DIGITAL MODULATIONS Chapter 8

1
DIGITAL MODULATIONS(Chapter 8)
2
Why digital modulation?

• If our goal was to design a digital baseband
  communication system, we have done that
• Problem is baseband communication wont takes us
  far, literally and figuratively
• Digital modulation to a square pulse is what
  analog modulation was to messages

3
A block diagram
1011
Messsage source
Source coder
Line coder
Pulse shaping
modulator
channel
demodulator
detector
decision
4
GEOMETRIC REPRESENTATION OF SIGNALS
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

6
Have we seen this before?

• Why yes! Remember the beloved ej2pfct which can
  be written as
• ej2pfctcos(2pfct)jsin(2pfct)

quadrature
inphase
7
Expressing signals as a weighted sum

• Suppose a signal set consists of M signals
  si(t),I1,,M. Each signal can be represented by
  a linear sum of basis functions

8
Conditions on basis functions

• For the expansion to hold, basis functions must
  be orthonormal to each other
• Mathematically
• Geometrically

?j
?i
?k
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)

si
?j
sij
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
• NltM

11
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

12
Example four signal set

• A communication system sends one of 4 possible
  signals. Expand each signal in terms of two given
  basis functions

1
2
1
-0.5
1
1
1
1 2
13
Components of s1(t)

• This is a 2-Dsignal space. Therefore, each signal
  can be represented by a pair of numbers. Lets
  find them
• For s1(t)

s1(t)
1
t
-0.5
?1
1
s(1,-0.5)
t
1 2
14
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
15
Signal constellation

• Finding individual components of each signal
  along the two dimensions gets us the constellation

?2
s4
s2
?1
-0.5
0.5
s1
-0.5
s3
16
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, lt s1, s4gt
• lt s1, s4gts11xs41s12xs42(1)(0.5)(1)(-0.5)0

17
Finding the energy from the constellation

• This is a simple matter. Remember,
• Replace the signal by its expansion

18
Exploiting the orthogonality of basis functions

• Expanding the summation, all cross product terms
  integrate to zero. What remains are N terms where
  jk

19
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

2
E9413
3
20
Constrained energy signals

• Lets 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 )

21
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
?
22
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?

23
Distance between two signals

• The closer signals are together the more chances
  of detection error. Here is how we can find their
  separation

2
1
1 2
24
Constellation building using correlator banks

• We can decompose the signal into its components
  as follows

s1
?1
s2
N components
s(t)
?2
sN
?N
25
Detection in the constellation space

• Received signal is put through the filter bank
  below and mapped to a point

s1
?1
s2
s(t)
components mapped to a single point
?2
sN
?N
26
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

27
Actual example

• Here is a 16-level constellation which is
  reconstructed in the presence of noise

28
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
29
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
30
Defining decision regions

• An easy detection method, is to compute decision
  regions offline. Here are a few examples

decide s2
decide s1
decide s1
s2
s1
decide s1
measurement
s2
s1
s1
s4
s3
decide s2
decide s3
decide s4
31
More formally...

• Partition the decision space into M decision
  regions Zi, i1,,M. Let X be the measurement
  vector extracted from the received signal. Then
• if X?Zi ?si was transmitted

32
How does detection error occur?

• Detection error occurs when X lands in Zi but it
  wasnt si that was transmitted. Noise, among
  others, may be the culprit

X
si
departure from transmitted position due to noise
33
Error probability

• we can write an expression for error like this
• PerrorsiPX does not lie in Zisi was
  transmitted
• Generally

34
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 s1Acos(2pfct)
• 0s2 - Acos(2pfct)
• 1s and 0s are identified by 180 degree phase
  reversal at bit transitions

35
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
36
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 ---gtAsqrt(2Eb/Tb)
• We can write the two bits as follows

37
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

38
Formulating BER

• BPSK constellation looks like this

received
if noise is negative enough, it will push X to
the left of the boundary, deciding 0 instead
X1vEbn,n
noise
vEb
-vEb
noise
transmitted
39
Finding BER

• Lets rewrite BER
• But n is gaussian with mean 0 and variance No/2

-sqrt(Eb)
40
BER for BPSK

• Using the trick to find the area under a gaussian
  density(after normalization with respect to
  variance)
• BERQ(2Eb/No)0.5
• or
• BER0.5erfc(Eb/No)0.5

41
BPSK Example

• Data is transmitted at Rb106 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. 1s and 0s are
  sent by

42
Solving for Eb

• Since bit rate is 106, bit length must be
  1/Rb10-6
• Therefore,
• Eb20x10-620 ?w-sec
• Remember, this is the received energy. What was
  transmitted are probably several orders of
  magnitude bigger

43
Solving for BER

• Noise PSD is No/2 10-6. We know for BPSK
• BER0.5erfc(Eb/No)0.5
• What we have is then
• Finish this using erf tables

44
Binary FSK(Frequency Shift Keying)

• Another method to transmit 1s and 0s 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?

