1. DIGITAL MODULATIONS
1999 Bijan 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
©2000 Bijan Mobasseri

3. A block diagram Messsage source Source coder 1011 Line coder Pulse
shaping
modulator
channel
decision
detector
demodulator
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

6. Have we seen this before?
Why yes! Remember the beloved ej2πfct which can be written as
ej2πfct=cos(2πfct)+jsin(2πfct)
quadrature
inphase
©2000 Bijan 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
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8. Conditions on basis functions
For the expansion to hold, basis functions must be orthonormal to each other
Mathematically:
Geometrically: j i k
©2000 Bijan 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) si j
sij
©2000 Bijan 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
©2000 Bijan Mobasseri

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
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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
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13. Components of s1(t) s=(1,-0.5)
This is a 2-Dsignal space. Therefore, each signal can be represented by a pair of numbers. Let’s find them
For s1(t)
s1(t) 1 t
-0.5 1 1
s=(1,-0.5) t
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

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
-0.5 s1 s3
©2000 Bijan Mobasseri

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, < s1, s4>
< s1, s4>=s11xs41+s12xs42(1)(0.5)+(1)(-0.5)=0
©2000 Bijan Mobasseri

17. Finding the energy from the constellation
This is a simple matter. Remember,
Replace the signal by its expansion
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18. Exploiting the orthogonality of basis functions
Expanding the summation, all cross product terms integrate to zero. What remains are N terms where j=k
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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
E=9+4=13 3
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20. 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 )
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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 
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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?
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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
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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
©2000 Bijan Mobasseri

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) 2
components
mapped to a single point sN N
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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
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27. Actual example
Here is a 16-level constellation which is reconstructed in the presence of noise
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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
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

31. if XZi si was transmitted
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
©2000 Bijan Mobasseri

32. 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
©2000 Bijan Mobasseri

33. P{error|si}=P{X does not lie in Zi|si was transmitted}
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
©2000 Bijan Mobasseri

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: s1=Acos(2πfct)
0:s2= - Acos(2πfct)
1’s and 0’s are identified by 180 degree phase reversal at bit transitions
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

36. Eb= A2Tb/2 --->A=sqrt(2Eb/Tb)
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
©2000 Bijan Mobasseri

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
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38. Formulating BER BPSK constellation looks like this -√Eb √Eb received
if noise is negative enough, it will push
X to the left of the boundary, deciding 0
instead
X|1=[√Eb+n,n]
noise
-√Eb
√Eb
noise
transmitted
©2000 Bijan Mobasseri

39. Finding BER Let’s rewrite BER
But n is gaussian with mean 0 and variance No/2
-sqrt(Eb)
©2000 Bijan Mobasseri

40. 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]
©2000 Bijan Mobasseri

41. 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
©2000 Bijan Mobasseri

42. 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 is probably several orders of magnitude bigger
©2000 Bijan Mobasseri

43. 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
©2000 Bijan Mobasseri

44. 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?
©2000 Bijan Mobasseri

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.
©2000 Bijan Mobasseri

46. Example tones
Let’s say we are sending data at the rate of 1 Mb/sec in BFSK. What are a few orthogonal tones?
Bit length is 10-6 sec. Therefore, possible tones are (use nc=0)
f1=1/Tb=1 MHz
f2=2/Tb=2MHz
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

48. BFSK constellation
There are two dimensions. Find the components of signals along each dimension using
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

50. 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 s2
Pe1=P{x2>x1|s1 was sent} x2
X=received s1 x1
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

52. 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
©2000 Bijan Mobasseri

53. BER for BFSK Skipping the details of derivation, we get
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54. 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; a 3dB disadvantage
©2000 Bijan Mobasseri

55. 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
©2000 Bijan Mobasseri

56. Differential PSK Concept of differential encoding is very powerful
Take the the bit sequence
Differentially encoding of this stream means that we start we a reference bit and then record changes
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57. Differential encoding example
Data to be encoded
Set the reference bit to 1, then use the following rule
Generate a 1 if no change
Generate a 0 if change
©2000 Bijan Mobasseri

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
? ? ? ? ? ? ? ?
unknown transmitted
bits
Encoded received
bits
©2000 Bijan Mobasseri

59. 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
0 0 π π
Differentially encoded data
Phase encoded(BPSK)
©2000 Bijan Mobasseri

