communications ii lecture 7: performance of digital modulationkkleung/communications2...psk can be...

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.