Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ =...

76
Meixia Tao @ SJTU Principles of Communications Meixia Tao Shanghai Jiao Tong University Chapter 8: Digital Modulation Techniques Selected from Chapter 10.1 – 10.5 of Fundamentals of Communications Systems, Pearson Prentice Hall 2005, by Proakis & Salehi 1

Transcript of Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ =...

Page 1: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Principles of Communications

Meixia TaoShanghai Jiao Tong University

Chapter 8: Digital Modulation Techniques

Selected from Chapter 10.1 – 10.5 of Fundamentals of Communications Systems, Pearson Prentice Hall

2005, by Proakis & Salehi

1

Page 2: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Topics to be Covered

Binary digital modulation

M-ary digital modulation Comparison study

Source A/D converter

Source encoder

Channelencoder Modulator

Channel

DetectorChannel decoder

Source decoder

D/A converter

User

Absent if source is

digitalNoise

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Page 3: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Digital Modulation

In digital communications, the modulation process corresponds to switching or keying the amplitude, frequency, or phase of a sinusoidal carrier wave according to incoming digital data

Three basic digital modulation techniques Amplitude-shift keying (ASK) - special case of AM Frequency-shift keying (FSK) - special case of FM Phase-shift keying (PSK) - special case of PM

Will use signal space approach in receiver design and performance analysis

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Page 4: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

101101001

Binary Modulation In binary signaling, the modulator produces one of two

distinct signals in response to one bit of source data at a time.

Binary modulation types

Binary PSK Binary FSK Binary ASK

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Page 5: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

: transmitted signal energy per bit, i.e.

, bit duration

Binary Phase-Shift Keying (BPSK)

Modulation 1 1 1 1

“1”

“0”

: carrier frequency, chosen to be for some fixedinteger or 1/c bf T>>

The pair of signals differ only in a 180-degree phase shift

0 1 1 0 1 0 0 1

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Page 6: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Then and

There is one basis function

Signal Space Representation for BPSK

s1s20

A binary PSK system is characterized by a signal space that is one-dimensional (i.e. N=1), and has two message points (i.e. M =2)

with

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Page 7: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Decision rule:

Guess signal (or binary 0) was transmitted otherwise

Guess signal (or binary 1) was transmitted if the received signal point r falls in region R1

Decision Rule of BPSK Assume that the two signals are equally likely, i.e.

s1s2

Region R1Region R2

0

Then the optimum decision boundary is the midpoint of the line joining these two message points

rr

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Page 8: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

0

Probability of Error for BPSK

The conditional probability of the receiver deciding in favor of given that is transmitted is

Due to symmetry

r

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Page 9: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Since the signals and are equally likely to be transmitted, the average probability of error is

Pe for BPSK (cont’d)

This ratio is normally called bit energy to noise density ratio (or SNR/bit)

depends on ratio

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Page 10: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

BPSK Transmitter

Product modulator

Binary PSKwave s(t)

Input binary data

Carrier wave

Spectrum shaping filter

Rectangular pulse

m(t)

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Page 11: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

BPSK Receiver

is the carrier-phase offset, due to propagation delay or oscillators at transmitter and receiver are not synchronous

The detection is coherent in the sense of Phase synchronization Timing synchronization

Decision device∫

bTdt

0

Threshold

Say 1 if threshold is exceeded

Say 0, otherwise

demodulator detector

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Page 12: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Binary FSK

Modulation

: transmitted signal energy per bit

“1”

“0”

fi : transmitted frequency with separation 1 0f f f∆ = −

(Example?)

