[PPT] - EE456 Digital Communications Professor Ha Nguyen September 2015 PowerPoint Presentation

SLIDE 1

Chapter 7: Basic Digital Passband Modulation

EE456 – Digital Communications

Professor Ha Nguyen September 2015

EE456 – Digital Communications 1

SLIDE 2

Chapter 7: Basic Digital Passband Modulation

Introduction to Basic Digital Passband Modulation

Baseband transmission is conducted at low frequencies. Passband transmission happens in a frequency band toward the high end of the spectrum. Satellite communication is in the 6–8 GHz band, while mobile phones systems are in the 800 MHz–2.0 GHz band. Bits are encoded as a variation of the amplitude, phase or frequency, or some combination of these parameters of a sinusoidal carrier. The carrier frequency is much higher than the highest frequency of the modulating signals (or messages). Shall consider binary amplitude-shift keying (BASK), binary phase-shift keying (BPSK) and binary frequency-shift keying (BFSK): Error performance, optimum receivers, spectra. Extensions to quadrature phase-shift keying (QPSK), offset QPSK (OQPSK) and minimum shift keying (MSK).

EE456 – Digital Communications 2

SLIDE 3

Chapter 7: Basic Digital Passband Modulation

Examples of Binary Passband Modulated Signals

1

b

T

b

T 2

b

T 3

b

T 4

b

T 5

b

T 6

b

T 7

b

T 8

b

T 9 1 1 1 1 1 (a) Binary data V V − (b) Modulating signal ( ) m t (c) BASK signal V V − (d) BPSK signal V V − (e) BFSK signal t t t t

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SLIDE 4

Chapter 7: Basic Digital Passband Modulation

Binary Amplitude-Shift Keying (BASK)

s1(t) = 0,

“0T ” s2(t) = V cos(2πfct), “1T ” , 0 < t ≤ Tb, fc = n/Tb

) (

1 t

φ ) (

1 t

s ) (

2 t

s

BASK

E

✁ ✂ ✄

Comparator

( )

∫ •

b

T

t d

1

r ) (

1 t

φ

b

T t = ( ) t r

1 1

1

h D h D

T T ≥ ⇒ < ⇒ r r

BASK 1 2 BASK

ln 2 2

h

E N P T P E   = +     (b) dt t t r r

b

T

) ( ) (

1 1

φ

=

) (

1 t

s ) (

2 t

s

T T

1 Choose Choose

☎

⇐

BASK 2

E

✆ ✝ ✞

P [error]BASK = Q

EBASK

2N0

= Q
Eb

N0

,

where Eb = 0.5 × 0 + 0.5 × EBASK = EBASK 2 is the energy per bit.

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SLIDE 5

Chapter 7: Basic Digital Passband Modulation

PSD of BASK

SBASK(f) = V 2 16

δ(f − fc) + δ(f + fc) + sin2[πTb(f + fc)]

π2Tb(f + fc)2 + sin2[πTb(f − fc)] π2Tb(f − fc)2

.

BASK( )

S f f

b c

T f 1 −

b c

T f 1 +

c

f

2

16 V

Approximately 95% of the total transmitted power lies in a band of 3/Tb (Hz), centered at fc. The carrier component could be helpful for frequency and phase synchronization at the receiver.

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SLIDE 6

Chapter 7: Basic Digital Passband Modulation

Binary Phase-Shift Keying (BPSK)

s1(t) = −V cos(2πfct), if “0T ” s2(t) = +V cos(2πfct), if “1T ” , 0 < t ≤ Tb, ) (

1 t

s ) (

2 t

s

BPSK

E

BPSK

E

1

2 ( ) cos(2 )

c b

t f t T φ π = P [error]BPSK = Q

2EBPSK

N0

= Q
2Eb

N0

,

where Eb = 0.5 × EBPSK + 0.5 × EBPSK = EBPSK is the energy per bit. SBPSK(f) = V 2 4 sin2[π(f − fc)Tb] π2(f − fc)2)Tb + sin2[π(f + fc)Tb] π2(f + fc)2Tb

.

Similar to that of BASK, but no impulse functions at ±fc.

