1. EE445S Real-Time Digital Signal Processing Lab Fall 2012
Quadrature Amplitude Modulation
(QAM) Transmitter
Prof. Brian L. Evans
Dept. of Electrical and Computer Engineering
The University of Texas at Austin
Lecture 15

2. Introduction
• Digital Pulse Amplitude Modulation (PAM)
Modulates digital information onto amplitude of pulse
May be later upconverted (e.g. to radio frequency)
• Digital Quadrature Amplitude Modulation (QAM)
Two-dimensional extension of digital PAM
Baseband signal requires sinusoidal amplitude modulation
May be later upconverted (e.g. to radio frequency)
• 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
15 - 2

3. Review
Amplitude Modulation by Cosine
• Example: y(t) = f(t) cos(ωc t)
Assume f(t) is an ideal lowpass signal with bandwidth ω1
Assume ω1 < ωc
Y(ω) is real-valued if F(ω) is real-valued
F(ω) 1 ½F(ω + ωc) Y(ω) ½F(ω − ωc)
½
ω ω
-ω1 0 ω1 -ωc - ω1 -ωc + ω1 0 ωc - ω1 ωc + ω1
−ωc ωc
Baseband signal Upconverted signal
• Demodulation: modulation then lowpass filtering
• Similar for modulation with sin(ω0 t) shown next
15 - 3

4. Review
Amplitude Modulation by Sine
• Example: y(t) = f(t) sin(ωc t)
Assume f(t) is an ideal lowpass signal with bandwidth ω1
Assume ω1 < ωc
Y(ω) is imaginary-valued if F(ω) is real-valued
F(ω) 1 j ½F(ω + ωc) Y(ω) -j ½F(ω − ωc)
j ½
ωc
ωc - ω1 ωc + ω1
ω ω
-ω1 0 ω1 -ωc - ω1 -ωc + ω1
−ωc
-j ½
Baseband signal Upconverted signal
• Demodulation: modulation then lowpass filtering
15 - 4

5. Baseband Digital QAM Transmitter
• Continuous-time filtering and upconversion
Impulse gT(t)
i[n] modulator
Index Pulse shapers
Bits (FIR filters) Delay s(t)
Serial/ Local
Map to 2-D
1
parallel
J constellation Oscillator +
converter 90o
Q q[n] Impulse gT(t)
modulator
d
-d d Delay matches delay through 90o phase shifter
I
-d Delay required but often omitted in diagrams
4-level QAM
15 - 5
Constellation

6. Phase Shift by 90 Degrees
• 90o phase shift performed by Hilbert transformer
1 1
cosine => sine cos(2π f 0 t ) ⇒ δ ( f + f 0 ) + δ ( f − f 0 )
2 2
j j
sine => – cosine sin( 2π f 0 t ) ⇒ δ ( f + f 0 ) − δ ( f − f 0 )
2 2
• Frequency response H ( f ) = − j sgn( f )
Magnitude Response Phase Response
| H( f )| ∠H ( f )
90o
f f
-90o
All-pass except at origin
15 - 6

7. Hilbert Transformer
• Continuous-time ideal • Discrete-time ideal
Hilbert transformer Hilbert transformer
H ( f ) = − j sgn( f ) H (ω ) = − j sgn(ω )
1/(π t) if t ≠ 0 2 sin 2 (πn / 2) if n≠0
h(t) = π n
h[n] =
0 if t = 0 0 if n=0
h(t) h[n]
t n
Even-indexed
samples are7 zero
15 -

8. 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
15 - 8

9. Baseband Digital QAM Transmitter
Impulse gT(t)
i[n] modulator
Index Pulse shapers
Bits (FIR filters) Delay s(t)
Serial/ Local
Map to 2-D
1
parallel
J constellation Oscillator +
converter 90o
q[n] Impulse gT(t)
modulator
100% discrete time i[n]
L gT[m]
Index
Bits s[m]
Serial/
Map to 2-D Pulse shapers s(t)
parallel cos(ω0 m) + D/A
1 J constellation (FIR filters)
converter sin(ω0 m)
L samples/symbol q[n] L gT[m]
(upsampling factor) 15 - 9

10. Performance Analysis of PAM
• If we sample matched filter output at correct time
instances, nTsym, without any ISI, received signal
x(nTsym ) = s (nTsym ) + v(nTsym ) v(nT) ~ N(0; σ 2/Tsym)
where the signal component is
3d
s (nTsym ) = an = (2i − 1)d for i = -M/2+1, …, M/2
d
v(t) output of matched filter Gr(ω) for input of
−d
channel additive white Gaussian noise N(0; σ2)
Gr(ω) passes frequencies from -ωsym/2 to ωsym/2 , −3 d
where ωsym = 2 π fsym = 2π / Tsym 4-level PAM
Constellation
• Matched filter has impulse response gr(t)
15 - 10

11. Performance Analysis of PAM
• Decision error d 
PI (e) = P ( v(nTsym ) > d ) = 2 Q Tsym 
for inner points σ 
• Decision error d 
PO− (e) = P (v(nTsym ) > d ) = Q Tsym 
for outer points σ 
d 
PO+ (e) = P(v(nTsym ) < −d ) = P(v(nTsym ) > d ) = Q Tsym 
σ 
• Symbol error probability
M −2 1 1 2( M − 1)  d 
P (e ) = PI (e) + PO+ (e) + PO− (e) = Q Tsym 
M M M M σ 
O- I I I I I I O+
8-level PAM
Constellation -7d -5d -3d -d d 3d 5d 7d
15 - 11

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

13. Performance Analysis of 16-QAM
Q
• Type 1 correct detection 3 2 2 3
P (c) = P( vI (nTsym ) < d & vQ (nTsym ) < d )
1 2
1 1
2
I
( ) (
= P vI (nTsym ) < d P vQ (nTsym ) < d ) 2
1 1
2
( ( )) ( (
= 1 − P vI (nTsym ) > d 1 − P vQ (nTsym ) > d )) 3 2 2 3
16-QAM
d d
2Q ( T) 2Q ( T)
σ σ
2
 d 
= 1 − 2Q
 Tsym  

 σ  1 - interior decision region
2 - edge region but not corner
3 - corner -region
15 13

14. Performance Analysis of 16-QAM
Q
• Type 2 correct detection 3 2 2 3
P2 (c) = P(vI (nTsym ) < d & vQ (nTsym ) < d ) 1 1
2 2
I
= P (vI (nTsym ) < d ) P ( vQ (nTsym ) < d )
2 2
 d   d  1 1
= 1 − Q 
 Tsym  1 − 2Q
 Tsym  

 σ   σ  3 3
2 2
• Type 3 correct detection 16-QAM
P3 (c) = P(vI (nTsym ) < d & vQ (nTsym ) > −d )
= P(vI (nTsym ) < d ) P (vQ (nTsym ) > − d )
2
 d  1 - interior decision region
= 1 − Q 
 Tsym  
 2 - edge region but not corner
 σ 
3 - corner -region
15 14

15. Performance Analysis of 16-QAM
• Probability of correct detection
2 2
4 d  4 d 
P(c) = 1 − 2Q Tsym   + 1 − Q Tsym  
16 
 σ
 
  16  σ 

8 d   d 
+ 1 − Q  Tsym  1 − 2Q Tsym  
16 
 σ

  σ 

d  9 d 
= 1 − 3Q Tsym  + Q 2  Tsym 
σ  4 σ 
• Symbol error probability
d  9 d 
P (e) = 1 − P (c) = 3Q Tsym  − Q 2  Tsym 
σ  4 σ 
• What about other QAM constellations?
15 - 15

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