24 Apr'06CS3282 Sectn 81 University of Manchester CS3282: Digital Communications Section 8: Carrier Modulated Transmission Convert binary data into form.

24 Apr'06 CS3282 Sectn 8 1

University of ManchesterCS3282: Digital Communications

Section 8: Carrier Modulated Transmission

•Convert binary data into form suited to channel characteristics; i.e. usable frequency band, gain & phase distortion within usable band anticipated noise characteristics frequency (e.g. Doppler) shifts

ChannelTrans Rec10110 10111

24 Apr'06 CS3282 Sectn 8 2

Band-pass modulation

• Up to now, we have assumed a “base-band” channel.

• Frequency range from zero to B Hz.

• Suitably shaped ‘pulses’ are symbols.

• Need transmission over channels which are not base-band:

e.g. channel of bandwidth 200 kHz centred on 900 MHz.

• Requires carrier modulated digital modulation.

• Approaches for base-band may be adapted to carrier modulated.

• Based on modulation techniques as used in radio.

24 Apr'06 CS3282 Sectn 8 3

8.1.1 Modulation of sine-wave carriers

• Pure sine-wave exists at just 1 frequency.• Infinitessimally narrow bandwidth• Some aspect varied in sympathy with baseband

e.g. amplitude or frequency• Detectable at receiver• Spreads energy about the nominal frequency.• No longer infinitessimally narrow bandwidth

24 Apr'06 CS3282 Sectn 8 4

8.1.2 Spread-spectrum modulation

• Use pseudo-random signal as carrier

• Wide bandwidth.

• Intended receiver knows the pseudo-random sequence.

• Has ‘matched filter’ tuned to it.

• To other receivers the pseudo-random carrier is just noise.

• Increases their bit-error rate a little.

• More users allowed until accumulated noise gets too much.

• Known as DS-SSMA & CDMA.

24 Apr'06 CS3282 Sectn 8 5

8.1.3 Multi-carrier modulation

• Use set of sub-carriers instead of 1 carrier• Currently sinusoidal• Good for frequency selective fading in radio• OFDM• Used for DTV, DAB, WLAN, ADSL• 64, 1024 or more sub-carriers• OFDM based on FFT

24 Apr'06 CS3282 Sectn 8 6

8.2. Modulation

• 8.2.1 Introduction to ‘am’ and ‘fm’

• Most well known modulation techniques are ‘am’ and ‘fm’ as used for radio & TV.

• For ‘am’, multiply sine-wave by baseband signal..

• For ‘fm’ cause frequency to be modified by baseband.

• Baseband may be speech, music, or just a sine wave.

• With digital, baseband will be pulse sequence.

24 Apr'06 CS3282 Sectn 8 7

24 Apr'06 CS3282 Sectn 8 8

Frequency modulation (fm) by sine-wave

Modulatefrequency

t

volts

t

24 Apr'06 CS3282 Sectn 8 9

Effect of modulation on frequency spectrum

carrier

frequency

Power spectral density

24 Apr'06 CS3282 Sectn 8 10

carrier * message

A cos(Ct) * cos(Mt)

= 0.5A cos(Ct + Mt) + 0.5A cos(Ct - Mt)

= 0.5A cos( (C + M) t ) + 0.5 A cos( (C - M) t )

upper sideband lower sideband

C = 2fC , etc

Where do we get ‘side-bands’ from?

24 Apr'06 CS3282 Sectn 8 11

Amplitude modulation

• Amplitude of sinewave can’t be ve.• Make bb purely +ve by adding constant.• Always done with broadcast ‘am’ radio stations

• Instead of cos(Mt) use [1 + cos(Mt)]

A cos(C t ) 0.5A [cos( (C + M) t ) + cos((C - M)t) ]

• Large carrier DSP ‘am’. • More easily demodulated (by envelope detector)

24 Apr'06 CS3282 Sectn 8 12

Large carrier DSP ‘am’ modulator

•

V

t

V

Multiply

V

t

t

1+cos(Mt)

24 Apr'06 CS3282 Sectn 8 13

‘Envelope detector’ for LC-DSB ‘am’

RectifyLow-pass filter

t

V

t

V

V

t

24 Apr'06 CS3282 Sectn 8 14

Coherent demodulation

• Envelope detection is ‘non-coherent’.• ‘Coherent’ demod needs local carrier at receiver.• Exact in freq & phase. • Derived from received signal.

24 Apr'06 CS3282 Sectn 8 15

Coherent demodulation of ‘am’V

t

V

Mult

V

t

t1+cos(Mt)

Lowpass filter

Local carrier

Received signal

Derivelocalcarrier

24 Apr'06 CS3282 Sectn 8 16

Proof that coherent demodulation works

• Let received signal be A cos(Ct) .(1+cos(Mt) )

• Multiplying by local carrier gives

A cos2 (Ct) . ( 1+cos(Mt) )

= 0.5A(1 + cos(2Ct)) .(1 + cos(Mt) )

= 0.5A(1+cos(Mt)) + 0.5A cos(2Ct)(1+cos(Mt) )

Low-pass filter removes this

24 Apr'06 CS3282 Sectn 8 17

Coherent demodulation again

• No longer requires modulating signal to be purely +ve

• Works with cos(Mt) just as well as with 1+cos(Mt)

• No longer ‘large carrier & envelope detectn no good.

• When cos(Mt) becomes ve, carrier amplitude remains +ve, but phase changes by 180o

• With digital, modulating signal no longer sinewaves or music

24 Apr'06 CS3282 Sectn 8 18

8.2.2 Vector modulator & complex baseband

• Independently modulate cos(2fCt) & sin(2fCt) and sum.

• Coherent demodulatr for ‘cos’ transmission blind to ‘sin’ trans.

• And vice-versa.

Mult

Mult

ADD

Cos(2fCt)

Sin(2fCt)

bR(t)

bI(t)

• “2 channels for price of 1”

• Still single carrier

• Complex baseband:

b(t) = bI(t) + jbR(t)

• More about this later

24 Apr'06 CS3282 Sectn 8 19

Vector demodulator

Mult

Mult

Cos(2fCt)

Sin(2fCt)

bR(t)

bI(t)

Derive local carrier(cos & sin)

Lowpassfilter

Lowpassfilter

bR(t)cos(2fCt) + bI(t)sin(2fCt)

24 Apr'06 CS3282 Sectn 8 20

8.2.3 Modulation for digital transmission

• Generate base-band symbols from bit-stream (map to b_b)• Use these symbols to modulate ‘carrier’.• Modulation shifts b_b symbols up in frequency to transmission band of channel. • Various forms of modulation may be used,

e.g. amplitude modulation (“am”) frequency modulation (“fm”).

• Doubles bandwidth of base-band signal.

24 Apr'06 CS3282 Sectn 8 21

Mapping bit stream to base-band

Pulse-shaping filter

..1 1 0 1 0 ...Generate impulses

t

V V

t

‘Map to base-band’

• Stream of impulses produced according to bits & approach

e.g. for unipolar: unit impulse for ‘1’ & zero for ‘0’.

• Pass impulse stream through pulse shaping filter.

• Impulses & filter may be analogue or digital (generally digital)

24 Apr'06 CS3282 Sectn 8 22

Techniques for digital transmission

• Can modulate amplitude, frequency &/or phase of cos(2Ct).

• These 3 forms of modulation when used independently give us

(a) amplitude shift keying (ASK) (b) frequency shift keying (FSK)(c) phase shift keying (PSK).

• There are many versions of each of these.

• Possible to use a combination of more than one form.

• Consider simplest binary forms first.

