demodulation
DESCRIPTION
modulationTRANSCRIPT
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
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A block diagram
Messsagesource
Source coder
Line coderPulse
shaping
demodulatordetector
channel
modulator
decision
1011
GEOMETRIC REPRESENTATION OF SIGNALS
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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)
inphase
quadrature
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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
si t( )= sijφj t( )j=1
N
∑ i =1,...,M
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:
Geometrically: φi t( )φj t( )∫ dt=0 i ≠j
1 i =j⎧ ⎨ ⎩
i
j
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:
sij means projection of si (t)on j(t)
sij = si(t)φj t( )0
T
∫ dtsij
si
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
cos2πnfot( )To
∫ cos2πmfot( )dt=0,m≠n
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
1 1 2
1
-0.5
21
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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)
s11 = s1(t)φ1 t( )0
2
∫ dt= 1( ) 1( )0
1
∫ dt+0=1
s12 = s1(t)φ2 t( )0
2
∫ dt=0+ −0.5( ) 1( )0
1
∫ dt=−0.5t
t
1 2
1
1
-0.5
s1(t)
1
s=(1,-0.5)
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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
1
-0.5
1
2
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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
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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,
Replace the signal by its expansion
Ei = si
2(t)dt0
T
∫
Ei = sijφj (t)j=1
N
∑⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
T
∫ sikφk(t)k=1
N
∑⎡
⎣ ⎢ ⎤
⎦ ⎥ dt
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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
Ei = sij
2φj
2 t( )j=1
N
∑⎡
⎣ ⎢ ⎤
⎦ ⎥ 0
T
∫ dt= sij2φ
j
2 t( )dt0
T
∫⎡
⎣ ⎢ ⎤
⎦ ⎥ j=1
N
∑
sij2 φ
j
2 t( )dt0
T
∫=1
1 2 4 3 4 j=1
N
∑ = sij2
j=1
N
∑
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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
3
2 E=9+4=13
E
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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
s1
s2 cosθ12( ) =s
1
Ts2s1 ×s2
transpose
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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?
s1Ts2 = 1 2[ ]
2
1⎡
⎣ ⎢ ⎤
⎦ ⎥ =2+2=4
s1 = 1+4= 5
s2 = 4+1= 5
cosθ12( ) =4
5× 5=
45
⇒ θ12 =36.9o
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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
d12
2 = s1 −s22
= s1j −s2j( )2
j=1
N
∑
=(1)2 +(1)2 =2
⇒ d12 = 21 2
1
2
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Constellation building using correlator banks
We can decompose the signal into its components as follows
s(t)
1
2
N
dt0
T
∫
dt0
T
∫
dt0
T
∫
s1
s2
sN
N components
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Detection in the constellation space
Received signal is put through the filter bank below and mapped to a point
s(t)
1
2
N
dt0
T
∫
dt0
T
∫
dt0
T
∫
s1
s2
sN
componentsmapped to a single point
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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
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Eb/No=5 dB
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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 signalwhich 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 s1
decide s2
s1s2
measurement
decide s1decide s2
decide s3 decide s4
s1s2
s3 s4
decide s1
s1
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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 XZi 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
departure from transmittedposition due to noise
X
si
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Error probability
we can write an expression for error like thisP{error|si}=P{X does not lie in Zi|si was
transmitted} Generally
Pe = P X∉Zi |si{ }P{i=1
M
∑ si}
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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
A-Acos(2pifct)
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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
s1 t( ) =2EbTb
cos2πfct( )
s2 t( )=−2EbTb
cos2πfct( )
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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
E=1φ1 t( )=2Tb
cos2πfct( )
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Formulating BER
BPSK constellation looks like this
√Eb-√Eb
X|1=[√Eb+n,n]
transmitted
received
noise
nois
e
Pe1 =P Eb +n<0|1 is transmitted{ }
if noise is negative enough, it will pushX to the left of the boundary, deciding 0instead
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Finding BER
Let’s rewrite BER
But n is gaussian with mean 0 and variance No/2
Pe1 =P Eb +n<0|1 { }=P n<− Eb{ }
-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]
orBER=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
Tbcos(2πfct)⇒
2EbTb
=0.2
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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
Finish this using erf tables
BER=12erfc
Eb
No
⎛
⎝ ⎜
⎞
⎠ ⎟ =
12erfc
2×10−7
2×10−6
⎛
⎝ ⎜
⎞
⎠ ⎟
=12erfc( 0.1) =
12erfc(0.316)
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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
But what is the requirements on the tones? Can they be any tones?
si t( )=
2EbTb
cos2πfit( ),0≤t≤Tb
0
⎧ ⎨ ⎪
⎩ ⎪
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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.
f i=nc +iTb
,i =1,2
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:
Notice f1 and f2 are selected to make them orthogonal
φi t( )=2Tb
cos2πfit( ),0≤t≤Tb
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BFKS constellation
There are two dimensions. Find the components of signals along each dimension using
s11 = s1 t( )0
Tb
∫ φ1 t( )dt= Eb
s12 = s1 t( )0
Tb
∫ φ2 t( )dt=0
s1 =( Eb,0)
Eb
Eb
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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
45 d
egre
e lin
e
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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?
x2>x1 puts X in the
s2 partition s1
s2
X=received
x1
x2Pe1=P{x2>x1|s1 was sent}
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Where do (x1,x2) come from?