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 π π
Detected data
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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 π π 0 π
Incoming phases
Detected bits
Transmitted bits
Back on track:no errors
2 errors
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

64. M-ary signaling
Binary communications sends one of only 2 levels; 0 or 1
There is another way: combine several bits into symbols
Combining two bits at a time gives rise to 4 symbols; a 4-ary signaling
©2000 Bijan Mobasseri

65. 8-level PAM Here is an example of 8-level signaling binary 7 5 3 2 1 -1 -3 -5 -7
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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 Tb T
©2000 Bijan Mobasseri

67. 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
©2000 Bijan Mobasseri

68. 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
©2000 Bijan Mobasseri

69. R=symbol rate=Rb/n=Rb/logM
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
©2000 Bijan Mobasseri

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.
©2000 Bijan Mobasseri

71. 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 baud
©2000 Bijan Mobasseri

72. 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
©2000 Bijan Mobasseri

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, BT=Rb/2--> =2 bits/sec/Hz
©2000 Bijan Mobasseri

74. 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
©2000 Bijan Mobasseri

75. 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
©2000 Bijan Mobasseri

76. BM-ary=Bbinary/logM=9600/log8=3200 Hz
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
©2000 Bijan Mobasseri

77. Bandwidth efficiency numbers
Here are some numbers
n(bits/symbol) M(levels) (bits/sec/Hz)
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

79. 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
©2000 Bijan Mobasseri

80. QPSK constellation 00 01 45o √E 10 11 Basis functions
S=[0.7 √E,- 0.7 √E]
©2000 Bijan Mobasseri

81. QPSK decision regions 00 01 10 11 Decision regions re color-coded
©2000 Bijan Mobasseri

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!
©2000 Bijan Mobasseri

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 s2 s1 11 00
©2000 Bijan Mobasseri

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 00 10 10 11 10
Sym.error=1/10
Bit error=1/20 00 11 10 11 10
10 symbols = 20 bits
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

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 E=2Eb
©2000 Bijan Mobasseri

87. QPSK vs. BPSK Let’s compare the two based on BER and bandwidth
BER Bandwidth
BPSK QPSK BPSK QPSK
Rb Rb/2
EQUAL
©2000 Bijan Mobasseri

88. 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
©2000 Bijan Mobasseri

89. 8-PSK constellation
Distribute 8 phasors uniformly around a circle of radius √E
45o
Decision region
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90. 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
©2000 Bijan Mobasseri

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 M=2n symbols
©2000 Bijan Mobasseri

92. 16-QAM constellation using Gray coding
16-QAM has the following constellation
Note gray coding
where adjacent symbols
differ by only 1 bit
0000
0001
0011
0010
1000
1001
1011
1010
1100
1101
1111
1110
0100
0101
0111
0110
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

95. Eavg for 16-QAM
Using the [ai,bi] matrix and using E=ai^2+bi^2 we get one energy per signal
Eavg=10
©2000 Bijan Mobasseri

96. Symbol error for M-ary QAM
With the definition of energy in mind, symbol error is approximated by
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

99. MFSK constellation: 3-dimensions
MFSK is different from MPSK in that each signal sits on an orthogonal axis(basis) 3 s3 √E
s1=[√E ,0, 0]
s2=[0,√E, 0]
s3=[0,0,√E] √E s1 1 √E s2 2
©2000 Bijan Mobasseri

100. 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
©2000 Bijan Mobasseri

101. How to pick the tones?
Orthogonal FSK requires tones that are orthogonal.
Two carrier frequencies separated by integer multiples of period are orthogonal
©2000 Bijan Mobasseri

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 f1=1000 and T=1/1000. Then
if f2=2000 , the two are orthogonal
so will f2=3000,4000 etc
©2000 Bijan Mobasseri

103. MFSK symbol error
Here is the error expression with the usual notations
©2000 Bijan Mobasseri

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
Let’s first look at binary modulation bandwidth
©2000 Bijan Mobasseri

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 BT=Rb.
It doesn’t take much to realize that carrier modulation doubles this bandwidth
©2000 Bijan Mobasseri

106. Illustrating BPSK bandwidth
The expression for baseband BPSK (polar) bandwidth is
SB(f)=2Ebsinc2(Tbf)
BT=2Rb
2/Tb=2Rb
BPSK
1/Tb f
fc-/Tb fc
fc+/Tb
©2000 Bijan Mobasseri

