Upload
others
View
8
Download
0
Embed Size (px)
Citation preview
Doc. no. ESS140/ext:10, rev. A
Notes on Quadrature Amplitude Modulation
Erik Strom
October 22, 2002
1 Introduction
This document should be viewed as a complement to Sections 7.3.3, 7.5.6, and7.6.5 of Proakis and Salehi’s book [1]. In particular, an error in the book’sequation (7.6.71) is pointed out, and a corrected formula is presented.
2 Quadrature Modulation
A block diagram for a generic inphase quadrature phase modulator (IQ-modulator)is found in Figure 1. We see that the transmitted signal can be viewed as thesum of two PAM processes with different pulse shapes: ψI(t) in the top (in-phase)branch and ψQ(t) in the lower (quadrature) branch, where
ψI =
√2
EggT (t) cos(2πfct)
ψQ =
√2
EggT (t) sin(2πfct),
and where Eg is the energy of the pulse shape gT (t). The pulse shape andthe carrier frequency fc is chosen such that the power spectral density of thetransmitted signal will fit the frequency response of the channel.
As implied by the IQ-modulator block diagram, the signals ψI(t) and ψQ(t)spans a two-dimensional signal space. In fact, it can be shown that signals areapproximately orthonormal if the carrier frequency fc is much larger than thebandwidth of gT (t). The approximation is very good for many practical choicesof fc and gT (t), and we will from now consider ψI(t) and ψQ(t) to be exactly or-thonormal and to make up the standard basis for the IQ signal space. The readershould here be cautioned that some texts use −ψQ(t) as one of the basis func-tions. Hence, we need to be a bit careful when using material on IQ modulationfrom different sources.
1
2 (12) Notes on Quadrature Amplitude Modulation
bits/
symbol
symbol/
vector
PAM
1( )T
g
g tE
PAM
1( )T
g
g tE
2 cos(2 )cf tπ
2 sin(2 )cf tπ
,I ms
,Q ms
transmitted vectorI
Q
s
s
=
s
( )s t
PAM with ( )I tψ
PAM with ( )Q tψ
2-dimensional
constellation
mbits
Figure 1: Generic IQ-modulator
A generic demodulator is shown in Figure 2. We will assume that the channelnoise is white and Gaussian with power spectral density N0/2. It then followsthat the noise vector elements, nI and nQ, are independent Gaussian randomvariables with zero mean and variance N0/2.
Let us consider the transmission of the kth signal vector an M-ary constel-
lation s1, s2, . . . , sM. The transmitted vector is sk =[sI,k sQ,k
]T, and the
transmitted signal is therefore
sk(t) = sI,kψI(t) + sQ,kψQ(t)
= sI,k
√2
EggT (t) cos(2πfct) + sQ,k
√2
EggT (t) sin(2πfct)
If we define Ak and θk as (see Figure 3 for a geometrical interpretation)
Ak = ‖sk‖ =√s2
I,k + s2Q,k,
θk = − tansQ,k
sI,k.
According to the choice of definition of the angle θk, we have that
sI,k = Ak cos(−θk) = Ak cos(θk)
sQ,k = Ak sin(−θk) = −Ak sin(θk),
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
Notes on Quadrature Amplitude Modulation 3 (12)
2 cos(2 )cf tπ
2 sin(2 )cf tπ
Matched filter with respect to ( )I tψ
0
1( )T
g
g T tE
−
0
1( )T
g
g T tE
−
Matched filter with respect to ( )Q tψ
( )s t
( )n t
( )r t
0t nT T= +
Decision
rulem
0t nT T= +
I I
Q Q
s n
s n
= + = +
r s n
Figure 2: Generic IQ-demodulator
,
,
I k
kQ k
s
s
=
s
( )Q tψ
( )I tψ
kθ
,Q ks
,I ks
kA
Figure 3: Geometric interpretation of Ak and θk. Note that θ taken as the anglefrom the vector to the ψI(t) axis, not the other way around. In this figure,θk ≈ −0.12π.
and we can rewrite the transmitted signal as
sk(t) = Ak cos(θk)
√2
Eg
gT (t) cos(2πfct) −Ak sin(θk)
√2
Eg
gT (t) sin(2πfct)
= Ak
√2
Eg
gT (t) cos(2πfct+ θk),
where we have used the trigonometric identity cos(α) cos(β) − sin(α) sin(β) =cos(α+ β).