45
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.

46
Example tones

• Lets 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 nc0)
• f11/Tb1 MHz
• f22/Tb2MHz

47
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

48
BFKS constellation

• There are two dimensions. Find the components of
  signals along each dimension using

49
Decision regions in BFSK

• Decisions are made based on distances. Signals
  closer to s1 will be classified as s1 and vice
  versa

45 degree line
50
Detection error in BFSK

• Let the received signal land where shown.
• Assume s1 is sent. How would a detection error
  occur?
• x2gtx1 puts X in the
• s2 partition

s2
Pe1Px2gtx1s1 was sent
Xreceived
x2
s1
x1
51
Where do (x1,x2) come from?

• Use the correlator bank to extract signal
  components

x1(gaussian)
?1
x s1(t)noise
x2(gaussian)
?2
52
Finding BER

• We have to answer this question what is the
  probability of one random variable exceeding
  another random variable?
• To cast P(x2gtx1) into like of P(xgt2), rewrite
• P(x2gtx1x1)
• x1 is now treated as constant. Then, integrate
  out x1 to eliminate it

53
BER for BFSK

• Skipping the details of derivation, we get

54
BPSK and BFSK comparisonenergy efficiency

• Lets compare their BERs

• 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

55
Comparison in the constellation space

• Distances determine BERs. Lets compare
• Both have the same Eb, but BPSKs are farther
  apart, hence lower BER

56
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

57
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

58
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

unknown transmitted bits
Encoded received bits
59
DPSK generation

• Once data is differentially encoded, carrier
  modulation can be carried out by a straight BPSK
  encoding
• Digit 1phase 0
• Digit 0phase 180
• 1 1 0 1 1 0 1 1 1
• 0 0 p 0 0 p 0 0 0

Differentially encoded data
Phase encoded(BPSK)
60
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 p 0 0 p 0 0 0

Detected data
1 0 0 1 0 0 1 1
61
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 p p 0 p 0 0 0

Incoming phases
1 0 1 0 0 0 1 1 1
0 0 1 0 0 1 1
Detected bits
Transmitted bits
Back on trackno errors
2 errors
62
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

63
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

64
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

65
8-level PAM

• Here is an example of 8-level signaling

binary
0 1 0 1 0 0 0 0 0 0 0 1
1 1 0 1 0 0 1 1 1
7 5 3 2 1
-1 -3 -5 -7
66
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
Tb
T
67
Bit length-symbol length relationship

• When we combine n bits into one symbol the
  following relationships hold
• TnTb- symbol length
• nlogM bits/symbol
• TTbxlogM- symbol length
• All logarithms are base 2

68
Example

• If 8 bits are combined into one symbol, the
  resulting symbol is 8 times wider
• Using n8, we have M28256 symbols to pick from
• Symbol length TnTb8Tb

69
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
• Rsymbol rateRb/nRb/logM
• For 8-level signaling, baud rate is 1/3 of bit
  rate

70
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.

71
Issues in transmitting 9600 bits/sec

• Want to transmit 9600 bits/sec. Options
• Nyquists minimum bandwidth9600/24800 Hz
• Full roll off raised cosine9600 Hz
• None of them fit inside the 4 KHz wide phone
  lines
• Go to a 16 - level signaling, M16. Pulse rate is
  reduced to
• RRb/logM9600/42400 Hz

72
Using 16-level signaling

• Go to a 16-level signaling, M16. Pulse rate is
  then cut down to
• RRb/logM9600/42400 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

73
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, BTRb/2--gt
  ?2 bits/sec/Hz

74
M-ary bandwidth efficiency

• In M-ary signaling , pulse rate is given by
  RRb/logM. Full roll-off raised cosine bandwidth
  is BTR Rb/logM.
• Bandwidth efficiency is then given by
• ?Rb/BTlogM bits/sec/Hz
• For M2, binary we have 1 bit/sec/Hz. For M16,
  we have 4 bits/sec/Hz

75
M-ary bandwidth

• Summarizing, M-ary and binary bandwidth are
  related by
• BM-aryBbinary/logM
• Clearly , M-ary bandwidth is reduced by a factor
  of logM compared to the binary bandwidth

76
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-aryBbinary/logM9600/log83200 Hz

77
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

78
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
79
QPSKquadrature 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
80
QPSK constellation
00
01
45o
vE
10
11
Basis functions
S0.7 vE,- 0.7 vE
81
QPSK decision regions
00
01
10
11
Decision regions re color-coded
82
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!

