communications ii lecture 7: performance of digital modulationkkleung/communications2...psk can be...
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Communications II
Lecture 7: Performance of digital modulation
Professor Kin K. LeungEEE and Computing Departments
Imperial College London© Copyright reserved
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Outline
• Digital modulation and demodulation
• Error probability of ASK
• Error probability of FSK
• Error probability of PSK
• Noncoherent demodulation
• Reference: Lathi, Chap. 13.
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Bandpass Data Transmission
How do we transmit symbols 1 and 0?
• Amplitude Shift Keying (ASK)
• Frequency Shift Keying (FSK)
• Phase Shift Keying (PSK)
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ASK, FSK and PSK
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How do we recover the transmitted symbol?
• Coherent (synchronous) detection Use a BPF to reject out-of-band noiseMultiply the incoming waveform with a cosine of the carrier frequency
Use a LPF Requires carrier regeneration (both frequency and phase synchronization by using a phase-lock loop)
• Noncoherent detection (envelope detection etc.) Makes no explicit efforts to estimate the phase
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Amplitude shift keying (ASK) = on-off keying (OOK)
s0(t) = 0 s1(t) = A cos(2 fct)
Ors(t) = A(t) cos(2 fct), A(t) ∈ {0, A}
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Coherent Detection
Assume an ideal band-pass filter with unit gain on [fc −W, fc +W ]. For a practical band-pass filter, 2W should be interpreted as the equivalent bandwidth.
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Pre-detection signal:
)2sin()()2cos()]()([)2sin()()2cos()()2cos(
)()()(
tftntftntAtftntftntfA
tntstx
CSCC
CSCCC
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• After multiplication with 2 cos(2 fct):
• After low-pass filtering:
)4sin()())4cos(1)](()([)2cos()2sin(2)()2(cos2)]()([)( 2
tftntftntAtftftntftntAty
CSCC
CCSCC
)()()(~ tntAty C (166)
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Reminder: The in-phase noise component nc(t) has the same variance as the original band-pass noise n(t) The received signal (166) is identical to the received signal (147) for baseband digital transmission
The sample values of will have PDFs that are identical to those of the baseband case
For ASK the statistics of the receiver signal are identical to those of a baseband system
)(~ ty
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The probability of error for ASK is the same as for the baseband case
Assume equiprobable transmission of 0s and 1s.Then the decision threshold must be A/2 and the probability of error is given by:
2,AQP ASKe
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Phase shift keying (PSK)
s(t) = A(t) cos(2 fct), A(t) ∈ {−A, A}
Use coherent detection again, to eventually get the detection signal:
)()()(~ tntAty C
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Probability density functions for PSK for equiprobable 0s and 1s in noise:
(a): symbol 0 transmitted(b): symbol 1 transmitted
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Use threshold 0 for detection
Conditional error probabilities:
dnAnPe
0 2
2
0 2)(exp
21
dnAnPe
0
2
2
1 2)(exp
21
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In the first set
In the second set
AnnnddnAnn ~,0whenand~~
ndnPAe
~2
~exp
21
2
2
0
:~,,~,0whenand~)(~
nnwhenandAnn
nddnAnAnn
ndn
ndnP
A
A
e
~2
~exp
21
~)1(2
~exp
21
2
2
2
2
1
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So:
Change variable of integration to z ≡ n/σ ⇒ dn = σdz and when n = A, z = A/σ. Then:
Remember that
dnnPPPAPSKeee
2
2
,10 2exp
21
AQdzeP Az
PSKe2/
,
2
21
dttxQx
)2/exp(21 2
(172)
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Frequency Shift Keying (FSK)
s0(t) = A cos(2 f0t), if symbol 0 is transmitted s1(t) = A cos(2 f1t), if symbol 1 is transmitted
Symbol recovery:Use two sets of coherent detectors, one operating at a frequency f0 and the other at f1.
