DIGITAL MODULATIONS
(Chapter 8)
Why digital modulation?



If our goal was to design a digital
baseband communication system, we have
done that
Problem is baseband communication won’t
takes us far, literally and figuratively
Digital modulation to a square pulse is
what analog modulation was to messages
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A block diagram
Messsage
source
Source
coder
1011
Line coder
Pulse
shaping
modulator
channel
decision
detector
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demodulator
3
GEOMETRIC
REPRESENTATION OF
SIGNALS
The idea



We are used to seeing signals expressed
either in time or frequency domain
There is another representation space that
portrays signals in more intuitive format
In this section we develop the idea of
signals as multidimensional vectors
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Have we seen this before?

Why yes! Remember the beloved ej2πfct
which can be written as
ej2πfct=cos(2πfct)+jsin(2πfct)
quadrature
inphase
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Expressing signals as a
weighted sum

Suppose a signal set consists of M signals
si(t),I=1,…,M. Each signal can be
represented by a linear sum of basis
functions
N
si t    sij j t  i  1,..., M
j 1
0t T
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Conditions on basis functions



For the expansion to hold, basis functions
must be orthonormal to each other
Mathematically:
0 i  j
Geometrically:
 i t  j t dt  
1 i  j
j
i
k
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Components of the signal
vector

Each signal needs N numbers to be
represented by a vector. These N numbers
are given by projecting each signal onto
the individual basis functions:
si
T

sij   si (t) j t dt
0
sij means
projection of si (t)ons j(t)
ij
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j
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Signal space dimension



How many basis functions does it take to
express a signal? It depends on the
dimensionality of the signal
Some need just 1 some need an infinite
number.
The number of dimensions is N and is
always less than the number of signals in
the set
N<=M
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Example: Fourier series


Remember Fouirer series? A signal was
expanded as a linear sum of sines and
cosines of different frequencies. Sounds
familiar?
Sines and cosines are the basis functions
and are in fact orthogonal to each other
 cos2nfot cos2mfot dt  0,m  n
To
fo  1/ To
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Example: four signal set
A communication system sends one of 4
possible signals. Expand each signal in
terms of two given basis functions

1
1
2
-0.5
1
1
1
1
2
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Components of s1(t)


This is a 2-Dsignal space. Therefore, each
signal can be represented by a pair of
numbers. Let’s find them
For s1(t)
s (t)
2
1
0
0
1
s11   s1 (t)1 t dt   11dt  0  1
2
1
s12   s1 (t)2 t dt  0   0.51dt  0.5
0
1
0
t
-0.5
1
1
s=(1,-0.5)
t
1
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Interpretation


s1(t) is now condensed into just two
numbers. We can “reconstruct” s1(t) like
this
s1(t)=(1)1(t)+(-0.5)2(t)
Another way of looking at it is this
2
1
1
-0.5
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Signal constellation

Finding individual components of each
signal along the two dimensions gets us
the constellation 2
s4
s2
-0.5
1
0.5
s1
-0.5
s3
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Learning from the
constellation


So many signal properties can be inferred
by simple visual inspection or simple math
Orthogonality:
• s1 and s4 or orthogonal. To show that, simply find
their inner product, < s1, s4>
< s1, s4>=s11xs41+s12xs42(1)(0.5)+(1)(-0.5)=0
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Finding the energy from the
constellation

This is a simple matter. Remember,
T
Ei   s 2i (t)dt
0

Replace the signal by its expansion
 N
 N

Ei    sij  j (t)  sik k (t) dt
j 1




k 1
0
T
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Exploiting the orthogonality
of basis functions

Expanding the summation, all cross
product terms integrate to zero. What
remains are N terms where j=k
N
 N 2 2 
T 2 2

Ei    s ij  j t  dt    sij j t dt
j 1



j1




0
0
T
T
N
N
2
s


t

dt

s
  j
 ij
j 1
2
ij
2
j 1
0
1
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Energy in simple language

What we just saw says that the energy of a
signal is simply the square of the length of
its corresponding constellation vector
E
2
E=9+4=13
3
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Constrained energy signals

Let’s say you are under peak energy Ep
constraint in your application. Just make
sure all your signals are inside a circle of
radius sqrt(Ep )
Ep
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Correlation of two signals

A very desirable situation in is to have
signals that are mutually orthogonal. How
do we test this? Find the angle between
them
transpose
s1
s2
cos12  
s1T s2
s1  s2

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Find the angle between s1 and
s2

Given that s1=(1,2)T and s2=(2,1)T, what is
the angle between the two?
2 
s s  1 2  2  2  4

1 

s1  1  4  5
T
1 2
s2  4  1  5
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4

5 5 5
 12  36.9o
cos12  
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Distance between two signals

