Receivers, Antennas,
and Signals
Modulation and Coding
Professor David H. Staelin
Lec14.10-1
2/6/01
Multi-Phase-Shift Keying, “MPSK”
BPSK: Binary Phase-Shift Keying
S1(t)
T
S2(t)
M = 2 phases
M = 4 (2 bits)
M = 8 (3 bits)
M = 16
B/R
(4 bits/baud)
1.0
Pe for
Ideal
Receiver
-2
10
BER
(Bit Error
Rate) 10-4
10
Lec14.10-2
2/6/01
Bandwidth/bit = B/R
0.5 Small
Loss
0
-6
0
QPSK: Quadrature Phase-Shift Keying
Step
“Eye
Diagram”
10
20
[Hz] bits/sec
M=2
4
8
16 32
E/No(dB)
10
20
30
QPSK is bandwidth efficient,
but has little E/No penalty.
E/No(dB) (E is J/bit)
F1
Error Probabilities for Binary Signaling
Matched filter reception of unipolar
baseband, coherent OOK or coherent FSK
Pe = Probability of Bit Error
1.0
Pe 0.5
-1
10
10
-2
10
-3
10
-4
10
-5
10
-6
10
-7
–1 0
Q
Eb /No
Eb = average energy per limit
Matched filter
reception of
bipolar baseband,
BPSK or QPSK
Q
2(Eb /No )
2
4
3 dB
6
8 10 12 14
(Eb/No) (dB)
Source: Digital and Analog Communication Systems,
L.W. Couch II, (4th Edition), Page 351
Lec14.10-3
2/6/01
F2
Phasor Diagrams
S(t) = Re{Sejωt}
Im{S}
Equal-Noise
Contour
Im{S}
Re{S}
Decision
Region for H1
Decision
Region for H1
Phasor
Im{S}
M = 16, Amplitude
and Phase
Re{S}
H1
H1
Rectangular
Lec14.10-4
2/6/01
16-PSK
QPSK
Re{S}
Im{S}
M = 19
Re{S}
Superior to
Rectangular
Hexagonal
F3
Intersymbol Interference
MPSK Examples:
Boxcar
Envelope
Typical
Baud
Si(f)
0
Si(f)
Typical
Baud
Window
Function
0
Baud
Boundary Overlapping
S (f)
Window Functions i
0
Want symbols to be orthogonal
within and between windows
Lec14.10-5
2/6/01
Spectrum for
Channel i
f
fi Sidelobes Interfere
with Adjacent Channels
Broader Channel
Bandwidth
f
fi Lower Sidelobes
Narrowband with
Lower Sidelobes
f
fi
F4
Performance Degradation Due to Interchannel Interfence
N = 1 Interfering Channel
N=∞
“Tamed FM” (window overlap)
Boxcar (QPSK)
10
Signal degradation
(dB) same as required
signal boost for
constant Pe
1
0.1
0 0.5 1
2
Channel spacing (Hz)
∆f/R for given E/No
BPS
Closer channel spacing requires more signal power to maintain Pe
Recover by boosting signal power (works until No becomes negligible)
BER
10
-2
N = ∞ Interfering Channels
BER
(for a given E/No)
N=1
10
Lec14.10-6
2/6/01
“Tamed FM”
-8
0.4 0.5 0.6 0.7
∆f/R
F5
Error Reduction via Channel Coding
Shannon’s Channel Capacity Theorem
We want Pe → 0 (banking, etc.) ⇒ E/No → ∞ using prior methods
Theorem: Pe → 0 if channel capacity “C” not exceeded, in bits/sec
C = B LOG2(1 + S/N) bits/sec
[Hz]
Noise Power = NoB
Average Signal Power
(Shannon showed “can,” not “how”)
Examples: S/N = 10 yields C = 3B (3 bits/Hz), S/N = 127 yields C = 7B (7 bits/Hz)
∼21 dB
e.g. 3-kHz phone at 9600 bps requires S/N ≥~10 dB
Lec14.10-7
2/6/01
H1
Channel Codes
Definitions:
1. “Channel codes” reduce Pe
2. “Source codes” reduce redundancy
3. “Cryptographic” codes conceal
Solomon Golomb: A message with content and clarity
has gotten to be quite a rarity;
to combat the terror of serious error,
use bits of appropriate parity.
Channel codes are our principal approach to letting
R → C with acceptable Pe
Lec14.10-8
2/6/01
H2
Coding Delays Message and Increases Bandwidth
Can show: Pe ≤ 2–Tk(C,R), T = time delay in coding process
e.g. use M = 2RT possible messages in T sec.
