TUTORIAL ON DIGITAL MODULATIONS
Part 15: m-QAM
[last modified: 2010
2010--1111-25]
Roberto Garello
Garello,, Politecnico di Torino
Free download at: www.tlc.polito.it/
www.tlc.polito.it/garello
garello
(personal use only
only))
1
m-QAM: characteristics
1. Band-pass modulation
2. 2D signal set
3. Basis signals p(t)cos(2πf0t) e p(t)sin(2πf0t)
4. Constellation = m points on the plane (typically on a grid)
5. Information associated to both the amplitude and the carrier phase
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
2
m-QAM:
QAM: constellation
SIGNAL SET
NOT CONSTANT ENVELOPE
M = { si (t ) = Ai p (t ) cos(2π f 0t − ϕi ) }
m
i =1
Information associated to both the amplitude and the carrier phase
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
3
m-QAM:
QAM: constellation
M = { si (t ) = Ai p(t ) cos(2π f 0t − ϕi ) }im=1
We can write
si (t ) = ( Ai cos ϕi ) p(t ) cos(2π f 0t ) + ( Ai sin ϕi ) p (t ) sin(2π f 0t )
Clearly, we have two versors
b1 (t ) = p(t ) cos(2π f 0t )
b2 (t ) = p(t ) sin(2π f 0t )
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
4
m-QAM:
QAM: constellation
SIGNAL SET
VERSORS
M = {si (t ) = Ai p(t ) cos(2π f 0t − ϕi ) }im=1
b1 (t ) = p(t ) cos(2π f 0t )
b2 (t ) = p(t ) sin(2π f 0t )
VECTOR SET
M = {s i = (α i , βi ) }im=1 ⊆ R 2
α i = Ai cos ϕi
βi = Ai sin ϕi
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
5
We focus on grid QAM
If m is a square (m=q2)
we use square grid QAM with q points on each face
[ 16-QAM = 4 x 4 ]
[ 64-QAM = 8 x 8 ]
[ 256-QAM = 16 x 16 ]
If m is a not a square (m≠q2)
we use a subset of the following square grid QAM
[ 8-QAM ⊆ 16-QAM ]
[ 32-QAM ⊆ 64-QAM]
[ 128-QAM ⊆ 256-QAM ]
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
6
Example: 1616-QAM
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
7
Example: 6464-QAM
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
8
Example 88-QAM
8-QAM ⊆ 16-QAM
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
9
Example 88-QAM
8-QAM ⊆ 16-QAM (not unique choice)
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
10
Example 3232-QAM
32-QAM ⊆ 64-QAM
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
11
m-QAM:
QAM: constellation
The constellation envelope is not constant (problems with power
amplifiers)
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
12
m-QAM:
QAM: binary labeling
e : Hk ↔ M
For grid QAM it is always possible to build Gray labelings
1001/ s 3
1000 / s 2
0001/ s 7
0000 / s 6
−3a
−a
0101/ s11
b1 (t )
3a
a
1010 / s1
0010 / s 5
a
0100 / s10
1011/ s 0
0011/ s 4
b0 (t )
3a
0110 / s 9
0111/ s8
1110 / s13
1111/ s12
−a
1101/ s15
1100 / s14
−3a
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
13
m-QAM:
QAM: transmitted waveform
k = log 2 m
T = kTb
R=
Rb
k
Each symbol has duration T
Each symbol component (α and β) lasts for T second
Transmitted waveform




s (t ) =  ∑ α [n]p(t − nT )  cos(2π f 0t ) +  ∑ β [n]p(t − nT )  sin(2π f 0t )
 n

 n

i(t )
I component (in
phase)
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
q(t )
Q component (in
quadrature)
14
Example: 1616-QAM transmitted waveform
16- QAM
1
p(t ) =
PT (t )
T
f 0 = 2 Rb
1001/ s 3
1000 / s 2
0001/ s 7
0000 / s 6
−3a
−a
0101/ s11
b1 (t )
3a
a
1010 / s1
0010 / s 5
a
0100 / s10
1011/ s 0
0011/ s 4
b0 (t )
3a
0110 / s 9
0111/ s 8
1110 / s13
1111/ s12
−a
u T = 0010001111010110
1101/ s15
1100 / s14
−3a
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
15
Example
0011
0010
0110
1101
6
4
2
0
-2
-4
-6
0
1
2
3
4
5
6
7
8
9
t/T
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
10
11
12
13
14
15
16
16
16
Example
1001/ s 3
1000 / s 2
0001/ s 7
0000 / s 6
−3a
−a
0101/ s11
b1 (t )
3a
a
1010 / s1
0010 / s 5
a
0100 / s10
1011/ s 0
0011/ s 4
b0 (t )
3a
0110 / s 9
0111/ s8
1110 / s13
1111/ s12
−a
1101/ s15
1100 / s14
−3a
π

vT [0] = 0010 
→ sT [0] = a cos(2π f 0t ) + a sin(2π f 0t ) = 2a cos  2π f 0t − 
4

