above
calculation is for a QPSK symbol. To compute the bit error rate, need to consider Gray coding. Copyright 2002 Charan Langton www.complextoreal.com 13 All About Modulation – Part II Bit and Symbol error rates for QPSK
Example: Imagine two independent BPSK streams. We transmit 1000 bits and see two errors in one and
three in the other. The bit errors rate individually is .002 and .003 and is .0025 on the average. Figure 13 – QPSK can be represented by two independent BPSK streams. Its BER is the sum of bit
errors of both streams vs. total bits.
Now imagine the same thing but as a QPSK signal, which is what it is when you have two BPSK streams
together. Figure 14 – Each bit error represents a symbol mistake in QPSK.
Each time a bit is decoded incorrectly, the whole symbol is wrong. Now we have 5 symbol errors.
(φ = 90) (φ = 90) 01 01 1 1 (φ = 180)
10 1 2 (φ = 0) QPSK
Natural
Ordering 00 (φ = 180)
11 2
11 (φ = 270) (φ = 0) QPSK
Gray
Coding 00 1
10 (φ = 27 0) Figure 15 a – On the average if a symbol is mistaken, there will be 1.33 bit errors. If we consider only
the adjacent symbol errors, (ignore the red link) then on the average, there will be 1.5 errors per
symbol for a QPSK that is numbered according to natural numbering and one bit per symbol for
Gray coded QPSK.
If Symbol 1 is misdecoded, then we can have on the average 1+1+2/3 =1.33 bits in error per miscoded
symbol If we ignore the possibility that symbol 1 will be mistaken for symbol 3 – they are 180 degrees
apart in phase – the average number of bits decoded incorrectly is 1.5.
But we can reduce this BER by utilizing Gray coding, which will reduce average miscoded bits down to
one per symbol. Copyright 2002 Charan Langton www.complextoreal.com 14 All About Modulation – Part II If we organized the symbols such that adjacent symbols are only 1 bit apart then, decoding a symbol
incorrectly will lead to making only a 1 bit error. If adjacent symbols are organized in a binary fashion, this
is not the case and 50% of the time, a miscoded symbol will mean two bit errors. Gray coding helps us to
organize our signaling such that adjacent symbols are different by exactly one bit and hence the Bit Error
rate for QPSK is onehalf the symbol error rate.
Since there are two bits per symbol, the BER of a QPSK when it is Gray coded can be written as P(e)QPSK = 1
E
erfc b
2
N0 33 This is exactly the same as BPSK.
Gray coding and 8PSK modulation
In 8PSK modulation we transmit 8 symbols. Each one of these symbols represents 3 bits. How do we
decide which symbol should represents which bit pattern. The first choice is the binary order, called natural
order here in Column 2. We can alternately also number the symbols by a different ordering, Gray coding
as in Column 4.
Table I – Natural vs. Gray coding of symbols
Symbol Natural
Number Ordering
1
2
3
4
5
6
7
8 000
001
010
011
100
101
110
111 Bit Diff. to
Next
Neighbor
3
1
2
1
3
1
2
1 Gray
Coding Bit Diff. to
Next Neighbor 000
001
011
010
110
111
101
100 1
1
1
1
1
1
1
1 011 Gray code ordering 010
010 001
0 01 011 Natural ordering 110 100 000 111 101
Copyright 2002 Charan Langton 111 000 110
101 100 www.complextoreal.com 15 All About Modulation – Part II Figure 16  Mapping bits to symbols, natural and Gray Coding
Natural Ordering
Notice that symbol 1 differs from symbol 2 by only 45 degrees of phase. So is it likely that when an error is
made, one symbol will be mistaken for one that looks similar to it. And that means we may confuse symbol
2 for either 3 or symbol 1. It is more unlikely that we will mistake symbol 2 for symbol 6 as these two
differ by 180 degrees. When a symbol is misidentified then the bit pattern that it stands for is also misdecoded.
•
•
• Symbol 4 is mistaken for Symbol 5; the bit pattern of symbol 4, 011 is decoded as 100. Three bit
errors are made in the final information signal.
Symbol 4 is mistaken for Symbol 3; the bit pattern of symbol 4, 011 is decoded as 010. One bit
error is made in the final information signal.
