m ary signalling
TRANSCRIPT
1999 BG Mobasseri 1
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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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) 2E
Tcos 2fct (2i 1)
4
, i 1,2,3,4
0
, 0 t T
Symbol energy
Symbol width
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QPSK constellation
45o
0001
11 10
√E
1 t 2T
cos2fct
2 t 2
Tsin 2fct
Basis functions S=[0.7 √E,- 0.7 √E]
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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
BERPe
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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)
BER1
2erfc(
E
2No)
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QPSK vs. BPSK
Let’s compare the two based on BER and bandwidth
BER BandwidthBPSK QPSK BPSK QPSK
1
2erfc
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) 2E
Tcos 2fct (i 1)
4
, 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 erfcE
Nosin
M
,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 1
Mai
2 bi2
i1
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
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Symbol error for M-ary QAM
With the definition of energy in mind, symbol error is approximated by
Pe 2 1 1
M
erfc
2Eavg2 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 2E
Tcos 2fit ,0 t T
i 1,...,M
1999 BG Mobasseri 35
MFSK constellation:3-dimensions
MFSK is different from MPSK in that each signal sits on an orthogonal axis(basis)
s1
s2
s3
1
2
3i t 2
Tcos 2fit ,
0 t Ti 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
2 cos 2f1t cos 2f2t dt cos2 f1 f2 dt0
T
averages to zero
0
T
cos2 f1 f2 dt0
T
averages to zero if T=i/(f1-f2); i=integer
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 1
2M 1 erfc
E
2No
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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
2
Tbcos(2f1t)
0
Tb
cos(2f2t)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
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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
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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?
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Power-bandwidth tradeoff
The goal is to keep BER fixed as we increase M. Consider an 8PSK set.
What happens if you go to 16PSK? Signals get closer hence higher BER
Solution: go to a larger circle-->higher energy
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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
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for MPSK
At a bit rate of Rb, BPSK bandwidth is 2Rb When we go to MPSK, bandwidth goes down
by a factor of log2M
BT=2Rb/ log2M Then
=Rb/BT= log2M/2 bits/sec/Hz
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Some numbers
Let’s evaluate vs. M for 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 66
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 MTs
bits/ sec
Ts symbol duration
Rs symbol rate
Tb 1R
Tsm
1mRs
bit length
Bandwidth efficiency measure
RW
log2 MWTs
1WTb
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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.
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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 MM
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
EbNoR
EbNo
(dB) SNo
(dB Hz) R(dB bits / sec
13.2dB
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Is BER met? Yes
The symbol error probability in 8-PSK is
Solve for Es/No
Solve for PE
PE M 2Q2EsNo
sinM
BERPE
log2 M
2.2 10 5
37.3 10 6
EsNo
log2 M EbN0
3 20.89 62.67
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Power-limited uncoded system
Same bit rate and BER Available bandwidth W=45 KHz Available S/No=48-dBHz Choose a modulation scheme that yields the
required performance
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Binary vs. M-ary Model
M-ary ModulatorR bits/s
Rs R
log2 Msymbols / s
M-ary demodulator
SNo
EbNoR
EsNo
Rs
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Choice of Modulation
With R=9600 bits/sec and W=45 KHz, the channel is not bandwidth limited
Let’s find the available Eb/No
EbNo
(dB) SNodB Hz R(dB bit / s)
EbNo
(dB) 48dB Hz
(10 log 9600)dB bits / s
8.2dB
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Choose MFSK
We have a lot of bandwidth but little power ->orthogonal modulation(MFSK)
The larger the M, the more power efficiency but more bandwidth is needed
Pick the largest M without going beyond the 45 KHz bandwidth.
1999 BG Mobasseri 84
MFSK Parameters
From Table 1, M=16 for an MFSK modulation requires a bandwidth of 38.4 KHz for 9600 bits/sec data rate
We also wanted to have a BER<10^-5. Question is if this is met for a 16FSK modulation.
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16-FSK
Again from Table 1, to achieve BER of 10^-5 we need Eb/No of 8.1dB.
We solved for the available Eb/No and that came to 8.2dB
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Symbol error for MFSK
For noncoherent orthogonal MFSK, symbol error probability is
PE M M 12
exp Es
2No
Es Eb log2 M
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BER for MFSK
We found out that Eb/No=8.2dB or 6.61 Relating Es/No and Eb/No
BER and symbol error are related by
EsNo
log2 M EbNo
PB 2m 1
2m 1PE
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Example
Let’s look at the 16FSK case. With 16 levels, we are talking about m=4 bits per symbol. Therefore,
With Es/No=26.44, symbol error prob. PE=1.4x10^-5-->PB=7.3x10^-6
PB 23
24 1PE
815PE