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SLIDE 7

Chapter 7: Basic Digital Passband Modulation

Binary Frequency-Shift Keying (BFSK)

✟ ✠ ✡ ☛ ☞ ☞ ✌ ✍ ✎ ✏ ✑ ✒ ✓ ✔ ✕ ✟ ✠ ✡ ☛ ☞ ☞ ✌ ✍ ✎ ✏ ✖ ✒ ✓ ✗ ✕ ✘ ✙✘ ✘ ✑ ✘

) (t s

✒ ✌ ✕

s1(t) = V cos(2πf1t + θ1),

if “0T ” s2(t) = V cos(2πf2t + θ2), if “1T ” , 0 < t ≤ Tb. (i) Minimum frequency separation for coherent orthogonality (θ1 = θ2): (∆f)[coherent]

min

= 1 2Tb . (ii) Minimum frequency separation for noncoherent orthogonality (θ1 = θ2): (∆f)[noncoherent]

min

= 1 Tb .

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SLIDE 8

Chapter 7: Basic Digital Passband Modulation

φ1(t) = s1(t) √EBFSK , φ2(t) = s2(t) √EBFSK . ) (

1 t

s ) (

2 t

s

T

1 Choose

T

Choose ) ( 2

2

t r φ ) (

1 1

t r φ

[ ]

2 ) ( ) (

1 2

t t φ φ −

( )

BFSK

0, E

( )

BFSK ,0

E

1 2

when boundary Decision P P = P [error]BFSK = Q

EBFSK

N0

= Q
Eb

N0

,

where Eb = 0.5 × EBFSK + 0.5 × EBFSK = EBFSK is the energy per bit.

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SLIDE 9

Chapter 7: Basic Digital Passband Modulation

PSD of BFSK

SBFSK(f) = V 2 16

δ(f − f2) + δ(f + f2) +

sin2[πTb(f + f2)] π2Tb(f + f2)2 + sin2[πTb(f − f2)] π2Tb(f − f2)2

+

V 2 16

δ(f − f1) + δ(f + f1) +

sin2[πTb(f + f1)] π2Tb(f + f1)2 + sin2[πTb(f − f1)] π2Tb(f − f1)2

.

✚

2 1

Bandwidth ( ) 3/

b

W f f T = − +

1

f

2

f

2

1

b

f T −

2

1.5

b

f T +

1

1.5

b

f T −

1

b

f T +

2

16 V

2

16 V

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SLIDE 10

Chapter 7: Basic Digital Passband Modulation

Performance Comparison of BASK, BPSK and BFSK

2 4 6 8 10 12 14 10

−6

10

−5

10

−4

10

−3

10

−2

10

−1

Eb/N0 (dB) P[error] BASK and BFSK BPSK P [error]BPSK = Q

2Eb

N0

, P [error]BASK = P [error]BFSK = Q
Eb

N0

.

Question: You have designed a BPSK system that achieves P [error] = 10−6. What needs to be changed if you want to double the transmission bit rate and still meet P [error] = 10−6?

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SLIDE 11

Chapter 7: Basic Digital Passband Modulation

Comparison Summary of BASK, BPSK and BFSK

The BER performance curves of BASK, BPSK and BFSK shown in the previous slide can only be realized with coherent receivers, i.e., when perfect carrier frequency and phase synchronization can be established at the receivers. A receiver that does not require phase synchronization is called a non-coherent receiver, which is simpler (hence less expensive) than a coherent receiver. Using the peaks of the first side lobes on both sides of the carrier frequency (or 2 frequencies in BFSK), the transmission bandwidths for BASK, BPSK and BFSK are approximated as: WBASK = WBPSK ≈ 3 Tb = 3rb WBFSK ≈ |f2 − f1| + 3 Tb = 3.5rb, coherently orthogonal 4rb, non-coherently orthogonal Taking into account both bandwidth and power, BPSK is the best scheme if a coherent receiver can be afforded! This is followed by BASK and BFSK. The question is when would one use BASK or BFSK? The answers are as follows:

If the carrier frequency can be synchronized, but not the phase, then BPSK does not work! On the contrary one can still use BASK but the receiver has to be redesigned to be a non-coherent receiver. See Assignment 7 (Problem 7.7), and the next few slides. Similarly, BFSK can be detected non-coherently without phase synchronization.

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SLIDE 12

Chapter 7: Basic Digital Passband Modulation

Non-Coherent Detection of BASK

r(t) = sR

1 (t) + w(t) = 0 + w(t)

: “0T ” sR

2 (t) + w(t) =

√ E

2

Tb cos(2πfct + θ) + w(t) : “1T ” Without the noise, the receiver sees one of the two signals sR

1 (t) = 0

: “0T ” sR

2 (t) =

√ E

2

Tb cos(2πfct + θ)

: “1T ” Write sR

2 (t) as follows:

sR

2 (t) =

√ E cos θ 2 Tb cos(2πfct)

φ1(t)

+ √ E sin θ 2 Tb sin(2πfct)

φ2(t)

. Thus, for an arbitrary θ two basis functions are required to represent sR

1 (t) and sR 2 (t).