We conclude that the transmitted signal can be seen as the pulse gT (t) mul-tiplied with a cosine-carrier, where the amplitude and phase of the carrier is
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
4 (12) Notes on Quadrature Amplitude Modulation
determined by sI,k and sQ,k. Hence, the transmitted information can affect boththe amplitude and phase of the transmitted signal.
We can choose the signal constellation such that the amplitude is the same forall signal alternatives, by placing the signal vectors on a circle in the signal space.This implies that A1 = A2 = · · · = AM , and the transmitted information is thencarried by the phase of the carrier. Phase-shift keying (PSK) is an example ofsuch a modulation scheme.
Conversely, if the signal vectors are placed on a straight line that crosses theorigin in the signal space, then the carrier phase will be the same for all signalalternatives. (Which follows from the fact that sI,k/sQ,k is the same for all k.)Amplitude-shift keying (ASK) is an example of such a modulation scheme.
The general case when both amplitude and phase is allowed to change betweensignal alternatives is called quadrature amplitude modulation (QAM). There ex-ists (of course) many possible QAM constellations, but we will limit the discus-sion here to when the signal points are placed on a regular rectangular grid inthe signal space, see Figure 4 for two examples. Moreover, we will only considerconstellation sizes such that M = 2k, where k is an integer. That is, each symbolrepresents k bits.
( )ψ I t
( )ψ Q t
−A3− A A 3A
A
3A
−A
3− A
16-QAM
( )ψ I t
( )ψ Q t
−A3− A A 3A
A
3A
−A
3− A
8-QAM
Figure 4: Rectangular QAM constellations
As seen from Figure 4, the signal vectors are spaced with the distance 2Aalong the axes. Hence, the minimum distance of the constellation is dmin = 2A.When M is an even square, e.g., when M = 16 = 42, there will be
√M possible
amplitudes for both sI and sQ. In fact,
sI ∈ ±A,±3A, . . . ,±A(√M − 1)
sQ ∈ ±A,±3A, . . . ,±A(√M − 1).
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
Notes on Quadrature Amplitude Modulation 5 (12)
When M is not an even square, e.g., M = 8, the signal points are more spreadalong one axis than the other.
Regardless if M is an even square or not, the maximum likelihood decisionregions will be rectangular-shaped. The regions can be squares (type a), squareswith one open side (type b) or squares with two open sides (type c), see Figure 5.
( )ψ I t
( )ψQ t
−A3− A A 3A
A
3A
−A
3− A
type a type b
type c
type a
type b
type c
Figure 5: ML decision regions for 16-QAM
To compute the symbol error probability, we therefore need only to computethe conditional error probabilities for the three types of decision regions. Con-ditioned on that we send a symbol that has a decision region of type a, theprobability of wrong decision is denoted Pe|a and the probability of correct de-cision is denoted Pc|a. Since the elements on the noise vector, nI and nQ, areindependent Gaussian random variables with zero mean and variance N0/2, wecan express
Pc|a = Pr−A < nI < A,−A < nQ < A= Pr−A < nI < APr−A < nQ < A= [Pr−A < nI < A]2.
From Figure 6, we see that Pr−A < nI < A can be written in terms of theQ-function as
Pr−A < nI < A = Ω2 = 1 − 2Q
(√2A2
N0
).
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
6 (12) Notes on Quadrature Amplitude Modulation
Hence,
Pe|a = 1 −[1 − 2Q
(√2A2
N0
)]2
= 4Q
(√2A2
N0
)− 4Q2
(√2A2
N0
).
3
0 / 2
AQ
N
Ω =
A
2
00
1( ) exp
In
xf x
NNπ
= −
A−
1
0 / 2
AQ
N
Ω =
2
0
1 2/ 2
AQ
N
Ω = −
x
Figure 6: The probability density function for nI and nQ is a zero mean Gaussiandistribution with variance N0/2. The areas Ω1, Ω2, and Ω3 represents the prob-abilities Ω1 = PrnI < −A, Ω2 = Pr−A < nI < A, and Ω3 = PrnI > A.
Proceeding with the type b region, we note that
Pc|b = PrnI > −A,−A < nQ < A= PrnI > −APr−A < nQ < A= (1 − Ω1)Ω2
=
[1 −Q
(√2A2
N0
)][1 − 2Q
(√2A2
N0
)]
= 1 − 3Q
(√2A2
N0
)+ 2Q2
(√2A2
N0
),
and
Pe|b = 3Q
(√2A2
N0
)− 2Q2
(√2A2
N0
).