The closer signals are together the more
chances of detection error. Here is how we
can find their separation
N
d 12  s1  s2   s1 j  s2 j 
2
2
2
2
j 1
1
 (1)  (1)  2
2
2
1
2
 d12  2
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Constellation building using
correlator banks

We can decompose the signal into its
components as follows
T
 dt
0
s1
1
T
 dt
0
s2
N components
s(t)
2
T
 dt
N
0
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sN
24
Detection in the constellation
space

Received signal is put through the filter
bank below and mapped to a point
T
 dt
0
s1
1
T
 dt
0
s2
s(t)
components
mapped to a single point
2
T
 dt
N
0
sN
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Constellation recovery in
noise

Assume signal is contaminated with noise.
All N components will also be affected.
The original position of si(t) will be
disturbed
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Actual example

Here is a 16-level constellation which is
reconstructed in the presence of noise
Eb/No=5 dB
2
1.5
1
0.5
0
-0 .5
-1
-1 .5
-2
-2
-1 .5
-1
-0 .5
0
0.5
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1
1.5
2
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Detection in signal space

One of the M allowable signals is
transmitted, processed through the bank
of correlators and mapped onto
constellation question is based on what we
see , what was the transmitted signal?
received signal
which of the four did it
come from
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Minimum distance decision
rule

It can be shown that the optimum decision,
in the sense of lowest BER, is to pick the
signal that is closest to the received
vector. This is called maximum likelihood
decision making
this is the most likely
transmitted signal
received
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Defining decision regions

An easy detection method, is to compute
“decision regions” offline. Here are a few
examples
decide s2
decide s1
s2
decide s1
s1
decide s1
measurement
s2
decide s2
s1
s1
s3
decide s3
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s4
decide s4
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More formally...

Partition the decision space into M
decision regions Zi, i=1,…,M. Let X be the
measurement vector extracted from the
received signal. Then
if XZi si was transmitted
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How does detection error
occur?

Detection error occurs when X lands in Zi
but it wasn’t si that was transmitted.
Noise, among others, may be the culprit
X
si
departure from transmitted
position due to noise
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Error probability


we can write an expression for error like
this
P{error|si}=P{X does not lie in Zi|si was
transmitted}
Generally
Pe   PX  Zi | si P{si}
M
i 1
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Example: BPSK
(binary phase shift keying)
BPSK is a well known digital modulation
obtained by carrier modulating a polar NRZ
signal. The rule is
1: s1=Acos(2πfct)
0:s2= - Acos(2πfct)
1’s and 0’s are identified by 180 degree
phase reversal at bit transitions

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Signal space for BPSK

Look at s1 and s2. What is the basis
function for them? Both signals can be
uniquely written as a scalar multiple of a
cosine. So a single cosine is the sole basis
function. We have a 1-D constellation
cos(2pifct)
-A
A
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Bringing in Eb


We want each bit to have an energy Eb.
Bits in BPSK are RF pulses of amplitude A
and duration Tb. Their energy is A2Tb/2 .
Therefore
Eb= A2Tb/2 --->A=sqrt(2Eb/Tb)
We can write the two bits as follows
2Eb
s1 t  
cos2fc t 
Tb
s2 t   
2Eb
cos2fc t 
Tb
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BPSK basis function

As a 1-D signal, there is one basis function.
We also know that basis functions must
have unit energy. Using a normalization
factor
2
1 t  
cos2fct 
Tb
E=1
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Formulating BER

BPSK constellation looks like this
received
X|1=[√Eb+n,n]
-√Eb
noise
if noise is negative enough, it will push
X to the left of the boundary, deciding 0
instead
√Eb
transmitted
Pe1  P Eb  n  0 |1 is transmitted 
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Finding BER

Let’s rewrite BER
Pe1  P Eb  n  0 |1  Pn   Eb 

But n is gaussian with mean 0 and
variance No/2
-sqrt(Eb)
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BER for BPSK

Using the trick to find the area under a
gaussian density(after normalization with
respect to variance)
BER=Q[(2Eb/No)0.5]
or
BER=0.5erfc[(Eb/No)0.5]
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BPSK Example


Data is transmitted at Rb=106 b/s. Noise
PSD is 10-6 and pulses are rectangular with
amplitude 0.2 volt. What is the BER?
First we need energy per bit, Eb. 1’s and 0’s
are sent by

2Eb
2Eb
cos(2fct) 
 0.2
Tb
Tb
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Solving for Eb



Since bit rate is 106, bit length must be
1/Rb=10-6
Therefore,
Eb=20x10-6=20 w-sec
Remember, this is the received energy.
What was transmitted are probably several
orders of magnitude bigger
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Solving for BER


Noise PSD is No/2 =10-6. We know for BPSK
BER=0.5erfc[(Eb/No)0.5]
What we have is then
 Eb  1
 2  10 7 
1
  erfc 

BER  erfc
6
2
 No  2
 2  10 

1
1
 erfc( 0.1)  erfc(0.316)
2
Finish this2 using erf tables
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Binary FSK
(Frequency Shift Keying)


