Digital-modulation.ppt

DIGITAL MODULATIONS
(Chapter 8)

1999 BG Mobasseri 2
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

1999 BG Mobasseri 3
A block diagram
Messsage
source
Source
coder
Line coder
Pulse
shaping
demodulator
detector
channel
modulator
decision
1011

GEOMETRIC
REPRESENTATION OF
SIGNALS

1999 BG Mobasseri 5
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

1999 BG Mobasseri 6
Have we seen this before?
 Why yes! Remember the beloved ej2πfct
which can be written as
ej2πfct=cos(2πfct)+jsin(2πfct)
inphase
quadrature

1999 BG Mobasseri 7
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

1999 BG Mobasseri 8
Conditions on basis functions
 For the expansion to hold, basis functions
must be orthonormal to each other
 Mathematically:
 Geometrically:
i
j
k

1999 BG Mobasseri 9
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:
 sij means projection of si (t)on j(t)
sij
si
j

1999 BG Mobasseri 10
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

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

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
1 1 2
1
-0.5
2
1

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)
t
t
1 2
1
1
-0.5
s1(t)
1
s=(1,-0.5)

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
1
-0.5
1
2

Signal constellation
 Finding individual components of each
signal along the two dimensions gets us
the constellation
s4
s1
s2
s3
1
2
-0.5
-0.5 0.5

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

Finding the energy from the
constellation
 This is a simple matter. Remember,
 Replace the signal by its expansion

Exploiting the orthogonality
of basis functions
 Expanding the summation, all cross
product terms integrate to zero. What
remains are N terms where j=k

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
3
2
E=9+4=13

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 )

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

s1
s2
transpose

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?

Distance between two signals
 The closer signals are together the more
chances of detection error. Here is how we
can find their separation
1 2
1
2

Constellation building using
correlator banks
 We can decompose the signal into its
components as follows
s(t)
1
2
N
s1
s2
sN
N components

Detection in the constellation
space
 Received signal is put through the filter
bank below and mapped to a point
s(t)
1
2
N
s1
s2
sN
components
mapped to a single point

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

Actual example
 Here is a 16-level constellation which is
reconstructed in the presence of noise

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

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

Defining decision regions
 An easy detection method, is to compute
“decision regions” offline. Here are a few
examples
decide s1
decide s2
s1
s2
measurement
decide s1
decide s2
decide s3 decide s4
s1
s2
s3 s4
decide s1
s1

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

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
departure from transmitted
position due to noise
X
si

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

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

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
A
-A
cos(2pifct)

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

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
E=1

Formulating BER
 BPSK constellation looks like this
√Eb
-√Eb
X|1=[√Eb+n,n]
transmitted
received
noise
if noise is negative enough, it will push
X to the left of the boundary, deciding 0
instead

Finding BER
 Let’s rewrite BER
 But n is gaussian with mean 0 and
variance No/2
-sqrt(Eb)

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]

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

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

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
 Finish this using erf tables

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
 But what is the requirements on the tones?
Can they be any tones?

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.

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

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:
 Notice f1 and f2 are selected to make them
orthogonal

BFKS constellation
 There are two dimensions. Find the
components of signals along each
dimension using

Decision regions in BFSK
 Decisions are made based on distances.
Signals closer to s1 will be classified as s1
and vice versa

Detection error in BFSK
 Let the received signal land where shown.
 Assume s1 is sent. How would a detection
error occur?
 x2>x1 puts X in the
s2 partition s1
s2
X=received
x1
x2
Pe1=P{x2>x1|s1 was sent}

Where do (x1,x2) come from?
 Use the correlator bank to extract signal
components
x=
s1(t)+noise
1
2
x1(gaussian)
x2(gaussian)

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

BER for BFSK
 Skipping the details of derivation, we get

BPSK and BFSK comparison:
energy efficiency
 Let’s compare their
BER’s
 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

Comparison in the
constellation space
 Distances determine BER’s. Let’s compare
 Both have the same Eb, but BPSK’s are
farther apart, hence lower BER

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

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

Detection logic
 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 1 0 1 1 0 1 1 1
Encoded received
bits
unknown transmitted
bits

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)

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 0 0 0
1 0 0 1 0 0 1 1
Detected data

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
1 0 1 0 0 0 1 1
1 0 0 1 0 0 1 1
Back on track:no errors
Transmitted bits
Incoming phases
Detected bits
2 errors

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

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

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

8-level PAM
 Here is an example of 8-level signaling
0 1 0 1 0 0 0 0 0 0 0 1 1 1 0 1 0 0 1 1 1
binary
7
5
3
2
1
-1
-3
-5
-7

A few definitions
 We used to work with bit length Tb. Now
we have a new parameter which we call
symbol length,T
1 1
0
T
Tb

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

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

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

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.

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

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

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

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

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

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

Bandwidth efficiency
numbers
 Here are some numbers
n(bits/symbol) M(levels) (bits/sec/Hz)
1 2 1
2 4 2
3 8 3
4 16 4
8 256 8

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

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
Symbol width

QPSK constellation
45o
00
01
11 10
√E
Basis functions S=[0.7 √E,- 0.7 √E]

QPSK decision regions
00
01
11 10
Decision regions re color-coded

QPSK error rate
 Symbol error rate for QPSK is given by
 This brings up the distinction between
symbol error and bit error. They are not the
same!