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
Notes on Quadrature Amplitude Modulation 7 (12)
Finally,
Pc|c = Pr−A < nI ,−A < nQ= PrnI > −APrnQ > −A= (1 − Ω1)(1 − Ω1)
=
[1 −Q
(√2A2
N0
)]2
= 1 − 2Q
(√2A2
N0
)+Q2
(√2A2
N0
),
Pe|c = 2Q
(√2A2
N0
)−Q2
(√2A2
N0
).
Suppose that all symbols are transmitted with the same probability 1/M . Ifna denotes the number of type a regions in the constellation then na/M is theprobability that a symbol with a type a decision region is transmitted. Hence,we can compute the (average) symbol error probability as
Pe =1
M(naPe|a + nbPe|b + nbPe|b). (1)
We always have nc = 4 (there is one type c region per corner, and there arealways four corners), but na and nb depends on M . For 16-QAM, we have thatna = 4 and nb = 8, and for 8-QAM na = 0 and nb = 4. If M is an even square,we have na = (
√M − 2)2 and nb = 4(
√M − 2).
It is desirable to put the final result in terms of Es/N0, where Es is the averageenergy per symbol. Assuming equally likely transmitted symbols, it is possibleto compute Es for any QAM constellation by just considering the first quadrantin the signal space. For 16-QAM,
Es =1
4[(A2 + A2) + 2(A2 + 9A2) + (9A2 + 9A2)] = 10A2,
and for 8-QAM, we have
Es =1
2[(A2 + A2) + (A2 + 9A2)] = 6A2.
It can be shown (see Section A) that when M is an even square, we have that
Es =A22(M − 1)
3⇒ 2A2
N0
=3Es
(M − 1)N0
. (2)
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
8 (12) Notes on Quadrature Amplitude Modulation
We can now write the symbol error probability for QAM when M is an evensquare (assuming equally likely transmitted symbols, ML detection, and AWGNchannel with noise power spectral density N0/2) as
Pe =1
M(naPe|a + nbPe|b + ncPe|c)
=1
M
(√M − 2)2
[4Q
(√3Es
(M − 1)N0
)− 4Q2
(√3Es
(M − 1)N0
)]
+ 4(√M − 2)
[3Q
(√3Es
(M − 1)N0
)− 2Q2
(√3Es
(M − 1)N0
)]
+ 4
[2Q
(√3Es
(M − 1)N0
)−Q2
(√3Es
(M − 1)N0
)]
=4
M(M −
√M)Q
(√3Es
(M − 1)N0
)
− 4
M(M − 2
√M + 1)Q2
(√3Es
(M − 1)N0
)(3)
We can form a simple upper bound on the symbol error probability by notingthat Pe|a > Pe|b > Pe|c. Hence, by replacing Pe|b and Pe|c in (1) with Pe|a andsince na + nb + nc = M , we can write
Pe =1
M(naPe|a + nbPe|b + ncPe|c)
<1
M(naPe|a + nbPe|a + ncPe|a)
= Pe|a
= 4Q
(√2A2
N0
)− 4Q2
(√2A2
N0
)(4)
< 4Q
(√2A2
N0
). (5)
The bounds (4) and (5) are valid for any M (assuming equally likely transmittedsymbols, rectangular constellation, ML detection, and AWGN channel with noisepower spectral density N0/2). The bound presented in Proakis and Salehi [1],equation (7.6.71) is not correct1.
1A counterexample can be found, e.g., by computing the exact symbol error probabilityaccording to (1) for the M = 8 constellation in Figure 4. For Es/N0 slightly larger than 9.5 dB,(7.6.71) is actually smaller than the exact symbol error probability. Hence, (7.6.71) is not aupper bound.
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
Notes on Quadrature Amplitude Modulation 9 (12)
Finally, we recall that the minimum distance of a QAM constellation is dmin =2A, and a standard union bound therefore yields
Pe ≤ (M − 1)Q
√d2
min
2N0
= (M − 1)Q
(√4A2
2N0
)
= (M − 1)Q
(√2A2
N0
), (6)
which is not as tight as (4) or (5).For (4), (5), or (6) to be really useful, we usually need to relate A2 to Es. For
a rectangular constellation, there are MI possible values of sI and MQ possible
values of sQ and M = MIMQ, where s =[sI sQ
]T. Due to symmetry, we can
compute Es by only considering the signal points in the first quadrant, which areof the form
sI
sQ
= A
2i− 1
2j − 1
, for i = 1, 2, . . . ,MI/2 and j = 1, 2, . . . ,MQ/2,
Hence, sI = A(2i − 1) and sQ = A(2j − 1) and the energy of the correspondingsignal alternative is
s2I + s2
Q = A2[(2i− 1)2 + (2j − 1)2].