Another method to transmit 1’s and 0’s is
to use two distinct tones, f1 and f2 of the
form below
2Eb


cos2fi t , 0  t  Tb
si t    Tb

0
But what is the requirements on the tones?
Can they be any tones?
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Picking the right tones


It is desirable to keep the tones orthogonal
Since tones are sinusoids, it is sufficient
for the tones to be separated by an integer
multiple of inverse duration, i.e.
nc  i
f i
,i  1,2
Tb
nc  some integer
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Example tones


Let’s say we are sending data at the rate
of 1 Mb/sec in BFSK, What are some
typical tones?
Bit length is 10-6 sec. Therefore, possible
tones are (use nc=0)
f1=1/Tb=1 MHz
f2=2/Tb=2MHz
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BFSK dimensionality



What does the constellation of BFSK look
like? We first have to find its dimension
s1 and s2 can be represented by two
orthonormal basis functions:
2
cos2fi t ,0  t  Tb
Tb selected to make them
Notice f1 and f2 are
orthogonal
i t  
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BFKS constellation

There are two dimensions. Find the
components of signals along each
dimension using
Eb
Tb
s11   s1 t 1 t dt  Eb
0
Eb
Tb
s12   s1 t 2 t dt  0
0
s1  ( Eb ,0)
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Decision regions in BFSK

Decisions are made based on distances.
Signals closer to s1 will be classified as s1
and vice versa
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Detection error in BFSK


Let the received signal land where shown.
Assume s1 is sent. How would a detection
error occur?
s2
Pe1=P{x2>x1|s1 was sent}
x2>x1 puts X in the
s2 partition

X=received
x2
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s1
x1
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Where do (x1,x2) come from?

Use the correlator bank to extract signal
components
Tb
 dt
0
x=
s1(t)+noise
x1(gaussian)
1
Tb
 dt
0
x2(gaussian)
2
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Finding BER



We have to answer this question: what is
the probability of one random variable
exceeding another random variable?
To cast P(x2>x1) into like of P(x>2), rewrite
P(x2>x1|x1)
x1 is now treated as constant. Then,
integrate out x1 to eliminate it
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BER for BFSK

Skipping the details of derivation, we get
 Eb 
1

Pe  BER  erfc
2
 2N o 
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BPSK and BFSK comparison:
energy efficiency

Let’s compare their
BER’s

 Eb 
1
, BPSK
Pe  erfc 
2
 No 
 Eb 
1
, BFSK
Pe  erfc 
2
 2No 


What does it take to
have the same BER?
Eb in BFSK must be
twice as big as BPSK
Conclusion: energy per
bit must be twice as
large in BFSK to
achieve the same BER
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Comparison in the
constellation space

Distances determine BER’s. Let’s compare
Eb
2 Eb

1.4  Eb
 Eb have theEsame
b
Both
Eb, but BPSK’s areEb
farther apart, hence lower BER
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Differential PSK



Concept of differential encoding is very
powerful
Take the the bit sequence 11001001
Differentially encoding of this stream
means that we start we a reference bit and
then record changes
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Differential encoding example


Data to be encoded
1 0 0 1 0 0 1 1
Set the reference bit to 1, then use the
following rule
• Generate a 1 if no change
• Generate a 0 if change
1 0 0 1 0 0 1 1
1 1 0 1 1 0 1 1 1
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Detection logic


1
Detecting a differentially encoded signal is
based on the comparison of two adjacent
bits
If two coded bits are the same, that means
data bit must have been a 1, otherwise 0
?
?
?
?
?
?
?
?
1
0
1
1
0
1
1
unknown transmitted
bits
1
Encoded received
bits
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DPSK: generation

Once data is differentially encoded, carrier
modulation can be carried out by a straight
BPSK encoding
•
•
Digit 1:phase 0
Digit 0:phase 180
1 1 0 1 1 0 1 1 1
0 0 π 0 0 π 0 0 0
Differentially encoded data
Phase encoded(BPSK)
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DPSK detection

Data is detected by a phase comparison of
two adjacent pulses
• No phase change: data bit is 1
• Phase change: data bit is 0
0 0 π 0 0 π
Detected data
1
0
0
1
0
0 0 0
0
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1
1
60
Bit errors in DPSK



Bit errors happen in an interesting way
Since detection is done by comparing
adjacent bits, errors have the potential of
propagating
Allow a single detection error in DPSK
0 0 π π 0 π 0 0 0
Incoming phases
Detected bits
Transmitted bits
1
1
0
0
1
0
0
1
2 errors
0 0
0 0
1
1
1
1
Back on track:no errors
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Conclusion


In DPSK, if the phase of the RF pulse is
detected in error, error propagates
However, error propagation stops quickly.
Only two bit errors are misdetected. The
rest are correctly recovered
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Why DPSK?