R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
17
Example
0011
0010
0110
1101
6
4
2
0
-2
-4
-6
0
1
2
3
4
5
6
7
8
9
t/T
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
10
11
12
13
14
15
18
16
18
Example
1001/ s 3
1000 / s 2
0001/ s 7
0000 / s 6
−3a
−a
0101/ s11
b1 (t )
3a
a
1010 / s1
0010 / s 5
a
0100 / s10
1011/ s 0
0011/ s 4
b0 (t )
3a
0110 / s 9
0111/ s 8
1110 / s13
1111/ s12
−a
1101/ s15
1100 / s14
−3a
π 

vT [0] = 0011 
→ sT [0] = 3a cos(2π f 0t ) + a sin(2π f 0t ) = 10a cos  2π f 0t − 
10 

R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
19
Example
0011
0010
0110
1101
6
4
2
0
-2
-4
-6
0
1
2
3
4
5
6
7
8
9
t/T
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
10
11
12
13
14
15
20
16
20
Example
1001/ s 3
1000 / s 2
0001/ s 7
0000 / s 6
−3a
−a
0101/ s11
b1 (t )
3a
a
1010 / s1
0010 / s 5
a
0100 / s10
1011/ s 0
0011/ s 4
b0 (t )
3a
0110 / s 9
0111/ s8
1110 / s13
1111/ s12
−a
1101/ s15
1100 / s14
−3a
π

vT [0] = 1101 
→ sT [0] = −3a cos(2π f 0t ) − 3a sin(2π f 0t ) = 3 2a cos  2π f 0t + 3 
4

R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
21
Example
0011
0010
0110
1101
6
4
2
0
-2
-4
-6
0
1
2
3
4
5
6
7
8
9
t/T
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
10
11
12
13
14
15
22
16
22
Example
1001/ s 3
1000 / s 2
0001/ s 7
0000 / s 6
−3a
−a
0101/ s11
b1 (t )
3a
a
1010 / s1
0010 / s 5
a
0100 / s10
1011/ s 0
0011/ s 4
b0 (t )
3a
0110 / s 9
0111/ s8
1110 / s13
1111/ s12
−a
1101/ s15
1100 / s14
−3a
π

vT [0] = 0110 
→ sT [0] = a cos(2π f 0t ) + −a sin(2π f 0t ) = 2a cos  2π f 0t + 
4

R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
23
Example
0011
0010
0110
1101
6
4
2
0
-2
-4
-6
0
1
2
3
4
5
6
7
8
9
t/T
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
10
11
12
13
14
15
24
16
24
m-QAM:
QAM: analytic signal




s (t ) =  ∑ α [n]p(t − nT )  cos(2π f 0t ) +  ∑ β [n]p(t − nT )  sin(2π f 0t )
 n

 n

i(t )
q (t )
s(t ) = Re [ s&(t )] = Re  s%(t )e j 2π f0t 
s% (t ) = i (t ) − jq(t ) = ∑ γ [n] p(t − nT )
γ [n] = α[n] − j β [n]
n
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
25
m-QAM:
QAM: bandwidth and spectral efficiency
Transmitted waveform




s (t ) =  ∑ α [n]p(t − nT )  cos(2π f 0t ) +  ∑ β [n]p(t − nT )  sin(2π f 0t )
 n

 n

Gs ( f ) = z  P( f − f 0 ) + P( f + f 0 )

