Quadrature Amplitude Modulation. QAM Transmitter.

1. EE445S Real-Time Digital Signal Processing Lab Spring 2011
Lecture 15
Quadrature Amplitude Modulation
(QAM) Transmitter
Prof. Brian L. Evans
Dept. of Electrical and Computer Engineering
The University of Texas at Austin

2. 15 - 2
Introduction
• Digital Pulse Amplitude Modulation (PAM)
Modulates digital information onto amplitude of pulse
May be later upconverted (e.g. by sinusoidal amplitude
modulation)
• Digital Quadrature Amplitude Modulation (QAM)
Two-dimensional extension of digital PAM
Requires sinusoidal amplitude modulation
• Digital QAM modulates digital information onto
pulses that are modulated onto
Amplitudes of a sine and a cosine, or equivalently
Amplitude and phase of single sinusoid

3. 15 - 3
Amplitude Modulation by Cosine
• Example: y(t) = f(t) cos(wc t)
Assume f(t) is an ideal lowpass signal with bandwidth w1
Assume w1 < wc
Y(w) is real-valued if F(w) is real-valued
• Demodulation: modulation then lowpass filtering
• Similar for modulation with sin(w0 t) shown next
w
0
1
w1
-w1
F(w)
w
0
Y(w)
½
-wc - w1 -wc + w1
-wc
wc - w1 wc + w1
wc
½F(w - wc)
½F(w + wc)
Review
Baseband signal Upconverted signal

4. 15 - 4
Amplitude Modulation by Sine
• Example: y(t) = f(t) sin(wc t)
Assume f(t) is an ideal lowpass signal with bandwidth w1
Assume w1 < wc
Y(w) is imaginary-valued if F(w) is real-valued
• Demodulation: modulation then lowpass filtering
w
Y(w)
j ½
-wc - w1 -wc + w1
-wc
wc - w1 wc + w1
wc
-j ½F(w - wc)
j ½F(w + wc)
-j ½
w
0
1
w1
-w1
F(w)
Review
Baseband signal Upconverted signal

5. 15 - 5
Digital QAM Modulator
Matched Delay matches delay through 90o phase shifter
(required but often omitted on block diagrams)
Serial/
Parallel
Map to 2-D
constellation
Impulse modulator
Impulse modulator
Pulse shaping gT(t)
Local
Oscillator
+
90o
Pulse shaping gT(t)
d[n]
bn
an
b*(t)
a*(t)
s(t)
1 J
Matched
Delay
bit
stream J bits/symbol
+
–
in-phase (I)
component
quadrature (Q)
component
4-level QAM
Constellation
I
Q
d
d
-d
-d

6. 15 - 6
Phase Shift by 90 Degrees
• 90o phase shift performed by Hilbert transformer
cosine => sine
sine => – cosine
• Frequency response of ideal
Hilbert transformer
)
(
2
1
)
(
2
1
)
2
cos( 0
0
0 f
f
f
f
t
f -
+
+
 


)
(
2
)
(
2
)
2
sin( 0
0
0 f
f
j
f
f
j
t
f -
-
+
 


)
sgn(
)
( f
j
f
H -









-



0
if
1
0
if
0
0
if
1
)
sgn(
x
x
x
x

7. 15 - 7
Hilbert Transformer
• Magnitude response
All pass except at origin
• For fc > 0
• Phase response
Piecewise constant
• For fc < 0
)
2
sin(
)
2
2
cos( t
f
t
f c
c 

 
+
)
)
(
2
sin(
)
2
)
(
2
cos(
))
2
2
(
cos(
)
2
2
cos(
t
f
t
f
t
f
t
f
c
c
c
c
-

+
-

+
-

-







f
)
( f
H

-90o
90o
f
|
)
(
| f
H
)
sgn(
)
( f
j
f
H -


8. 15 - 8
Hilbert Transformer
• Continuous-time ideal
Hilbert transformer
• Discrete-time ideal
Hilbert transformer
h(t) =
1/( t) if t  0
0 if t = 0
h[n] =
if n0
0 if n=0
n
n )
2
/
(
sin
2 2


Even-indexed
samples are zero
t
h(t)
)
sgn(
)
( f
j
f
H -
 )
sgn(
)
( w
w j
H -

n
h[n]

9. 15 - 9
Discrete-Time Hilbert Transformer
• Approximate by odd-length linear phase FIR filter
Truncate response to 2 L + 1 samples: L samples left of
origin, L samples right of origin, and origin
Shift truncated impulse response by L samples to right to
make it causal
L is odd because every other sample of impulse response is 0
• Linear phase FIR filter of length N has same phase
response as an ideal delay of length (N-1)/2
(N-1)/2 is an integer when N is odd (here N = 2 L + 1)
• Matched delay block on slide 15-5 would be an
ideal delay of L samples