The signal sR

1 (t) always lies at the origin, while sR 2 (t) could be anywhere on the circle

f radius

√ E, depending on θ.

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SLIDE 13

Chapter 7: Basic Digital Passband Modulation

θ ) (

1 t

φ ) (

1 t

s ) (

2 t

sR E ) (

2 t

φ ) (

f

locus

2 t

sR assuming boundary decision = θ θ ) (

1 t

φ ) (

1 t

s ) (

2 t

sR D ) (

2 t

φ account into y uncertaint phase taking boundary decision (a) (b) 2 E

(a) If the receiver assumes θ = 0, the optimum decision boundary is a line perpendicular to φ1(t) and at distance √ E/2 away from sR

1 (t). The error is:

P [error] = 0.5Q

E

2N0

+ 0.5Q
2E

N0 [cos θ − 0.5]

(b) Without phase synchronization, the decision boundary is a circle centered at sR

1 (t) and with

some diameter D. It can be shown that the optimum value of D is: D = √ E √ 2

1 + 4N0

E

≈
E

2 (for high SNR) The error probability is well approximated as P [error] ≈ 1

2 e − E 4N0 . EE456 – Digital Communications 13

SLIDE 14

Chapter 7: Basic Digital Passband Modulation

2

r ) (

2 t

φ ) (

1 t

φ

2 2 2 2 1

to compare and Compute D r r + ( ) t r Decision

( )

∫ •

b

T

t d

( )

∫ •

b

T

t d

b

T t =

1

r

b

T t =

5 10 15 10

−4

10

−3

10

−2

10

−1

10 Eb/N0 (dB) Pr[error] θ=0 θ=π/6 θ=π/3 θ=π/2 Noncoherent EE456 – Digital Communications 14

SLIDE 15

Chapter 7: Basic Digital Passband Modulation

Quadrature Phase Shift Keying (QPSK)

Basic idea behind QPSK: cos(2πfct) and sin(2πfct) are orthogonal over [0, Tb] when fc = k/Tb, k integer ⇒ Can transmit two different bits over the same frequency band at the same time. The symbol signaling rate (i.e., the baud rate) is rs = 1/Ts = 1/(2Tb) = rb/2 (symbols/sec), i.e., halved. Bit Pattern Message Signal Transmitted 00 m1 s1(t) = V cos(2πfct), 0 ≤ t ≤ Ts = 2Tb 01 m2 s2(t) = V sin(2πfct), 0 ≤ t ≤ Ts = 2Tb 11 m3 s3(t) = −V cos(2πfct), 0 ≤ t ≤ Ts = 2Tb 10 m4 s4(t) = −V sin(2πfct), 0 ≤ t ≤ Ts = 2Tb V V − t

b

T 2

b

T 4

b

T 6

b

T 8

2

01 m =

3

11 m =

1

00 m =

4

10 m =

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SLIDE 16

Chapter 7: Basic Digital Passband Modulation

Signal Space Representation of QPSK

Ts s2

i (t)dt = V 2

2 Ts = V 2Tb = Es, φ1(t) = s1(t) √Es , φ2(t) = s2(t) √Es . ) (

1 t

φ ) (

1 t

s ) (

2 t

s

E ) (

2 t

φ ) (

3 t

s ) (

4 t

s

EE456 – Digital Communications 16

SLIDE 17

Chapter 7: Basic Digital Passband Modulation

Optimum Receiver for QPSK

1 1 1

r

) ( Choose m t s ℜ

2 2 2

r

) ( Choose m t s ℜ

3 3 3

r

) ( Choose m t s ℜ

4 4 4

r

) ( Choose m t s ℜ

1 2

dimensional observation space

( , , , )

m

m r = r r r

✛ ✜

P [correct] =

ℜ1

P1f( r|s1(t))d r +

ℜ2

P2f( r|s2(t))d r +

ℜ3

P3f( r|s3(t))d r +

ℜ4

P4f( r|s4(t))d r. Choose si(t) if Pif( r|si(t)) > Pjf( r|sj(t)), j = 1, 2, 3, 4; j = i.