24 Apr'06 CS3282 Sectn 8 23

Binary frequency shift keying (B-FSK)

Modulatecarrier

Map to base-band

10110

tvolts

t

24 Apr'06 CS3282 Sectn 8 24

Binary amplitude shift keying (B-ASK)

Map to base-band

10110

tvolts

Multiply

24 Apr'06 CS3282 Sectn 8 25

Binary phase shift keying (B-PSK)

Map to base-band

10110

t

volts

Multiply

t

24 Apr'06 CS3282 Sectn 8 26

4-ary amplitude shift keying (ASK)

Map to base-band

10110

t

volts

Multiply

tvolts

volts

24 Apr'06 CS3282 Sectn 8 27

Combined multi-level ASK & PSK

Map to base-band

10110

t

volts

Multiply

24 Apr'06 CS3282 Sectn 8 28

8.3. Amplitude shift keying

r(t)

cos(2ct)

b(t)

t

b(t)r(t)

t

24 Apr'06 CS3282 Sectn 8 29

8.3.2. Non-coherent detection of ASK • Detection carried out without local carrier locked in frequency &

phase with received carrier. • A possible method is 'envelope detector’. • Diode & resistor produce 'half-wave rectified' voltage waveform

when input voltage is ASK waveform. • Smoothed by low-pass filter (or simple capacitor). • Produces voltage waveform shown on next slide.• Sampled at appropriate points in time to recover the bit-stream.

24 Apr'06 CS3282 Sectn 8 30

Coherent demodulation of ASK

10110

tvolts

Multiply Threshold detector

Lowpass

24 Apr'06 CS3282 Sectn 8 31

Non-coherent detection of ASK

Rectify & smooth

Thresholddetector

t

10110

24 Apr'06 CS3282 Sectn 8 32

Low-pass filter(smoother)

t t t

V V

SampleDiode

Resistor

Envelope detector for ASK

24 Apr'06 CS3282 Sectn 8 33

8.3.3. Constellation diagrams

Show “in phase” and “quadrature” components as a graph as illustrated below for two examples:

Binary ASK with symbols 0 & Acos(..)

In phase with carrier

Quadrature to carrier

Q

I

4-ary ASK with symbols 0, Acos(..), 2Acos(..), 3Acos(..)

0 A 2A 3A

24 Apr'06 CS3282 Sectn 8 34

8.3.4. Coherent demodulation of ASK

Multiply by local carrier locked in frequency & phase with carrier received.

Lowpassfilter

cos(2ct)

s(t)cos(2ct)

Generate local carrier

Thresholddetector

)4cos()( )(5.0 )(cos)( 2cc ftststfts

Removed by lowpass filtercos2= 2cos2 - 1

24 Apr'06 CS3282 Sectn 8 35

8.3.5. Coherent versus non-coherent detectionLet the signal be: b(t)cos(2ct).Noise is: N(t)cos(2ct + (t)) where N(t) is random envelope & (t) is random phase. This equals:

Half noise power in phase with cos(2ct ) & half with sin(2ct ).Non-coherent detection measures envelope of signal plus noise & is affected by full power of noise. Coherent detection multiplies by cos(2ct ) low-pass filters & thus eliminates half the noise power 3dB reduction in effective noise power as seen by detector. coherent detection tolerates 3dB more noise than non-coherent to achieve same BER.

)2sin()(sin)( )2cos()(cos)( tfttNtfttN cc

24 Apr'06 CS3282 Sectn 8 36

8.4 Complx baseband & vector-modulator/demodulatr

8.4.1 Vector modulator:

..11010.. Map

sin(2fCt)

cos(2fCt)

bI(t)

bR(t)

bR(t)cos(2fCt) +bI(t)sin(2fCt)

Map..10010..

24 Apr'06 CS3282 Sectn 8 37

Complex notation for vector-modulator

• bR(t) is ‘in-phase’ component & bI(t) is ‘quadrature’ component. • Complex base-band signal is bR(t) + jbI(t) where j = (-1). • Output is real part of: [ bR(t) + jbI(t)] . exp(-2jfC t) since [ bR(t) + jbI(t)] . [cos(2fC t) jsin(2fC t) ] = [ bR(t) cos(2fC t) + bI(t)sin(2fC t) ] + j(..)

MultMap

exp(-2 j fCt)

b(t)10110

11011

Complexsignal. Take real part.

Complx base-band

24 Apr'06 CS3282 Sectn 8 38

8.4.2. Vector-demodulator

• Receives bR(t)cos(2fC t) + bI(t)sin(2fC t)

• Recovers bR(t) & bI(t) separately.

• bR(t) & bI(t) may be considered independent channels.

• If each transmits at 1 b/s/Hz, we get 2 b/s per Hz.• “Two channels for price of one”. • Constellation diagrams becomes more interesting:

24 Apr'06 CS3282 Sectn 8 39

Vector demodulator (cont)

Mult

Mult ThresholdDetector

ThresholdDetector

Cos(2fCt)

Sin(2fCt)

bR(t)

bI(t)Lowpass

Lowpass

..11010..

Derive localcarrier(cos & sin)

Receivedsignal r(t)

..10010..

24 Apr'06 CS3282 Sectn 8 40

Show why this works for cosine modulation

Let r(t) = bR(t) cos(2 fC t) + bI(t) sin(2fC t) )

Then r(t) cos(2 fC t)

= bR(t)cos2(2 fC t) + bI(t) sin(2fC t) )cos(2 fC t)

= 0.5 bR(t)[1 + cos(4 fC t)] + 0.5 bI(t) sin(4fC t) )

= 0.5bR(t) + 0.5bR(t) cos(4 fC t) + 0.5 bI(t) sin(4fC t) )

Hence cosine demodulator recovers bR(t) & is blind to bI(t)

Removed by lowpass filter

24 Apr'06 CS3282 Sectn 8 41

Similarly for sine modulation

r(t)sin(2 fC t)

= bR(t) cos(2 fC t)sin(2fC t) + bI(t) sin2(2fC t) )

= 0.5 bR(t) sin(4 fC t) + 0.5 bI(t) [1 - cos(4fC t) ]

= 0.5 bR(t) sin(4 fC t) + 0.5 bI(t) - 0.5bI(t)cos(4fC t)

Removed by lowpass filter

Sine demodulator recovers bI(t) & is blind to bR(t)

24 Apr'06 CS3282 Sectn 8 42

Trig formulae

This works because cos2() & sin2() have a constant (or DC) component 0.5 whereas sin()cos() does not.

Relevant formulae are:

• cos 2 () = 0.5 + 0.5 cos(2)

• sin 2 () = 0.5 - 0.5 cos(2)

• sin() cos () = 0.5sin(2)

24 Apr'06 CS3282 Sectn 8 43

8.4.3. Constellation diags for ASK with complx baseband

In phase with carrier

Quadrature to

carrier

0 A 2A 3A

Binary ASK for bR(t) & bI(t) 4-ary ASK

for bR(t) & bI(t)

In quadrature

Inphas

A

A 3A

A

0

24 Apr'06 CS3282 Sectn 8 44

Symbol allocation tables for binary & 4-ary ASK

Bits bR bI

0 0 0 00 1 0 A1 0 A 01 1 A A

Bits bR bI

0 0 0 0 0 00 0 0 1 0 A0 0 1 0 0 2A0 0 1 1 0 3A0 1 0 0 A 00 1 0 1 A A0 1 1 0 A 2A0 1 1 1 A 3A1 0 0 0 2A 01 0 0 1 2A A.....1 1 1 1 3A 3A

24 Apr'06 CS3282 Sectn 8 45

8.5 Frequency Shift Keying (FSK)

• Can be straightforward form of digital modulation. • Simple to generate and detect,• Constant amplitude, insensitive to fluctuations of channel attenuation. • Based on frequency modulation (fm)• Uses set of distinct frequencies to represent symbols. • Transmit constant amplitude sine-wave whose frequency varies between the frequencies assigned to each symbol. •For binary signalling there are 2 frequencies, 0 & 1 say. •Consider 3 generation methods.