Detecting regular BPSK needs a coherent
detector, requiring a phase reference
DPSK needs no such thing. The only
reference is the previous bit which is
readily available
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M-ary signaling



Binary communications sends one of only
2 levels; 0 or 1
There is another way: combine several bits
into symbols
1 0 1 1 0 1 1 0 1 1 1 0 0 1 1
Combining two bits at a time gives rise to 4
symbols; a 4-ary signaling
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8-level PAM

Here is an example of 8-level signaling
binary
7
5
3
2
1
0 1 0 1 0 0 0 0 0 0 0 1 1
1 0 1 0 0 1 1 1
-1
-3
-5
-7
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A few definitions

We used to work with bit length Tb. Now
we have a new parameter which we call
symbol length,T
1
0
1
Tb
T
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Bit length-symbol length
relationship


When we combine n bits into one symbol;
the following relationships hold
T=nTb- symbol length
n=logM bits/symbol
T=TbxlogM- symbol length
All logarithms are base 2
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Example



If 8 bits are combined into one symbol, the
resulting symbol is 8 times wider
Using n=8, we have M=28=256 symbols to
pick from
Symbol length T=nTb=8Tb
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Defining baud




When we combine n bits into one symbol,
numerical data rate goes down by a factor
of n
We define baud as the number of
symbols/sec
Symbol rate is a fraction of bit rate
R=symbol rate=Rb/n=Rb/logM
For 8-level signaling, baud rate is 1/3 of bit
rate
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Why M-ary?



Remember Nyquist bandwidth? It takes a
minimum of R/2 Hz to transmit R
pulses/sec.
If we can reduce the pulse rate, required
bandwidth goes down too
M-ary does just that. It takes Rb bits/sec
and turns it into Rb/logM pulses sec.
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Issues in transmitting 9600
bits/sec

Want to transmit 9600 bits/sec. Options:
• Nyquist’s minimum bandwidth:9600/2=4800 Hz
• Full roll off raised cosine:9600 Hz


None of them fit inside the 4 KHz wide
phone lines
Go to a 16 - level signaling, M=16. Pulse
rate is reduced to
R=Rb/logM=9600/4=2400 Hz
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Using 16-level signaling



Go to a 16-level signaling, M=16. Pulse rate
is then cut down to
R=Rb/logM=9600/4=2400 pulses/sec
To accommodate 2400 pulses /sec, we
have several options. Using sinc we need
only 1200 Hz. Full roll-off needs 2400Hz
Both fit within the 4 KHz phone line
bandwidth
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Bandwidth efficiency


Bandwidth efficiency is defined as the
number of bits that can be transmitted
within 1 Hz of bandwidth
=Rb/BT bits/sec/Hz
In binary communication using sincs,
BT=Rb/2--> =2 bits/sec/Hz
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M-ary bandwidth efficiency



In M-ary signaling , pulse rate is given by
R=Rb/logM. Full roll-off raised cosine
bandwidth is BT=R= Rb/logM.
Bandwidth efficiency is then given by
=Rb/BT=logM bits/sec/Hz
For M=2, binary we have 1 bit/sec/Hz. For
M=16, we have 4 bits/sec/Hz
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M-ary bandwidth


Summarizing, M-ary and binary bandwidth
are related by
BM-ary=Bbinary/logM
Clearly , M-ary bandwidth is reduced by a
factor of logM compared to the binary
bandwidth
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8-ary bandwidth


Let the bit rate be 9600 bits/sec. Binary
bandwidth is nominally equal to the bit
rate, 9600 Hz
We then go to 8-level modulation (3
bits/symbol) M-ary bandwidth is given by
BM-ary=Bbinary/logM=9600/log8=3200 Hz
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76
Bandwidth efficiency
numbers

Here are some numbers
n(bits/symbol) M(levels)
1
2
2
4
3
8
4
16
8
256
1999 BG Mobasseri
(bits/sec/Hz)
1
2
3
4
8
77
Symbol energy vs. bit energy

Each symbol is made up of n bits. It is not
therefore surprising for a symbol to have n
times the energy of a bit
E(symbol)=nEb
Eb
E
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78
QPSK
quadrature phase shift keying



This is a 4 level modulation.
Every two bits is combined and mapped to
one of 4 phases of an RF signal
These phases are 45o,135o,225o,315o
Symbol energy

 2E cos2fct  (2i  1)  ,i  1,2,3, 4
si (t)   T

,0  t  T
4 

0
Symbol width
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QPSK constellation
01
00
45o
√E
11
2
1 t  
cos2fc t
T
2
2 t  
sin 2fc t
T
Basis functions
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10
S=[0.7 √E,- 0.7 √E]
80
QPSK decision regions
01
00
10
11
Decision regions re color-coded
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81
QPSK error rate