(RT = #bits in T sec; “block coding”)
use M = 2RT frequencies spaced at ∼1/T Hz
then B = 2RT/T (can → ∞!)
Lec14.10-9
2/6/01
H3
Minimum S/No for Pe → 0
∆
Can show : C ∞ = lim C = S (No ln 2 ) ≥ R( P →0 ) bits/sec
e
B→∞
Therefore
S/No ≥ 0.69 R for Pe → 0 as B → ∞
Pe{Bit Error}
1
∞ = log2 M
-1
10
-2
10
-3
10
-4
10
-5
10
0.1
Lec14.10-10
2/6/01
Typically 2 ≤ M ≤ 64
20 = log2 M (M ≅ 10 )
6
M≅2
0.69
1
10
E/No →
(J/bit)
(W/Hz)
M = number of
frequencies
H4
Error Detection K + R Code
Blocks:
message
check bits
K bits
R bits
Simple parity check –
xxx…x P
K bits
where P ∋ Σ 1’s =
even
or (2 standards)
odd
R = 1 bit
i.e. = A single bit error transforms its block to “illegal” message set
(half are illegal here).
Lec14.10-11
2/6/01
H5
Error Correction Code
Message = m1 m2…mK
Checks = mK + 1….mK + R
Any of these K + R bits can be erroneous
Receive:
ˆ1
m
ˆ 2……………m
ˆK+R
m
Correct:
m1
m2……………mK + R
Sum (modulo 2) = 0’s if no error → 0
0
0
Consider locations of “1”s in K + R slots of Sum
If we wish to detect and correct 0 or 1 bit error in the block of K + R bits,
we need
K + R ≥ K + LOG2 (K + R + 1) bits/block
Original
Information
Number of
Slots for
1-bit error
Lec14.10-12
2/6/01
“No Error”
Message
H6
Single-Bit Error Correction
If we wish to detect and correct 0 or 1 bit error in the block of K + R bits,
we need
K + R ≥ K + LOG2 (K + R + 1) bits/block
Original
Information
Number of
Slots for
1-bit error
K=
1
2
3
4
5
100
3
10
6
10
Lec14.10-13
2/6/01
R≥
2
3
3
3
4
7
~10
~20
R/(R +K)
0.67
0.6
Not too
0.5
efficient
0.4
0.4
0.07
0.01
0.002
“No Error”
Message
R = Check bits needed
to detect and fix ≤ 1 error
in a block of K + R
H7
Two-Bit Error Correction
If we wish to correct two errors:
We need
K + R ≥ K + LOG2 1 + K + R +
(K + R)(K + R - 1)
“No Error” 1 Error
Message
K=
R≥
5
7
0.6
3
∼20
0.02
6
∼40
0.004
10
10
Lec14.10-14
2/6/01
2
2 Errors
R/(R +K)
R = Check bits needed
to detect and fix ≤ 2 errors
in a block of K + R bits
J1
Implementation: Single-Error Correction
∆
Block =
m1
m2
m3
m4
C1
C2
C3
(K = 4, R = 3)
(4 message bits, 3 check bits)
∆
Let C1 = m1 ⊕ m2 ⊕ m3
∆
∆
C2 = m1 ⊕ m2 ⊕ m4 C3 = m1 ⊕ m3 ⊕ m4
∆
(Note: C1 ⊕ C1 = m1 ⊕ m2 ⊕ m3 ⊕ C1 ≡ 0)
Modulo-2 1
matrix 1
multiply 1
1
1
0
1
0
1
0
1
1
∆
=H
(Note: m1 ⊕ m2 ⊕ m3 = C1)
1
0
0
0
1
0
∆
Q=
0
0
1
⊕
0
1
0
0
1
1
1
0
“Sum, modulo-2”
Truth Table
m1
m2
m3
m4
C1
C2
C3
∆
=
0
0
0
If no errors
HQ = 0 defines legal codewords Q
Lec14.10-15
2/6/01
J2
Implementation: Single-Error Correction
HQ = 0 defines legal codewords Q
Only 1/8 of all 7-bit words are legal because C1, C2, and C3 are each
3
correct only half the time and (0.5) = 1/8
Suppose transmitted Q is legal and received R = Q + E then HR = HQ + HE
≡0 ≠0
Interpret to yield error-free Q from R
1
1
Say HR = 0 ⇒ Error in m3 Note that Hi3 = 0
1
1
Can even rearrange transmitted word so:
0 0 0 1 1 1 1 C3
0 1 1 0 0 1 1 C2 = E = Binary representation
of error location “L”
1 0 1 0 1 0 1 m4
L = 1 2 3 4 5 6 7 C1
m3
m2
m1
Lec14.10-16
2/6/01
J3
Pe Benefits of Channel Coding
Suppose Pe = 10 , then P{error in 4-bit word} = 1 – (1 – 10 ) ≅ 4 × 10
(no-coding case)
P{no errors}
-5
-5 4
-5
If we add 3 bits to block (4 + 3 = 7) for single-error correction,
and send it in the same time ⇒ 4 less E/No (2.4 dB loss)
7
-4
Pe → 6 × 10 (per bit; depends on modulation)
p{2 errors in 7 bits @ 6 × 10 } = p{no error} • p{error}
-4
5
2
7 ≅ 8 × 10-6
2
(6 × 10 ) 7 • 6/2!
