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QAM

Quadrature amplitude modulation (QAM) is a modulation scheme in which two sinusoidal car-

riers, 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 pha-

se 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 de-

gradation in the bit error ratio (BER).

Mathematically two independent signals and , each bandlimited by are trans-

mitted. One signal is modulated by a cosine the other signal by a sinus carrier. Thus the signal

(1)

is transmitted. For simplicity back to back transmission is concerned. Thus the received signal

is equal to (1). The in-phase part of the signal can be demodulated by multiplication of

(2)

and lowpass filtering

. (3)

* is the convolution operator and an ideal lowpass filter with cut-off frequency .

Equation (3) only holds for .

Task 1

Show, that the quadrature part of the received signal (1) can be demodulated by multipli-

cation of . (click for solution)

With complex notation equation (1) can be written as

. (4)

In case of transmission of digital information

t() x

t() ω

t()cos⋅ x

t() ω

t()sin⋅–

t()

t()cos

t() ω

t()cos⋅ x

t() ω

t()sin⋅–[]ω

t()cos⋅

t() cos

t()⋅ x

t() ω

t()sin ω

t()cos⋅⋅–=

t()

-----------

12ω

t()cos+[]

t()

-------------

2ω

t()sin⋅–=

t()

-----------

12ω

t()cos+[]

t()

-------------

2ω

t()sin⋅–*r

t()

-----------=

t() ω

2ω

qt()

t()sin–

Re x

t() jx

t()+[]e

jω







t() a

I n,

gt nT–()

n ∞–=

∞

∑

t() a

Q n,

gt nT–()

n ∞–=

∞

∑

. (5)

is the complex symbol sequence at the output of the mapper. Thus equation (4) can be writ-

ten as

. (6)

Task 2

Show, that with complex notation the in-pase and quadrature component of the modulated si-

gnal (6) can be demodulated by multiplication of and low-pass filtering. (click for solu-

tion)

In the following is assumed to be a Nyquist impulse with equidistant zero crossings at

, thus

(7)

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

(8)

with discrete additive gaussian noise .

In the following the impulse is chosen suchlike, that the average signal energy

(9)

and the average noise energy

. (10)

The decision device has to estimate out of . should be the same sequence as . The

influence of noise should be as low as possible. and are assumed to be statistically

independent. Thus the decision problem can be separated. In case of AWGN the noise and

follows the probability density function (pdf)

, (11)

respectively. Thus the noise process is normally distributed with zero-mean and variance .

In total because real and imaginary part are statistically independent in complex notation

I n,

Q n,

Re a

gt nT–()e

jω

n ∞–=

∞

∑







jω

t–

gt()

tnT=

g 0()Re a

jω

n ∞–=

∞

∑







⋅

jω

j– ω

nt()



























transmitter tx AWGN

receiver rx

mapper

Re{}

demapper

sampler

low-pass

filter

Fig. 1: principle communication system

decision

device

gt()

g 0()

-----------

I n,

Q n,

gt()

-----

k 1=

∑

NEr

I n,

{}Er

Q n,

{}σ

== =

I n,

Q n,

I n,

Q n,

r() p

r()

2πσ

--------------

---

-----

–

(12)

is the pdf of .

For example a four quadrature pulse shift keying (4-QPSK) system is analyzed. Thus

. Fig. 2 shows this signal constellati-

on.

Task 3

Calculate the average signal energy of the 4-QPSK. (click for solution)

To this symbols AWGN is added. All four signal points are assumed to occure with the same

probability. For simplicity is assumed to be 2.

Task 4

Calculate the pdf for . (click for solution)

Fig. 3 shows 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 (eq. (23); eg.

Fig. 4) over the region in the complex -plane where received symbols would not be deci-

ded to the appropriate sent symbol , respectively. This has to be done for all signal

points. The SER is the arithmetic average over all.

pr() p

()p

()⋅

2πσ

-------------

---–

-------

1j+= A

1j–= A

1–j+= A

1–j–=,, ,{}∈

Re A

[]

Im A

[]

Fig. 2: 4-QPSK constellation

4-QPSK

g 0()

r() a

r()

r a

= k

(13)

0.15

0.1

0.0432

0.02

0.01

-2

-1

-2

-1

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

r()

Re r[]

Im r[]

Fig. 3: probability density function of fora

σ 0.5=

-2

-1

-2

-1

0.1

0.2

0.3

0.4

0.5

0.6

0.7

Im r[]

Re r[]

r()

Fig. 4: probability density function of if fora

= σ 0.5=

r()

4-QPSK

---

r()

∞–

∞

∫

∞–

∫







r()

∞–

∫

∞

∫

r()

∞–

∞

∫

∞–

∫

++=

r()

∞

∫

∞

∫

r()

∞–

∞

∫

∞

∫

r()

∞–

∫

∞–

∫

++ +

r()

∞–

∞

∫

∞

∫

r()

∞

∫

∞–

∫







(14)

The simplification in (14) can be done because of symmetry. To calculate this integral, the se-

paration of (12) can be used.

(15)

In equation (15) the substitutions

and (16)

are used. With some routine the notation with the Q-function is very efficient. The Gaussian er-

ror function is defined as

. (17)

Task 5

Calculate as a function of SNR (average symbol energy per average noise energy) and

as a function of (SNR in decibel).

Hint: Scale the signal points by ( ). adapt the result in (15). Normalize adequa-

tely. (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 as a function of .

Calculate the numerical values of for .

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 . (click for solution)

Task 8

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)

4-QPSK

r()

∞–

∞

∫

∞–

∫

r()dr

∞–

∫

∞

∫

+=1p

r()

∞

∫

∞

∫

–==

4-QPSK

2πσ

-------------

---–

–

-------------------

∞

∫

∞

∫

–=

2πσ

--------------

---

I1,

–()

--------------------------

–

∞

∫

2πσ

--------------

---

Q1,

–()

-----------------------------

–

∞

∫

⋅–=

-------–e

–

2σ

-----------

∞–

∫

–=1

---

erf

2σ

-----------





---+

–2Q

---





---





–==

I/Q

1–

2σ

-----------------–= Qx()

---

1 erf

-------





–= erf x()⇔ 12Q 2x()–=

erf x()

-------

–

∫

4-QPSK

SNR

2a=

16-QAM

SNR

16-QAM

SNR

01234567891011,,,,,,,,,, ,=

SNR

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