Since there are M/4 signal vectors in the first quadrant, we can compute Es as
Es =4
MA2
MI/2∑i=1
MQ/2∑j=1
(2i− 1)2 + (2j + 1)2.
This can be simplified to (see Section A)
Es =A2
3(M2
I +M2Q − 2). (7)
By combining (4), (5), or (6) with (7), we can form a bound on the symbol errorprobability in terms of Es/N0.
As an example, we see that for the M = 8 constellation in Figure 4, MI = 4and MQ = 2, and
Es =A2
3(16 + 4 − 2) = 6A2,
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
10 (12) Notes on Quadrature Amplitude Modulation
and the bound (5) evaluates to
Pe < Q
(√2A2
N0
)= 4Q
(√Es
3N0
).
Plots of the exact error probability and the bounds (4), (5), and (6) are found inFigure 7.
0 2 4 6 8 10 12 14 16 18 2010
−6
10−5
10−4
10−3
10−2
10−1
100
Symbol error probability for rectangular 8−QAM
10 log10
Es/N
0
Standard Union BoundBound 1Bound 2Exact
Figure 7: Plots of the exact expression and some upper bounds on the symbolerror probability for rectangular 8-QAM (see Figure 4). Bound 1 and 2 are definedby (4) and (5), respectively. The Standard Union Bound is defined by (6).
3 Summary
In these notes, we have developed a method for computing the exact symbol errorprobability for rectangular QAM (assuming ML detection, AWGN channel, andequally likely transmitted symbols). The general expression, found in (1), canbe further simplified for the case when M is an even square, see (3). A numberof upper bounds on the symbol error probability has also been presented in (4),(5), and (6), and it was noted that (7.6.71) in Proakis and Salehi [1] is in error.
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
Notes on Quadrature Amplitude Modulation 11 (12)
To compute the symbol energy, we can use (7). The result can be combinedwith the symbol error formulas to form yield expressions that depend only onEs/N0.
A Derivation of (2) and (7)
First of all, we recall the following formulas (see, e.g., [2, p. 189])
n∑j=0
j =n(n + 1)
2
n∑j=0
j2 =n(n + 1)(2n+ 1)
6
Hence,
n∑j=1
(2i− 1)2 =
n∑j=1
4i2 − 4i+ 1
= 4n(n+ 1)(2n+ 1)
6− 4
n(n+ 1)
2+ n
=2
3n(n + 1)(2n+ 1) − 2n(n+ 1) + n
=n
3[2(n+ 1)(2n+ 1) − 6(n+ 1) + 3]
=n
3[4n2 + 6n + 2 − 6n− 6 + 3]
=n
3(4n2 − 1).
In particular,
MQ/2∑j=1
(2j + 1)2 =1
3
MQ
2
(4M2
Q
22− 1
)=MQ
6(M2
Q − 1).
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
12 (12) Notes on Quadrature Amplitude Modulation
The symbol energy is
Es =4
MA2
MI/2∑i=1
MQ/2∑j=1
(2i− 1)2 + (2j + 1)2
=4
MA2
MI/2∑i=1
MQ/2∑
j=1
(2i− 1)2
+
MQ/2∑
j=1
(2j + 1)2
=4
MA2
MI/2∑i=1
[MQ
2(2i− 1)2 +
MQ
6(M2
Q − 1)
]
=4
MA2
MI/2∑
i=1
MQ
2(2i− 1)2
+
MI/2∑
i=1
MQ
6(M2
Q − 1)
=4
MA2
[MQ
2
MI
6(M2
I − 1) +MI
2
MQ
6(M2
Q − 1)
]
=4
MA2MIMQ
12[M2
I +M2Q − 1]
=A2
3[M2
I +M2Q − 2],
where we have used that MIMQ = M . This completes the derivation of (7).The formula holds for any rectangular QAM; in particular, if M is an even
square and MI = MQ =√M , the formula reduces to
Es =A2
3[M2
I +M2Q − 2] =
A2
3[M +M − 2] =
A2
32(M − 1),
and this justifies (2).
B References
[1] John G. Proakis and Masoud Salehi. Communication Systems Engineering.Prentice-Hall, second edition, 2002.
[2] Lennart Rade and Bertil Westergren. Beta, Mathematics Handbook. Stu-dentlitteratur, fourth edition, 1998.
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex