2/6/01 Receivers, Antennas, and Signals Modulation and Coding Professor David H. Staelin Lec14.10-1
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2/6/01
Receivers, Antennas, and Signals
Modulation and Coding Professor David H. Staelin
Lec14.10-1
2/6/01
Multi-Phase-Shift Keying, “MPSK”
F1
ing Step
S2(t)T
“EyeDiagram”
S1(t)
E/No(dB)
B/R
Small Loss
M = 2 4
8 3216
QPSK is bandwidth efficient, but has little E/No penalty.
0
Bandwidth/bit = B/R
[Hz] bits/sec
1.0 0.5
BPSK: Binary Phase-Shift Keying
E/No(dB) (E is J/bit)
M = 16 (4 bits/baud)
10-2
10-4
10-6
0
BER (Bit Error Rate)
Pe for Ideal
Receiver
Lec14.10-2
QPSK: Quadrature Phase-Shift Key
10 20 30
M = 2 phases M = 4 (2 bits) M = 8 (3 bits)
10 20
2/6/01
Error Probabilities for Binary Signaling
3 dB
1.0 0.5
10-1
10-2
10-3
10-4
10-5
10-6
10-7
2 4 6 8 10 12 14 (Eb/No) (dB)
Pe
P e =
Pro
babi
lity
of B
it E
rror
, L.W. Couch II, (4th Edition), Page 351
Matched filter reception of unipolar baseband, coherent OOK or coherent FSK
Matched filter reception of bipolar baseband, BPSK or QPSK
Q Eb /No
Q 2(Eb /No )
F2
bE = average energy per limit
Lec14.10-3
–1 0
Source: Digital and Analog Communication Systems
2/6/01
Phasor Diagrams S(t) = Re{Sejωt}
Equal-NoiseContour QPSK
Re{S}
Decision Region for H1
Phasor
Im{S} Im{S}
16-PSK
Re{S} Decision Region for H1
M = 16, Amplitudeand Phase
Im{S}
Re{S}
Rectangular
H1
Re{S}
Hexagonal
Superior toRectangular
M = 19 Im{S}
H1
F3Lec14.10-4
2/6/01
Baud Boundary
Want symbols to be orthogonalwithin and between windows
Overlapping Window Functions
Si(f)
0 fi
f
Narrowband with Lower Sidelobes
Lower Sidelobes
Window FunctionTypical
Baud
Si(f)
0 fi
f
Broader Channel Bandwidth
F4
Intersymbol Interference MPSK Examples:
Boxcar Envelope
Typical Baud
Si(f)
0 fi
f Sidelobes Interfere with Adjacent Channels
Spectrum for Channel i
Lec14.10-5
2/6/01
Performance Degradation Due to Interchannel Interfence
10 Signal degradation (dB) same as required signal boost for constant Pe
1
0.1 2
∆f/R for given E/No
N = 1 Interfering Channel
“Tamed FM” (window overlap) Boxcar (QPSK)
e
o becomes negligible)
BPS
N = ∞
N = ∞
∆f/R10-8
10-2
BER
N = 1
“Tamed FM”
BER o)
F5
Channel spacing (Hz)
Lec14.10-6
0 0.5 1 Closer channel spacing requires more signal power to maintain PRecover by boosting signal power (works until N
Interfering Channels
0.4 0.5 0.6 0.7
(for a given E/N
2/6/01
Error Reduction via Channel Coding
We want Pe → ⇒ o → ∞ using prior methods
Theorem: Pe →
∼21 dB e.g. 3-kHz phone at 9600 bps requires S/N ≥ 10 dB ~
Shannon’s Channel Capacity Theorem
C = B LOG2 /N) bits/sec
[Hz] oB
Average Signal Power
H1Lec14.10-7
0 (banking, etc.) E/N
0 if channel capacity “C” not exceeded, in bits/sec
Examples: S/N = 10 yields C = 3B (3 bits/Hz), S/N = 127 yields C = 7B (7 bits/Hz)
(1 + S
Noise Power = N
(Shannon showed “can,” not “how”)
2/6/01
Channel Codes
Channel codes are our principal approach to letting R → C with acceptable Pe
e
Solomon Golomb: has gotten to be quite a rarity; to combat the terror of serious error, use bits of appropriate parity.