Use the correlator bank to extract signal components
x=s1(t)+noise
1
2
dt0
Tb
∫
dt0
Tb
∫
x1(gaussian)
x2(gaussian)
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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
Pe =BER=12erfc
Eb2No
⎛
⎝ ⎜
⎞
⎠ ⎟
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BPSK and BFSK comparison:energy efficiency
Let’s compare their BER’s
Pe =12erfc
Eb2No
⎛
⎝ ⎜
⎞
⎠ ⎟ ,BFSK
Pe =12erfc
Eb
No
⎛
⎝ ⎜
⎞
⎠ ⎟ ,BPSK
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
Both have the same Eb, but BPSK’s are farther apart, hence lower BER
Eb− Eb
2 Eb
Eb
Eb 1.4× Eb
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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 encoded1 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 11 1 0 1 1 0 1 1 1
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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 transmittedbits
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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 0 0 0
1 0 0 1 0 0 1 1Detected data
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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 11 0 0 1 0 0 1 1
Back on track:no errors
Transmitted bits
Incoming phases
Detected bits
2 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
0 1 0 1 0 0 0 0 0 0 0 1 1 1 0 1 0 0 1 1 1binary
75321
-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 10
T
Tb
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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/symbolT=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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Bandwidth efficiency numbers
Here are some numbersn(bits/symbol) M(levels) (bits/sec/Hz)
1 2 12 4 23 8 34 16 48 256 8
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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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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
si(t)=2ET
cos2πfct+(2i−1)π4
⎡ ⎣
⎤ ⎦ ,i =1,2,3,4
0
⎧ ⎨ ⎪
⎩ ⎪ ,0≤t≤T
Symbol energy
Symbol width
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QPSK constellation
45o
0001
11 10
√E
φ1 t( )=2T
cos2πfct
φ2 t( ) =2T
sin2πfctBasis functions S=[0.7 √E,- 0.7 √E]
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QPSK decision regions
0001
11 10
Decision regions re color-coded
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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!
Pe =erfc(E
2No
)
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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
0011s1s2
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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
10 10 11 1000
11 10 11 1000
10 symbols = 20 bits
Sym.error=1/10Bit error=1/20
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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
PelogM
≤BER≤Pe
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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
Pe =erfc(E
2No
)
BER=12erfc(
E2No
)
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QPSK vs. BPSK
Let’s compare the two based on BER and bandwidth
BER BandwidthBPSK QPSK BPSK QPSK
12erfc
EbNo
⎛
⎝ ⎜
⎞
⎠ ⎟ 1
2erfc
EbNo
⎛
⎝ ⎜
⎞
⎠ ⎟ 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
si(t)=2ET
cos2πfct+(i−1)π4
⎡ ⎣
⎤ ⎦ ,i =1,...,8
0
⎧ ⎨ ⎪
⎩ ⎪ ,0≤t≤T
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8-PSK constellation
Distribute 8 phasors uniformly around a circle of radius √E
45o
Decision region
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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
Pe ≅erfcENo
sinπM
⎛ ⎝
⎞ ⎠
⎛
⎝ ⎜
⎞
⎠ ⎟ ,M ≥4
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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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16-QAM constellation using Gray coding
16-QAM has the following constellation Note gray codingwhere adjacent symbolsdiffer by only 1 bit
0010001100010000
1010
1110
0110
1011
1111
0111
1001
1101
0101
1000
1100
0100
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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
[ai,bi ]=
−3,3( ) −1,3( ) 1,3( ) 3,3( )
−3,1( ) −1,1( ) 1,1( ) 3,1( )
−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=1M
ai2 +bi
2( )i=1
M
∑
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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
E =
18 10 10 18
10 2 2 10
10 2 2 10
18 10 10 18
⎡
⎣
⎢ ⎢ ⎢
⎤
⎦
⎥ ⎥ ⎥
Eavg=10
1999 BG Mobasseri 96
Symbol error for M-ary QAM
With the definition of energy in mind, symbol error is approximated by
Pe ≅2 1−1M
⎛ ⎝
⎞ ⎠ erfc
2Eavg
2 M−1( )No
⎛
⎝ ⎜
⎞
⎠ ⎟
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Familiar constellations
Here are a few golden oldies
V.22 600 baud1200 bps
V.22 bis600 baud2400 bps
V.32 bis2400 baud9600 bps
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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
si t( )=2ET
cos2πfit( ),0≤t≤T
i =1,...,M
1999 BG Mobasseri 99
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
φi t( ) =2T
cos2πfit( ),
0≤t≤T
i =1,...,M
s1=[√E ,0, 0]s2=[0,√E, 0]s3=[0,0,√E]
√E
√E
√E
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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
2cos2πf1t( )cos2πf2t( )dt= cos2π f1 + f2( )dt0
T
∫averages to zero
1 2 4 4 4 3 4 4 4 0
T
∫
+ cos2π f1 −f2( )dt0
T
∫averages to zero if T=i/(f1-f2); i=integer
1 2 4 4 4 3 4 4 4
Take f1=1000 and T=1/1000. Thenif 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
Pe ≤12M−1( )erfc
E2No
⎛
⎝ ⎜
⎞
⎠ ⎟
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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)
BT=2Rbf1/Tb
BPSK
fc+/Tbfc-/Tbfc
2/Tb=2Rb
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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
0 1000 2000 3000 4000 5000 6000 7000 8000-1
-0.5
0
0.5
1
0 1000 2000 3000 4000 5000 6000 7000 8000-1
-0.5
0
0.5
1
0 1000 2000 3000 4000 5000 6000 7000 8000-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1BFSK for 1 0 0 1 0 1 1
+
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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
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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?