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SLIDE 18

Chapter 7: Basic Digital Passband Modulation

Minimum-Distance Receiver

When Pi = 0.25, i = 1, . . . , 4, the decision rule is the minimum-distance rule: Choose si(t) if (r1 − si1)2 + (r2 − si2)2 is the smallest

✢ ✣✤✤✥ ✦ ✧ ★ ✩ ✪✫ ✢ ✣✤✤✥ ✦ ✧ ✬ ✩ ✫✪ ✭ ✮ ✯ ✯ ✰ ✱ ✲ ✳ ✴ ✵ ✵ ✶ ✷ ✸ ✸ ✹ ✺ ✻ ✼ ✽ ✾ ✾

1 1

( ), t r φ ) (

1 t

s ) (

2 t

s

2 2

( ), t r φ ) (

3 t

s ) (

4 t

s / 4 π

EE456 – Digital Communications 18

SLIDE 19

Chapter 7: Basic Digital Passband Modulation

Simplified Decision Rule and Receiver Implementation

Choose si(t) if N0 2 ln Pi + r1si1 + r2si2 > N0 2 ln Pj + r1sj1 + r2sj2 j = 1, 2, 3, 4; j = i.

2

r ) (

2 t

φ

1

r ) (

1 t

φ 2

s b

t T T = =

1 1 2 2

Compute ln( ) 2 for 1,2,3,4 and choose the

j j j

N r s r s P j largest + + = ( ) t r Decision

( )

d

s

T

t

✿

2

s b

t T T = =

( )

d

s

T

t

✿

EE456 – Digital Communications 19

SLIDE 20

Chapter 7: Basic Digital Passband Modulation

Symbol (Message) Error Probability of QPSK

❀ ❁❂❂❃ ❄ ❅ ❆ ❇❈❉ ❀ ❁❂❂❃ ❄ ❅ ❊ ❇❉❈ ❋

❍

❍ ■ ❏ ❑ ▲ ▼ ◆ ◆ ❖ P ◗ ◗ ❘ ❙ ❚ ❯ ❱ ❲ ❲

1

r ) (

1 t

s ) (

2 t

s

2

r ) (

3 t

s ) (

4 t

s / 4 π / 2

s

E / 2

s

E −

1

ˆ r

2

ˆ r

P [error] =

4

i=1

P [error|si(t)]P [si(t)] = 1 4

4

i=1

P [error|si(t)] = P [error|si(t)] = 1 − P [correct|si(t)] P [correct|s1(t)] = P [(r1, r2) ∈ shaded quadrant|s1(t)] = P [(ˆ r1, ˆ r2) ∈ shaded quadrant|s1(t)] = P [(ˆ r1 ≥ 0) AND (ˆ r2 ≤ 0)|s1(t)] = P [ˆ r1 ≥ 0|s1(t)] × P [ˆ r2 ≤ 0|s1(t)] =

1 − Q
Es/N0
×
1 − Q
Es/N0
=
1 − Q
Es/N0

2 P [error] = 1 −

1 − Q
Es/N0

2 = 2Q

Es/N0
− Q2
Es/N0
≈ 2Q
Es/N0
.

EE456 – Digital Communications 20

SLIDE 21

Chapter 7: Basic Digital Passband Modulation

Illustration of Symbol Error Probability Analysis for QPSK

−4 −2 2 4 −4 −2 2 4 0.05 0.1 0.15 Observation ˆ r1 Observation ˆ r2 Joint pdf f(ˆ r1, ˆ r2) Observation ˆ r1 Observation ˆ r2 −2 −1 1 2 −2 −1 1 2

EE456 – Digital Communications 21

SLIDE 22

Chapter 7: Basic Digital Passband Modulation

Bit Error Probability of QPSK

❳ ❨❩❩❬ ❭ ❪ ❫ ❴❵❛ ❳ ❨❩❩❬ ❭ ❪ ❜ ❴❛❵ ❝ ❞ ❡ ❡ ❢ ❣ ❤ ✐ ❥ ❦ ❦ ❧ ♠ ♥ ♥ ♦ ♣ q r s t t

1

r ) (

1 t

s ) (

2 t

s

2

r ) (

3 t

s ) (

4 t

s / 4 π / 2

s

E / 2

s

E −

1

ˆ r

2

ˆ r P [m2|m1] = Q  

Es

N0    1 − Q  

Es

N0     , P [m3|m1] = Q2  

Es

N0   , P [m4|m1] = Q  

Es

N0    1 − Q  

Es

N0     .