H2
Definitions:
Lec14.10-8
1. “Channel codes” reduce P
2. “Source codes” reduce redundancy
3. “Cryptographic” codes conceal
A message with content and clarity
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 H32/6/01
2/6/01
Minimum S/No for Pe → 0
Therefore
S/No ≥ e → 0 as B → ∞
H4
( ) ( ) bits/secRNS 0ePo →∞ ≥= ∆ = B→∞
Pe{Bit Error}
0.1 0.69
1 E/No →
(J/bit) (W/Hz)
∞ = log2 M1 10-1
10-2
10-3
10-4
10-5
20 = log2 M (M ≅ 106)
M ≅ 2
Typically 2 ≤ M ≤ 64
frequencies
Lec14.10-10
0.69 R for P
2 ln C lim C : show Can
10 M = number of
2/6/01
Error Detection K + R Code
Blocks: message check bits
K bits R bits
Simple parity check – xxx…x P where P ∋ Σ 1’s = even or odd
(2 standards)
(half are illegal here).
K bits R = 1 bit
H5Lec14.10-11
i.e. = A single bit error transforms its block to “illegal” message set
2/6/01
Error Correction Code
1 m2…mK ….m
Any of these K + R bits can be erroneous
Receive: m1 m2……………mCorrect: m1 m2……………m
Sum (modulo 2) = 0’s if no error → 0 0 0
Consider locations of “1”s in K + R slots of Sum
we need K + R ≥ K + LOG2 its/block
Original
Number of Slots for
1-bit error
“No Error” Message
ˆ ˆ ˆ
H6Lec14.10-12
Message = m Checks = mK + 1 K + R
K + R
K + R
If we wish to detect and correct 0 or 1 bit error in the block of K + R bits, (K + R + 1) b
Information
2/6/01
Single-Bit Error Correction
K = R ≥ R/(R +K) 1 2 0.67 2 3 3 3 4 3 5 4 100 7 103 ~10 106 ~20
Not too efficient
R = Check bits needed ≤ 1 error
H7
we need K + R ≥ K + LOG2 its/block
Original
Number of Slots for
1-bit error
“No Error” Message
Lec14.10-13
0.6 0.5 0.4 0.4 0.07 0.01 0.002
to detect and fix in a block of K + R
If we wish to detect and correct 0 or 1 bit error in the block of K + R bits, (K + R + 1) b
Information
2/6/01
Two-Bit Error Correction
) 2
“No Error” Message
K = R ≥ R/(R +K) 5 7 103 ∼20 0.02 106 ∼40 0.004
R = Check bits needed ≤ 2 errors
We need K + R ≥ K + LOG2 1 + K + R +
1 Error
J1Lec14.10-14
(K + R)(K + R - 1
0.6 to detect and fix in a block of K + R bits
If we wish to correct two errors:
2 Errors
2/6/01
Implementation: Single-Error Correction ∆
(Note: C1 ⊕ C1 = m1 ⊕ m2 ⊕ m3 ⊕ C1 ≡ 0)∆
C2 = m1 ⊕ m2 ⊕ m4 ∆ C3 = m1 ⊕ m3 ⊕ m4
∆
J2
Block = m1 m2 m3 m4 C1 C2 C3
(K = 4, R = 3) (4 message bits, 3 check bits)
If no errors (Note: m1 ⊕ m2 ⊕ m3 = C1)
1 1 1 0 1 0 0 m1 0 1 1 0 1 0 1 0 m2 = 0 1 0 1 1 0 0 1 m3 0
m4 C1 C2 C3
∆
= H∆
Modulo-2 matrix
multiply
Q =∆
1 = m1 ⊕ m2 ⊕ m3 ⊕ 0 1 0 0 1 1 1 0
“Sum, modulo-2” Truth Table
∆
Lec14.10-15
HQ = 0 defines legal codewords Q
Let C
2/6/01
Implementation: Single-Error Correction
Can even rearrange transmitted word so: 0 0 0 1 1 1 1 C3 0 1 1 0 0 1 1 C2 1 0 1 0 1 0 1 m4 1 2 3 4 5 6 7 C1
m3 m2 m1
of error location “L”
J3
1, C2, and C3 3
1 1 ⇒ Error in m3 Note that Hi3 = 0
1 1
≡ 0 ≠ 0
Interpret to yield error-free Q from R
Lec14.10-16
HQ = 0 defines legal codewords Q
L =
= E = Binary representation
Only 1/8 of all 7-bit words are legal because C are each correct only half the time and (0.5) = 1/8
Say HR = 0
Suppose transmitted Q is legal and received R = Q + E then HR = HQ + HE
2/6/01
Pe Benefits of Channel Coding
⇒ less E/No
Pe → 6 × 10-4 (per bit; depends on modulation)
4 7
e
Benefits depend on Pe(E/No) relation
7 2p{2 errors in 7 bits @ 6 × 10-4} = p{no error}5 2 ≅ 8 × 10-6
(6 × 10-4)2 !