24 Apr'06 CS3282 Sectn 8 46

FM Modulator

(VCO)1

0 0 1 0

Better to have smoothly changing pulse for gradual transition. This is “continuous phase form of FSK i.e. CPFSK.

2. “Switched oscillator” method of generating FSK.

1

0

FSK

1. “Voltage controlled oscillator(VCO)”method.

Clearly this may not produce a continuous phase output.

8.5.1 Methods for generating FSK

24 Apr'06 CS3282 Sectn 8 47

3. “Vector-modulator” method: For binary FSK with c+1 & c-1, apply cos (21t) to ‘Q’ and sin(21t) to ‘I’ . Sign determines the symbol.

“Q” input

“I” input

Sin(2ct)

Cos(2ct)

cos (21t)

sin(21t)

24 Apr'06 CS3282 Sectn 8 48

Exercise 8.1: Check that this works.

Solution:

When I=+sin(21t), output is:sin(21t)cos(2ct)+cos(21t)sin(2ct)

=sin(2(c+1)t)

When I=-sin(2f1t) the output is:-sin(21t)cos(2ct)+cos(21t)sin(2ct)

=sin(2(c–1)t)

24 Apr'06 CS3282 Sectn 8 49

8.5.2. Non-coherent detection of FSK at receiver (low bit-rates)

Consider 3 methods

1. Set of band-pass filters with envelope-detectors;

BPF (f0)

BPF (f1)

Decide

24 Apr'06 CS3282 Sectn 8 50

Discriminator

Low-pass filter (smoother)

t t t

V

t

f

GainResistor

f1 f0

f1 f0

2. Discriminator followed by envelope-detector.

Turns FSK into ASK for easier detection

24 Apr'06 CS3282 Sectn 8 51

PLL

t

Vt

t

VCO input(Voltage input frequency)

VCO output

Frequency modulated input

3. Phase Locked Loop detector for FSK.

PLL is 'black box' with one input & 2 useful outputs:

24 Apr'06 CS3282 Sectn 8 52

PLL has VCO with frequency adapted to match that of FSK

signal.

VCO controlled by voltage generated by measuring phase

difference between VCO output & incoming FSK signal.

Voltage input frequency & can be used for detecting data bits

Low-pass filter

VCO

VCO input voltage

VCO output voltage

tt

V

V

8.5.3. Phase-locked loop (PLL)

24 Apr'06 CS3282 Sectn 8 53

8.5.4 Non-coherent FSK detector for higher data rates: “Zero crossing counter” type of detector

LimitingAmplifier

Clock

DecideCounter

Reset

DataFSK

and

24 Apr'06 CS3282 Sectn 8 54

8.5.5 Coherent FSK detection:

Similar to coherent ASK detection. Must have local carrier sine-waves at receiver.Must match exactly in frequency & phase the FSK symbols being received. For binary transmission there would be two locally generated sine-waves of frequency 0 and 1 respectively. The incoming signal is multiplied by both sine waves and the two signals which result are low-pass filtered. A comparator then has to decide which frequency 0 or 1 produced the larger output, and that determines the symbol.

24 Apr'06 CS3282 Sectn 8 55

8.5.6 Spectrum of FSK:At 1/T symbols/s, base-band signal has spectrum which is non-zero for –1/T<<1/T if 100% RC spectral shaping is applied Non-zero for –1/(2T)<<1/(2T) with 0% RC spectral shaping. When base-band signal is modulated to form FSK with signalling frequencies 1 & 0,

‘one’s form a DSB spectrum centred on 1 ‘zero’s form a DSB spectrum centred on 0.

Resulting spectrum is sum of these two spectra.PSD

0-1/T 0 0+1/T

PSD

1-1/T 1 1+1/T

24 Apr'06 CS3282 Sectn 8 56

PSD

0-1/T 0 0+1/T

PSD

1-1/T 1 1+1/T

PSD

0-1/T 0 1

1+1/T

+=

24 Apr'06 CS3282 Sectn 8 57

Sunde’s FSK method

Place 0 at 11/T & 1 at o1/T.

24 Apr'06 CS3282 Sectn 8 58

8.5.7. Minimum shift keying (MSK)•Form of FSK where difference between 0 & 1 is 1/(2T) Hz. •Makes MSK very efficient in its spectral utilisation.•Price is increased complexity in generation & detection process. •Non-coherent detection is difficult for MSK. •The detection is recommended to be coherent (Sklar p152).

Pulse-shaping filter: e.g. 100r % RRC, controls FSK spectrum. •Placed just before the FSK modulator.•Controls how frequency changes from 0 to 1 and vice-versa. •In GSM phone systems the shaping is root-Gaussian filter. •This form of binary FSK is known as “Gaussian MSK”.

24 Apr'06 CS3282 Sectn 8 59

FIR Gaussian shaping filter

VCO

Map to impulses

..10110

..

GMSK

GMSK transmitter

24 Apr'06 CS3282 Sectn 8 60

Gaussian minimum shift keying (GMSK)

• Spectrally efficient form of binary FSK with ‘Gaussian’ pulse shaping.

• 2 bits/s /Hz

• Spectrum similar to ASK

• Used for GSM

24 Apr'06 CS3282 Sectn 8 61

8.5.8. Advantages & disadvantages of FSK Advantages:

1. Constant envelope hence not too sensitive to varying attenuation on the channel.

2. Detection based on frequency changes, so not very sensitive to frequency shifts of channel, (Doppler shifts etc).

3. Simple implementations possible for low bit-rates.

Disadvantages of FSK:

1. Less bandwidth efficient than ASK or PSK (except MSK)

2. Bit-error rate performance in AWGN worse than PSK.

24 Apr'06 CS3282 Sectn 8 62

8.6. Phase shift keying (PSK)

cos(2ct)

t

cos(2ct)

b(t)

Map

..1010010..

• Send sinusoidal carrier with phase changes determined by bits

• Consider binary PSK with 1 bit/cycle, 00 & 1800 phase shifts & rectangular pulse shaping

24 Apr'06 CS3282 Sectn 8 63

A binary PSK waveform

t

V

1 1 0 0 1 1 0

Assuming 1 bit per cycle.

24 Apr'06 CS3282 Sectn 8 64

8.6.2 Coherent Detector for binary PSK

Lowpassfilter

cos(2C t)

cos2(2ct) = 0.5(1+cos4ct)

ThresholdDetector

Data+1/2:”1”-1/2:”0”

1/2

cos(2Ct)

Generatelocal carrier

24 Apr'06 CS3282 Sectn 8 65

•Low-pass filter eliminates cos(4C t). •Matched filter will achieve this because of orthogonality of cos(4ct) to sin(2ct). •Local carrier must be generated from received signal. (Square incoming signal & divide frequency of result by 2).•Spectrum of PSK similar to that of ASK.•PSK multiplies carrier by bipolar base-band: ASK by unipolar. •Shifts up base-band spectrum producing DSB spectrum centred on carrier frequency.