The average number of bits that are decoded incorrectly for this symbol is 2 bits. Gray Coding
Now examine a different type of mapping where adjacent bit mappings differ by just one bit. Now when a
mistake is made in identifying a symbol, then the maximum number of bits that will be wrong is one and
not two as was the case for natural ordering.
This way of mapping the bit groups to the symbols is called Gray coding. Gray coding is far more natural
then binary system. In nature such as in DNA and chemical structures we find that Mother Nature prefers
Gray coding and uses it liberally where errors can be devastating. Gray coding is used for nearly all digital
modulations starting with QPSK
Relationship of Bit Error Rate with Symbol Error Rate
If a symbol consists of N bits, and if it is decoded incorrectly then the net effect is that anywhere from 1 to
N bits could be decoded incorrectly. Let's take a look at QPSK which has 4 symbols. Since only one is
correct, the probability of a bit error comes from
Symbol 2  1 bit error out of 2
Symbol 3  2 bit error out of 2
Symbol 4  1 bit error out of 2
So now we can write the bit error as
1
(1 bit error ) + 1 (1 bit error ) + 2 (2 bit error )
2
2
2
Pb =
3 There are three symbols that have errors, and each adds to the total bit errors as shown above.
Pb = 2
Ps
3 Copyright 2002 Charan Langton 34 www.complextoreal.com 16 All About Modulation – Part II And if the signal is gray coded this improves by factor of N which is the number of bits per symbol.
Pb = M /2
Ps
M −1 35 The following table gives a list of BER relationships. One of the outstanding things is that BER is a
function of only one variable, Eb/N0. However, don’t forget that these are theoretical equations under ideal
conditions, that is no ISI, no distortions, no filtering effects, no amplitude nonlinearities, no fading. In fact
when we incorporate all these effects, the required Eb/N0 to achieve a certain BER goes up considerably.
What these numbers do provide is a baseline from which we can judge improvements and degradations. Intersymbol distance concept
Constellation diagrams are very important in developing intuitive understanding of BER. The constellation
lets us see the signals as population of objects from which the receiver must pick out the correct one. How
densely packed are the symbols? This is obviously the key to how good the bit error rate of a signal will be.
We want a lot of signals but we also want them to be “different” from each other so mistakes are not
frequent.
Note that the bit energy of a modulated signal is equal to Eb = PTb
c 36 Where Pc is carrier power and Tb the bit time of the information signal.
The carrier power is defined as A2
Pc =
2 The symbol energy is defined as Es = PTs
c 37 The bit time and symbol time of BPSK is the same, so its bit and symbol energies are the same. For QPSK,
which has two independent I and Q channels, we have A2Ts
Es = PcTs =
2 38 This energy is divided equally between the I and Q channels, so we have A2Ts
A2Tb
→ Eb =
A2Ts
2
Es = PcTs =
→4
2
2
2
A Ts
A Tb
→ Eb =
4
2 39 Bit times for QPSK are related by Copyright 2002 Charan Langton www.complextoreal.com 17 All About Modulation – Part II Tb = 2Ts from which we get that
A2Tb
2 Eb( BPSK ) =
Eb(QPSK ) = 40 2 A Tb
2 Now lets’ examine the constellation of a BPSK signal. Each point is located at an amplitude equal to
from the origin.  Es Eb Es 1 0 2 Es Figure 17 – The distance between the two BPSK symbols is the largest that is possible and that means
that these two symbols are least like each other.
We define distance as the degree of dissimilarity between signals. The distance of BPSK symbols to the
origin is E b and the distance between the two points is d = 2 PTb = 2 Eb
c 41 This distance although expressed as a scalar quantity is the property of the signal space. It has no physical
meaning but is very very helpful in understanding how different modulations result in different bit error
rates.
Let's rewrite the BER equation of a BPSK signal by setting
Eb = d2
2 Now substitute d for Eb, E
1
P(e) = erfc b
2
N0
1
d2
= erfc
2
4 N0 42 This equation tells us that the BER is directly related to intersymbol distance d. Let's see how this applies to
other PSK modulations.