Compare new p{block error} 8 × 10-6 to 4 × 10-5 without coding
J4
Suppose Pe = 10-5, then P{error in 4-bit word} = 1 – ( -5)4 ≅ 4 × 10-5
P{no errors}(no-coding case)
Lec14.10-17
If we add 3 bits to block (4 + 3 = 7) for single-error correction, and send it in the same time (2.4 dB loss)
Alternatively, reduce power and maintain P
• p{error}
7 • 6/2
Coding reduced block errors by a factor of 5 with same transmitter power
1 – 10
2/6/01
Benefits of Soft Decisions Soft decisions can yield ∼2 dB SNR improvement for same Pe
incorrectly. Choose the one bit for which the decision was least clear.
e.g.
v A B C G H
i = 1
i = 2
“Soft Decision:” Say 8 Levels
J5Lec14.10-18
Example: Parity bit implies one of n bits was received
D E F
“Hard Decision:” 2 Alternatives
2/6/01
Constraint Length
employ overlappingblocks (sliding window)
One advantage: accommodates soft decisions Here each message bit impacts 3 output bits and thereforeimpacts decoder decisions impacting 3 or more reconstructedbits, so soft decisions help identify erroneous bits.
J6
2R bits/sec output
3-bit shift register Sum modulo 2
Example:
InR (bits/sec)
Lec14.10-19
Convolutional Codes
Convolutional codes
This is a “rate 1/2, constraint-length-3 convolutional coder”
2/6/01
Receivers, Antennas, and Signals
Modulation and Coding Professor David H. Staelin
Lec14.10-20
2/6/01
e.g. Fading from deep vigorous multipath
Σ xi cos ωt + yi sin ωt (sum of N phasors, one per path)
N
i = 1
Im{z} z2 (t)(filtered)
Deep Fadet00 Re{z}
i = 1z
K1
z
i and yi are independent g.r.v.z.m. Im{z} P{|z|}
0 0Re{z} |z|eσ2 1
σ
Lec14.10-21
Rayleigh Fading Channels
Consider multipath with output signal z(t) =
Rayleigh fading: x
2/6/01
Variance of Re{z}, Im{z} ≡ σ2
⟨|z|⟩ = σ π/2
⟨|z|2⟩ = σ2(2 – π/2)
|z| – ⟨|z|⟩ 2 ≅ 2σ/3 ≠ f(N)
P{|z| > zo} = e–(zo/σ)2/2
K2
i and yi are independent g.r.v.z.m.
Im{z} P{|z|}
0 0 z
Re{z} |z|eσ2
1
σ
Lec14.10-22
Rayleigh Fading Channels Rayleigh fading: x
2/6/01
Effect of Fading on Pe(Eb/No) Pe curve increases and flattens when there is fading
Pe
Eb/No(dB)
Pe New Pe{Eb/No} Relation After Fading
Eb(t) Fading history, increases Pe(t)
t
Deep fades produce bursts of errors (error clusters)
p{Eb/No}
K3Lec14.10-23
2/6/01
Remedies for Error Bursts
AAA…A AAA…A
(Tolerate adjacent errors better than random ones)
so then block error-correct the symbols A:
≥ 2 independent paths
K4
Error burst, hits only,one bit per block
Burst hits fewer bits per block
t
Lec14.10-24
2. General error-correcting codes
A…
4. Reed-Solomon codes
e.g. multivalue symbols A (say 4 bits each, 16 possibilities)
1. Diversity – Space – Frequency – Polarization
3. Same, plus interleaving:
2/6/01
Remedies for Error Bursts Fading flattens Pe(Eb/No) curve, so potential coding gain can exceed 10 dB sometimes
10 dB 10-8
10-2
Pe Increase in Eb required to accommodate coding
Reduction in Pe using coding
Eb/No (dB)
New Pe for coded fading channel
Note: Coding gain greater for flatter Pe(Eb/No)
Coding Gain
K5
Flatter Pe(Eb/No) for fading channel
Lec14.10-25
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