Details of coherent PSK demodulator/detector

24 Apr'06 CS3282 Sectn 8 66

900 & 2700 phase shifts often preferred with binary DPSK:

t

V

1 1 0 1 1 0Discontinuities tell receiver when next symbol starts.Makes bit-synchronisation easier when symbol rate not fully synchronised with carrier (not exact no. of cycles/bit)..

1 bit/cycle

24 Apr'06 CS3282 Sectn 8 67

8.6.4 Differential detection of binary DPSK•Consider case where phase shifts are 00 & 1800 & there is an integer number (e.g. 1) of cycles per bit.•Instead of generating local carrier, take previous symbol delayed as required carrier segment. •Small penalty compared with a fully coherent technique.

Lowpassfilter

cos(2Ct)

cos2(2ct) = 0.5(1+cos4ct)

Thresholddetector

Delay by T(Delay for 1 bit)

0.5

24 Apr'06 CS3282 Sectn 8 68

Lowpass filter output is +0.5 if carrier has been subject to 00 phase shift (logic 1 say) and –1/2 for 1800 (logic ‘0’).

Channel noise affects both data & delayed data used as carrier.

Was used for modem data over telephone lines, 1200 b/s being possible over worst case lines.

Increased to 2400bits/s using quaternary PSK (QPSK).

24 Apr'06 CS3282 Sectn 8 69

8.6.5 Detector for binary DSPK with 90O & 270O phase shifts rather than 0 and 180O.

LPF Detect

Delay by T(Delay for 1

bit)

900 phase shift

24 Apr'06 CS3282 Sectn 8 70

8.6.6 Quaternary PSK (QPSK)

• Consider a vector modulator where bR(t) & bI(t) are bipolar

• Then bR(t)cos(2fCt) & bI(t) sin(2fCt) are both binary PSK.

• ‘2-channel’ modulation process is QPSK or 4-PSK.

Mult

Mult

ADD

Map

Map

Cos(2fCt)

Sin(2fCt)

bR(t)

bI(t)10110

11011

24 Apr'06 CS3282 Sectn 8 71

QPSK de-modulator

Mult

Mult Detect

Detect

Cos(2fCt)

Sin(2fCt)

bR(t)

bI(t) 10110

11011

Lowpass

Lowpass

Detectcarrier

24 Apr'06 CS3282 Sectn 8 72

Two ways of looking at QPSK

• One way is ‘vector modulation’ approach where cos(2fCt) & sin(2fCt) are binary PSK modulated independently.

• At receiver, coherent PSK detector for cos(2fCt) channel is blind to transmission on sin(2fCt) & vice-versa.• Refer to bR(t) + j bI(t) as 'complex base-band' signal b(t). • Transmitted QPSK signal is Re{ [bR(t) +j bI(t)] exp(-j2fCt) }.

MultMap

exp(-2j fCt)

b(t)10110

11011

Transmit real partComplx

base-band

24 Apr'06 CS3282 Sectn 8 73

Another way to look at QPSK

• QPSK sends 2 bits at once , using bipolar bR(t) & bI(t)

• Let bR(t) & bI(t) be rect pulses of amplitude -A or +A.

• Mapping to base-band may then be as follows (C=2fC)

Bit1 bit2 bR(t) bI(t) QPSK symbol transmitted

0 0 A A Acos(Ct) A sin(Ct) = Acos(Ct1350)

0 1 A +A Acos(Ct) + A sin(Ct) = Acos(Ct+1350)

1 0 +A A Acos(Ct) A sin(Ct) = Acos(Ct 450)

1 1 +A +A Acos(Ct) + A sin(Ct) = Acos(Ct +450)

Looking at a constellation diag for this mapping makes it clear why Acos(Ct) + A sin(Ct) = Acos(Ct +450) etc.

24 Apr'06 CS3282 Sectn 8 74

Constellation diagram for 45o, 135o QPSK

In phase with cos

(real pt)

1,1

1,0

0,1

0,0

45o

V

V

-V

In quadrature with cosSymbol allocation table:

Bit1 bit2 bR(t) bI(t) 0 0 A A 0 1 A +A 1 0 +A A 1 1 +A +A

24 Apr'06 CS3282 Sectn 8 75

Real

Alternative constellation diag ( 0o,90,180,270o QPSK)

Symbol allocation table:

Bit1 bit2 bR(t) bI(t)

0 0 A 0

0 1 0 +A

1 0 -A 0

1 1 -A -A

Imag pt

0,1

1,1

0,01,0

24 Apr'06 CS3282 Sectn 8 76

ReReal pt

8-PSK16-PSK

Imag pt

QPSK is 4-PSK. What about 8-PSK & 16-PSK?

Can have 8-PSK (3 bits/symbol) & 16-PSK (4 bits/symbol). Constellation diagrams for shown below.

Differential forms of QPSK & M-PSK often used where changes in phase signify the data. Principle similar to DPSK .

24 Apr'06 CS3282 Sectn 8 77

Exercise 8.6: Consider how symbols for 8-PSK & 16-PSK may be associated with sequences of 3 or 4 bits, i.e. label the constellation diagrams. Use a form of 'Gray coding'.

000

001011

010

110

111

101

100With Gray coding, a symbol error generally causes just one bit-error

24 Apr'06 CS3282 Sectn 8 78

Exercise 8.6 (cont): What happens if we don’t use Gray coding?

000

001010

011

100

101

110

111If symbol 111 mistaken for 000 get 3 bit-errors

24 Apr'06 CS3282 Sectn 8 79

Advantage of Gray coding

•With Gray coding of multi-level symbols,

bit-error rate may be assumed to be:

symbol-error rate no. of bits/symbol

except when the noise is exceptionally high.

(We assume a symbol error just takes us to a nearby symbol which differs in just one bit with Gray coding)

• Repeat the labeling now for 16-PSK.

24 Apr'06 CS3282 Sectn 8 80

Exercise 8.7: Show how a vector-modulator may be used to generate the 8 or 16 symbols of 8-PSK & 16-PSK.

000

001011

010

110

111

101

100

Symbol bR(t) bI(t)

000 V 0

001 V/1.4 V/1.4

010 -V/1.4 V/1.4

011 0 V

100 V/1.4 -V/1.4

101 0 -V

110 -V 0

111 -V/1.4 -V/1.4

V

24 Apr'06 CS3282 Sectn 8 81

Example 8.7 (cont) How would you detect 8-PSK with a vector demodulator & threshold detectors?

Exercise 8.8: If radius of constellation diagram circle is V volts for QPSK, 8-PSK & 16-PSK calculate energy per bit for each of these schemes assuming rectangular pulses.

Take 'noise immunity' as distance between each symbol on constellation diagram & nearest one to it, Estimate noise immunity for QPSK, 8-PSK & 16-PSK when radius is V in each case.

24 Apr'06 CS3282 Sectn 8 82

Exercise 8.9:

How will pulse-shaping be applied to QPSK, 8-PSK and 16-PSK?

With 100% RRC pulse shaping & symbol duration T, what is band-with efficiency (in b/s / Hz) for each of these techniques.

What is theoretical maximum bandwidth efficiency in each case?

24 Apr'06 CS3282 Sectn 8 83

Single carrier digital modulation schemes

•ASK, FSK, PSK, DPSK, QPSK

•Differential QPSK

•Gaussian FSK & MSK

•Combined ASK & PSK (QAM, APK)

•etc.

24 Apr'06 CS3282 Sectn 8 84

Other modulation techniques

•Direct sequence spread spectrum techniques (DSSS)

•Frequency hopping (FHSS)

•Complementary code keying (CCK)

24 Apr'06 CS3282 Sectn 8 85

8.7. Introduction to multi-carrier modulation & OFDM

• Introduces concept of multi-carrier modulation

• Compares with single carrier modulation to determine some advantages & disadvantages.