Symbol error rate for QPSK is given by

E
Pe  erfc(
)
2No
This brings up the distinction between
symbol error and bit error. They are not the
same!
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Symbol error

Symbol error occurs when received vector
is assigned to the wrong partition in the
constellation
11

s2
s1
00
When s1 is mistaken for s2, 00 is mistaken
for 11
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Symbol error vs. bit error


When a symbol error occurs, we might
suffer more than one bit error such as
mistaking 00 for 11.
It is however unlikely to have more than
one bit error when a symbol error occurs
00
10
10
11
10
00
11
10
11
10
Sym.error=1/10
Bit error=1/20
10 symbols = 20 bits
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84
Interpreting symbol error


Numerically, symbol error is larger than bit
error but in fact they are describing the
same situation; 1 error in 20 bits
In general, if Pe is symbol error
Pe
 BER  Pe
log M
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85
Symbol error and bit error for
QPSK



We saw that symbol error for QPSK was
E
Pe  erfc(
)
2No
Assuming no more than 1 bit error for each
symbol error, BER is half of symbol error
1
E
BER  erfc(
)
2
2NE=2E
Remember symbol energy
o
b
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86
QPSK vs. BPSK
Let’s compare the two based on BER and
bandwidth
BER
Bandwidth
BPSK
QPSK
BPSK
QPSK

 Eb  1
1
Eb 
 erfc 
erfc 


2
 N o  2
 N o 
Rb
Rb/2
EQUAL
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M-phase PSK (MPSK)

If you combine 3 bits into one symbol, we
have to realize 23=8 states. We can
accomplish this with a single RF pulse
taking 8 different phases 45o apart

 2E cos2fct  (i  1)  ,i  1,...,8
si (t)   T

,0  t  T
4 

0
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88
8-PSK constellation

Distribute 8 phasors uniformly around a
circle of radius √E
45o
Decision region
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89
Symbol error for MPSK


We can have M phases around the circle
separated by 2π/M radians.
It can be shown that symbol error
probability is approximately given by
 E
 

, M  4
Pe  erfc
sin


M 
 No
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Quadrature Amplitude
Modulation (QAM)



MPSK was a phase modulation scheme. All
amplitudes are the same
QAM is described by a constellation
consisting of combination of phase and
amplitudes
The rule governing bits-to-symbols are the
same, i.e. n bits are mapped to M=2n
symbols
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91
16-QAM constellation using
Gray coding
16-QAM has the following constellation
 Note gray coding
0000
0001
0011
where adjacent symbols
differ by only 1 bit
1000
1001
1011

0010
1010
1100
1101
1111
1110
0100
0101
0111
0110
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Vector representation
of 16-QAM

There are 16 vectors, each defined by a
pair of coordinates. The following 4x4
matrix describes the 16-QAM constellation
 3,3 1,3 1,3 3,3 
 3,1
1,1
1,1
3,1 
[ai ,bi ]  
3,1 1, 1 1, 1 3,1


3, 3 1,3 1,3 3, 3
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What is energy per symbol in
QAM?


We had no trouble defining energy per
symbol E for MPSK. For QAM, there is no
single symbol energy. There are many
We therefore need to define average
symbol energy Eavg
Eavg
1 M 2
  ai  bi2 
M i 1
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Eavg for 16-QAM

Using the [ai,bi] matrix and using
E=ai^2+bi^2 we get one energy per signal
18 10 10
10 2 2
E  
10 2 2

18 10 10
18 
10 
10 

18 
Eavg=10
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Symbol error for M-ary QAM

With the definition of energy in mind,
symbol error is approximated by
2Eavg 
1  


Pe  2 1 
erfc


M
 2 M  1No 
1999 BG Mobasseri
96
Familiar constellations

Here are a few golden oldies
V.22
600 baud
1200 bps
V.22 bis
600 baud
2400 bps
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V.32 bis
2400 baud
9600 bps
97
M-ary FSK


Using M tones, instead of M
phases/amplitudes is a fundamentally
different way of M-ary modulation
The idea is to use M RF pulses. The
frequencies chosen must be orthogonal
2E
si t  
cos2fi t ,0  t  T
T
i  1,..., M
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MFSK constellation:
3-dimensions

MFSK is different from MPSK in that each
signal sits on an orthogonal axis(basis)
2
i t  
cos2fi t ,
T
0t T
i  1,..., M
3
s3
√E
s1=[√E ,0, 0]
s2=[0,√E, 0]
s3=[0,0,√E]
√E
s1
√E
2
1
s2
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Orthogonal signals:
How many dimensions, how many
signals?



We just saw that in a 3 dimensional space,
we can have no more than 3 orthogonal
signals
Equivalently, 3 orthogonal signals don’t
need more than 3 dimensions because
each can sit on one dimension
Therefore, number of dimensions is always
less than or equal to number of signals
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How to pick the tones?