BER and QPSK intersymbol distance Copyright 2002 Charan Langton www.complextoreal.com 18 All About Modulation – Part II Eb
11 01 2 Eb 10 00 Eb Figure 18 – QPSK intersymbol distance is same as BPSK
The QPSK intersymbol distance is equal to 2 Eb 43 when we substitute this into Eq. 42, we get 1
d2
P(e) = erfc
2
4 N0
1
4 Eb
= erfc
2
4N0 44 E
1
= erfc b
2
N0
This is same as BPSK, again.
BER and Intersymbol distance for 6PSK
Now examine a MPSK signal space where M = 8. The signal space for a MPSK is always a circle, and
each bit sequence of N bits is represented by one of these M symbols.
M = 2N and M = 8, N = 3, for 8PSK, where three bits can be represented by one symbol. Each signal is this
circular space is located at a distance Ps Ts from the origin. Each symbol is 360/M degrees apart. So 8PSK signals are 360/8 = 45 degrees apart. For small M, the distance d between symbols is given by
π d 2 = 4 NE b sin 2 M 45 for 8PSK, M = 8 and N=3, we get
d 2 = 12E b 46
Plug this into equation 42, we get BER of 8PSK, Copyright 2002 Charan Langton www.complextoreal.com 19 All About Modulation – Part II E
1
P(e) = erfc b
2
N0
1
d2
= erfc
2
4 N0
12 Eb
1
= N erfc
2
4 N0 47 3
3Eb
= erfc
2
N0
Of course, we can extrapolate to this any MPSK by using consistent values of M and N in equation.
BER and Intersymbol distance of 16QAM
Now let's do MQAM, specifically 16QAM. We will compute its BER using the concept of the distance.
Now we have 16 symbols to transmit, with each symbol standing for 4 bits. Figure 11 – 16_QAM has three
bit sequence.
N = 4, and M = 16
Computing its BER is different from 8 Figure 19 – 16QAM constellation. Its intersymbol distance is smaller than 16PSK.
In PSK, all signals were at the same distance from its neighbor. There was only one intersymbol distance
to be computed. Here we have three different types of symbols, those that are in the middle and have
neighbors on all sides, and those that are on the sides so have neighbors on only one side and those in the
corners.
We assume that the inner most symbol is at (a, a) volts. First we will compute the signal energy. Energy of
a symbol is equal to the sum the energy of all the individual symbols. Another way to look at symbol
energy is to see it as the average energy of all different types of symbols. In case of PSK, all symbols have
the same energy (since all the amplitudes are the same), PAM and others where the amplitudes vary would
have different energies for the symbols. So we write the average energy as
E av = 1
M M ∑E m 48 m =1 Copyright 2002 Charan Langton www.complextoreal.com 20 All About Modulation – Part II The symbol energy is just the sum of all the individual energies of the different looking symbols.
Eb = E av
E av
=
N
log 2 M 49 Using a as the distance, we can express the average energy of the symbol as
Es = [ 12
( a + a 2 ) + (9a 2 + a 2 ) + ( a 2 + 9a 2 ) + (9 a 2 + 9a 2 )
4 = 10 a 2 Now we have
a = .1 E s 50 d = 2 .1 E s And since in this modulation, each symbol represents 4 bits, we have
E s = 4 Eb 51 d = 2 .4 Eb Compare this to the intersymbol distance of QPSK which is
d = 2 Eb And 16PSK, which is
d = 2 16Eb sin 2 π
16 = 2 .15Eb 52 The distance we just computed for 16QAM is less than QPSK but more than 16PSK. And what this says
is that 16QAM has a BER that is in between those two other schemes.
