ee456 – digital communications · satellite communication is in the 6–8 ghz band, while mobile...

Chapter 7: Basic Digital Passband Modulation

EE456 – Digital CommunicationsProfessor Ha Nguyen

September 2016

EE456 – Digital Communications 1


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 thespectrum.

Satellite communication is in the 6–8 GHz band, while mobile phones systems arein the 800 MHz–2.0 GHz band.

Bits are encoded as a variation of the amplitude, phase or frequency, or somecombination of these parameters of a sinusoidal carrier.

The carrier frequency is much higher than the highest frequency of themodulating signals (or messages).

Shall consider binary amplitude-shift keying (BASK), binary phase-shift keying

(BPSK) and binary frequency-shift keying (BFSK): Error performance, optimumreceivers, spectra.

Extensions to quadrature phase-shift keying (QPSK), offset QPSK (OQPSK) andminimum shift keying (MSK).



Examples of Binary Passband Modulated Signals

0

1

0 bT bT2 bT3 bT4 bT5 bT6 bT7 bT8 bT9

1 0 00 01 11 1(a) Binary data

0

V

V−

(b) Modulating signal

( )m t

(c) BASK signal

0

V

V−

(d) BPSK signal

0

V

V−

(e) BFSK signal

t

t

t

t



Binary Amplitude-Shift Keying (BASK){

s1(t) = 0, “0T ”s2(t) = V cos(2πfct), “1T ”

, 0 < t ≤ Tb, fc = n/Tb

)(1 tφ0

)(1 ts )(2 ts

BASKE

��

Comparator( )∫ •bT

t0

d1r

)(1 tφ

bTt =( )tr 1

1

1

0

h D

h D

T

T

≥ ⇒

< ⇒

r

r

BASK0 1

2BASK

ln22

h

EN PT

PE

= +

(b)

dtttrrbT

)()( 101 φ�=)(1 ts )(2 ts

TT 1 Choose 0 Choose �⇐

0BASK 2E

��

P [error]BASK = Q

(√

EBASK

2N0

)

= Q

(√

Eb

N0

)

,

where Eb = 0.5 × 0 + 0.5 × EBASK =EBASK

2is the energy per bit.



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

bc T

f1−

bc T

f1+cf0

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 synchronizationat the receiver.



Binary Phase-Shift Keying (BPSK)

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

, 0 < t ≤ Tb,

0)(1 ts )(2 ts

BPSKE BPSKE

1

2( ) cos(2 )c

b

t f tT

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.



Binary Frequency-Shift Keying (BFSK)

��

��

��

��

)(ts

� �{

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 6= θ2):

(∆f)[noncoherent]min =

1

Tb

.



φ1(t) =s1(t)√EBFSK

, φ2(t) =s2(t)√EBFSK

.

0 )(1 ts

)(2 ts

T1 Choose

T0 Choose

)( 22 tr φ

)(1

1

t

r

φ

[ ] 2)()( 12 tt φφ − ( )BFSK0, E

( )BFSK ,0E

12 when

boundary Decision

PP =

P [error]BFSK = Q

(√

EBFSK

N0

)

= Q

(√

Eb

N0

)

,

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



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 1Bandwidth ( ) 3/ bW f f T= − +

1f 2f2

1

b

fT

− 2

1.5

b

fT

+1

1.5

b

fT

− 1

1

b

fT

+

2

16

V 2

16

V



Performance Comparison of BASK, BPSK and BFSK

0 2 4 6 8 10 12 1410

−6

10−5

10−4

10−3

10−2

10−1

Eb/N

0 (dB)

P[e

rror

]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?



Comparison Summary of BASK, BPSK and BFSK

The BER performance curves of BASK, BPSK and BFSK shown in the previousslide can only be realized with coherent receivers, i.e., when perfect carrierfrequency 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 2frequencies in BFSK), the transmission bandwidths for BASK, BPSK and BFSKare approximated as:

WBASK = WBPSK ≈ 3

Tb

= 3rb

WBFSK ≈ |f2 − f1|+3

Tb

=

{3.5rb, coherently orthogonal4rb, non-coherently orthogonal

Taking into account both bandwidth and power, BPSK is the best scheme if acoherent 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 notwork! On the contrary one can still use BASK but the receiver has to be redesigned tobe 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.