10. 15 - 10
Performance Analysis of PAM
• If we sample matched filter output at correct time
instances, nTsym, without any ISI, received signal
where the signal component is
v(t) output of matched filter Gr(w) for input of
channel additive white Gaussian noise N(0; 2)
Gr(w) passes frequencies from -wsym/2 to wsym/2 ,
where wsym = 2  fsym = 2 / Tsym
• Matched filter has impulse response gr(t)
)
(
)
(
)
( sym
sym
sym nT
v
nT
s
nT
x +

d
i
a
nT
s n
sym )
1
2
(
)
( -

 for i = -M/2+1, …, M/2
v(nT) ~ N(0; 2/Tsym)
Review
4-level PAM
Constellation
d
-d
-3 d
3 d

11. 15 - 11
Performance Analysis of PAM
• Decision error
for inner points
• Decision error
for outer points
• Symbol error probability
-7d -5d -3d -d d 3d 5d 7d
O- I I I I I I O+








 sym
sym
I T
d
Q
d
nT
v
P
e
P

2
)
)
(
(
)
(









- sym
sym
O T
d
Q
d
nT
v
P
e
P

)
)
(
(
)
(









-


+ sym
sym
sym
O T
d
Q
d
nT
v
P
d
nT
v
P
e
P

)
)
(
(
)
)
(
(
)
(






-

+
+
-
 -
+ sym
O
O
I T
d
Q
M
M
e
P
M
e
P
M
e
P
M
M
e
P

)
1
(
2
)
(
1
)
(
1
)
(
2
)
(
8-level PAM
Constellation
Review

12. 15 - 12
Performance Analysis of QAM
• Received QAM signal
• Information signal s(nTsym)
where i,k  { -1, 0, 1, 2 } for 16-QAM
• Noise, vI(nTsym) and vQ(nTsym) are independent
Gaussian random variables ~ N(0; 2/Tsym)
• Decision regions must span entire I-Q plane
)
(
)
(
)
( sym
sym
sym nT
v
nT
s
nT
x +

d
k
j
d
i
b
j
a
nT
s n
n
sym )
1
2
(
)
1
2
(
)
( -
+
-

+

)
(
)
(
)
( sym
Q
sym
I
sym nT
v
j
nT
v
nT
v +

4-level QAM
Constellation
I
Q
d
d
-d
-d

13. 15 - 13
Performance Analysis of 16-QAM
• Type 1 correct detection
)
)
(
&
)
(
(
)
(
1 d
nT
v
d
nT
v
P
c
P sym
Q
sym
I 


( ) ( )
d
nT
v
P
d
nT
v
P sym
Q
sym
I 

 )
(
)
(
( )
( ) ( )
( )
d
nT
v
P
d
nT
v
P sym
Q
sym
I 
-

-
 )
(
1
)
(
1
2
2
1 













-
 sym
T
d
Q

)
(
2 T
d
Q

)
(
2 T
d
Q

3
3
3
3
2 2
2
2
2 2
2
2
1
1
1 1
I
Q
16-QAM
1 - interior decision region
2 - edge region but not corner
3 - corner region

14. 15 - 14
Performance Analysis of 16-QAM
• Type 2 correct detection
• Type 3 correct detection
)
)
(
&
)
(
(
)
(
2 d
nT
v
d
nT
v
P
c
P sym
Q
sym
I 


)
)
(
(
)
)
(
( d
nT
v
P
d
nT
v
P sym
Q
sym
I 
















-














-
 sym
sym T
d
Q
T
d
Q


2
1
1
)
)
(
&
)
(
(
)
(
3 d
nT
v
d
nT
v
P
c
P sym
Q
sym
I -



)
)
(
(
)
)
(
( d
nT
v
P
d
nT
v
P sym
Q
sym
I -



2
1 













-
 sym
T
d
Q

3
3
3
3
2 2
2
2
2 2
2
2
1
1
1 1
I
Q
16-QAM
1 - interior decision region
2 - edge region but not corner
3 - corner region

15. 15 - 15
Performance Analysis of 16-QAM
• Probability of correct detection
• Symbol error probability
• What about other QAM constellations?
2
2
1
16
4
2
1
16
4
)
( 













-
+














-
 sym
sym T
d
Q
T
d
Q
c
P


2
1
1
16
8














-














-
+ sym
sym T
d
Q
T
d
Q








+






-
 sym
sym T
d
Q
T
d
Q


2
4
9
3
1






-







-
 sym
sym T
d
Q
T
d
Q
c
P
e
P


2
4
9
3
)
(
1
)
(

16. 15 - 16
Average Power Analysis
• Assume each symbol is equally likely
• Assume energy in pulse shape is 1
• 4-PAM constellation
Amplitudes are in set { -3d, -d, d, 3d }
Total power 9 d2 + d2 + d2 + 9 d2 = 20 d2
Average power per symbol 5 d2
• 4-QAM constellation points
Points are in set { -d – jd, -d + jd, d + jd, d – jd }
Total power 2d2 + 2d2 + 2d2 + 2d2 = 8d2
Average power per symbol 2d2
4-level PAM
Constellation
d
-d
-3 d
3 d
4-level QAM
Constellation
I
Q
d
d
-d
-d