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Coherent detector for FSK
The two BPF’s are non-overlapping in frequency spectrum
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Each branch = an ASK detector
presentnot symbol if noise
present symbol if noiseA branch each onoutput LPF
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n0(t): the noise output of the top branchn1(t): the noise output of the bottom branch
Each of these noise terms has identical statistics to n(t).
Output if a symbol 1were transmittedy1(t) = A + [n1(t) − n0(t)]
Output if a symbol 0 were transmittedy0(t) = -A + [n1(t) − n0(t)]
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Set detection threshold to 0
Difference from PSK: the noise term is now n1(t) − n0(t).
The noises in the two channels are independent because their spectra are non-overlapping.
the variances add
the noise variance has doubled!
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Probability of error for FSK:
Replace σ2 in (172) by 2σ2 (or σ by 2 σ )
2,AQP FSKe
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Noise of the sum or difference of two independent zero mean random variables:
x1: a random variable with variance σ12
x2: a random variables with variance σ22
What is the variance of y ≡ x1 ± x2?
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By definition, the variance of y ≡ x1 ± x2 is
For independent variables: E{x1x2} = E{x1}E{x2} For zero-mean random variables:E{x1} = E{x2} = 0 ⇒ E{x1x2} = 0
So
}{}{}{}2{
}){(
}{}{
2221
21
2221
21
221
222
xExxExExxxxE
xxE
yEyEy
22
21
22
21
2 }{}{
xExEy
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Comparison of the three schemes
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To achieve the same error probability (fixed Pe):
PSK can be reduced by 6 dB compared with a baseband or ASK system (a factor of 2 reduction in amplitude)
FSK can be reduced by 3 dB compared with a baseband or ASK (a factor of 2 reduction in amplitude)
Caution: The comparison is based on peak SNR. In terms of average SNR, PSK only has a 3 dB improvement over ASK, and FSK has the same performance as ASK
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Properties of the Q-function
Upper bounds and good approximations:
Becomes tighter for large x.
A better upper bound for small x.
0,21)( 2/2
xex
xQ x
0,21)( 2/2
xexQ x
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Non-coherent detection
Accurate phase synchronization may be difficult in a dynamic channel.
Phase synchronization error is due to frequency drift, instability of the local oscillator, effects of strong noise ...
When the carrier phase is unknown, one must rely on non-coherent detection.
The phase θ is assumed to be uniformly distributed on [0, 2 ].
Circuitry is simpler, but analysis is more difficult!
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ASK
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Error analysis
When symbol 0 is sent, the envelope (noise alone) has Rayleigh distribution
When symbol 1 is sent, the envelope (signal + noise) has Riciandistribution
(180) dominates the error probability when A/σ >> 1.
0,)( )2/(2
22
rerrf r
0,)( 20)2/()(
2
222
rArIerrf Ar
(180)
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Let the threshold be A/2 for simplicity.
The error probability (dominated by symbol 0) is given by
Cf. coherent demodulation
)8/(
2/
)2/(2
22
22
21
0,21
A
A
re
e
rdrerP
)8/(,
22
21
2
A
ASKe eAQP
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FSK
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Error probability
When a symbol 1 is sent, one branch has Rayleigh distribution, the other has Rice distribution.Error occurs if Rice < Rayleigh, and it can be shown that
Cf. coherent demodulation
)4/( 22
21 A
e eP
)4/(,
22
21
2
A
FSKe eAQP
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DPSK: Differential PSK
It is impossible to demodulate PSK with an envelop detector, since PSK signals have the same frequency and amplitude.
We can demodulate PSK differentially, where phase reference is provided by a delayed version of the signal in the previous interval.
Differential encoding is essential: bn = bn−1 an, where an , bn ∈ 1.
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Differential detection
Error probability
Cf. coherent demodulation
)2/(,
22
21 A
DPSKe eP
)2/(,
22
21
A
PSKe eAQP
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Conclusions
Non-coherent demodulation retains the hierarchy of performance.
Non-coherent demodulation has error performance slightly worse than coherent demodulation, but approaches coherent performance at high SNR.
Non-coherent demodulators are considerably easier to build.