Description of the 16-QAM simulation.

QAM
Quadrature amplitude modulation (QAM) is a modulation scheme in which two sinusoidal carriers,
one exactly 90 degrees out of phase with respect to the other, are used to transmit data over
a given physical channel. One signal is called the I signal, and the other is called the Q signal.
Because the orthogonal carriers occupy the same frequency band and differ by a 90 degree phase
shift, each can be modulated independently, transmitted over the same frequency band, and
separated by demodulation at the receiver. For a given available bandwidth, QAM enables data
transmission at twice the rate of standard pulse amplitude modulation (PAM) without any degradation
in the bit error ratio (BER).
Mathematically two independent signals x I () t and x Q () t , each bandlimited by ω g are transmitted.
One signal is modulated by a cosine the other signal by a sinus carrier. Thus the signal
x I () t ⋅ cos( ω 0 t)
– x Q () t ⋅ sin( ω 0 t)
(1)
is transmitted. For simplicity back to back transmission is concerned. Thus the received signal
is equal to (1). The in-phase part x I
() t of the signal can be demodulated by multiplication of
cos( ω 0
t)
[ x I
() t ⋅ cos( ω 0
t)
– x Q
() t ⋅ sin( ω 0
t)
] ⋅ cos( ω 0
t)
= x I () t ⋅ cos 2 ( ω 0 t)
– x Q () t ⋅ sin( ω 0 t)
⋅ cos( ω 0 t)
(2)
=
x I
() t
x
---------- Q
() t
[ 1 + cos( 2ω
2
0
t)
] – ------------ ⋅ sin( 2ω
2
0
t)
and lowpass filtering
x I () t
x
---------- Q () t
[ 1+
cos( 2ω
2
0
t)
]– ------------ ⋅ sin( 2ω
2
0
t)
* r ωg
() t
x I () t
= ----------
2
. (3)
* is the convolution operator and r ωg
() t an ideal lowpass filter with cut-off frequency ω g .
Equation (3) only holds for 2ω g
< ω 0 .
Task 1
Show, that the quadrature part qt () of the received signal (1) can be demodulated by multiplication
of –sin( ω 0
t)
. (click for solution)
With complex notation equation (1) can be written as
⎧
Re [ x I () t + jx Q () t ]e jω 0t ⎫
⎨
⎬. (4)
⎩
⎭
In case of transmission of digital information
∞
∑
x I () t = a I, n gt ( – nT)
n = – ∞
∞
∑
x Q () t = a Q, n gt ( – nT)
n = – ∞

a n
= a I, n + ja Q,
n . (5)
a n
is the complex symbol sequence at the output of the mapper. Thus equation (4) can be written
as
∞
⎧
Re a n
gt ( – nT)e jω 0t ⎫
⎨ ∑
⎬. (6)
⎩
⎭
n = – ∞
Task 2
Show, that with complex notation the in-pase and quadrature component of the modulated signal
(6) can be demodulated by multiplication of e jω 0t –
and low-pass filtering. (click for solution)
In the following
t = nT, thus
∞
gt ()
is assumed to be a Nyquist impulse with equidistant zero crossings at
⎧
g( 0) Re a n
e jω 0nT ⎫
⋅ ⎨ ∑ ⎬
(7)
⎩n
= – ∞ ⎭
is the received signal (6) after sampling. Now additive white gaussian noise (AWGN) is added.
This is depicted in Fig. 1. As sampling and demodulation with ideal low-pass filtering are linear
processes, this will lead to the received sequence
g( 0)
ã n
= ----------a (8)
2 n
+ r n
with discrete additive gaussian noise r n
= r I, n + jr Q,
n .
In the following the impulse gt () is chosen suchlike, that the average signal energy
M
1
S = ---- A
2
(9)
M ∑ k
k = 1
and the average noise energy
N = E { r 2 } I,
n
= E { r 2
. (10)
Q ,
} = σ 2 n
The decision device has to estimate â n
out of ã n
. â n
should be the same sequence as a n
. The
influence of noise r n
should be as low as possible. r I, n and r Q, n are assumed to be statistically
independent. Thus the decision problem can be separated. In case of AWGN the noise r I,
n and
r Q, n follows the probability density function (pdf)
1
1 –-- -----
r2
p I ( r) = p Q ( r)
= -------------- e
2σ 2
, (11)
2πσ
respectively. Thus the noise process is normally distributed with zero-mean and variance σ 2 .
In total because real and imaginary part are statistically independent in complex notation
e jω 0t
nt ()
e – jω
0t
low-pass
decision
demapper
b m
mapper a sampler
n
filter device
ã â
+
n bˆ
gt ()
m
× Re{}
×
n
⎧ ⎪⎪⎪
⎪
⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩
⎧ ⎨⎩
⎧ ⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩
transmitter tx AWGN receiver rx
Fig. 1: principle communication system