1 1 1 10 1 1 0 1 0 0 1

is selected so that and are orthogonal i.e. f∆

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Page 13: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Signal Space for BFSK

Two orthogonal basis functions are required

Signal space representation

Message point

Message point

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Page 14: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Decision Regions of Binary FSK

The receiver decides in favor of s1 if the observation vector rfalls inside region R1. This occurs when r1 > r2

Decision boundary

Message point

Message point

R1

R2

Observation vector

When r1 < r2 , r falls inside region R2 and the receiver decides in favor of s2

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Page 15: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Thus, n is also Gaussian with

Since n1 and n2 are i.i.d with

Define a new random variable

Since the condition r1 < r2 corresponds to the receiver making a decision in favor of symbol s2, the conditional probability of error given s1 is transmitted is given by

Probability of Error for Binary FSK Given that s1 is transmitted,

and

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Page 16: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

By symmetry

3 dB worse than BPSK

To achieve the same Pe, BFSK needs 3dB more transmission power than BPSK

Pe for BFSK (cont’d)

Since the two signals are equally likely to be transmitted, the average probability of error for coherent binary FSK is

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Page 17: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Binary FSK Transmitter

On-off signalling form

0

1

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Page 18: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Coherent Binary FSK Receiver

∫bT

dt0

∫bT

dt0

Decision Device+ Choose 1 if l>0

Choose 0 otherwise

+

-

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Page 19: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Average energy per bit

Binary ASK

Modulation 1 1 1 1

(On-off signaling)

“1”

“0”

s1s2

Region R1Region R2

0

0 1 1 0 1 0 0 1

Decision Region

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Page 20: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Probability of Error for Binary ASK

Average probability of error is

Exercise: Prove Pe

Identical to that of coherent binary FSK

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Page 21: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Probability of Error and the Distance Between Signals

In general,

BPSK BFSK BASK

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Page 22: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

babi

lity

of B

it E

rror

PSK

ASK/FSK

3dB

Probability of Error for BPSK and FSK/ASK

e.g. 22

Page 23: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Example

Binary data are transmitted over a microwave link at the rate of 106 bits/sec and the PSD of the noise at the receiver input is 10-10 watts/Hz.

a) Find the average carrier power required to maintain an average probability of error for coherent binary FSK.

b) Repeat the calculation in a) for noncoherent binary FSK

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Page 24: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU 24

We have discussed Coherent modulation schemes, .e.g.

BPSK, BFSK, BASK They needs coherent detection,

assuming that the receiver is able to detect and track the carrier wave’s phase

Update

We now consider: Non-coherent detection on binary FSK Differential phase-shift keying (DPSK)

In many practical situations, strict phase synchronization is not possible. In these situations, non-coherent reception is required.

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Page 25: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Non-coherent scheme: BFSK

Consider a binary FSK system, the two signals are

, : unknown random phases with uniform distribution

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Page 26: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Signal Space Representation

Since

Choose four basis functions as

Signal space representation

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Page 27: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

The vector representation of the received signal

Detector

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Page 28: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Decision Rule for Non-coherent FSK

ML criterion:

Conditional pdf

Similarly,

Choose s1><

Choose s2

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Page 29: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

For ML decision, we need to evaluate

i.e.

Removing the constant terms

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Page 30: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

We have the inequality

By definition

where I0(.) is a modified Bessel function of the zeroth order

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Page 31: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Decision Rule (cont’d) Thus, the decision rule becomes: choose s1 if

Noting that this Bessel function is monotonically increasing. Therefore we choose s1 if

Interpretation: compare the energy in the two frequencies and pick the larger => envelop detector

Carrier phase is irrelevant in decision making

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Page 32: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Structure of Non-Coherent Receiver for Binary FSK

It can be shown that

Comparator(select

thelargest)

(For detailed proof, see Section 10.4.2 in the textbook )32

Page 33: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Performance Comparison Between coherent FSK and Non-Coherent FSK

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

babi

lity

of B

it E

rror

BPSK

ASK/FSK

NC FSK

DPSK

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Page 34: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Differential PSK (DPSK)

Non-coherent version of PSK

Phase synchronization is eliminated using differential encoding Encode the information in phase difference between successive

signal transmission. In effect, to send “0”, advance the phase of the current signal by 1800 ; to send “1”, leave the phase unchanged

Provided that the unknown phase contained in the received wave varies slowly (constant over two bit intervals), the phase difference between waveforms received in two successive bit intervals will be independent of .

