signal space analysis of bask, bfsk, bpsk, and qam on mac

Signal Space Analysis of BASK, BFSK, BPSK, and QAM on Mac

30. Signal Space Analysis of BASK, BFSK, BPSK, and QAM

The vector-space representation of signals and the optimum detection process whichchooses the signal closest to the received signal is particularly useful in signal design and inprobability-of-error calculations. The material presented herein includes the errorperformance of binary amplitude-shift keying (BASK), binary frequency-shift keying(BFSK), binary phase-shift keying (BPSK), and M-ary quadrature amplitude modulation(QAM) signals.

Error Performance of Binary ASK

A binary amplitude-shift keying (BASK) signal can be defined by

s(t) = A fct t T

elsewhere

cos( ),

,

2 0

0

π ≤ ≤

(30.1)

where A is a constant, fc is the carrier frequency, and T is the bit duration. It has a

power P = A2/2, so that A = 2P . Thus equation (30.1) can be written as

s(t) = 2 2 0

0

P fct t T

elsewhere

cos( ),

,

π ≤ ≤

= PT

Tfct t T

elsewhere

22 0

0

cos( ),

,

π ≤ ≤

= E

Tfct t T

elsewhere

22 0

0

cos( ),

,

π ≤ ≤

(30.2)

where E = PT is the energy contained in a bit duration. Figure 30.1 shows the signalconstellation diagram of BASK signals and the conditional probability density functionsassociated with the signals.

Figure 30.1 BASK signal constellation diagram and the conditional probability densityfunctions associated with the signals.

The energy contained in a bit duration is

E = d2 (30.3)

and so

30.1


d = E (30.4)

It is assumed that the noise is additive white Gaussian noise (AWGN) with a two-sided

power spectral density of n0/2, zero mean and fixed variance σ2 = n0/2. In the

presence of AWGN, the conditional probability density function of φ1 assuming that s0is transmitted is

f(φ1 / 0) = 1

2 2

12

2 2

πσ

φσe

−

and the probability of error given that s0 is transmitted is

Pe0 =

φ1 2=

∞∫d /

f(φ1/0)dφ1 =

φ1 2=

∞∫d /

1

2 2

12

2 21

πσ

φ

σ φe d−

(30.5)

Similarly, the probability of error given that s1 is transmitted is

Pe1 =

−∞

=

∫φ1 2d /

f(φ1/1)dφ1 = 1

2 2

12

2 21

1 2

πσ

φσ φ

φe

d

dd − −

−∞

=

∫( )/

= Pe1

where

f(φ1 /1) = 1

2 2

12

2 2

πσ

φσe

d− −( )

(30.6)

Let p0 be the probability of sending s0 and p1 be the probability of sending s1. For

equally likely transmission of binary signals, we have p0 = p1 = 0.5. The average

probability of error is given by

Pe = p0 Pe0 + p1 Pe1= Pe0

=

φ1 2=

∞∫d /

1

2 2

12

2 21

πσ

φ

σ φe d−

(30.7)

30.2


Let y = φσ1

2. Then y2 =

φ

σ12

2 2 and d y = dφ

σ1

2 . Substituting y and dy into

equation (30.7), we get

Pe = 1

2 2

πσ

y d=

∞∫ e-y2 dy

= 12 [

2π

y

∞∫ e-y2 dy ]

= 12 erfc( d

2 2σ)

= 12 erfc( d

n2 0)

= 12 erfc(

E

n4 0) (30.8)

where the complementary error function is

erfc (x) = 2π

x

∞∫ e-α2 dα (30.9)

σ2 = n0/2 for AWGN, and d = E .

Error Performance of Binary PSK

A binary phase-shift keying (BPSK) signal can be defined by

s(t) = +A cos 2πfct, 0 < t < T (30.10)

where A is a constant, fc is the carrier frequency, and T is the bit duration. It has a

power P = A2/2, so that A = 2P . Thus equation (30.10) can be written as

s(t) = + 2P cos 2πfct

= + PT2T

cos 2πfct

= + E2T

cos 2πfct (30.11)

30.3


where E = PT is the energy contained in a bit duration. Figure 30.2 shows the signalconstellation diagram of BPSK signals and the conditional probability density functionsassociated with the signals.

Figure 30.2 BPSK signal constellation diagram and the conditional probability densityfunctions associated with the signals.