Non-Coherent Detection of BASK

r(t) =

{ sR1 (t) +w(t) = 0 +w(t) : “0T ”

sR2 (t) +w(t) =√E

√

2

Tb

cos(2πfct + θ) +w(t) : “1T ”

Without the noise, the receiver sees one of the two signals

{ sR1 (t) = 0 : “0T ”

sR2 (t) =√E√

2Tb

cos(2πfct + θ) : “1T ”

Write sR2 (t) as follows:

sR2 (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 sR1 (t) and sR2 (t).The signal sR1 (t) always lies at the origin, while sR2 (t) could be anywhere on the circle

of radius√E, depending on θ.



θ)(1 tφ

0

)(1 ts

)(2 tsR

E

)(2 tφ

)( of locus 2 tsR

0 assuming

boundary decision

=θ

θ)(1 tφ

0

)(1 ts

)(2 tsR

D

)(2 tφ

account into

yuncertaint 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 sR1 (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 sR1 (t) and withsome 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] ≈ 12 e

− E4N0 .



2r

)(2 tφ

)(1 tφ2

22

21

tocompare and

Compute

D

rr +( )tr Decision

( )∫ •bT

t0

d

( )∫ •bT

t0

d

bTt =

1rbTt =

0 5 10 1510

−4

10−3

10−2

10−1

100

Eb/N

0 (dB)

Pr[

erro

r]

θ=0 θ=π/6θ=π/3θ=π/2

Noncoherent



Quadrature Phase Shift Keying (QPSK)

Basic idea behind QPSK: cos(2πfct) and sin(2πfct) are orthogonal over [0, Tb] whenfc = k/Tb, k integer ⇒ Can transmit two different bits over the same frequency band atthe same time.

The symbol signaling rate (i.e., the baud rate) is rs = 1/Ts = 1/(2Tb) = rb/2(symbols/sec), i.e., halved. Thus, the required bandwidth is halved compared to BPSK.

An alternative view is that, for the same bandwidth as in BPSK, the bit rate can be doubled.

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

0

V

V−

t

bT2 bT4 bT6 bT8

2 01m =

0

3 11m =1 00m = 4 10m =



Comparison of BPSK and QPSK Waveforms

0 1 2 3 4 5 6 7 8−1.5

−1

−0.5

0

0.5

1

1.5

t/Tb

BPSK

0 1 2 3 4 5 6 7 8−1.5

−1

−0.5

0

0.5

1

1.5

t/Tb

QPSK Approach 1: Same rb with BPSK, same power to achieve the same Eb, hence the same P [bit error], half the bandwidth since the baud rate is halved (Ts = 2Tb)

0 1 2 3 4 5 6 7 8−1.5

−1

−0.5

0

0.5

1

1.5

t/Tb

QPSK Approach 2: Twice bit rate compared to BPSK, twice the power to achieve the same Eb, hence the same P [bit error], same bandwidth since the baud rate is unchanged (Ts = Tb)

0 0 1 1 01 10

00

00

11

11

01

01

10

10 0001 1011



Signal Space Representation of QPSK

∫

Ts

0

s2i (t)dt =

V 2

2Ts = V

2Tb = Es,

φ1(t) =s1(t)√Es

, φ2(t) =s2(t)√Es

.

)(1 tφ0

)(1 ts

)(2 ts

sE

)(2 tφ

)(3 ts

)(4 ts



Optimum Receiver for QPSK

11

1

or )( Choose mts

ℜ

22

2

or )( Choose mts

ℜ

33

3

or )( Choose mts

ℜ

44

4

or )( Choose mts

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 6= i.



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

�� ! "#$%&

�� ! "'$&%

()**+,-./00

12334567899

1 1( ), t rφ0

)(1 ts

)(2 ts

2 2( ), t rφ

)(3 ts

)(4 ts

/ 4π



Simplified Decision Rule and Receiver Implementation

Choose si(t) ifN02 lnPi + r1si1 + r2si2 > N0

2 lnPj + r1sj1 + r2sj2j = 1, 2, 3, 4; j 6= i.