The BER of a 16QAM is computed by plugging in this value of intersymbol distance d in Eq 42. And we
get,
P (e ) =
= Eb
1
erfc
2
N0 1
d2
erfc
2
4N 0 .4 Eb
1
= 4 erfc
2
N0
= 2erfc 53 .4 Eb
N0 BER and intersymbol distance of a FSK signal Copyright 2002 Charan Langton www.complextoreal.com 21 All About Modulation – Part II Eb
1 2Eb
Eb 0 Figure 20 – FSK with two signals of two different frequencies
In a FSK system, we have symbol signals given by
u1 (t ) = 2
cos 2π m f c t
Tb u 2 (t ) = 2
cos 2π n f c t
Tb 54 These are the FSK basis functions and are harmonics. The symbol signals are now given by (after scaling
with Eb)
s1 (t ) = Eb u1 (t )
s 2 (t ) = E b u 2 (t ) 55 The signals are orthogonal when n and m are integers. Now imagine just a two dimensional system. The
signals lie on the two axes and the distance between these two is easily computed as
d = 2E b 56 This distance is larger than 4PSK but smaller than 16QAM. Now the BER can be written as
Pe = 1
d2
erfc
2
N0 Pe = Eb
1
erfc
2
2N0 57 Relative Intersymbol distances Copyright 2002 Charan Langton www.complextoreal.com 22 All About Modulation – Part II 01 Gray c ode or der ing 0 1 BPSK 00 11 QPSK 10 011
010 001
Gray code order ing 000 110 111 100
101 8PSK 16QAM s1 Figure 20  Intersymbol distance decreases as number of symbols in a signal space goes up.
The above figure shows that for the same power signal, one with more symbols will have smaller intersymbol distance. This leads to larger bit error rates. This problem is excerbated by other distortions making
the power required to operate these modulations nonlinearly related to the linear case. Other techniques
can be applied to control the distortions but there is no way to overcome the minimum required Eb/N0 for
these modulations. Only with coding can we reduce the requirements down but then we do pay the price of
a smaller throughput.
“No pain, no gain.” Copyright 2002 Charan Langton www.complextoreal.com 23 All About Modulation – Part II Eb/N0 vs. Bit Error Rate
Threoretical Bounds
1.E01 BPSK, QPSK
32QAM 1.E02 8PSK
16 PSK 1.E03 32PSK
FSK 1.E04
Bit Error Rate QPR
1.E05
Summary
1.E06
1.E07
1.E08
1.E09
1 .E10
0 5 10 15 20 25 30 Eb/N0, dB Figure 21 – BER curves for various modulations
Table II  BER Equations
Modulation
BPSK M = 2N, N
2, 1 QPSK 4, 2 SQPSK 4, 2 MSK 4, 2 MPSK M, N NE erfc b sin 2 π / M N0 16QAM 16, 4 E
2 erfc 0.4 b N0 Copyright 2002 Charan Langton Pe Bandwidth
Rb 1/ 2 1
erfc
2 Eb N0 1
erfc
2 Eb N0 Rb /2 1/ 2 Rb /2 1/ 2 E 1
erfc b 2 N0 1
1 erfc
2 Eb N0 Rb /2 1/ 2 1/ 2 1/ 2 Rb /N
Rb /4 www.complextoreal.com
24 All About Modulation – Part II MQAM M, N 3 Eb 1 21 − erfc 2 N N − 1 N 0 QPR L levels π 2 Eb 1
erfc 2 16 N 0 LQPR L levels MFSK M, N Rb /N 1/ 2 Rb /4 1/ 2 1/ 2
π
1 6 Eb 21 − 2 erfc (log 2 L )1 / 2 2 L −1 N0 L
4 M −1
erfc
2 N Eb 2N 0 1/ 2 Rb /L M Rb /N 1/ 2 http://www.vmsk.org/BERMeas.pdf
http://www.ctr.kcl.ac.uk/lectures/Reza/Lec3354/slides6.pdf
________________________________________________________
Charan Langton
Errors, corrections, comments, please email me: [email protected] Symbols Bits
1 00 2
3
4 01
10
11 AM
A1 sin(ω t + φ ) A2 sin (ω t + φ )
A3 sin (ω t + φ )
A4 sin (ω t + φ ) Copyright 2002 Charan Langton FM
A sin (ω1 t + φ ) A sin (ω 2 t + φ )
A sin (ω 3 t + φ )
A sin (ω 4 t + φ ) PM ( ) A sin ω t + φ1 A sin (ω t + φ 2)
A sin (ω t + φ 3)
A sin (ω t + φ 4) www.complextoreal.com ...
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This note was uploaded on 02/07/2011 for the course EEE EE567 taught by Professor Tutorials during the Spring '11 term at Birla Institute of Technology & Science, Pilani  Hyderabad.
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