1 -- ------
pr ( ) = p I
( r I
) ⋅ p Q
( r Q
) = ------------e
2 σ 2
(12)
2πσ 2
is the pdf of r n
.
For example a four quadrature pulse shift keying (4-QPSK) system is analyzed. Thus
a n
∈ { A 1
= 1 + j, A 2
= 1–
j, A 3
= – 1 + j,
A 4
= – 1 – j}
. Fig. 2 shows this signal constellation.
Task 3
1
– r 2
Calculate the average signal energy S 4-QPSK
of the 4-QPSK. (click for solution)
To this symbols a n
AWGN is added. All four signal points are assumed to occure with the same
probability. For simplicity g( 0)
is assumed to be 2.
Task 4
Calculate the pdf p ãn
( r) for ã n
. (click for solution)
Fig. 3 shows p ãn
( r)
in a 3-dimensional sketch. This is similar to the frequency of occurence in
the 16-QAM animation, where the 3rd dimension is equivalent to the density of received signal
points, which are plotted in green and red colour. Green stands for correct decision and red for
symbol error.
The decision device has to choose the most probable signal point, wich may be sent. If the a-
priori probability for the signal points is equal as mentioned, the decision thresholds consist of
perpendicular bisectors of the sides of adjacent signal points. That means the signal point, which
is most close to the received point will be decided to. In this example the axes in Fig. 2 are also
the decision thresholds.
The probability of symbol errors (SER) can be calculated by integration of p Ak
( r)
(eq. (23); eg.
Fig. 4) over the region in the complex r -plane where received symbols ã n
would not be decided
to the appropriate sent symbol a n
= A k
, respectively. This has to be done for all k signal
points. The SER is the arithmetic average over all.
Im[ A k
]
A 3
1
A 1
1 Re[ A k
]
A 4
A 2
Fig. 2: 4-QPSK constellation

0
∞
1⎛
P 4-QPSK
= -- ⎜ p
4 ∫ ∫ A1
( r)
dr Q
dr I
+ ∫ ∫ p A1
( r)
dr Q
dr I
+ ∫ ∫ p A2
( r)
dr Q
dr I
⎝
–∞ –∞
∞ ∞
∞
0
0 –∞
0
∞
–∞ –∞
+ ∫ ∫ p ( A2
r ) dr Q
dr I
+ ∫ ∫ p A3
( r)
dr Q
dr I
+ ∫ ∫ p A3
( r)
dr Q
dr I
0 0
∞ ∞
∞ ∞
0 –∞
∫ ∫ p ( A4
r ) dr Q dr I ∫ ∫ p A4
r
+ +
0 –∞
0
∞
–∞ 0
⎞
( ) dr Q dr I ⎟
⎠
0
0
–∞ –∞
(13)
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
p ãn
( r)
p ãn
( r)
0.15
0.1
0.0432
0.02
0.01
-2
-1
0
Re[ r]
1
2
-2
-1
0
1
Im[ r]
2
Fig. 3: probability density function of ã n
for σ = 0.5
P A1
( r)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
P A1
( r)
-2
-1
0
Re[ r]
1
2
-2
-1
0
1
Im[ r]
2
Fig. 4: probability density function of ã n
if a n
= A 1
for σ = 0.5