Orthogonal FSK requires tones that are
orthogonal.
Two carrier frequencies separated by
integer multiples of period are orthogonal
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Example

Take two tones one at f1 the other at f2. T
must cover one or more periods for the
integral to be zero
T
T
 2 cos2f t cos2f t dt   cos 2  f
1
2
0
1
 f2 dt
0
averages to zero
T
  cos 2  f1  f2 dt
0
averages to zero if T =i/(f1-f2)
; i=integer
Take f1=1000 and T=1/1000. Then
if f2=2000 , the two are orthogonal
so will f2=3000,4000 etc
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MFSK symbol error

Here is the error expression with the usual
notations
 E 
1

Pe   M  1erfc
2
 2N o 
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Spectrum of M-ary signals



So far Eb/No, i.e. power, has been our main
concern. The flip side of the coin is
bandwidth.
Frequently the two move in opposite
directions
Let’s first look at binary modulation
bandwidth
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BPSK bandwidth



Remember BPSK was obtained from a
polar signal by carrier modulation
We know the bandwidth of polar NRZ using
square pulses was BT=Rb.
It doesn’t take much to realize that carrier
modulation doubles this bandwidth
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Illustrating BPSK bandwidth
The expression for baseband BPSK (polar)
bandwidth is
SB(f)=2Ebsinc2(Tbf)

2/Tb=2Rb
BPSK
1/Tb
BT=2Rb
f
fc-/Tb
1999 BG Mobasseri
fc
fc+/Tb
106
BFSK as a sum of two RF
streams
BFSK can be thought of superposition of
two unipolar signals, one at f1 and the
other at f2

1
0.5
BFSK for 1 0 0 1 0 1 1
1
0
0.8
0.6
-0.5
0.4
-1
0.2
0
1000
2000
3000
4000
5000
6000
7000
8000
0
1000
2000
3000
4000
5000
6000
7000
8000
0
1
-0.2
-0.4
+
0.5
-0.6
0
-0.8
-1
0
1000
2000
3000
4000
5000
6000
7000
8000
-0.5
-1
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Modeling of BFSK bandwidth

Each stream is just a carrier modulated
unipolar signal. Each has a sinc spectrum
1/Tb=Rb
f
BT=2 f+2Rb
f= (f2-f1)/2
f1
fc
f2
fc=(f1+f2)/2
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Example: 1200 bps bandwidth



The old 1200 bps standard used BFSK
modulation using 1200 Hz for mark and
2200 Hz for space. What is the bandwidth?
Use
BT=2f+2Rb
f=(f2-f1)/2=(2200-1200)/2=500 Hz
BT=2x500+2x1200=3400 Hz
This is more than BPSK of 2Rb=2400 Hz
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Sunde’s FSK

We might have to pick tones f1 and f2 that
are not orthogonal. In such a case there
will be a finite correlation between the
tones

T
2 b
   cos(2f1t) cos(2f2 t)dt
Tb 0
Good points,zero correlation
1
1999 BG Mobasseri
2
3
2(f2-f1)Tb
110
Picking the 2nd zero crossing:
Sunde’s FSK
If we pick the second zc term (the first
term puts the tones too close) we get
2(f2-f1)Tb=2--> f=1/2Tb=Rb/2
remember f is (f2-f1)/2
 Sunde’s FSK bandwidth is then given by
BT=2f+2Rb=Rb+2Rb=3Rb
 The practical bandwidth is a lot smaller

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Sunde’s FSK bandwidth

Due to sidelobe cancellation, practical
bandwidth is just BT=2f=Rb
1/Tb=Rb
f
f
BT=2 f+2Rb
f= (f2-f1)/2
f1
fc
f2
fc=(f1+f2)/2
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B FSK example




A BFSK system operates at the 3rd zero
crossing of -Tb plane. If the bit rate is 1
Mbps, what is the frequency separation of
the tones?
The 3rd zc is for 2(f2-f1)Tb=3. Recalling that
f=(f2-f1)/2 then f =0.75/Tb
Then f =0.75/Tb=0.75x106=750 KHz
And BT=2(f +Rb)=2(0.75+1)106=3.5 MHz
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Point to remember



FSK is not a particularly bandwidthfriendly modulation. In this example, to
transmit 1 Mbps, we needed 3.5 MHz.
Of course, it is working at the 3rd zero
crossing that is responsible
Original Sunde’s FSK requires BT=Rb=1 MHz
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Bandwidth of MPSK
modulation
MPSK bandwidth review


In MPSK we used pulses that are log2M
times wider tan binary hence bandwidth
goes down by the same factor.
T=symbol width=Tblog2M
For example, in a 16-phase modulation,
M=16, T=4Tb.
Bqpsk=Bbpsk/log2M= Bbpsk/4
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MPSK bandwidth

