Doc. no. ESS140/ext:10, rev. A
Notes on Quadrature Amplitude Modulation Erik Str¨om October 22, 2002
1
Introduction
This document should be viewed as a complement to Sections 7.3.3, 7.5.6, and 7.6.5 of Proakis and Salehi’s book [1]. In particular, an error in the book’s equation (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 the sum 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 2 gT (t) cos(2πfc t) ψI = Eg 2 gT (t) sin(2πfc t), ψQ = Eg and where Eg is the energy of the pulse shape gT (t). The pulse shape and the carrier frequency fc is chosen such that the power spectral density of the transmitted 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 are approximately orthonormal if the carrier frequency fc is much larger than the bandwidth of gT (t). The approximation is very good for many practical choices of fc and gT (t), and we will from now consider ψI (t) and ψQ (t) to be exactly orthonormal and to make up the standard basis for the IQ signal space. The reader should here be cautioned that some texts use −ψQ (t) as one of the basis functions. Hence, we need to be a bit careful when using material on IQ modulation from different sources. 1
2 (12)
Notes on Quadrature Amplitude Modulation PAM with ψ I (t )
sI , m
2-dimensional constellation bits
bits/ symbol
PAM 1 Eg
gT ( t ) 2 cos(2π fc t )
symbol/ vector
m
sI sQ
s (t )
PAM
transmitted vector s =
1 sQ ,m
Eg
gT ( t )
2 sin(2π f c t ) PAM with ψ Q (t )
Figure 1: Generic IQ-modulator A generic demodulator is shown in Figure 2. We will assume that the channel noise is white and Gaussian with power spectral density N0 /2. It then follows that the noise vector elements, nI and nQ , are independent Gaussian random variables with zero mean and variance N0 /2. Let us consider the transmission of the kth signal vector an M-ary constelT lation {s1 , s2 , . . . , sM }. The transmitted vector is sk = sI,k sQ,k , and the transmitted signal is therefore sk (t) = sI,k ψI (t) + sQ,k ψQ (t) 2 2 gT (t) cos(2πfc t) + sQ,k gT (t) sin(2πfc t) = sI,k Eg Eg
If we define Ak and θk as (see Figure 3 for a geometrical interpretation) Ak = sk = s2I,k + s2Q,k , sQ,k . θk = − tan 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
3 (12)
Notes on Quadrature Amplitude Modulation
Matched filter with respect to ψ I (t )
t = nT + T0
1 gT (T0 − t ) Eg s (t )
r (t )
2 cos(2π fc t )
t = nT + T0
Decision rule
mˆ
1 gT (T0 − t ) Eg n(t )
2 sin(2π f c t )
sI nI + sQ nQ
r =s+n =
Matched filter with respect to ψ Q (t )
Figure 2: Generic IQ-demodulator
ψ Q (t ) sQ ,k
sI ,k sQ ,k
Ak
sk =
θk sI ,k
ψ I (t )
Figure 3: Geometric interpretation of Ak and θk . Note that θ taken as the angle from 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 2 2 sk (t) = Ak cos(θk ) gT (t) cos(2πfc t) − Ak sin(θk ) gT (t) sin(2πfc t) Eg Eg 2 gT (t) cos(2πfc t + θk ), = Ak Eg
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) multiplied 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 both the amplitude and phase of the transmitted signal. We can choose the signal constellation such that the amplitude is the same for all 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 then carried by the phase of the carrier. Phase-shift keying (PSK) is an example of such a modulation scheme. Conversely, if the signal vectors are placed on a straight line that crosses the origin in the signal space, then the carrier phase will be the same for all signal alternatives. (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 between signal alternatives is called quadrature amplitude modulation (QAM). There exists (of course) many possible QAM constellations, but we will limit the discussion here to when the signal points are placed on a regular rectangular grid in the signal space, see Figure 4 for two examples. Moreover, we will only consider constellation sizes such that M = 2k , where k is an integer. That is, each symbol represents k bits. 16-QAM
8-QAM
ψ Q (t )
−3 A
ψ Q (t )
3A
3A
A
A
−A
A
3A
ψ I (t )
−3 A
−A
A
−A
−A
−3 A
−3 A
3A
ψ I (t )
Figure 4: Rectangular QAM constellations As seen from Figure 4, the signal vectors are spaced with the distance 2A along the axes. Hence, the minimum distance of the constellation is √ dmin = 2A. 2 When M is an even square, e.g., when M = 16 = 4 , 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
5 (12)
Notes on Quadrature Amplitude Modulation
When M is not an even square, e.g., M = 8, the signal points are more spread along one axis than the other. Regardless if M is an even square or not, the maximum likelihood decision regions will be rectangular-shaped. The regions can be squares (type a), squares with one open side (type b) or squares with two open sides (type c), see Figure 5.