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Page 35: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Generation of DPSK signal Generate DPSK signals in two steps

Differential encoding of the information binary bits Phase shift keying

Differential encoding starts with an arbitrary reference bit

1 0 0 1 0 0 1 1Information sequence

1 1 0 1 1 0 1 1 1

0 0 π 0 0 π 0 0 0

Initial bit

Differentially encoded sequence

Transmitted Phase

___________

1 iii mdd ⊕= −

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Page 36: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

DPSK Transmitter Diagram

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Page 37: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Differential Detection of DPSK Signals

Output of integrator (assume noise free)

The unknown phase becomes irrelevant If = 0 (bit 1), then y > 0 if =π (bit 0), then y < 0

Error performance

Decision device∫

bTdt

0

Threshold of zero volts

Choose 1 if l > 0

Otherwise choose 0

DelayTb

37

Page 38: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Summary of Pe for Different Binary Modulations

Coherent PSK

Coherent ASK

Coherent FSK

Non-Coherent FSK

DPSK

38

Page 39: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

0 2 4 6 8 10 12 1410

-7

10-6

10-5

10-4

10-3

10-2

10-1

100

Eb/No in [dB]

Pro

babi

lity

of B

it E

rror

BPSK(QPSK)

ASK/FSK

NC FSK

DPSK

Pe Plots for Different Binary Modulations

39

Page 40: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Topics to be Covered

Binary digital modulation

M-ary digital modulation Comparison study

Source A/D converter

Source encoder

Channelencoder Modulator

Channel

DetectorChannel decoder

Source decoder

D/A converter

User

Absent if source is

digitalNoise

40

Page 41: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

M-ary Modulation (多进制调制)

64QAM

128 QAM

256 QAM

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Page 42: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

M-ary Modulation Techniques In binary data transmission, send only one of two possible

signals during each bit interval Tb

In M-ary data transmission, send one of M possible signals during each signaling interval T

In almost all applications, M = 2n and T = nTb, where n is an integer

Each of the M signals is called a symbol These signals are generated by changing the amplitude,

phase, frequency, or combined forms of a carrier in M discrete steps.

Thus, we have: MASK MPSK MFSK MQAM

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Page 43: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

M-ary Phase-Shift Keying (MPSK)

The phase of the carrier takes on M possible values:

Signal set:

= Energy per symbol

Basis functions

1cf T>>

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Page 44: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

MPSK (cont’d)

Signal space representation

44

Page 45: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

MPSK Signal Constellations

BPSK QPSK 8PSK 16PSK

45

Page 46: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Euclidean distance

The minimum Euclidean distance is

plays an important role in determining error performance as discussed previously (union bound)

In the case of PSK modulation, the error probability is dominated by the erroneous selection of either one of the two signal points adjacent to the transmitted signal point.

Consequently, an approximation to the symbol error probability is

2 ( )2 1 cosmn m n sm nd EM

π − = − = −

s s

min22 1 cos 2 sins sd E EM Mπ π = − =

min

0

/ 22 2 2 sin/ 2MPSK s

dP Q Q EMNπ ≈ =

46

Page 47: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Exercise

Consider the M=2, 4, 8 PSK signal constellations. All have the same transmitted signal energy Es.

Determine the minimum distance between adjacent signal points

For M=8, determine by how many dB the transmitted signal energy Es must be increased to achieve the same as M =4.

mind

mind

47

Page 48: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Error Performance of MPSK

For large M, doubling the number of phases requires an additional 6dB/bit to achieve the same performance

4dB 5dB 6dB

48

Page 49: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

M-ary Quadrature Amplitude Modulation (MQAM 正交幅度调制)

In MPSK, in-phase and quadrature components are interrelated in such a way that the envelope is constant (circular constellation)

If we relax this constraint, we get M-ary QAM

16PSK 16QAM

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Page 50: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

MQAM

Signal set:

is the energy of the signal with the lowest amplitude are a pair of independent integers

Basis functions:

Signal space representation

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Page 51: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

MQAM Signal Constellation

Square lattice

1 3 5 7

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Page 52: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Error Performance of MQAM

Upper bound of the symbol error probability

Exercise: Determine the increase in Eb required to maintain the same error performance if the number of bits per symbol is increased from k to k+1, where k is large.