The energy contained in a bit duration is

E = ( )d

22 (30.12)

and so

d = 4E (30.13)

In the presence of AWGN, the conditional probability density function of φ1 assuming

that s0 is sent is

f(φ1/0) = 1

2 2

1 22

2 2

πσ

φ

σe

d

−+( )

(30.14)

and the probability of error given that s0 is transmitted is

Pe0 =

0

∞∫ f(φ1/0)dφ1 =

0

∞∫ 1

2 2

1 22

2 21

πσ

φ

σ φe

d

d−

+( )

(30.15)

Similarly, the probability of error given that s1 is sent is

Pe1 =

−∞∫0

f(φ1 /1) dφ1 =

−∞∫0

1

2 2

1 22

2 21

πσ

φ

σ φe

d

d−

−( )

= Pe0 (30.16)

where

30.4


f(φ1 /2) = 1

2 2

1 22

2 2

πσ

φ

σe

d

−−( )

(30.17)

Let p0 be the probability of sending s0 and p1 be the probability of sending s1. For

equally likely transmission of binary signals, we have p0 = p1 = 0.5. The average

probability of error is

Pe = p0 Pe0 + p1 Pe1= Pe0

=

0

∞∫ 1

2 2

1 22

2 21

πσ

φ

σ φe

d

d−

+( )

(30.18)

Let y = φ

σ1 22

+ d

. Then y2 =( )φ

σ

1 22

2 2

+ d

and dy = dφ

σ1



Pe =

y

d

=+

∞∫

φ

σ1 22

1π

e -y2 dy

= 12

2π

y

d

=+

∞∫

φ

σ1 22

e -y2 dy

= 12

erfc (φ

σ1 2

2

+ d

)

= 12

erfc (d

2 2σ)

= 12 erfc( d

n2 0)

= 12 erfc(

E

n0) (30.19)

where φ1 = 0 is the threshold value, the complementary error function is

30.5


erfc (x) = 2π

x

∞∫ e-α2 dα (30.20)

σ2 = n0/2 for AWGN, and d = 4E .

Error Performance of Orthogonal Binary FSK

A binary frequency-shift keying (BFSK) signal can be defined by

s(t) = A fc t t T

A fc t elsewhere

cos( ),

cos( ),

2 1 0

2 2

π

π

≤ ≤

(30.21)

where A is a constant, fc1 and fc2

are the carrier frequencies, and T is the bit duration.

It has a power P = A2/2, so that A = 2P . Thus equation (30.21) can be written as

s(t) = 2 2 1 0

2 2 2

P fc t t T

P fc t elsewhere

cos( ),

cos( ),

π

π

≤ ≤

=

PTT

fc t t T

PTT

fc t elsewhere

22 1 0

22 2

cos( ),

cos( ),

π

π

≤ ≤

=

ET

fc t t T

ET

fc t elsewhere

22 1 0

22 2

cos( ),

cos( ),

π

π

≤ ≤

(30.22)

where E = PT is the energy contained in a bit duration. Figure 30.3 shows the signalconstellation diagram of orthogonal BFSK signals and the conditional probability densityfunctions associated with the signals.

Figure 30.3 Orthogonal BFSK signal constellation diagram and the conditionalprobability density functions associated with the signals.

The energy in a bit duration is

E = ( )d

22 (30.23)

and

30.6


d = 2E (30.24)

If we rotate the two orthonormal axes φ1 and φ2 to u1 and u2, we end up with a signal

constellation diagram for BPSK signals with a separation distance of d. Thus, the averageprobability of error is

Pe = 12 erfc( d

2 2σ)

= 12 erfc( d

n2 0)

= 12 erfc(

E

n2 0) (30.25)

σ2 = n0/2 for AWGN, and d = 2E .

Table 30.1 summarises the performance of BPSK, orthogonal BFSK, and BASK signalsin the presenec of AWGN.

________________________________________________________________________

Modulation d Pe = 12 erfc( d

2 2σ) Relative (dB)

= 12 erfc( d

n2 0)

________________________________________________________________________

BPSK 4E12 erfc(

E

n0) 0

Orthogonal BFSK 2E12 erfc(

E

n2 0) -3

BASK E12 erfc(

E

n4 0) -6

________________________________________________________________________

Table 30.1 Performance of various modulation systems.

Also, we can express the performance of these modulation systems in terms of the average

signal energy E , where

E = E/2 (30.26)

for BASK signals,

30.7


E = E (30.27)

for orthogonal BFSK signals, and

E = E (30.28)

for BPSK signals. This is summarised in Table 30.2.

________________________________________________________________________

Modulation E Pe Relative (dB)

________________________________________________________________________

BPSK E12 erfc(

E

n0) =

12 erfc(

E

n0) 0

Orthogonal BFSK E12 erfc(

E

n2 0) =

12 erfc(

E

n2 0) -3

BASK E/212 erfc(

E

n4 0) =

12 erfc(

E

n2 0) -3

________________________________________________________________________

Table 30.2 Performance of various binary modulation systems in terms of the averagesignal energy.