P 4-QPSK
= = ∫ ∫ p A1
( r)
dr Q
dr I
+ ∫ ∫ p A1
( r)dr Q
dr I
= 1–
∫ ∫ p A1
( r)
dr Q
dr I
The simplification in (14) can be done because of symmetry. To calculate this integral, the separation
of (12) can be used.
P 4-QPSK
= 1 –
(14)
In equation (15) the substitutions
r
t I/Q
– 1
1 x
= –---------------- and Qx ( ) = -- 1 – erf⎛------
⎞ ⇔ erf( x)
= 1 – 2Q( 2x)
(16)
2σ
2 ⎝ 2⎠
are used. With some routine the notation with the Q-function is very efficient. The Gaussian error
function is defined as
erf( x)
Task 5
Calculate P 4-QPSK as a function of SNR (average symbol energy per average noise energy) and
as a function of SNRdB (SNR in decibel).
Hint: Scale the signal points by a ( A k
= 2a ). adapt the result in (15). Normalize adequately.
(click for solution)
Task 6
In Fig. 5 the constellation diagram of a 16-QAM is depicted. All signal points occure with the
same probability. Calculate P 16-QAM as a function of SNR dB .
Calculate the numerical values of P 16-QAM for SNR dB = 01234567891011
, , , , , , , , , , , .
Compare your results with the calculated values in the 16-QAM animation.
Hint: There are three types of signal points with different error rate. (click for solution)
Task 7
Give the limit of SER of the 16-QAM for very low SNR . (click for solution)
Task 8
0
= 1 –
∞
–∞ –∞
∞ ∞
∫ ∫
0 0
∞
∫
0
1
------------e
2πσ 2
1
-------------- e
2πσ
–∞
– 2
1
–--
r A 1
-------------------
2
σ 2
1( -- r I – A I1 , ) 2
– -------------------------
2
σ 2
∞
0
0 –∞
dr Q
dr I
dr I
⋅
∞
∫
0
1
-------------- e
2πσ
1( -- r Q – A Q1 , ) 2
– -----------------------------
2
Fig. 6 shows a screenshot of the 16-QAM animation. Only one symbol error occured (red
point). Which signal point was most probably sent? (click for solution)
σ 2
∞ ∞
0 0
dr Q
1
1 –------
e – 1 1
= – ∫
t2 dt 1 --erf ----------
π
2 ⎝
⎛ 2σ⎠
⎞ 1 2
= – + -- = 2Q⎛ 1
2 ⎝σ -- ⎞ – Q
⎠
2 ⎛ 1 ⎝σ -- ⎞
⎠
x
0
1
----------
2σ
2
2
= ------
∫e – x2 dx . (17)
π
(15)

Task 9
Start the 16-QAM animation with SNR dB ≈ 20dB . Wait until more than 10 errors (red points)
occured. Make sure, they agglomerate at the decision thresholds.
Gray coding is a coding scheme where adjacent signal points differ only by one bit in coding.
Why is this coding scheme useful for QAM modulation?
What about low SNR ? (click for solution)
Im[ A k
]
3a
A 1
A 2
A 3
A 4
a
A 5
A 6
A 7 A 8
a
3a
Re[ A k
]
A 9
A 10
A 11
A 12
A 13
A 14
A 15
A 16
Fig. 5: 16-QAM constellation
Fig. 6: Screenshot of the 16-QAM animation

Solution 1 (click to go back)
–[ x I () t ⋅ cos( ω 0 t)
– x Q () t ⋅ sin( ω 0 t)
] ⋅ sin( ω 0 t)
= – x I () t ⋅ sin( ω 0 t)
⋅ cos( ω 0 t)
+ x Q () t ⋅ sin 2 ( ω 0 t)
x I
() t
x
---------- Q
() t
= – ⋅ sin( 2ω
2
0
t)
+ ------------ [ 1 – cos( 2ω
2
0
t)
]
After lowpass filtering (* is the convolution operator):
x I
() t
x
---------- Q
() t
– ⋅ sin( 2ω
2
0
t)
+ ------------ [ 1 – cos( 2ω
2
0
t)
] * r ωg
() t
x Q
() t
= ------------
2
(click to go back)
(18)
(19)

Solution 2 (click to go back)
1
With Re{ z}
= -- ( z + z)
, whereby z is the conjugate complex number of z , after mutliplication
– 2
of e jω 0t
∞
⎧
Re a n
gt ( – nT)e jω 0t ⎫ – jω 0 t
⎨ ∑
⎬⋅
e
⎩
⎭
n = – ∞
∞
1
-- a
2 n e jω 0t –
a n e jω 0t
–
( + )gt ( – nT) e jω 0t
= ∑
⋅
n = – ∞
∞
1
–
-- a
2 n
a n
e j2ω 0t
= ∑ ( + )gt ( – nT)
n = – ∞
and after lowpass filtering (20) gets
∞
1
–
-- a
2 n
a n
e j2ω 0t
∑ ( + )gt ( – nT )
1
* r ωg
() t = -- a
2 ∑ n
gt ( – nT)
n = – ∞
(click to go back)
∞
n = – ∞
(20)
(21)

Solution 3 (click to go back)
4
1
S 4-QPSK
-- A 2 1
=
4 ∑ k
= -- ( 1 + j
4
+ 1 – j
+ – 1 + j
+ – 1 – j
) =
k = 1
2 . (22)
The signal and noise power do not have units, because of standardization!
(click to go back)

Solution 4 (click to go back)
p ãn
4
1
( r)
= -- p
4 ∑ Ak
( r)
=
k = 1
(click to go back)
1
--
4
4
∑
k = 1
1
------------e
2πσ 2
– 2
1
–--
r A k
------------------
2
σ 2
(23)