2
Each symbol
2


α[n] and β[n] has time
z∈R
T = kTb
duration
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
26
m-QAM:
QAM: bandwidth and spectral efficiency
Case 1: p(t) = ideal low pass filter
− f0
1/RT
Bid = R =
f0
1/RT
Rb
k
Total bandwidth
(ideal case)
ηid =
Rb
= k bps / Hz
Bid
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
27
m-QAM:
QAM: bandwidth and spectral efficiency
Case 2: p(t) = RRC filter with roll off α
Rb
B = R (1 + α ) =
(1 + α )
k
− f0
f0
1 R(1+α)
(1 + α )
T
1R(1+α)
(1 + α )
T
Total bandwidth
η=
Rb
k
=
bps / Hz
B (1 + α )
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
28
Exercize
Given a bandpass channel with bandwidth B = 4000 Hz, centred around
f0=2 GHz, compute the maximum bit rate Rb we can transmit over it with
an 16-QAM constellation or a 64-QAM constellation in the two cases:
•
Ideal low pass filter
•
RRC filter with α=0.25
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
29
m-QAM:
QAM: interpretation
Square grid QAM constellations (m=q2):
it can be view as the Cartesian product of two independent q-ASK
constellations
(one for I channel/cosine and the other for Q channel/sine)
( 3α )
( A, A)
(α )
( −α )
( −3α )
( −3α ) ( −α )
(α )
( 3α )
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
30
m-QAM: modulator for square constellations
Rb / 2
p(t )
11100001
ee
(⋅)
i (t )
cos
uT [i ]
1101101100010110
cos ( 2π f 0t )
S/P
Rb
01110110
s (t )
90
sin
ee
(⋅)
q (t )
p(t )
Rb / 2
Example: 16-QAM = 4-ASK x 4-ASK
One symbol = 4 bits, two on I channel and two on Q channel
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
31
m-QAM: demodulator for square constellations
p (t )
t0 + nT
ρ1[n]
11100001
ML
CRITERION
e-1
cos ( 2π f 0t )
SYMBOL
SYNCHRONIZATION
1101101100010110
P/S
R
90
uR[i]
sin ( 2π f 0t )
CARRIER
SYNCHRONIZATION
p (t )
t0 + nT
ρ2[n]
ML
CRITERION
e-1
01110110
cos ( 2π f 0t )
Decision is taken separately on I channel (based on ρ1[n] )
and on Q channel (based on ρ2[n] )
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
32
m-QAM:
QAM: eye diagram
16-QAM constellation with RRC filter (α=0.5)
[
α and β components =
-3 , -1 , +1 , +3 ]
I and Q channel
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
33
m-QAM:
QAM: eye diagram
64-QAM constellation with RRC filter (α=0.5)
[
α and β components = +7,+5,+3,+1,-1,-3,-5,-7]
I and Q channel
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
34
m-QAM: modulator for nonnon-square constellations
Non-Square grid QAM constellations (m≠q2):
cannot be view as the Cartesian product of two independent q-ASK:
modulator and demodulator do not work independently on I and Q
channels.
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
35
m-QAM: modulator for non
non--square constellations
p(t )
α [ n]
k
cos
sT [n] cos ( 2π f 0t )
vT [ n ]
uT [i]
i (t )
e(⋅)
s (t )
90
sin
β [ n]
p(t )
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
q (t )
36
m-QAM: demodulator for non
non--square constellations
p (t )
t0 + nT
ρ1[n]
cos ( 2π f 0t )
SYMBOL
SYNCHRONIZATION
ML
CRITERION
R
sR [n]
vR [n]
e-1
90
uR [i]
k
sin ( 2π f 0t )
CARRIER
SYNCHRONIZATION
p (t )
t0 + nT
ρ2[n]
cos ( 2π f 0t )
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
37
m-QAM: error probability
Square grid m=q2 -QAM
q2-QAM
( 3α )
= Cartesian product of
( A, A)
(α )
two independent q-ASK (q-PAM)
( −α )
( −3α )
q2-QAM has the same error
performance of q-PAM
( −3α ) ( −α )
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
(α )
( 3α )
38
m-QAM: error probability
4-QAM = 2-PAM
01
00
 Eb 
1
Pb (e) = erfc 


2
 N0 
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
11
10
0
1
39
m-QAM: error probability
16-QAM = 4-PAM
 2 Eb 
3
Pb (e) ≈ erfc 


8
 5 N0 
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
40
m-QAM: error probability
64-QAM = 8-PAM
 1 Eb 
7
Pb (e) ≈ erfc 

24
 7 N0 
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
41
m-QAM: error probability
q2-QAM = q-PAM
q2-QAM
General expression
derived for m-PAM constellations