Error Performance of M -ary QAM

Consider the (M=16)-ary quadrature amplitude modulation (QAM) signal constellationdiagram of Figure 30.4 as an example.

Figure 30.4 16-ary QAM signal constellation diagram.

The average signal energy of an M-ary QAM signal is

E = 1

0

1

Mi

M

=

−∑ Ei (30.29)

where Ei is the energy of si. For M = 16, we have

E = 116

[

42

4

2

4

5 6 9 10

× +

d d

s , , ,1 244 344

+

89 2

4

2

4

1 2 4 111314 7 8

× +

d d

s , , , , , , ,1 244 344

+

49 2

49 2

4

0 312 15

× +

d d

s , , ,1 2444 3444

]

30.8


= 5d2

2(30.30)

and so

d = 25

E (30.31)

To find the probability of error, it is simpler to first find the probability Pc of correct

detection. It is apparent from Figure 30.4 that we can group signal points into 3 classes andcompute Pc based on the 3 decision regions as shown in Figure 30.5.

Figure 30.5 Decision regions.

Let n1 and n2 be additive white Gaussian noise samples with two-sided power spectral

density of n0/2, zero mean and fixed variance σ2 = n0/2 along the φ1-axis and φ2-

axis, respectively.

Case 1 - four inner signal points (s5,6,9,10) :

In the presence of AWGN, the probability of correct detection given that si, i = 5, 6, 9,

10, is transmitted is

P(C/si) = P(-d/2 < n1 < d/2) x P(-d/2 < n2 < d/2)

= P(-d/2 < n1 < d/2) x P(-d/2 < n1 < d/2)

= p x p (30.32)

where

p = P(-d/2 < n1 < d/2)

=

−∫

d

d

/

/

2

21

2 2

12

2 2

πσ

φ

σe−

dφ1

= 2

0

2d /

∫ 1

2 2

12

2 2

πσ

φ

σe−

dφ1 (30.33)

30.9


Let y = φσ1

2. Then y2 =

φ

σ12

2 2 and d y = dφ

σ1



p = 2

0

2 2d

σ∫ 1

πe -y2 dy

= erf(d

2 2σ)

= erf(d

n2 0) (30.34)

where the error function is

erf (x) = 2π

0

x

∫ e-α2 dα (30.35)

and σ2 = n0/2 for AWGN.

Case 2 - four corner signal points (s0,3,12,15) :


15, is transmitted is

P(C/si) = P( −∞ < n1 < d/2) x P(-d/2 < n2 < ∞) (30.36)

= P(-d/2 < n1 < ∞) x P(-d/2 < n2 < ∞)

= P(-d/2 < n1 < ∞) x P(-d/2 < n1 < ∞)

= r x r

where

r = P(-d/2 < n1 < ∞)

=

−

∞∫

d / 2

1

2 2

12

2 2

πσ

φ

σe−

dφ1

30.10


=

−∫

d / 2

01

2 2

12

2 2

πσ

φ

σe−

dφ1 +

0

∞∫ 1

2 2

12

2 2

πσ

φ

σe−

dφ1

=

0

2d /

∫ 1

2 2

12

2 2

πσ

φ

σe−

dφ1 +

0

∞∫ 1

2 2

12

2 2

πσ

φ

σe−

dφ1

= 0.5p + 0.5 (30.37)

Case 3 - eight edge signal points (s1,2,4,7,8,11,13,14) :


7, 8, 11, 13, 14, is transmitted is

P(C/si) = P(-d/2 < n1 < d/2) x P(-d/2 < n2 < ∞)

= P(-d/2 < n1 < d/2) x P(-d/2 < n1 < ∞)

= p x r (30.38)

Let P(si) be the probability of sending si, for i = 0, 1, ..., 15. Thus, the total probability of

correct detection is

Pc = i

M

=

− =∑

0

1 15

P sia priori probability

( )123

P(C/si) (30.39)

With equally likely transmission of si, we have

Pc = 4 (116

p x p) + 4 (116

r x r) + 8(116

p x r)

= 116

[ 4 p2 + 4(p2

+12

)2 + 8p (p2

+12

)]

= ( 3p + 1)

2

16(30.40)

The probability of error is

Pe = (1 - Pc)

= 1 - ( 3p + 1)

2

16 =

42

− ( 3p + 1)2

16

= 3(1 −p)(5 + 3p)

16

30.11


For normal operation, p approaches 1 and Pe << 1, and then

Pe ≈ 3(1 −p)8

16

= 32

(1 - p) (30.41)

Substituting p = erf (d

2 n0

) and d = 25

E into equation (30.41), we have

Pe ≈ 32

[1 −erf (E

10n0

)]

= 32

erfc (E

10n0

) (30.42)

Example 30.1

Consider the 4-ary QAM signal constellation diagram as shown in Figure 30.6.