-4 2
Compare new p{block error} 8 × 10 to 4 × 10 without coding
-6
-5
Coding reduced block errors by a factor of 5 with same transmitter power
Alternatively, reduce power and maintain Pe
Benefits depend on Pe(E/No) relation
Lec14.10-17
2/6/01
J4
Benefits of Soft Decisions
Soft decisions can yield ∼2 dB SNR improvement for same Pe
e.g.
i=1
“Hard Decision:” 2 Alternatives
i=2
v
A B C D E F G H “Soft Decision:”
Say 8 Levels
Example: Parity bit implies one of n bits was received
incorrectly. Choose the one bit for which the
decision was least clear.
Lec14.10-18
2/6/01
J5
Convolutional Codes
Constraint
Length
Convolutional codes
employ overlapping
blocks (sliding window)
Example:
R (bits/sec)
In
2R bits/sec output
3-bit shift register
Sum modulo 2
This is a “rate 1/2, constraint-length-3 convolutional coder”
One advantage: accommodates soft decisions
Here each message bit impacts 3 output bits and therefore
impacts decoder decisions impacting 3 or more reconstructed
bits, so soft decisions help identify erroneous bits.
Lec14.10-19
2/6/01
J6
Receivers, Antennas,
and Signals
Modulation and Coding
Professor David H. Staelin
Lec14.10-20
2/6/01
Rayleigh Fading Channels
e.g. Fading from deep vigorous multipath
N
Consider multipath with output signal z(t) = Σ xi cos ωt + yi sin ωt
i=1
(sum of N phasors, one per path)
Im{z}
z2 (t)(filtered)
z
i=1
Re{z}
0
0
Deep Fade
t
Rayleigh fading: xi and yi are independent g.r.v.z.m.
Im{z}
P{|z|}
1
2
Re{z}
0
eσ
0
σ
|z|
z
Lec14.10-21
2/6/01
K1
Rayleigh Fading Channels
Rayleigh fading: xi and yi are independent g.r.v.z.m.
Im{z}
1
eσ2
Re{z}
0
P{|z|}
σ
0
|z|
z
Variance of Re{z}, Im{z} ≡ σ2
⟨|z|⟩
=
σ π/2
2
⟨|z| ⟩
=
σ2(2 – π/2)
≅
2σ/3 ≠ f(N)
|z| – ⟨|z|⟩
P{|z| > zo}
Lec14.10-22
2/6/01
2
=
–(zo/σ)2/2
e
K2
Effect of Fading on Pe(Eb/No)
Pe curve increases and flattens when there is fading
Pe
Pe
New Pe{Eb/No} Relation
After Fading
p{Eb/No}
Eb(t)
Deep fades produce
bursts of errors
(error clusters)
Lec14.10-23
2/6/01
Eb/No(dB)
Fading history,
increases Pe(t)
t
K3
Remedies for Error Bursts
1. Diversity – Space
– Frequency
– Polarization
≥ 2 independent paths
2. General error-correcting codes
3. Same, plus interleaving:
Error burst, hits only,
one bit per block
t
Burst hits fewer
bits per block
4. Reed-Solomon codes
(Tolerate adjacent errors better than random ones)
e.g. multivalue symbols A (say 4 bits each, 16 possibilities)
so then block error-correct the symbols A:
AAA…A AAA…A A…
Lec14.10-24
2/6/01
K4
Remedies for Error Bursts
Fading flattens Pe(Eb/No) curve, so potential coding gain
can exceed 10 dB sometimes
Pe
Increase in Eb required to accommodate coding
New Pe for coded fading channel
10-2
Reduction in Pe using coding
Flatter Pe(Eb/No) for
fading channel
Coding Gain
-8
10
10 dB
Eb/No (dB)
Note: Coding gain greater for flatter Pe(Eb/No)
Lec14.10-25
2/6/01
K5