83
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

s1
s2
00
11
84
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
Sym.error1/10 Bit error1/20
11
10
11
10
00
10 symbols 20 bits
85
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

86
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 E2Eb

87
QPSK vs. BPSK

• Lets compare the two based on BER and bandwidth
• BER Bandwidth
• BPSK QPSK BPSK QPSK

Rb Rb/2
EQUAL
88
M-phase PSK (MPSK)

• If you combine 3 bits into one symbol, we have to
  realize 238 states. We can accomplish this with
  a single RF pulse taking 8 different phases 45o
  apart

89
8-PSK constellation

• Distribute 8 phasors uniformly around a circle of
  radius vE

45o
Decision region
90
Symbol error for MPSK

• We can have M phases around the circle separated
  by 2p/M radians.
• It can be shown that symbol error probability is
  approximately given by

91
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 M2n symbols

92
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
1011
1001
1000
1110
1111
1101
1100
0110
0111
0101
0100
93
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

94
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

95
Eavg for 16-QAM

• Using the ai,bi matrix and using Eai2bi2 we
  get one energy per signal

Eavg10
96
Symbol error for M-ary QAM

• With the definition of energy in mind, symbol
  error is approximated by

97
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
98
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

99
MFSK constellation3-dimensions

• MFSK is different from MPSK in that each signal
  sits on an orthogonal axis(basis)

?3
s3
vE
s1vE ,0, 0 s20,vE, 0 s30,0,vE
vE
?1
s1
vE
s2
?2
100
Orthogonal signalsHow 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 dont 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

101
How to pick the tones?

• Orthogonal FSK requires tones that are
  orthogonal.
• Two carrier frequencies separated by integer
  multiples of period are orthogonal

102
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 f11000 and T1/1000. Then if f22000 , the
two are orthogonal so will f23000,4000 etc
103
MFSK symbol error

• Here is the error expression with the usual
  notations

104
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
• Lets first look at binary modulation bandwidth

105
BPSK bandwidth

• Remember BPSK was obtained from a polar signal by
  carrier modulation
• We know the bandwidth of polar NRZ using square
  pulses was BTRb.
• It doesnt take much to realize that carrier
  modulation doubles this bandwidth

106
Illustrating BPSK bandwidth

• The expression for baseband BPSK (polar)
  bandwidth is
• SB(f)2Ebsinc2(Tbf)
• BT2Rb

2/Tb2Rb
BPSK
f
1/Tb
fc
fc/Tb
fc-/Tb
107
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

108
Modeling of BFSK bandwidth

• Each stream is just a carrier modulated unipolar
  signal. Each has a sinc spectrum

1/TbRb
?f
BT2 ?f2Rb
?f (f2-f1)/2
f1
f2
fc
fc(f1f2)/2
109
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
• BT2?f2Rb
• ?f(f2-f1)/2(2200-1200)/2500 Hz
• BT2x5002x12003400 Hz
• This is more than BPSK of 2Rb2400 Hz

110
Sundes 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

?
Good points,zero correlation
1
2
3
2(f2-f1)Tb
111
Picking the 2nd zero crossingSundes FSK

• If we pick the second zc term (the first term
  puts the tones too close) we get
• 2(f2-f1)Tb2--gt ?f1/2TbRb/2
• remember ?f is (f2-f1)/2
• Sundes FSK bandwidth is then given by
• BT2?f2RbRb2Rb3Rb
• The practical bandwidth is a lot smaller

112
Sundes FSK bandwidth

• Due to sidelobe cancellation, practical bandwidth
  is just BT2?fRb

1/TbRb
?f
?f
BT2 ?f2Rb
?f (f2-f1)/2
f1
f2
fc
fc(f1f2)/2
113
BFSK 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)Tb3. Recalling that
  ?f(f2-f1)/2 then ?f 0.75/Tb
• Then ?f 0.75/Tb0.75x106750 KHz
• And BT2(?f Rb)2(0.751)1063.5 MHz

114
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 Sundes FSK requires BTRb1 MHz

115
Bandwidth of MPSK modulation
116
MPSK bandwidth review

• In MPSK we used pulses that are log2M times wider
  tan binary hence bandwidth goes down by the same
  factor.
• Tsymbol widthTblog2M
• For example, in a 16-phase modulation, M16,
  T4Tb.
• BqpskBbpsk/log2M Bbpsk/4

117
MPSK bandwidth

• MPSK spectrum is given by
• SB(f)(2Eblog2M)sinc2(Tbflog2M)

Set to 1 for zero crossing BW Tbflog2M1 --gtf1/
Tbflog2M Rb/log2M
BT Rb/log2M
f/Rb
1/logM
Notice normalized frequency
118
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,
• Bmpsk2Rb/log2M