The average signal energy is

E = 14

[ 42

4

2

4× +

d d] =

d2

2

and

d = 25

E .

From our previous analysis of 16-ary QAM signals, the probability of correct detection in4-ary QAM signals given that si, i = 0, 1, 2, 3, is transmitted is

P(C/si) = P( −∞ < n1 < d/2) x P(-d/2 < n2 < ∞)

= P(-d/2 < n1 < ∞) x P(-d/2 < n2 < ∞)

= P(-d/2 < n1 < ∞) x P(-d/2 < n1 < ∞)

= r x r

where

30.12


r = 0.5p + 0.5

p = erf(d

2 2σ) = erf(

d

n2 0)

and σ2 = n0/2 for AWGN.

The total probability of correct detection is

Pc = i

M

=

− =∑

0

1 3

P sia priori probability

( )123

P(C/si)

With equally likely transmission of si, we have

Pc = 4 (14

r x r)

= (p2

+12

)2

= ( )1 2

4+ p

The probability of error is

Pe = (1 - Pc) = 1 - ( )1 2

4+ p

= ( )( )1 3

4− +p p

For normal operation, p approaches 1 and Pe << 1, and then

Pe ≈ 4 1

4( )− p

= (1 - p) (30.43)

Substituting p = erf (d

2 n0

) and d = 2E into equation (30.43), we have

Pe ≈ 12 0

− erfE

n( ) = erfc

E

n( )

2 0.

30.13


References

[1] M. Schwartz, Information Transmission, Modulation, and Noise, 4/e, McGraw-Hill, 1990.

[2] H. Taub and D. L. Schilling, Principles of Communication Systems, 2/e,MaGraw-Hill, 1996.

30.14


Probability density function

d E=0 d/2φ

Pe0Pe1

1

f ( φ /1)1

2π σ 2e 2σ 2

-=1

(φ )21d2

f ( φ /0)1

2πσ 2e 2σ 2

φ 2-

= 1

1

s 1s 0

φ 2

= 2T

cos 2 π f c t

= 2T

sin 2 π f c t

Figure 30.1 BASK signal constellation diagram and the conditional probability densityfunctions associated with the signals.

d2

E=-d2

10φ

Pe0Pe1

2φ

0s

1s

1

2π σ 2e 2σ 2

-= f ( φ /0)1

(φ ) 21

d2

+

f ( φ /1)1

2π σ 2e 2σ 2

-=1

(φ )21d2

-

= 2T

cos 2 π f c t

= 2T

sin 2 π f c t

Figure 30.2 BPSK signal constellation diagram and the conditional probability densityfunctions associated with the signals.

f (u /1)1

2π σ 2e 2σ 2

-=1

(u - )21

d2

1

2π σ 2e 2σ 2

-=f (u /0)1

(u + )21

d2

0φ

1

u 1

u 2

φ2

0

E2d

=

E2d

=

s 0

s 1

d

= 2T

cos 2 πf c t1

=2T

cos 2 π f c t2

Figure 30.3 Orthogonal BFSK signal constellation diagram and the conditionalprobability density functions associated with the signals.

30.15


s 1 s 3

s 7 s 6 s 5 s 4

s 8 s 9 s 11

s 15 s 14 s 13 s 12

d

d

φ1

φ2

0

d/2

d/2

d/2 d/2 d/2

s 10

s 2s 0

Decision boundary

= 2T

cos 2 π f c t

= 2T

sin 2 π f c t


φ1

φ2

-d/2 d/2

s 10

0

d/2

-d/2

(a)

s 0φ

1

φ2

d/20

-d/2

(b)

s 2 φ1

φ2

d/20-d/2

-d/2

(c)

Figure 30.5 Decision regions.

30.16


s 0 s 1

s 3

φ1

φ2

0

d/2

d/2

d/2 d/2

s 2

Decision boundary

= 2T

cos 2 π f c t

= 2T

sin 2 π f c t


30.17

signal space analysis of bask, bfsk, bpsk, and qam on mac

Documents

signal space analysis

conditional probability densityfunctions

complementary error function

orthogonal bfsk signals

modulation systems

transmitted isp

dy intoequation

written ass