0

34( 1)

be

kEP QM N

≤ −

(for )2kM =

52

Page 53: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

where

M-ary Frequency-Shift Keying (MFSK) or Multitone Signaling

Signal set:

Correlation between two symbols

53

Page 54: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

MFSK (cont’d)

For orthogonality, the minimum frequency separation is

-0.217

54

Page 55: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

M-ary orthogonal FSK has a geometric presenation as M M-dim orthogonal vectors, given as

The basis functions are

( )0 ,0,0, ,0sE=s

( )1 0, ,0, ,0sE=s

( )1 0,0, ,0,M sE− =s

( )2 cos 2m cf m f tT

φ π= + ∆

55

Page 56: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Error Performance of MFSK

56

Page 57: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Notes on Error Probability Calculations

Pe is found by integrating conditional probability of error over the decision region Difficult for multi-dimensions Can be simplified using union bound (see ch07)

Pe depends only on the distance profile of signal constellation

57

Page 58: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Example

The 16-QAM signal constellation shown below is an international standard for telephone-line modems (called V.29).

a) Determine the optimum decision boundaries for the detector

b) Derive the union bound of the probability of symbol error assuming that the SNR is sufficiently high so that errors only occur between adjacent points

c) Specify a Gray code for this 16-QAM V.29 signal constellation

58

Page 59: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Symbol Error versus Bit Error

Symbol errors are different from bit errors

When a symbol error occurs, all k bits could be in error

In general, we can find BER using

is the number different bits between and

Gray coding is a bit-to-symbol mapping, where two adjacent symbols differ in only one bit out of the k bits

An error between adjacent symbol pairs results in one and only one bit error.

59

Page 60: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Example: Gray Code for QPSK

0001

11 10

60

Page 61: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Bit Error Rate for MPSK and MFSK For MPSK with Gray coding An error between adjacent symbols will most likely occur Thus, bit error probability can be approximated by

For MFSK When an error occurs anyone of the other symbols may result

equally likely. Thus, k/2 bits every k bits will on average be in error when there is

a symbol error Bit error rate is approximately half of the symbol error rate

eb PP21

≅

61

Page 62: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Think …

Why 4G LTE prefers MQAM over MPSK/MFSK?

256QAM

62

Page 63: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Topics to be Covered

Binary digital modulation

M-ary digital modulation Comparison study

Source A/D converter

Source encoder

Channelencoder Modulator

Channel

DetectorChannel decoder

Source decoder

D/A converter

User

Absent if source is

digitalNoise

63

Page 64: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Comparison of M-ary Modulation Techniques

Channel bandwidth and transmit power are two primary communication resources and have to be used as efficient as possible Power utilization efficiency (energy efficiency): measured by

the required Eb/No to achieve a certain bit error probability Spectrum utilization efficiency (bandwidth efficiency):

measured by the achievable data rate per unit bandwidth Rb/B

It is always desired to maximize bandwidth efficiency at a minimal required Eb/No

64

Page 65: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Example Suppose you are a system engineer in Huawei/ZTE, designing a part of the

communication systems. You are required to design a modulation scheme for three systems using MFSK, MPSK or MQAM only. State the modulation level M to be low, medium or high

An ultra-wideband system

Large amount of bandwidth

Band overlays with other systems

Purpose: high data rate

A wireless remote control system

Use unlicensed band

Purpose: control devices remotely

A fixed wireless system

Use licensed band Transmitter and

receiver fixed with power supply

Voice and data connections in rural areas

65

Page 66: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Energy Efficiency Comparison

MPSKMFSK

Picture from: Proakis J G, Salehi M. Fundamentals of communication systems[M]. Pearson Education India, 2007.

66

Page 67: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Energy Efficiency Comparison (cont’d)

MFSK: At fixed Eb/No, increase M can provide an improvement on Pb

At fixed Pb increase M can provide a reduction in the Eb/Norequirement

MPSK BPSK and QPSK have the same energy efficiency At fixed Eb/No, increase M degrades Pb At fixed Pb, increase M increases the Eb/No requirement

MFSK is more energy efficient than MPSK

67

Page 68: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Bandwidth Efficiency Comparison To compare bandwidth efficiency, we need to know the

power spectral density (power spectra) of a given modulation scheme

MPSK/MQAM

If is rectangular, the bandwidth of mainlope is If it has a raised cosine spectrum, the bandwidth is

Spectrum shaping filter

Input data

Signal point mapper

Spectrum shaping filter

+MPSK/MQAM

signal

Page 69: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Bandwidth Efficiency Comparison (cont’d)

In general, bandwidth required to pass MPSK/MQAM signal is approximately given by

But

Then bandwidth efficiency may be expressed as

= bit rate

(bits/sec/Hz)

69

Page 70: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

MFSK: Bandwidth required to transmit MFSK signal is

Bandwidth efficiency of MFSK signal

Bandwidth Efficiency Comparison (cont’d)

(Adjacent frequencies need to be separated by 1/2T to maintain orthogonality)

(bits/s/Hz)

M 2 4 8 16 32 64

1 1 0.75 0.5 0.3125 0.1875(bits/s/Hz)

ρ

As M increases, bandwidth efficiency of MPSK/MQAM increases, but bandwidth efficiency of MFSK decreases.

70

Page 71: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Fundamental Tradeoff :Bandwidth Efficiency and Energy Efficiency

To see the ultimate power-bandwidth tradeoff, we need to use Shannon’s channel capacity theorem: Channel Capacity is the theoretical upper bound for the maximum

rate at which information could be transmitted without error (Shannon 1948)

For a bandlimited channel corrupted by AWGN, the maximum rate achievable is given by

Note that

Thus

)1(log)1(log0

22 BNPBSNRBCR s+=+=≤

RBSNR

BRNBP

RNP

NTP

NE sssb ====

0000

)12( /

0

−= BRb

RB

NE

71

Page 72: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Power-Bandwidth Tradeoff

Capacity boundary with R = CUnachievable

Region with R > C

Shannon limit

Picture from: Proakis J G, Salehi M. Fundamentals of communication systems[M]. Pearson Education India, 2007.

72

Page 73: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Notes on the Fundamental Tradeoff In the limits as goes to 0, we get

This value is called the Shannon Limit

Received must be >-1.6dB to ensure reliable communications

BPSK and QPSK require the same of 9.6 dB to achieve Pe=10-5. However, QPSK has a better bandwidth efficiency

MQAM is superior to MPSK MPSK/MQAM increases bandwidth efficiency at the cost of lower

energy efficiency MFSK trades energy efficiency at reduced bandwidth efficiency.

73

Page 74: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

System Design TradeoffWhich Modulation to Use ?

Power Limited Systems:

Power scarceBandwidth available

Bandwidth Limited Systems:

Bandwidth scarcePower available

74

Page 75: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Example Suppose you are a system engineer in Huawei/ZTE, designing a part of the

communication systems. You are required to design a modulation scheme for three systems using MFSK, MPSK or MQAM only. State the modulation level M to be low, medium or high

An ultra-wideband system

Large amount of bandwidth

Band overlays with other systems

Purpose: high data rate

A wireless remote control system

Use unlicensed band

Purpose: control devices remotely

A fixed wireless system

Use licensed band Transmitter and

receiver fixed with power supply

Voice and data connections in rural areas

75

Page 76: Principles of Communications - SJTUdE M π − = −= − ss min 2 d E E2 1 cos 2 sin ss MM ππ = −= min 0 /2 2 2 2 sin MPSK /2 s d P Q Q E N M π ≈= 46 Meixia Tao @ SJTU Exercise

Meixia Tao @ SJTU

Practical Applications BPSK:

WLAN IEEE802.11b (1 Mbps) QPSK:

WLAN IEEE802.11b (2 Mbps, 5.5 Mbps, 11 Mbps) 3G WDMA DVB-T (with OFDM)

QAM Telephone modem (16QAM) Downstream of Cable modem (64QAM, 256QAM) WLAN IEEE802.11a/g (16QAM for 24Mbps, 36Mbps; 64QAM for 38Mbps

and 54 Mbps) LTE Cellular Systems

FSK: Cordless telephone Paging system

76


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