119
QPSK bandwidth

• QPSK is a special case of MPSK with M4 phases.
  Its baseband spectrum is given by
• SB(f)2Esinc2(2Tbf)

B0.5Rb--gt half of BPSK
After modulation BqpskRb
f/Rb
0.5
1
120
Some numbers

• Take a 9600 bits/sec data stream
• Using BPSK B2Rb19,200 Hz (too much for 4KHz
  analog phone lines)
• QPSK B19200/log249600Hz, still high
• Use 8PSKB 19200/log286400Hz
• Use 16PSKB19200/ log2164800 Hz. This may
  barely fit

121
MPSK vs.BPSK

• Lets 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?

122
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--gthigher energy

123
Additional comparisons

• Take a 28.8 Kb/sec data rate and lets compare
  the required bandwidths
• BPSK BT2(Rb)57.6 KHz
• BFSK BT Rb 28.8 KHz ...Sundes FSK
• QPSK BThalf of BPSK28.8 KHz
• 16-PSK BTquarter of BPSK14.4 KHz
• 64-PSK BT1/6 of BPSK9.6 KHz

124
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

125
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

126
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 ?

127
? 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
• BT2Rb/ log2M
• Then
• ?Rb/BT log2M/2 bits/sec/Hz

128
Some numbers

• Lets 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 M2 and M64
• 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

129
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)

130
MFSK bandwidth

• Symbol duration in MFSK is M times longer than
  binary
• TTblog2M symbol length
• Each pair of tones are separated by 1/2T. If
  there are M of them,
• BTM/2TM/2Tblog2M
• --gtBTMRb/2log2M

131
Contrast with MPSK

• Variation of bandwidth with M differs drastically
  compared to MPSK
• MPSK MFSK
• BT2Rb/log2M BTMRb/2log2M
• As M goes up, MFSK eats up more bandwidth but
  MPSK save bandwidth

132
MFSK bandwidth efficiency

• Lets compute ?s for MFSK
• ?Rb/M2log2M/M bits/sec/HzMFSK
• 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

133
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

134
Notations

• Bandwidth efficiency measure

135
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 WTb0.3

136
Case of MPSK

• In MPSK, symbols are m times as wide as binary.
• Nyquist bandwidth is WRs/21/2Ts. However, the
  bandpass bandwidth is twice that, W1/Ts
• Then

137
Cost of Bandwidth Efficiency

• As M increases, modulation becomes more bandwidth
  efficient.
• Lets fix BER. To maintain this BER while
  increasing M requires an increase in Eb/No.

138
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

139
Case of MFSK

• MFSK is an orthogonal modulation scheme.
• Nyquist bandwidth is M-times the binary case
  because of using M orthogonal frequencies,
  WM/TsMRs
• Then

140
Select an Appropriate Modulation

• We have a channel of 4KHz with an available
  S/No53 dB-Hz
• Required data rate R9600 bits/sec.
• Required BER10-5.
• Choose a modulation scheme to meet these
  requirements

141
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, M8

142
What is the required Eb/No?
143
Is BER met? Yes

• The symbol error probability in 8-PSK is
• Solve for Es/No
• Solve for PE

144
Power-limited uncoded system

• Same bit rate and BER
• Available bandwidth W45 KHz
• Available S/No48-dBHz
• Choose a modulation scheme that yields the
  required performance

145
Binary vs. M-ary Model
R bits/s
M-ary Modulator
M-ary demodulator
146
Choice of Modulation

• With R9600 bits/sec and W45 KHz, the channel is
  not bandwidth limited
• Lets find the available Eb/No

147
Choose MFSK

• We have a lot of bandwidth but little power
  -gtorthogonal 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.

148
MFSK Parameters

• From Table 1, M16 for an MFSK modulation
  requires a bandwidth of 38.4 KHz for 9600
  bits/sec data rate
• We also wanted to have a BERlt10-5. Question is
  if this is met for a 16FSK modulation.

149
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

150
Symbol error for MFSK

• For noncoherent orthogonal MFSK, symbol error
  probability is

151
BER for MFSK

• We found out that Eb/No8.2dB or 6.61
• Relating Es/No and Eb/No
• BER and symbol error are related by

152
Example

• Lets look at the 16FSK case. With 16 levels, we
  are talking about m4 bits per symbol. Therefore,
• With Es/No26.44, symbol error prob.
  PE1.4x10-5--gtPB7.3x10-6

153
Summary

• Given
• R9600 bits/s
• BER10-5
• Channel bandwith45 KHz
• Eb/No8.2dB

• Solution
• 16-FSK
• required bw38.4khz
• required Eb/No8.1dB