Solution 5 (click to go back)
With the new signal points A 1
= a + ja ; A 2
= a – ja ; A 3
= – a + ja ; A 4
= – a – ja the
average signal energy
S 4-QPSK = 2a 2 . (24)
Thus the SER gets
P 4-QPSK
2Q⎛ a ⎝σ -- ⎞ Q
⎠
– 2 ⎛ a ⎝σ -- ⎞
⎠
2Q ⎛------
1 2a 2
2
-------- ⎞
⎝ σ 2 Q
⎠
2 1
= =
– ⎛------
⎝ 2
2a
--------
2 ⎞
⎠
SNR dB
SNR .
= 2Q
SNR ----------
⎝
⎛ 2 ⎠
⎞ – Q2 ⎛ SNR ---------- ⎞ 2Q 10 ---------------
dB
⎛
10
⎞ ⎛ ---------------
⎜ ------------------ ⎟ Q
⎝ 2 ⎠ ⎜ 2 ⎟
2 10
10
⎞
=
– ⎜ ------------------ ⎟
⎜ 2 ⎟
⎝ ⎠ ⎝ ⎠
(click to go back)
σ 2

Solution 6 (click to go back)
S 16-QAM = 10a 2
(25)
4
P 16-QAM = ----- 2Q⎛ a
16 ⎝σ -- ⎞ – Q
⎠
2 ⎛ a ⎝σ -- ⎞
⎠
⎧ ⎪⎪⎪⎨⎪⎪⎪⎩
A 1
, A 4
, A 13
, A 16
+
8
----- Q⎛ a
16 ⎝σ -- ⎞ ⎧
+ 1 – Q⎛ a
⎠ ⎝σ -- ⎞
⎨ ⎠
⎩
⎧ ⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩
⎫ ⎧
⎬⋅
⎨
⎭ ⎩
⎛ ⎞ Q⎛ a ⎝σ -- ⎞⎫
+
⎠⎬
⎭
Q a σ -- ⎝ ⎠
A 2
, A 3
, A 5
, A 8
, A 9
, A 12
, A 14
, A 15
+
4
----- 2Q⎛ a
16 ⎝σ -- ⎞ ⎧
+ 1–
2Q⎛ a
⎠ ⎝σ -- ⎞
⎨ ⎠
⎩
⎧ ⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩
A 6
, A 7
, A 10
, A 11
⎫ a
⎬⋅
2Q --
⎝ ⎛ ⎭ σ⎠
⎞
(26)
3Q⎛ a ⎝σ -- ⎞ 9
= – --Q
⎠ 4
2 ⎛ a ⎝σ -- ⎞ 3Q⎛ SNR ---------- ⎞ 9
=
– --Q
⎠ ⎝ 10 ⎠ 4
2 ⎛
⎝
SNR
---------- ⎞
10 ⎠
=
⎛
3Q⎜
⎜
⎝
SNR dB
10 ---------------
10
⎞
------------------ ⎟
10 ⎟
⎠
–
SNR dB
⎛ ---------------
9
--Q
4
2 10
10
⎞
⎜ ------------------ ⎟
⎜ 10 ⎟
⎝ ⎠
(27)
SNR dB
P 16-QAM
SNR dB
P 16-QAM
0 0.80979256 12 0.28773651
1 0.79027909 13 0.22268236
2 0.76759250 14 0.16230454
3 0.74125244 15 0.10984263
4 0.71074913 16 0.067830428
5 0.67557021 17 0.037404624
6 0.63524598 18 0.017932041
7 0.58941918 19 0.0072268271
8 0.53794439 20 0.0023467250
9 0.48101796 21 0.00058187175
10 0.41932991 22 0.00010290522
11 0.35421249 23 0.000011907710
(click to go back)

Solution 7 (click to go back)
15
In case of very low SNR all signal points are equal probable. Thus P 16-QAM → -----
16
(click to go back)
= 0.9375

Solution 8 (click to go back)
Signal point A 13
was most probably sent, because it is the second most close signal point to the
received signal point (red point).
(click to go back)

Solution 9 (click to go back)
In case of gray coding most symbol errors lead to only one bit error, because most wrong decided
symbols come from adjacent signal points and the coding of them differ only by one bit.
Thus approximately if M bits are transmitted per symbol the bit error ratio (BER) is equal to
P SER
P BER
≈ ----------- . (28)
M
This only holds for high SNR . In case of low SNR a lot of wrong decided symbols come from
signal points far away. The advatage of gray coding vanishs.
(click to go back)