Symbol error
 Symbol error occurs when received vector
is assigned to the wrong partition in the
constellation
 When s1 is mistaken for s2, 00 is mistaken
for 11
00
11
s1
s2

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
10 10 11 10
00
11 10 11 10
00
10 symbols = 20 bits
Sym.error=1/10
Bit error=1/20

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

Symbol error and bit error for
QPSK
 We saw that symbol error for QPSK was
 Assuming no more than 1 bit error for each
symbol error, BER is half of symbol error
 Remember symbol energy E=2Eb

QPSK vs. BPSK
 Let’s compare the two based on BER and
bandwidth
BER Bandwidth
BPSK QPSK BPSK QPSK
Rb Rb/2
EQUAL

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

8-PSK constellation
 Distribute 8 phasors uniformly around a
circle of radius √E
45o
Decision region

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

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

16-QAM constellation using
Gray coding
 16-QAM has the following constellation
 Note gray coding
where adjacent symbols
differ by only 1 bit
0010
0011
0001
0000
1010
1110
0110
1011
1111
0111
1001
1101
0101
1000
1100
0100

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

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 for 16-QAM
 Using the [ai,bi] matrix and using
E=ai^2+bi^2 we get one energy per signal
Eavg=10

Symbol error for M-ary QAM
 With the definition of energy in mind,
symbol error is approximated by

Familiar constellations
 Here are a few golden oldies
V.22
600 baud
1200 bps
V.22 bis
600 baud
2400 bps
V.32 bis
2400 baud
9600 bps

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

MFSK constellation:
3-dimensions
 MFSK is different from MPSK in that each
signal sits on an orthogonal axis(basis)
s1
s2
s3
1
2
3
s1=[√E ,0, 0]
s2=[0,√E, 0]
s3=[0,0,√E]
√E
√E
√E

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

How to pick the tones?
 Orthogonal FSK requires tones that are
orthogonal.
 Two carrier frequencies separated by
integer multiples of period are orthogonal

Example
 Take two tones one at f1 the other at f2. T
must cover one or more periods for the
integral to be zero
Take f1=1000 and T=1/1000. Then
if f2=2000 , the two are orthogonal
so will f2=3000,4000 etc

MFSK symbol error
 Here is the error expression with the usual
notations

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

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

Illustrating BPSK bandwidth
 The expression for baseband BPSK (polar)
bandwidth is
SB(f)=2Ebsinc2(Tbf)
BT=2Rb
f
1/Tb
BPSK
fc+/Tb
fc-/Tb
fc
2/Tb=2Rb

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
+

Modeling of BFSK bandwidth
 Each stream is just a carrier modulated
unipolar signal. Each has a sinc spectrum
f1 f2
1/Tb=Rb
fc
fc=(f1+f2)/2
f
BT=2 f+2Rb
f= (f2-f1)/2

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

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
1 2 3 2(f2-f1)Tb
Good points,zero correlation


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

Sunde’s FSK bandwidth
 Due to sidelobe cancellation, practical
bandwidth is just BT=2f=Rb
f1 f2
1/Tb=Rb
fc
fc=(f1+f2)/2
f
BT=2 f+2Rb
f= (f2-f1)/2
f

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

Point to remember
 FSK is not a particularly bandwidth-
friendly 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

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

MPSK bandwidth
 MPSK spectrum is given by
SB(f)=(2Eblog2M)sinc2(Tbflog2M)
f/Rb
Notice normalized frequency
1/logM
Set to 1 for zero crossing BW
Tbflog2M=1
-->f=1/ Tbflog2M
=Rb/log2M
BT= Rb/log2M

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

QPSK bandwidth
 QPSK is a special case of MPSK with M=4
phases. It’s baseband spectrum is given by
SB(f)=2Esinc2(2Tbf)
f/Rb
0.5
B=0.5Rb-->
half of BPSK
1
After modulation:
Bqpsk=Rb

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

MPSK vs.BPSK
 Let’s say we fix BER at some level. How do
bandwidth and power levels compare?
M Bm-ary/Bbinary (Avg.power)M/(Avg.power)bin
4 0.5 0.34 dB
8 1/3 3.91 dB
16 1/4 8.52 dB
32 1/5 13.52 dB
 Lesson: By going to multiphase modulation, we save
bandwidth but have to pay in increased power, But why?

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

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

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

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

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 

 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

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

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)

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

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

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

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
 Bandwidth efficiency
measure

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 time-
bandwidth product
 The GSM standard uses Gaussian minimum
shift keying(GMSK) with WTb=0.3

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

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.

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

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

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

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

What is the required Eb/No?

Is BER met? Yes
 The symbol error probability in 8-PSK is
 Solve for Es/No
 Solve for PE

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

Binary vs. M-ary Model
M-ary Modulator
R bits/s
M-ary demodulator

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

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.

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.

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

Symbol error for MFSK
 For noncoherent orthogonal MFSK, symbol
error probability is

BER for MFSK
 We found out that Eb/No=8.2dB or 6.61
 Relating Es/No and Eb/No
 BER and symbol error are related by

Example
 Let’s look at the 16FSK case. With 16
levels, we are talking about m=4 bits per
symbol. Therefore,
 With Es/No=26.44, symbol error prob.
PE=1.4x10^-5-->PB=7.3x10^-6

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

Digital-modulation.ppt

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