ψ Q (t ) type a
3A type c
type a
type b
A type b
−3 A
−A
A
3A
ψ I (t )
−A type c
−3 A
Figure 5: ML decision regions for 16-QAM To compute the symbol error probability, we therefore need only to compute the conditional error probabilities for the three types of decision regions. Conditioned on that we send a symbol that has a decision region of type a, the probability of wrong decision is denoted Pe|a and the probability of correct decision is denoted Pc|a. Since the elements on the noise vector, nI and nQ , are independent Gaussian random variables with zero mean and variance N0 /2, we can express Pc|a = Pr{−A < nI < A, −A < nQ < A} = Pr{−A < nI < A} Pr{−A < nQ < A} = [Pr{−A < nI < A}]2 . From Figure 6, we see that Pr{−A < nI < A} can be written in terms of the Q-function as Pr{−A < nI < A} = Ω2 = 1 − 2Q Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
2A2 N0
.
6 (12)
Notes on Quadrature Amplitude Modulation
Hence,
2 2A2 = 1 − 1 − 2Q N0 2 2 2A 2A = 4Q − 4Q2 . N0 N0
Pe|a
x2 1 exp − π N0 N0
f nI ( x ) =
A N 0 / 2
Ω 2 = 1 − 2Q
Ω1 = Q
A N 0 / 2
A N 0 / 2
Ω3 = Q
−A
x
A
Figure 6: The probability density function for nI and nQ is a zero mean Gaussian distribution with variance N0 /2. The areas Ω1 , Ω2 , and Ω3 represents the probabilities Ω1 = Pr{nI < −A}, Ω2 = Pr{−A < nI < A}, and Ω3 = Pr{nI > A}. Proceeding with the type b region, we note that Pc|b = Pr{nI > −A, −A < nQ < A} = Pr{nI > −A} Pr{−A < nQ < A} = (1 − Ω1 )Ω2 2A2 2A2 = 1−Q 1 − 2Q N0 N0 2A2 2A2 + 2Q2 , = 1 − 3Q N0 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
7 (12)
Notes on Quadrature Amplitude Modulation
Finally, Pc|c = Pr{−A < nI , −A < nQ } = Pr{nI > −A} Pr{nQ > −A} = (1 − Ω1 )(1 − Ω1 ) 2 2A2 = 1−Q N0 2 2A2 2A = 1 − 2Q + Q2 , N0 N0 Pe|c = 2Q
2A2 N0
− Q2
2A2 N0
.
Suppose that all symbols are transmitted with the same probability 1/M. If na denotes the number of type a regions in the constellation then na /M is the probability that a symbol with a type a decision region is transmitted. Hence, we can compute the (average) symbol error probability as Pe =
1 (na Pe|a + nb Pe|b + nb Pe|b ). M
(1)
We always have nc = 4 (there is one type c region per corner, and there are always four corners), but na and nb depends on M. For 16-QAM, we have that na = 4 and nb = a = 0 and nb = 4. If M is an even square, √ 8, and 2for 8-QAM n√ we have na = ( M − 2) and nb = 4( M − 2). It is desirable to put the final result in terms of Es /N0 , where Es is the average energy per symbol. Assuming equally likely transmitted symbols, it is possible to compute Es for any QAM constellation by just considering the first quadrant in the signal space. For 16-QAM, 1 Es = [(A2 + A2 ) + 2(A2 + 9A2 ) + (9A2 + 9A2 )] = 10A2 , 4 and for 8-QAM, we have 1 Es = [(A2 + A2 ) + (A2 + 9A2 )] = 6A2 . 2 It can be shown (see Section A) that when M is an even square, we have that Es =
2A2 A2 2(M − 1) 3Es ⇒ = . 3 N0 (M − 1)N0
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
(2)
8 (12)
Notes on Quadrature Amplitude Modulation
We can now write the symbol error probability for QAM when M is an even square (assuming equally likely transmitted symbols, ML detection, and AWGN channel with noise power spectral density N0 /2) as 1 (na Pe|a + nb Pe|b + nc Pe|c ) M √ 3E 3E 1 s s ( M − 2)2 4Q = − 4Q2 M (M − 1)N0 (M − 1)N0 √ 3Es 3E s + 4( M − 2) 3Q − 2Q2 (M − 1)N0 (M − 1)N0
3Es 3E s − Q2 + 4 2Q (M − 1)N0 (M − 1)N0 √ 3Es 4 (M − M)Q = M (M − 1)N0 √ 3E 4 s − (M − 2 M + 1)Q2 (3) M (M − 1)N0
Pe =
We can form a simple upper bound on the symbol error probability by noting that Pe|a > Pe|b > Pe|c . Hence, by replacing Pe|b and Pe|c in (1) with Pe|a and since na + nb + nc = M, we can write 1 (na Pe|a + nb Pe|b + nc Pe|c ) M 1 < (na Pe|a + nb Pe|a + nc Pe|a) M = Pe|a 2 2A2 2A = 4Q − 4Q2 N0 N0 2A2 . < 4Q N0
Pe =
(4)
(5)
The bounds (4) and (5) are valid for any M (assuming equally likely transmitted symbols, rectangular constellation, ML detection, and AWGN channel with noise power spectral density N0 /2). The bound presented in Proakis and Salehi [1], equation (7.6.71) is not correct1 . 1 A counterexample can be found, e.g., by computing the exact symbol error probability according 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 a upper bound.