UseBT=2f+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
ρ=2Tb
cos(2πf1t)0
Tb
∫ cos(2πf2t)dt
1 2 3 2(f2-f1)Tb
Good points,zero correlation
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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=2f+2Rb=Rb+2Rb=3Rb
The practical bandwidth is a lot smaller
1999 BG Mobasseri 112
Sunde’s FSK bandwidth
Due to sidelobe cancellation, practical bandwidth is just BT=2f=Rb
f1 f2
1/Tb=Rb
fc
fc=(f1+f2)/2
f
BT=2 f+2Rb
f= (f2-f1)/2
f
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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
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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
Bandwidth of MPSK modulation
1999 BG Mobasseri 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
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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 BWTbflog2M=1-->f=1/ Tbflog2M=Rb/log2M
BT= Rb/log2M
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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)
f/Rb0.5
B=0.5Rb-->half of BPSK
1
After modulation:Bqpsk=Rb
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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
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)bin4 0.5 0.34 dB8 1/3 3.91 dB16 1/4 8.52 dB32 1/5 13.52 dB Lesson: By going to multiphase modulation, we save
bandwidth but have to pay in increased power, But why?
1999 BG 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
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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 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
BT=2Rb/ log2M Then
=Rb/BT= log2M/2 bits/sec/Hz
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Some numbers
Let’s evaluate vs. M for MPSKM 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 130
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
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Contrast with MPSK
Variation of bandwidth with M differs drastically compared to MPSK
MPSK MFSKBT=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
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
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Notations
M =2m # of symbols
m=log2 M bits/symbol
R=mTs
=log2 M
Ts
bits/sec
Ts = symbol duration
Rs = symbol rate
Tb =1R=
Ts
m=
1mRs
bit length
Bandwidth efficiency measure
RW
=log2 MWTs
=1
WTb
1999 BG 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
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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
RW
=log2 MWTs
=log2 M / /bits sec Hz
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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.
1999 BG 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
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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
RW
=log2 MWTs
=log2 M
M / /bits sec Hz
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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?
SNo
=EbRNo
=Eb
No
R
Eb
No
(dB) =SNo
(dB−Hz)−R(dB−bits/sec
=13.2dB
1999 BG Mobasseri 143
Is BER met? Yes
The symbol error probability in 8-PSK is
Solve for Es/No
Solve for PE
PE M( ) =2Q2Es
No
sinπM ⎛ ⎝
⎞ ⎠
⎡
⎣ ⎢ ⎤
⎦ ⎥
BER=PE
log2 M=2.2 ×10−5
3=7.3×10−6
EsNo
= log2 M( )Eb
N0
=3×20.89 =62.67
1999 BG 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
1999 BG Mobasseri 145
Binary vs. M-ary Model
M-ary ModulatorR bits/s
Rs =R
log2 Msymbols/ s
M-ary demodulator
SNo
=Eb
No
R=Es
No
Rs
1999 BG 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
EbNo
(dB) =SNo
dB−Hz( )−R(dB−bit/ s) =
Eb
No
(dB) =48dB−Hz
=−(10 log9600)dB−bits/ s=8.2dB
1999 BG 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.
1999 BG 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.
1999 BG 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
1999 BG Mobasseri 150
Symbol error for MFSK
For noncoherent orthogonal MFSK, symbol error probability is
PE M( ) ≤M −12
exp−Es
2No
⎛ ⎝ ⎜ ⎞
⎠ ⎟
Es =Eb log2 M
1999 BG 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
EsNo
= log2 M( )Eb
No
PB =2m−1
2m −1PE
1999 BG 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
PB =23
24 −1PE =
815
PE
1999 BG Mobasseri 153
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