Eb 
m −1
3k
Pb (e) ≈ 2
erfc 

 2 ( m − 1) N 0 
mk


R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
q-PAM
42
m-QAM: error probability
4-QAM
16-QAM
64-QAM
256-QAM
1024-QAM
1
0.1
0.01
1E-3
1E-4
1E-5
BER
1E-6
1E-7
1E-8
1E-9
1E-10
1E-11
1E-12
1E-13
1E-14
0
2
4
6
8 10 12 14 16 18 20 22 24 26 28 30 32 34 36
Eb/N0 [dB]
For increasing m, the performance decrease
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
43
m-QAM: error probability (vs. ASK)
m=q2-QAM
1. COMPARISON QAM / ASK
An m-QAM constellation and an m-ASK constellation have:
the same spectral efficiency
m-QAM has better performance, because equal to q-PAM (remember
that, for increasing m, PAM performance decreases)
16-QAM = 4-PAM
16-ASK=16-PAM
64-QAM = 8-PAM
64-ASK=64-PAM
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
44
m-QAM: error probability (vs. PSK)
m=q2-QAM
2. COMPARISON QAM / PSK
An m-QAM constellation and an m-PSK constellation have:
the same spectral efficiency
m-QAM has better performance (better distribution of points on the
plane, larger minimum distance)
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
45
m-QAM: error probability (vs. PSK)
COMPARISON QAM / PSK
m − PSK
m − QAM


Eb
1
2π 
Pb (e) ≈ erfc  k
sin   


k
N
m


0



Eb 
m −1
3k
Pb (e) ≈ 2
erfc 



2
m
−
1
N
(
)
mk
0


Fixed a (sufficiently large) BER, PSK requires higher Eb/N0
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
46
m-QAM: error probability (vs. PSK)
Fixed a (sufficiently large) BER, PSK requires higher Eb/N0
 Eb 
 Eb 

 ≈

 N 0 PSK  N 0 QAM


3


2
 2(m − 1) sin (π / m ) 
m = 16
difference = 4.20 dB
m = 64
difference = 9.96 dB
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
47
m-QAM: error probability (vs. PSK)
Comparison 16-QAM vs. 16-PSK
1001/ s 3
1000 / s 2
b1 (t )
3a
1010 / s1
1011/ s 0
0100
0001/ s 7
0000 / s 6
a
0010 / s 5
0101
0011
0001
1100
0011/ s 4
0111
1101
−3a
0101/ s11
−a
a
0100 / s10
0110 / s 9
0000
b0 (t )
3a
1111
0111/ s 8
0010
1011
−a
0110
1001
1000 1010
1101/ s15
1100 / s14
1110 / s13
1110
1111/ s12
−3a
 2 Eb 
3
Pb (e) ≈ erfc 

8
 5 N0 

Eb 
1
Pb (e) ≈ erfc  0.152


4
N
0 

difference = 4.20 dB
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
48
m-QAM: error probability (vs. PSK)
Comparison 16-PSK vs. 16-QAM
16-PSK
16-QAM
1
0.1
0.01
difference = 4.20 dB
1E-3
1E-4
1E-5
BER
1E-6
1E-7
1E-8
1E-9
1E-10
1E-11
1E-12
1E-13
1E-14
0
2
4
6
8
10
12
14
16
18
20
22
24
26
Eb/N0 [dB]
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
49
m-QAM: error probability
Non-square grid m=q2 –QAM (8-QAM, 32-QAM, 128-QAM, 512-QAM,...)
Not yet a Cartesian product of
two independent ASK
The performance must be computed
case by case
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
50
m-QAM: error probability (vs. PSK)
Comparison 8-QAM vs. 8-PSK
001
000
010
011
011
111
001
101
101
100
110
111
000
010
100
110
 1 Eb 
5
Pb (e) ≈ erfc 


12
 2 N0 

Eb 
1
Pb (e) ≈ erfc  0.439


3
N0 

R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
51
m-QAM: error probability (vs. PSK)
Comparison 8-QAM vs. 8-PSK
8-PSK
8-QAM
1
0.1
0.01
1E-3
1E-4
1E-5
difference = 0.56 dB
BER
1E-6
1E-7
1E-8
1E-9
1E-10
1E-11
1E-12
1E-13
1E-14
0
2
4
6
8
10
12
14
16
18
20
Eb/N0 [dB]
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
52
Exercize
Given a baseband channel with bandwidth B = 4000 Hz, compute the
maximum bit rate Rb we can transmit over it by:
•
a 4-PAM constellation
•
a 16-QAM constellation (carrier frequency f0 = 2kHz)
when ideal low pass TX filters are supposed in both cases.
Which constellation has better performance?
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
53
m-QAM:
QAM: applications
•
Digital radio links (Up to 128-QAM)
•
Some satellite links (up to 16-QAM)
•
Internet modems (V90: 33600 bps in uplink, 1024-QAM)
•
ADSL modems (OFDM modulation, up to 256-QAM for each carrier)
•
DVB-T, DAB (OFDM)
•
…
R. Garello. “Tutorial on digital modulations - Part 15 m-QAM” [2010-11-25]
54