2r

)(2 tφ

1r

)(1 tφ

2s bt T T= =

01 1 2 2

Compute

ln( )2

for 1, 2,3, 4

and choose

the

j j j

Nr s r s P

j

largest

+ +

=( )tr Decision

( )0

dsT

t•:

2s bt T T= =

( )0

dsT

t•:



Symbol (Message) Error Probability of QPSK;<==>? @ABCD

;<==>? @EBDC

FGHHIJKLMNN

OPQQRSTUVWW

1r0

)(1 ts

)(2 ts

2r

)(3 ts

)(4 ts

/ 4π

/ 2sE

/ 2sE

−

1̂r2̂r

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 [(r̂1, r̂2) ∈ shaded quadrant|s1(t)]= P [(r̂1 ≥ 0) AND (r̂2 ≤ 0)|s1(t)] = P [r̂1 ≥ 0|s1(t)] × P [r̂2 ≤ 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

)

.



Illustration of Symbol Error Probability Analysis for QPSK

−4−2

02

4

−4−2

02

40

0.05

0.1

0.15

Observation r̂1Observation r̂2

Jointpdff(r̂

1,r̂

2)

Observation r̂1

Observationr̂ 2

−2 −1 0 1 2−2

−1

0

1

2



Bit Error Probability of QPSK

XYZZ[\ ]^_`a

XYZZ[\ ]b_a`

cdeefghijkk

lmnnopqrstt

1r0

)(1 ts

)(2 ts

2r

)(3 ts

)(4 ts

/ 4π

/ 2sE

/ 2sE

−

1̂r2̂r

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.



An Alternative Representation of QPSK

)(ts

)(taQ

)(ta I

( )tfV cπ2cos

uvwxyz{|yv}v~

( )tfV cπ2sin

( )a t

0

1

0 bT bT2 bT3 bT4 bT5 bT6 bT7 bT8 bT9

1 0 00 01 11 1sequenceBit

t( )a t

1−

0a 1a 2a 3a 4a 5a 6a 7a 8a

0

1

0 bT bT2 bT3 bT4 bT5 bT6 bT7 bT8 bT9t)(ta I

1−

0a 2a 4a 6a 8a

0

1

0 bT bT2 bT3 bT4 bT5 bT6 bT7 bT8 bT9t)(taQ

1−

1a 3a 5a 7a



0 2Tb 4Tb 6Tb 8Tb 10Tb

−V

0

V

aI(t)Vcos(2πf

ct)


−V

0

V

aQ

(t)Vsin(2πfct)


−V

0

V

sQPSK

(t)=aI(t)Vcos(2πf

ct)+a

Q(t)Vsin(2πf

ct)

t

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

=√

a2I(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)

.



Signal Space Representation of QPSK

φ1(t) =V cos(2πfct)√

V 2Tb

=√

2Ts

cos(2πfct)

φ2(t) =V sin(2πfct)√

V 2Tb

=√

2Ts

sin(2πfct), 0 < t < Ts = 2Tb,

1

2( ) cos(2 )c

s

t f tT

φ π=0

1, 1 ( ) / 4I Qa a tθ π= = � =

2sV T

1, 1 ( ) / 4I Qa a tθ π= = − � = −

1, 1 ( ) 3 / 4I Qa a tθ π= − = � =

1, 1 ( ) 3 / 4I Qa a tθ π= − = − � = −

/ 4π

2

2( ) sin(2 )c

s

t f tT

φ π=



Receiver Implementation of QPSK

1

2( ) cos(2 )c

s

t f tT

φ π=0

1, 1 ( ) / 4I Qa a tθ π= = � =

2sV T

1, 1 ( ) / 4I Qa a tθ π= = − � = −

1, 1 ( ) 3 / 4I Qa a tθ π= − = � =

1, 1 ( ) 3 / 4I Qa a tθ π= − = − � = −

/ 4π

2

2( ) sin(2 )c

s

t f tT

φ π=

Threshold = 01r

1( )tφ

( )0

dsT

t•∫

2r

2( )tφ

st T=

( )tr

Threshold = 0

Multiplexer

)(ta I

)(taQ

( )a t

( )0

dsT

t•∫

st T=

P [bit error] = Q

√

V 2Ts

N0

= · · · = Q

(√

2Eb

N0

)

.


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