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
9 (12)
Notes on Quadrature Amplitude Modulation
Finally, we recall that the minimum distance of a QAM constellation is dmin = 2A, and a standard union bound therefore yields 2 dmin Pe ≤ (M − 1)Q 2N0 4A2 = (M − 1)Q 2N0 2A2 = (M − 1)Q , (6) N0 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 T values of sQ and M = MI MQ , where s = sI sQ . Due to symmetry, we can compute Es by only considering the signal points in the first quadrant, which are of the form s 2i − 1 , I = A for i = 1, 2, . . . , MI /2 and j = 1, 2, . . . , MQ /2, 2j − 1 sQ Hence, sI = A(2i − 1) and sQ = A(2j − 1) and the energy of the corresponding signal alternative is s2I + s2Q = A2 [(2i − 1)2 + (2j − 1)2 ]. Since there are M/4 signal vectors in the first quadrant, we can compute Es as MI /2 MQ /2 4 2 A Es = (2i − 1)2 + (2j + 1)2 . M i=1 j=1
This can be simplified to (see Section A) Es =
A2 (MI2 + MQ2 − 2). 3
(7)
By combining (4), (5), or (6) with (7), we can form a bound on the symbol error probability in terms of Es /N0 . As an example, we see that for the M = 8 constellation in Figure 4, MI = 4 and MQ = 2, and A2 Es = (16 + 4 − 2) = 6A2 , 3 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 in Figure 7. Symbol error probability for rectangular 8−QAM
0
10
Standard Union Bound Bound 1 Bound 2 Exact
−1
10
−2
10
−3
10
−4
10
−5
10
−6
10
0
2
4
6
8
10 12 10 log10 Es/N0
14
16
18
20
Figure 7: Plots of the exact expression and some upper bounds on the symbol error probability for rectangular 8-QAM (see Figure 4). Bound 1 and 2 are defined by (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 error probability for rectangular QAM (assuming ML detection, AWGN channel, and equally likely transmitted symbols). The general expression, found in (1), can be further simplified for the case when M is an even square, see (3). A number of 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 combined with the symbol error formulas to form yield expressions that depend only on Es /N0 .
A
Derivation of (2) and (7)
First of all, we recall the following formulas (see, e.g., [2, p. 189]) n
j=
j=0 n
j2 =
j=0
n(n + 1) 2 n(n + 1)(2n + 1) 6
Hence, n n 2 (2i − 1) = 4i2 − 4i + 1 j=1
j=1
=4
n(n + 1)(2n + 1) n(n + 1) −4 +n 6 2
2 = n(n + 1)(2n + 1) − 2n(n + 1) + n 3 n = [2(n + 1)(2n + 1) − 6(n + 1) + 3] 3 n 2 = [4n + 6n + 2 − 6n − 6 + 3] 3 n = (4n2 − 1). 3 In particular,
MQ /2
j=1
1 MQ (2j + 1) = 3 2 2
MQ2 MQ 4 2 −1 = (MQ2 − 1). 2 6
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 MI /2 MQ /2 4 2 A Es = (2i − 1)2 + (2j + 1)2 M i=1 j=1 MQ /2 MQ /2 MI /2 4 2 = (2i − 1)2 + (2j + 1)2 A M i=1 j=1 j=1
MI /2 4 2 MQ MQ 2 2 A (2i − 1) + (MQ − 1) = M 2 6 i=1 MI /2 MI /2 MQ MQ 4 2 A (2i − 1)2 + (MQ2 − 1) = M 2 6 i=1 i=1 MI MQ 4 2 MQ MI 2 2 = A (MI − 1) + (MQ − 1) M 2 6 2 6 4 2 MI MQ 2 = A [MI + MQ2 − 1] M 12 A2 2 = [MI + MQ2 − 2], 3 where we have used that MI MQ = 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 A2 2 A2 A2 2 Es = [MI + MQ − 2] = [M + M − 2] = 2(M − 1), 3 3 3 and this justifies (2).
B
References
[1] John G. Proakis and Masoud Salehi. Communication Systems Engineering. Prentice-Hall, second edition, 2002. [2] Lennart R˚ ade and Bertil Westergren. Beta, Mathematics Handbook. Studentlitteratur, fourth edition, 1998.
Doc. no.: ESS140/ext:10, rev.: A, date: October 22, 2002, file: qam-notes.tex