• Orthogonal frequency division multiplexing (OFDM) introduced as highly efficient form of multi-carrier modulation widely used in broadcasting, ADSL & wireless LANs.

• Implementation of OFDM using FFT & inverse FFT.

• Parameters of 802.11 OFDM implementation investigated.

• First, revise some important aspects of single carrier modulation.

24 Apr'06 CS3282 Sectn 8 86

8.8 Matched filtering & equalization for single carrier

• ‘Map to base-band’ at transmitter has ‘pulse shaping filter’.

• Generates sinc-like pulses of correct amplitude & polarity at right time.

• Pulse added to previous pulses & modulated onto carrier.

• Diagram below illustrates generation & modulation of a single pulse

• With ASK, ‘sinc like’ pulse shape becomes ‘envelope’.

Pulseshapingfilter

ExcitePulse-s filter

b(t)..11101..

Multiply

tVolts

t

volts

Map to base-band

volts

t

24 Apr'06 CS3282 Sectn 8 87

sinc(x/T)(x)sinc 0 : 1

0 : x)(x)sin(

sinc(x)

T

x

x

T-3T

sincT(t)

t

1

-T

2T-2T

3T

4T-4T

Formulae for sinc(x) & sincT(x)

24 Apr'06 CS3282 Sectn 8 88

Single carrier PSK with pulse shaping

Pulseshapingfilter

ExcitePulse-s filter

b(t)..11101..

Multiply

tVolts

t

volts

Map to base-band

Volts

t

envelope

24 Apr'06 CS3282 Sectn 8 89

Output of transmitter with two PSK pulses

Volts

t

24 Apr'06 CS3282 Sectn 8 90

‘Single carrier’ receiver• Receiver must demodulate to obtain base-band b(t) .

• Pulse shapes distorted & affected by noise.

• Sample & detect for rectangular pulses discussed in last lecture.

• May work for low bit-rates over channels with little distortion or noise

• Performance can be improved by introduction of

– a matched filter optimally tuned to shape of transmitted pulses to minimise effect of noise (AWGN).

– a channel equaliser to cancel out distortion introduced by channel.

..1100..Matchedfilter

Demodulator Channelequaliser

Sample & detectb(t)

Channelsignal + AWGN

24 Apr'06 CS3282 Sectn 8 91

Matched filter & RRC pulses

• Matched filter & channel equaliser may have complex input signals.

• Multi-level pulses may be used instead of binary.

• Pulse shapes seen at input to ‘sample & detect’ block be Nyquist; i.e. centre of each pulse must coincide with zero-crossings of all others.

• e.g. R% ‘raised cosine’.

• Matched filter multiplies received pulse shape by a copy of itself.

• So transmitter must now send root raised cosine (RRC) pulses.

• Look very similar & ‘sinc-like’.

• Transmitted pulse is ‘squared’ by matched filter in receiver.

• If transmitter sent RC pulses, detector would see squared RC pulses.

• These would not have zero-crossings in the right places.

24 Apr'06 CS3282 Sectn 8 92

Channel equaliser

• Channel equaliser’ is an ‘adaptive filter’

• Programmed to correct any differences between pulses seen at output of matched filter & ideal RC pulses required by detector.

• Aims to cancel out effect of the channel,

• In particular the effects of frequency selective fading.

• Received amplitude reduced at some frequencies & reinforced at others.

• Equalizer must do opposite of this.

• Must adapt to changes in fading channel characteristics.

• A demanding filtering task, and it cannot always be successful.

• If there is a very deep fade, it will just not be possible to reverse it.

• Trying to do so will just emphasize noise at frequency of deep fade.

• Single carrier sine-wave modulation still widely used.

24 Apr'06 CS3282 Sectn 8 93

8.9. Spread spectrum modulation

• Use of a single sine-wave as a carrier is not the only possible choice.

• Could use a ‘pseudo-random’ carrier known at transmitter & receiver.

• Bandwidth much wider than that of a modulated sine-wave.

• This may appear very wasteful of bandwidth.

• It will appear as noise to receivers not tuned to its exact characteristics.

• Transmission is ‘coded’ by pseudo-random carrier & security is a bonus.

• Transmitter-receivers using different pseudo-random carriers can co-exist.

• This is direct sequence spread spectrum multiplexed access (DS-SSMA)

• Also referred to as ‘code division multiplexed access’ (CDMA).

• Basis of most 2G mobile phone systems in the USA.

• 3G mobile telephony will be based on enhanced form of CDMA.

24 Apr'06 CS3282 Sectn 8 94

8.10. Multi-carrier modulation

• Assume we have 20 MHz radio channel centred on 2.46 GHz.

• Could apply single carrier modulation to a sine-wave carrier at 2.46 GHz.

• With QPSK, max achievable bandwidth efficiency is 2 b/s per Hz

• Allows 40 Mb/s to be transmitted with 0% RRC pulses (pure sinc).

• 50% RRC pulses would reduce bandwidth efficiency to 1.33 bits/s per Hz.

• Only 26.7 Mbits/s now possible, but generating the pulses is much easier.

• In both cases, whole 20 MHz used by the single carrier modulated signal.

24 Apr'06 CS3282 Sectn 8 95

Alternative to single carrier modulation

• An alternative is to divide the 20 MHz band into sub-bands with a sinusoidal ‘sub-carrier’ in centre of each band.

• Instead of one carrier we now have many ‘sub-carriers’.

• IEEE802.11 divides 20 MHz into 64 sub-bands each of 312.5 kHz.

• Now 64 sub-carriers at frequencies F+f0, F+f1, …, F+f63 Hz.

• F is lowest frequency of the 20MHz band

• f0 = 156.25 Hz, f1 = 468.75 Hz, …, f63 = 19843.75 Hz.

• Modulating each sub-carrier with QPSK with 0% RRC pulse shaping would achieve 625 kb/s per sub-band.

• Total bit-rate = 625 x 64 = 40 Mb/s (same as with single carrier)

• But now the bits are divided into 64 parallel sub-streams.

• Bit-rate of each sub-stream is 1/64 of the total.

• This is multi-carrier modulation.

24 Apr'06 CS3282 Sectn 8 96

Pulse shaping again

• To see the main advantage of multi-carrier modulation, look again at the demands of pulse shaping

• For single carrier, it is necessary to have a band-limited spectrum.

• Use ‘sinc-like’ pulses with zero-crossings at t=T, 2T, etc.

• Pure sinc pulse has rectangular & strictly band-limited spectrum.

• Rectangular pulse of duration T would have a ‘sinc-like’ frequency spectrum with zero-crossings at f =1/T, 2/T, 3/T, etc.

• Unsuitable for single-carrier modulation.

• But (as we shall see) may be suitable for multi-carrier.

• Study the graphs on the next slide.

24 Apr'06 CS3282 Sectn 8 97

Spectra of rect & sinc pulsesT.sinc1/T(f)

t

T/2-T/2

1

rectT(t)

f

T

1/T-1/T

2/T-2/T

3/T-3/T

4/T-4/T

Fourier transform

Real part shownImag part = 0

f

1/(2T)-1/(2T)

T

T.rect1/T(f)sincT(t)

t

1

T-T

2T-2T

3T-3T

4T-4T

Fourier transform

Real pt shownImag pt = 0

24 Apr'06 CS3282 Sectn 8 98

Spectra of 50% RC pulses & spectraRC(f)

tT/2-T/2

1

rc(t)

f

T

1/T-1/T

2/T-2/T

3/T-3/T

4/T-4/T

Fourier transform

Real part shownImag part = 0

-3T/43T/4

f

1/(2T)-1/(2T)

T

RC(f)rc(t)

t

1

T-T

2T-2T

3T-3T

4T-4T

Fourier transform

Real pt shownImag pt = 0

3/(4T)

-3/(4T)

24 Apr'06 CS3282 Sectn 8 99

Sub-band spectral interference (ICI)

• With single carrier, R% RC (or RRC) pulses are used at expense of decreasing band-width efficiency.

• With multi-carrier, pulse shapes close to rectangular may be used.

• Their spectra are ‘sinc-like’ & of very wide bandwidth.

• With 64 adjacent sub-bands, there is clearly a danger of inter spectrum interference, or ‘inter-sub-carrier interference (ICI).

• Also a danger of spectrum leaking outside the 20 MHz band.

• Both these dangers may be avoided.

24 Apr'06 CS3282 Sectn 8 100

Eliminating ICI by OFDM

• Rectangular pulses may be used if peak of spectrum for each sub-band corresponds to zero crossings for all other modulated sub-carriers.

• Interference avoided in frequency-domain rather than time-domain.

• Looking at previous graphs, ICI is avoided if adjacent sub-carriers are spaced exactly 1/T Hz apart when sub-band bit-rate is 1/T b/s.

• This is orthogonal frequency division multiplexing (OFDM)

• Highly efficient because sub-carriers are as close together as they can possibly be without introducing spectral interference.

• Each modulated sub-carrier is ‘orthogonal’ to all others which means that they do not interfere with each other.

24 Apr'06 CS3282 Sectn 8 101

t

T/2-T/2

1

rectT(t)

Modulate F

tT/2-T/2

1

rectT(t)

Modulate F+2/T

tT/2-T/2

1

rectT(t)Modulate F+1/T

T.sinc1/T(..)

f

T

F+2/T

F

T.sinc1/T(f-F)

f

T

F+1/T

F

T.sinc1/T(..)

f

F+2/T

F+1/T F+3/T

Assume purely real spectrum

Combining OFDM sub-bands

24 Apr'06 CS3282 Sectn 8 102

OFDM spectrum

Fourier transformSUM

f

1/T 3/T

Combinerealspectra

Assume purely real spectra

24 Apr'06 CS3282 Sectn 8 103

Use of sub-carriers

• Bit-rate (1/T) for each sub-channel is 1/64 times total bit-rate • Zero-crossings of sinc spectra (at 1/T 2/T, ..) much closer together. • So the sinc spectra ‘die away’ must faster. • Ones in centre of 20 MHz band die away almost completely at edges. • Ones near edges not modulated. • Out of 64 sub-carriers, do not modulate first six, last five & no. 32. • Four other sub-carriers reserved as ‘pilots’, • Leaves 48 sub-carriers that can be modulated with data. • In IEEE802.11 standard, sub-carriers 0 & 27 to 37 not modulated

& 4 others are designated as pilots. • Again this leaves 48 sub-carriers for data. • Depending on processing, the 2 approaches are probably the same.

24 Apr'06 CS3282 Sectn 8 104

Modulation of sub-carriers

• With IEEE802.11, each OFDM sub-carrier modulated by choice of:– binary-PSK, (1 bit per pulse)

– QPSK, (2 bits per pulse)

– 16-QAM (4 bits per pulse)

– 64-QAM (6 bits per pulse)

• 16-QAM & 64-QAM are multi-level schemes.

• Implement by vector-modulator according to ‘constellations’.

• Illustrate for QPSK & 16-QAM

• ‘Gray coding’ for 16-QAM makes nearest dots differ in just 1 bit.

• Differential PSK, QPSK & QAM used where the difference between the current & previous pulse specifies the bit pattern.

24 Apr'06 CS3282 Sectn 8 105

Constellation for QPSK

modulating cos

0,0

0,1

1,0

1,1

Bit1 Bit2 bR bI

0 0 A A0 1 A -A1 0 -A A1 1 -A -A

Modulating sin

24 Apr'06 CS3282 Sectn 8 106

‘16_QAM’ constellation

A

3A

-A

-3A

A 3AReal

Imag

(0000)

-A

(0001)

(0010)

(0011)

(0100)(1000)

(1001)

(1010)

(1011)

(1100)

(1101)

(1110)

(1111)

(0110)

(0101)

(0111)

(modulates cos)

(modulates sin)

24 Apr'06 CS3282 Sectn 8 107

Vector-modulator as used for 16-QAM

Mult

Mult

ADDMap

Cos(2fCt)

Sin(2fCt)

3A,-3A,..

-3A,-A,..

1011 1101..

t

V

t

V

Re{..}

24 Apr'06 CS3282 Sectn 8 108

Vector modulator in complex notationTake b(t) + jq(t) as a complex b-b signal.

cos(2fCt).bR(t) + sin(2fCt).bI(t) = real { ( bR(t) + jbI(t) ) exp(-2jfCt) }

MultMap

exp(-2jfCt)

b(t)1011 1101..

Complxbase-band

Take real pt

Sometimes people make this exp(2jfCt).Makes little difference as long as they are consistent.

24 Apr'06 CS3282 Sectn 8 109

Fast Fourier Transform & its inverse

FFT : {x[n]}0,N-1 {X[k]}0,N-1

1 10for 1

0

/2 N-, ...,, k = enxkXN

n

Nknj

Inverse FFT: {X[k]}0,N-1 {x[n]}0,N-1

110for 1

1

0

/2 , ...,N-, k = eX[k]N

nxN

n

Nknj

Both are ‘fast’ in that they can be programmed or implemented in hardware very efficiently especially when N is a power of 2, e.g. 64, 512, 1024

24 Apr'06 CS3282 Sectn 8 110

• Take 64 sub-carrier frequencies over range F to F + 20 MHz: fC + 0, fC + fD, fC + 2fD, … , fC +63fD

with fD = 20MHz / 64 = 312.5 kHz

fC = F + 176.25 kHz

• For orthogonality (correct freq-domain zero crossings) sub-carriers must be 1/T Hz apart.

• So fD = 1/T & pulse duration T = 3.2 x 10-6 s = 3.2 s

• Could transmit 1/(3.2s) = 312.5 k pulses per second, but we don’t.• Extend each pulse to 4 s with a 0.8 s ‘guard-interval’.• Transmit 250 k ‘extended pulses’ per second.• Guard-interval’ extension could be 0.8 s of zero voltage. • But it’s not. Its a ‘cyclic extension’ as we will see later.

8.11 OFDM implementation

24 Apr'06 CS3282 Sectn 8 111

Bandwidth efficiency of IEEE802.11 OFDM

• Theoretical maximum is 1 pulse/s per Hz.

• Using only 48 out of 64 sub-channels loses 25% of total capacity.

• Lose another 20% (=0.8/4) because of guard-interval (cyclic extension)

• Max bandwidth efficiency is 60% (=3/4 x 4/5) of 1 pulse/s per Hz.

= 0.6 x 2 =1.2 b/s per Hz, if QPSK used for all 48 sub-carriers.

• With QPSK, bit-rate in 20 MHz will be 24 Mb/s.

• With 64-QAM, bit-rate achieved is 72 Mb/s.

• Reduced to 36 Mb/s by half rate convolutional coder.

• IEEE specifies ¾ rate ‘punctured coder’ for 64-QAM.

• Gives bit-rate of 72 x3/4 = 54 Mb/s.

• A ¾ rate punctured conv coder is half rate coder with 2 out of every 6 bits erased to reduce bit-rate to 4/3 times the original.

24 Apr'06 CS3282 Sectn 8 112

MultMap

exp(2jfCt)

X0(t)10110..

MultMap

exp(2j(fC+fD)t)

X1(t)11001..

MultMap

exp(2j(fC+63fD)t)

XN-1(t)11001..

Multi-carrier vector-modulation (in principle)

24 Apr'06 CS3282 Sectn 8 113

Multi-carrier modulation in practice:

Stage 1:Apply PSK, QPSK, QAM (or other) to obtain X0(t), X1(t), ..., X63(t)

which remain constant for a ‘pulse (symbol) period’ T. Then vector-modulate complex 'sub-carriers' of frequencies: 0 , fD, 2fD , …, 63fD

Stage 2: Vector-modulate exp(2jfCt) with output from Stage 1

24 Apr'06 CS3282 Sectn 8 114

Map X0(t)10110..

MultMap

exp(2jfDt)

X1(t)11001..

MultMap

exp(2j63fDt)

X63(t)11001..

Stage 1

63

0

2)(m

tjmfm

DetX

x(t)

t

X0(t)

24 Apr'06 CS3282 Sectn 8 115

exp(2jfCt)

63

0

2)(m

tjmfm

DetX

63

0

)(2)(m

tmffjm

DCetX

Stage 2

Complex

multiplication.

= x(t) (complex) (complex but need only

real part)

OFDM

24 Apr'06 CS3282 Sectn 8 116

Stage 1: 63 x(t) = Xm(t) exp (2jmfD t ) with fD = 1/T m=0• Take 64 samples of x(t) pulse of duration T• Let = T/64 & denote x(n) by x[n] for n = 0, 1, ..., 63.• Set Xm(n) =Xm : constant for 0<n<63 63 x(n) = x[n] = Xm exp (2jm n /T ) m=0 63 x[n] = Xm exp(jm(2/64)n) : 0 < n < 63 m=0• Generates a set {x[0], x[1], …, x[63]} of complex numbers.

24 Apr'06 CS3282 Sectn 8 117

Use of inverse FFT to generate x(t)

•Can now take 64 complex numbers {X0, X1, …, X63 } representing one symbol & generate 64 complex samples {x[0], x[1], …, x[63]} of x(t).

63 x[n] = Xm exp(jm(2/N)n) : 0<n<63 m=0

• This is ‘inverse FFT’ formula (apart from a factor 1/64). • Pulse is of duration T = 3.2 s.• It is sampled at T/64 = (1/20) s or 20 MHz (20 x 106 complex samples/second)• Real & imag pts of {x[n]}0,63 could be D to A converted & applied to analogue implementation of Stage 2. • Call {x[n]}0,63 ‘base-band OFDM pulse’• Repeat for next set of {X0, X1, ..., X63} to get another pulse & so on.

24 Apr'06 CS3282 Sectn 8 118

Stage 2• Real part of x(t) multiplies cos(2fCt) & imag part multiplies sin(2fCt). • Real part of output is OFDM symbol starting at fC Hz rather than zero. • More convenient to implement Stage 2 digitally• exp(2jfCt) must be sampled & x(t) ‘up-sampled’ to same sampling rate. • Assume fc = 100 MHz & cos(2fCt) & sin(2fCt) are sampled at 400 MHz. • Must increase sampling rate of x(t) by a factor of 20; i.e. 63 x[n] = Xm exp(jm(2/1280)n) : 0 < n < 1279 m=0 which is more conveniently written as 1279 x[n] = Ym exp(jm(2/1280)n) : 0 < n < 1279 m=0

127964 : 0

630 : X where m

m

mYm

24 Apr'06 CS3282 Sectn 8 119

Implementing Stage 2 digitally• The ‘up-sampling’ is achieved by increasing I-FFT order by factor 20.• Instead of 64 point I-FFT, we need a 1280 point I-FFT. • 1280 is not a power of 2, but there are fast algorithms for such an I-FFT.

• Applying 1280 point I-FFT to {Ym}0,1279 which is a ‘zero-padded’ version

of {Xm}0,63 gives a version of x(t) sampled at 400 MHz.

• Since exp(2jfCt) is also sampled at 400 MHz, we can now implement

‘Stage 2’ digitally by multiplying x(t) by exp(2jfCt) sample by sample.

• Taking the real part of the result we obtain 100 MHz sinusoidal carrier

modulated by a base-band OFDM signal. • The result is sampled at 40 MHz. • Converting to analogue & removing all frequencies above about 130

MHz leaves an analogue version of the required OFDM signal.• Up-sampling x(t) is useful even in analog implentations to simplify DAC.

24 Apr'06 CS3282 Sectn 8 120

• Shape of OFDM symbol conveys the bit-sequence.• With QPSK on 48 carriers, 296 1029 different symbol shapes.• With 16-QAM there are 1060 different pulse shapes

(with single carrier binary PSK there are just two!) • OFDM pulses must be accurately represented & processed by linear circuits.

(with a small number pulses, linearity is not so important)• Highly linear amplifiers (Class A) are very power inefficient.• Amplifiers used in 2G mobile phones (for GMSK - a form of binary FSK) are

not very linear but extremely power efficient.• GMSK is ‘constant envelope’ - OFDM is definitely not!

OFDM symbols (pulse shapes)

24 Apr'06 CS3282 Sectn 8 121

• Each 3.2 s pulse is extended to 4 s by prefixing a 0.8 s ‘guard time’• The prefix is made to be a copy of the final 0.8 us (16 samples) of the pulse.• It is called a ‘cyclic prefix’ or ‘cyclic extension’.• Generate 80 time-domain complex numbers for each ‘extended pulse’• Each extended pulse takes 4 us, so we send 250 k extended pulses/second.

Cyclic extension

Real{x[n]}

n

80 160-80

Similarly for imaginary part.

16

Cyclic prefix

3.2s pulse

Cyclic prefix

3.2s pulse

3.2s pulse

24 Apr'06 CS3282 Sectn 8 122

•Coherent demodulator with sampler, sync & symbol extraction.

•Apply FFT to recover {X0, X1, …, X63}.

•Channel distortion cancelled out by equaliser applied to FFT output.

OFDM receiver

exp(-2jfCt)

Complex multiplication.

OFDM

Derive local carrier

Sample & extract 4s ext- symbol

FFT

Detector

Detector

Detector

20 kHzlowpass filter

Equaliser

24 Apr'06 CS3282 Sectn 8 123

Detectors

• FFT of {x[n]}0,63 gets back to {X0, X1, …., X63}.

• Detect sequence of 1,2, 4 or 6 bits by finding nearest dot on the appropriate constellation diagram.

(B-PSK, QPSK, 60-QAM or 64-QAM),

• ‘Nearest dot’ detector for each complex number generated by FFT is required.

0,0

0,1

1,0

1,1

Re

Illustrate for QPSK

Im

24 Apr'06 CS3282 Sectn 8 124

Cyclic extension as ‘guard interval’

• Eliminates inter-symbol interference between 3.2s OFDM symbols.

• 0.8s is longer than any delay between a direct path & any reflected paths within a building.

• As speed of radio waves 300106 m/s, allows for a path-length difference of 0.8 300 = 250 m.

• Any reflected path up to 250 m longer than direct path will not cause one 3.2s OFDM symbol to interfere with the next.

• Multipath propagation may still distort structure of OFDM symbols.

• Equaliser required to reverse this distortion.

24 Apr'06 CS3282 Sectn 8 125

Cyclic extension for equalisation• A guard interval could be 0.8 s of zero voltage.

• Cyclic extension is more than just a guard interval.

• With the FFT, it greatly simplifies equalisation process.

• Multi-path propagation causes radio channel to act like a ‘filter’.

• Single carrier demodulator employs adaptive filter to cancel it.

• Filtering is computationally intensive.

• Filtering in time-domain becomes multiplication in frequency-domain.

• FFT is part of OFDM demodulator, so equalisation, using multiplication rather than filtering, can be applied to FFT output.

• Difference between ‘cyclic’ filtering with FFT & ‘linear’ filtering.

• Disappears when input to FFT is result of applying cyclically extended signal to channel.

• Cyclic extension to OFDM symbol allows equalisation by ‘cyclic’ filtering by FFT & complex multiplication.

24 Apr'06 CS3282 Sectn 8 126

Cyclic extension for synchronisation

• The cyclic extension is also useful for carrier & symbol synchronisation at the receiver since, if the first 16 samples of an extended pulse are the same as last 16, we are synchronised.

24 Apr'06 CS3282 Sectn 8 127

Exercise: generation of OFDM with 4 sub-carriers

•Given 8-bits, 00011011, show how one OFDM base-band symbol {x[n]} may be generated by a 4-point inverse FFT. Use QPSK to modulate the 4 sub-carriers.

•Extend to 6 samples {x[n]}0,6 by cyclic extension & explain how a high frequency carrier would be modulated by the samples of x.

•Show how original data can be recovered by 4-point FFT.

24 Apr'06 CS3282 Sectn 8 128

Solution:Data is: 00 01 10 11Then X0 =1+j, X1 = 1- j, X2 = -1+j, X3 = -1-j

X = [ 1+j 1-j -1+j -1-j ]; % array of 4 complex numbersPerform 4 point IFFT on X to obtain array xx=ifft(X) % This does it in MATLABArray x now contains the 4 samples of the required symbol: [ 0 0.5 + 0.5j j 0.5 - 0.5j ] Including the cyclic extension, this becomes: [ j 0.5 - 0.5 j 0 0.5 + 0.5j j 0.5 - 0.5j ]

24 Apr'06 CS3282 Sectn 8 129

8.12 Advantages of OFDM

Spectrally efficient because of orthogonality of the 64 carriers. Good for channels affected by frequency selective fading because:

(i) Effects of fading, affecting a small range of frequencies, can be spread out using ‘interleaving’ so that FEC can more easily correct any bit-errors.

(ii) Cyclic extension as a guard-interval, eliminates ISI caused by multi-path propagation. Simpler way of eliminating ISI than pulse-shaping as used in single carrier systems.

(iii) Equalisation is easier than with single carrier systems which use adaptive filtering. OFDM receiver can amplify real & imag parts of FFT outputs such that they have same amplitudes.

Possible because of the cyclic extension as explained earlier.

24 Apr'06 CS3282 Sectn 8 130

Disadvantages of OFDM• ‘Peak to mean’ ratio of symbols can be very large by nature of FFT & Inv-FFT. (Amplitudes can become very large in comparison to the mean)

• Shapes OFDM symbols very complex & must be sent & received accurately.

• With QPSK on each sub-carrier, 1029 shapes & even more with 64-QAM

• Transmitter & receiver must be linear to preserve shape.

• Definitely not "constant envelope".

• Need ‘class A’ amplifiers: less power efficient than those for constant envelope transmissions.

• Lot of power lost in the amplifiers.

• Not ideal for mobile phones, but fine for mobile computers with bigger batteries that are not sending data continuously.

• Sensitive to ‘Doppler’ frequency shifts.

24 Apr'06 CS3282 Sectn 8 131

8.13 Some more details about IEEE 802.11a/g OFDM

• With IEEE802.11a & g, OFDM symbols take 4 s; 250 k symbols/second.• Each symbol can carry 1-6 bits per carrier (BPSK, QPSK, 16- & 64-QAM).• Highest bit-rate with 64-QAM & 3/4 rate conv coder: 48 x 6 x (3/4) x 250 kb/s = 54 Mb/s.• Distances over which this bit-rate achievable will be restricted.• Lower bit-rates (48, 36, 24, 18, 12, 9 and 6 Mb/s) available.• Two lowest bit-rates (9 & 6 Mb/s) use binary PSK & 3/4 or 1/2 rate FEC : 48 x (3/4) x 250kb/s = 9 Mb/s 48 x (1/2) x 250 kb/s = 6 Mb/s. • For 18 & 12 Mb/s, QPSK is used on each of 48 data carriers.• For 36 & 24 Mb/s use 16-QAM.• With 1/2 rate coder 64-QAM would give 36 Mb/s, so use 2/3 rate for 48 Mb/s.

24 Apr'06 CS3282 Sectn 8 132

8.14. Conclusions and learning outcomes

• Matched filtering affects pulse-shaping in single carrier modulation.

• Channel equalisation, required to cancel effects of frequency selective fading, is a computationally expensive adaptive filtering task.

• OFDM is highly efficient form of multi-carrier modulation.

• Single carrier uses sinc pulses & eliminates inter-symbol interference

• OFDM uses rect pulses & eliminates inter-spectral interference.

• FFT & I- FFT implement OFDM directly.

• Channel equalisation much easier to implement - no adaptive filter needed.

• Need for highly linear amplification & wide range of peak-to-mean ratios cause practical problems especially for battery powered mobile equipment.

• Parameters of 802.11 OFDM implementation have been analysed.

24 Apr'06 CS3282 Sectn 8 133

8.15 Problems & discussion points

1. What is the max bit-rate that can be transmitted without ISI on a 1 MHz channel using (i) B-PSK, (ii) QPSK, (iii) 16-QAM.

2. What is the max bit-rate that can be transmitted with arbitrarily low bit-errors over a noise-less channel of 1 MHz bandwidth [Ans: ]

3. Repeat Q.2 for a noisy channel where the SNR is 30 dB.

4. How does spectrum of a 50% RC pulse differ from that of a pure sinc pulse.

5. Why are RRC rather than RC pulses used in single carrier transmissions.

6. How many different OFDM symbol shapes are there with 64-QAM?

7. Why are the first & last few sub-carriers left unmodulated?

8. With 16-QAM, why are the 4-bit numbers arranged in ‘Gray coder’ order?

9. Derive a constellation for 64-QAM.

10. Why are interleaving & FEC very important with OFDM?

24 Apr'06 CS3282 Sectn 8 134

Problems & discussion points (cont)

11.Given that their bandwidth was 30 kHz & in cities BC 30 kHz, why was an equaliser not needed in a ‘1G’ mobile phone. Why is an equaliser definitely needed in a WLAN receiver when single carrier modulation is used?

12. Explain why bandwidth efficiency of 802.11 OFDM is 0.6 symbols/s per Hz without FEC. What is bandwidth efficiency when ¾ rate convolutional coder is used?

13. If a single carrier modulation scheme is used with R% RRC pulse shaping, what value of R would give a bandwidth efficiency of 0.6 pulses (symbols) per Hz ?

14. How are 24 & 36 Mb/s achieved over an IEEE802.11g WLAN?

15. Some non-standard versions of 802.11 claim to achieve 108 Mb/s. How is this done?

16. 802.11g claims max bit-rate of 54Mb/s. But cost of sending sync preambles

& headers reduces this bit-rate even in ideal conditions. Assuming ideal conditions,

estimate max average bit-rate

(i) where close to max length packets ( 2000 byte payload) always sent ,

(ii) where packets contain only 160 bytes of payload (20 ms of G711 speech).