P [bit error] = 0.5P [m2|m1] + 0.5P [m4|m1] + 1.0P [m3|m1] = Q

Es

N0

= Q
2Eb

N0

,

because Es = 2Eb. Gray mapping: Nearest neighbors are mapped to the bit pairs that differ in only one bit.

EE456 – Digital Communications 22

SLIDE 23

Chapter 7: Basic Digital Passband Modulation

An Alternative Representation of QPSK

) (t s ) (t aQ ) (t aI

( )

t f V

c

π 2 cos

✉ ✈✇ ①② ③ ④ ⑤ ② ✈ ⑥✈ ⑦

( )

t f V

c

π 2 sin ( ) a t 1

b

T

b

T 2

b

T 3

b

T 4

b

T 5

b

T 6

b

T 7

b

T 8

b

T 9 1 1 1 1 1 sequence Bit t ( ) a t 1 − a

1

a

2

a

3

a

4

a

5

a

6

a

7

a

8

a 1

b

T

b

T 2

b

T 3

b

T 4

b

T 5

b

T 6

b

T 7

b

T 8

b

T 9 t ) (t aI 1 − a

2

a

4

a

6

a

8

a 1

b

T

b

T 2

b

T 3

b

T 4

b

T 5

b

T 6

b

T 7

b

T 8

b

T 9 t ) (t aQ 1 −

1

a

3

a

5

a

7

a EE456 – Digital Communications 23

SLIDE 24

Chapter 7: Basic Digital Passband Modulation 2Tb 4Tb 6Tb 8Tb 10Tb −V V aI(t)Vcos(2πfct) 2Tb 4Tb 6Tb 8Tb 10Tb −V V aQ(t)Vsin(2πfct) 2Tb 4Tb 6Tb 8Tb 10Tb −V V sQPSK(t)=aI(t)Vcos(2πfct)+aQ(t)Vsin(2πfct) t

s(t) = aI(t)V cos(2πfct) + aQ(t)V sin(2πfct) =

a2

I(t) + a2 Q(t)V cos

2πfct − tan−1

aQ(t) aI(t)

=

√ 2V cos[2πfct − θ(t)]. θ(t) =      π/4, if aI = +1, aQ = +1 (bits are 11) −π/4, if aI = +1, aQ = −1 (bits are 10) 3π/4, if aI = −1, aQ = +1 (bits are 01) −3π/4, if aI = −1, aQ = −1 (bits are 00) .

EE456 – Digital Communications 24

SLIDE 25

Chapter 7: Basic Digital Passband Modulation

Signal Space Representation of QPSK

     φ1(t) = V cos(2πfct) √

V 2Tb

=

2

Ts cos(2πfct)

φ2(t) = V sin(2πfct) √

V 2Tb

=

2

Ts sin(2πfct)

, 0 < t < Ts = 2Tb,

1

2 ( ) cos(2 )

c s

t f t T φ π = 1, 1 ( ) / 4

I Q

a a t θ π = =

⑧

=

2 s

V T 1, 1 ( ) / 4

I Q

a a t θ π = = −

⑨

= − 1, 1 ( ) 3 / 4

I Q

a a t θ π = − =

⑩

= 1, 1 ( ) 3 / 4

I Q

a a t θ π = − = −

❶

= − / 4 π

2

2 ( ) sin(2 )

c s

t f t T φ π =

EE456 – Digital Communications 25

SLIDE 26

Chapter 7: Basic Digital Passband Modulation

Receiver Implementation of QPSK

1

2 ( ) cos(2 )

c s

t f t T φ π = 1, 1 ( ) / 4

I Q

a a t θ π = =

❷

=

2 s

V T 1, 1 ( ) / 4

I Q

a a t θ π = = −

❸

= − 1, 1 ( ) 3 / 4

I Q

a a t θ π = − =

❹

= 1, 1 ( ) 3 / 4

I Q

a a t θ π = − = −

❺

= − / 4 π

2

2 ( ) sin(2 )

c s

t f t T φ π = Threshold = 0

1

r

1( )

t φ

( )

d

s

T

t

∫

2

r

2( )

t φ

s

t T = ( ) t r Threshold = 0 Multiplexer ) (t aI ) (t aQ ( ) a t

( )

d

s

T

t

∫

s

t T =

P [bit error] = Q  

V 2Ts

N0   = · · · = Q

2Eb

N0

.

EE456 – Digital Communications 26