MPSK spectrum is given by
SB(f)=(2Eblog2M)sinc2(Tbflog2M)
Set to 1 for zero crossing BW
Tbflog2M=1
-->f=1/ Tbflog2M
=Rb/log2M
1/logM
BT= Rb/log2M
f/Rb
Notice normalized frequency
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Bandwidth after carrier
modulation


What we just saw is MPSK bandwidth in
baseband
A true MPSK is carrier modulated. This will
only double the bandwidth. Therefore,
Bmpsk=2Rb/log2M
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QPSK bandwidth

QPSK is a special case of MPSK with M=4
phases. It’s baseband spectrum is given by
SB(f)=2Esinc2(2Tbf)
B=0.5Rb-->
half of BPSK
0.5
1
f/Rb
1999 BG Mobasseri
After modulation:
Bqpsk=Rb
119
Some numbers





Take a 9600 bits/sec data stream
Using BPSK: B=2Rb=19,200 Hz (too much
for 4KHz analog phone lines)
QPSK: B=19200/log24=9600Hz, still high
Use 8PSK:B= 19200/log28=6400Hz
Use 16PSK:B=19200/ log216=4800 Hz. This
may barely fit
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MPSK vs.BPSK

M
4
8
16
32

Let’s say we fix BER at some level. How do
bandwidth and power levels compare?
Bm-ary/Bbinary
(Avg.power)M/(Avg.power)bin
0.5
0.34 dB
1/3
3.91 dB
1/4
8.52 dB
1/5
13.52 dB
Lesson: By going to multiphase modulation, we save
bandwidth but have to pay in increased power, But why?
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Power-bandwidth tradeoff



The goal is to keep BER fixed as we
increase M. Consider an 8PSK set.
What happens if you go to 16PSK? Signals
get closer hence higher BER
Solution: go to a larger circle-->higher
energy
1999 BG Mobasseri
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Additional comparisons

Take a 28.8 Kb/sec data rate and let’s
compare the required bandwidths
•
•
•
•
•
BPSK: BT=2(Rb)=57.6 KHz
BFSK: BT = Rb =28.8 KHz ...Sunde’s FSK
QPSK: BT=half of BPSK=28.8 KHz
16-PSK: BT=quarter of BPSK=14.4 KHz
64-PSK: BT=1/6 of BPSK=9.6 KHz
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Power-limited systems

Modulations that are power-limited achieve
their goals with minimum expenditure of
power at the expense of bandwidth.
Examples are MFSK and other orthogonal
signaling
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Bandwidth-limited systems



Modulations that achieve error rates at a
minimum expenditure of bandwidth but
possibly at the expense of too high a
power are bandwidth-limited
Examples are variations of MPSK and
many QAM
Check BER rate curves for BFSK and
BPSK/QAM cases
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Bandwidth efficiency index


A while back we defined the following ratio
as a bandwidth efficiency measure in
bits/sec/HZ
=Rb/BT bits/sec/Hz
Every digital modulation has its own 
1999 BG Mobasseri
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 for MPSK



At a bit rate of Rb, BPSK bandwidth is 2Rb
When we go to MPSK, bandwidth goes
down by a factor of log2M
BT=2Rb/ log2M
Then
=Rb/BT= log2M/2 bits/sec/Hz
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Some numbers
Let’s evaluate  vs. M for MPSK
M
2
4
8
16
32
64

.5
1
1.5 2
2.5 3
 Notice that bits/sec/Hz goes up by a factor
of 6 from M=2 and M=64
 The price we pay is that if power level is
fixed (constellation radius fixed) BER will
go up. We need more power to keep BER
the same

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Defining MFSK:


In MFSK we transmit one of M frequencies
for every symbol duration T
These frequencies must be orthogonal.
One way to do that is to space them 1/2T
apart. They could also be spaced 1/T
apart. Following The textbook we choose
the former (this corresponds to using the
first zero crossing of correlation curve)
1999 BG Mobasseri
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MFSK bandwidth


Symbol duration in MFSK is M times longer
than binary
T=Tblog2M symbol length
Each pair of tones are separated by 1/2T. If
there are M of them,
BT=M/2T=M/2Tblog2M
-->BT=MRb/2log2M
1999 BG Mobasseri
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Contrast with MPSK


Variation of bandwidth with M differs
drastically compared to MPSK
MPSK
MFSK
BT=2Rb/log2M
BT=MRb/2log2M
As M goes up, MFSK eats up more
bandwidth but MPSK save bandwidth
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MFSK bandwidth efficiency
Let’s compute ’s for MFSK
=Rb/M=2log2M/M bits/sec/Hz…MFSK
M
2
4
8
16
32
64

1
1
.75 .5
.3
.18
 Notice bandwidth efficiency drop. We are
sending fewer and fewer bits per 1 Hz of
bandwidth

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COMPARISON OF DIGITAL
MODULATIONS*
*B. Sklar, “ Defining, Designing and Evaluating Digital Communication Systems,”
IEEE Communication Magazine, vol. 31, no.11, November 1993, pp. 92-101
Notations