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 +
©2000 Bijan Mobasseri

108. 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 f2 fc
fc=(f1+f2)/2
©2000 Bijan Mobasseri

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
BT=2f+2Rb
f=(f2-f1)/2=( )/2=500 Hz
BT=2x500+2x1200=3400 Hz
This is more than BPSK of 2Rb=2400 Hz
©2000 Bijan Mobasseri

110. Sunde’s FSK: f2-f1=Rb
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
Sunde’s
©2000 Bijan Mobasseri

111. 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
©2000 Bijan Mobasseri

112. 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 f2 fc
fc=(f1+f2)/2
©2000 Bijan Mobasseri

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)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
©2000 Bijan Mobasseri

114. Point to remember
FSK is not a particularly bandwidth-efficient modulation. In this example, to transmit 1 Mbps, we needed 3.5 MHz.
Of course, working at the 3rd zero crossing is responsible
Original Sunde’s FSK requires BT=Rb=1 MHz
©2000 Bijan Mobasseri

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.
T=symbol width=Tblog2M
For example, in a 16-phase modulation, M=16, T=4Tb.
Bqpsk=Bbpsk/log2M= Bbpsk/4
©2000 Bijan Mobasseri

117. SB(f)=(2Eblog2M)sinc2(Tbflog2M)
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
BT= Rb/log2M
1/logM
f/Rb
Notice normalized frequency
©2000 Bijan Mobasseri

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,
Bmpsk=2Rb/log2M
©2000 Bijan Mobasseri

119. 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
After modulation:
Bqpsk=Rb
0.5
f/Rb 1
©2000 Bijan Mobasseri

120. 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
©2000 Bijan Mobasseri

121. 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 dB
8 1/ dB
16 1/ dB
32 1/ dB
Lesson: By going to multiphase modulation, we save bandwidth but have to pay in increased power, But why?
©2000 Bijan Mobasseri

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-->higher energy
©2000 Bijan Mobasseri

123. 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
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

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 
©2000 Bijan Mobasseri

127. =Rb/BT= log2M/2 bits/sec/Hz
 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
©2000 Bijan Mobasseri

128. Some numbers Let’s evaluate  vs. M for MPSK M 2 4 8 16 32 64
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
©2000 Bijan Mobasseri

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 textbook we choose the former (this corresponds to using the first zero crossing of the correlation curve)
©2000 Bijan Mobasseri

130. T=Tblog2M symbol length
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
©2000 Bijan Mobasseri

131. BT=2Rb/log2M 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
©2000 Bijan Mobasseri

132. MFSK bandwidth efficiency
Let’s compute ’s for MFSK
=Rb/BT=2log2M/M bits/sec/Hz…MFSK M 
Notice bandwidth efficiency drop. We are sending fewer and fewer bits per 1 Hz of bandwidth
©2000 Bijan Mobasseri

133. COMPARISON OF DIGITAL MODULATIONS*
*B. Sklar, “ Defining, Designing and Evaluating Digital Communication Systems,”
IEEE Communication Magazine, vol. 31, no.11, November 1993, pp

134. Notations
Bandwidth efficiency measure
©2000 Bijan Mobasseri

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 WTb=0.3
©2000 Bijan Mobasseri

136. 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
©2000 Bijan Mobasseri

137. 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.
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

139. 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
©2000 Bijan Mobasseri

140. 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
©2000 Bijan Mobasseri

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, M=8
©2000 Bijan Mobasseri

142. What is the required Eb/No?
©2000 Bijan Mobasseri

143. Is BER met? Yes The symbol error probability in 8-PSK is
Solve for Es/No
Solve for PE
©2000 Bijan Mobasseri

144. 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
©2000 Bijan Mobasseri

145. Binary vs. M-ary Model R bits/s M-ary Modulator M-ary demodulator
©2000 Bijan Mobasseri

146. 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
©2000 Bijan Mobasseri

147. 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.
©2000 Bijan Mobasseri

148. 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.
©2000 Bijan Mobasseri

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
©2000 Bijan Mobasseri

150. Symbol error for MFSK
For noncoherent orthogonal MFSK, symbol error probability is
©2000 Bijan Mobasseri

151. 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
©2000 Bijan Mobasseri

152. 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
©2000 Bijan Mobasseri

153. Summary Given: Solution 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
©2000 Bijan Mobasseri