M  2 # of symbols
m = log2 M bits/symbol
m log2 M
R= 
bits/ sec
Ts
Ts
m
Bandwidth efficiency
measure
R log2 M
1


W
WTs
WTb
Ts  symbol duration
Rs  symbol rate
1 Ts
1
Tb   
bit length
R m mRs
1999 BG Mobasseri
134
Bandwidth-limited Systems



There are situations where bandwidth is at
a premium, therefore, we need
modulations with large R/W.
Hence we need standards with large timebandwidth product
The GSM standard uses Gaussian minimum
shift keying(GMSK) with WTb=0.3
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Case of MPSK



In MPSK, symbols are m times as wide as
binary.
Nyquist bandwidth is W=Rs/2=1/2Ts.
However, the bandpass bandwidth is twice
that, W=1/Ts
Then
R log2 M

 log 2 M bits/sec/Hz
W
WTs
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Cost of Bandwidth Efficiency


As M increases, modulation becomes more
bandwidth efficient.
Let’s fix BER. To maintain this BER while
increasing M requires an increase in Eb/No.
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137
Power-Limited Systems


There are cases that bandwidth is
available but power is limited
In these cases as M goes up, the
bandwidth increases but required power
levels to meet a specified BER remains
stable
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Case of MFSK



MFSK is an orthogonal modulation scheme.
Nyquist bandwidth is M-times the binary
case because of using M orthogonal
frequencies, W=M/Ts=MRs
Then
R log2 M log 2 M


bits/sec/Hz
W
WTs
M
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Select an Appropriate
Modulation




We have a channel of 4KHz with an
available S/No=53 dB-Hz
Required data rate R=9600 bits/sec.
Required BER=10-5.
Choose a modulation scheme to meet
these requirements
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Minimum Number of Phases



To conserve power, we should pick the
minimum number of phases that still meets
the 4KHz bandwidth
A 9600 bits/sec if encoded as 8-PSK
results in 3200 symbols/sec needing
3200Hz
So, M=8
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What is the required Eb/No?
S Eb R Eb


R
No
No
No
Eb
S
(dB) 
(dB  Hz)  R(dB  bits / sec
No
No
 13.2dB
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Is BER met? Yes

The symbol error probability in 8-PSK is

 2Es  
PE  M  2Q
sin  

No
M 


Solve for Es/No

E
E
Solve for PEs  log 2 M b  3  20.89  62.67
No
N0
PE
2.2 10 5
BER 

 7.3  10 6
log 2 M
3
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Power-limited uncoded
system




Same bit rate and BER
Available bandwidth W=45 KHz
Available S/No=48-dBHz
Choose a modulation scheme that yields
the required performance
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Binary vs. M-ary Model
R bits/s
M-ary Modulator
R
Rs 
symbols / s
log 2 M
M-ary demodulator
S
Eb
Es

R
Rs
N o No
No
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Choice of Modulation


With R=9600 bits/sec and W=45 KHz, the
channel is not bandwidth limited
Let’s find the available Eb/No
Eb
S
(dB)  dB  Hz   R(dB  bit / s) 
No
No
Eb
(dB)  48dB  Hz
No
 (10 log 9600)dB  bits / s
 8.2dB
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Choose MFSK



We have a lot of bandwidth but little power
->orthogonal modulation(MFSK)
The larger the M, the more power
efficiency but more bandwidth is needed
Pick the largest M without going beyond
the 45 KHz bandwidth.
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MFSK Parameters


From Table 1, M=16 for an MFSK
modulation requires a bandwidth of 38.4
KHz for 9600 bits/sec data rate
We also wanted to have a BER<10^-5.
Question is if this is met for a 16FSK
modulation.
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16-FSK


Again from Table 1, to achieve BER of 10^5 we need Eb/No of 8.1dB.
We solved for the available Eb/No and that
came to 8.2dB
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Symbol error for MFSK

For noncoherent orthogonal MFSK, symbol
error probability is
 Es 
M 1

PE  M  
exp 
2
 2 No 
Es  Eb log 2 M
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BER for MFSK

We found out that Eb/No=8.2dB or 6.61
Relating Es/No and Eb/No

Es
Eb
 log 2 M 
No
No
BER and symbol error are related by

2 m 1
PB  m
PE
2 1
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Example


Let’s look at the 16FSK case. With 16
levels, we are talking about m=4 bits per
symbol. Therefore,
23
8
PB  4
PE  PE
2 symbol
1
15 error prob.
With Es/No=26.44,
PE=1.4x10^-5-->PB=7.3x10^-6
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Summary

Given:

• R=9600 bits/s
• BER=10^-5
• Channel bandwith=45
KHz
• Eb/No=8.2dB
Solution
• 16-FSK
• required bw=38.4khz
• required Eb/No=8.1dB
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