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EITF25 Internet-‐-‐Techniques and Applica8ons Stefan Höst
L2 Physical layer
OSI model Open Systems Interconnec8on
• Developed by ISO, 1970~
2
TCP/IP model • Developed by DARPA, 1970~
3
Physical layer
• Analog vs digital signals • Sampling, quan8sa8on
• Modula8on • Represent digital data in a con8nuous world
• Transmission media • Cables and such
• Disturbances • Noise and distor8on
4
Data vs Signal
• Data: Sta8c representa8on of informa8on • For storage (oUen digital)
• Signal: Dynamic representa8on of informa8on • For transmission (oUen analog)
5
Analog vs digital
Analog • Con8nuous 8me and
amplitude signal • Electrical/op8cal domain
Digital • Discrete 8me and amplitude • Binary representa8on
6
t
s(t)
n
s[n]
Digitaliza8on of analog signals
Performed in three steps: 1. Sampling
Discre8za8on in 8me 2. Quan8za8on
Discre8za8on in amplitude 3. Encoding
Binary representa8on of amplitude levels
7
Sampling • The process of
discre8zing 8me of a con8nuous signal.
• Sampling 8me: • Sampling frequnecy:
• Loose iforma8on about 8me
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−4
−2
0
2
4
t
s(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−4
−2
0
2
4
ts(t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−4
−2
0
2
4
t
s(t)
0 5 10 15 20 25 30 35 40 45−4
−2
0
2
4
n
s[n]
8
s[n]= s(nTs )Ts
Fs = 1/Ts
Aliasing
9
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−1
−0.5
0
0.5
1y(t)=cos(14π t), ΩT=20π rad/s
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−1
−0.5
0
0.5
1yr(t)=cos(6π t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−1
−0.5
0
0.5
1y(t)=cos(14π t), ΩT=20π rad/s
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7−1
−0.5
0
0.5
1yr(t)=cos(6π t)
y(t)=cos(14πt) Fs=10 Hz
y(t)=cos(6πt)
Reconstruc8on to lowest possible frequency
Shannon-‐Nyquist Sampling Theorem If s(t) is a band limited signal with highest frequency component Fmax, then s(t) is uniquely determined by the samples s[n] = s(nT) if and only if The signal can be reconstructed with Fs/2 is the Nyquist frequency and 2Fmax the Nyquist rate
max21 FT
Fs ≥=
s(t) = s[n]sinc t − nTsTs
⎛⎝⎜
⎞⎠⎟n=−∞
∞
∑
10
Reconstruc8on Example
11
−2 0 2 4 6 8 10 12−1
−0.5
0
0.5
1
y(t)=sin(2π/10*t), Fs=1Hz
−2 0 2 4 6 8 10 12−1
−0.5
0
0.5
1
sum y[n]sinc((t−nT)/T)
−2 0 2 4 6 8 10 12−1
−0.5
0
0.5
1
y(t)=sin(2π/10*t), Fs=1Hz
−2 0 2 4 6 8 10 12−1
−0.5
0
0.5
1
sum y[n]sinc((t−nT)/T)
y(t)=sin(2/7πt), Fs=1Hz
Sampling theorem proof Two important transforms
12
-1/2 1/2
-1 1 2 3 -1 1 2 3
1 2 3-1 t f
t f
sinc(t)
F
p(f)
x(t) =P
n (t n)
Fx(f) =
Pn (f n)
+
+
1
Sampling theorem proof Mathema8cal descrip8on of sampling
13
-T T 2T
-1/T 1/T 2/T -1/T 1/T 2/T
-T T 2T 3T3Tt t t
f f f
F
s(t)
1Tx( t
T ) =P
n (t nT )
=
ss(t) = s(t) 1Tx( t
T )
=P
n s[n](t nT )
S(f) x(Tf) = 1T
Pn (f n
T ) Ss(f) = S(f) x(Tf)
= 1t
Pn S(f n
T ) =
+
+
2
Sampling theorem proof Reconstruc8on
14
-1/T 1/T 2/T -1/2T 1/2T
-T T 2T 3T T 2T 3T-T
T
f f f
t t t
F1
Ss(f) =1T
Pn S(f n
T )Tp(fT )
S(f)
=
ss(t) =P
n s[n](t nT )sinc( t
T )s(t) = ss(t) sinc( t
T )
=P
n s[n]sinc(tnTT )
=
+
+
3
Sampling theorem Aliasing
Let Fs<2Fmax
15
Sampling Reconstruction
-F F 2F -F/2 F/2f f f
S(f) Ss(f) S(f)
+
+
4
Example
16
y(t) = cos(2π 7t)→Y ( f ) = 12δ ( f + 7)+δ ( f − 7)( )
7-7
-10-20 10 20 30-7 -3 3 7 13 17 23 27-13-17
-5 5
-3 3
f
f
f
Sampling with Fs=10 Hz
Reconstruct in [-‐Fs/2, Fs/2]: Y ( f ) =12δ ( f + 3)+δ ( f − 3)( )→ y(t) = cos(2π 3t)
Quantization
Linear Quan8za8on for k bits • 2k equidistant
levels • Represent sample
with k bits
17
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34 10. Rate Distortion
x
xQ
D
D
M12 D
M12 D
M2 D M
2 D
Figure 10.7: A linear quantisation function.
used to form the quantiser for the statistics of the continuous source. However,in most practical implementations a linear quantiser is used. If the statisticsdiffer much from the uniform distribution the quantiser can be followed by asource code like the Huffman code.
In the figure a quantiser with M output levels is shown. Assuming a maximumlevel of the output mapping of D = M1
2 D, the granularity becomes D = 2DM1 .
The mth output level then corresponds to th evalue
xQ(m) =m M 1
2D= (2m M + 1)
D2
(10.58)
The mapping function shown in the figure is detrmined from finding an integerm such that
xQ(m) D2 x < xQ(m) D
2(10.59)
then the output index is y = m. If the input value x can exceed the intevallxQ(0) D
2 x < xQ(M 1) D2 the limits should be
y =
8><
>:
m, xQ(m) D2 x < xQ(m) D
2 , 1 m M 20, x < xQ(0) + D
2M 1, x xQ(M 1) D
2
(10.60)
Encoding Representa8on of quan8zed samples in bits
18
0 1 2−2
−1.5
−1
−0.5
0
0.5
1
1.5
2
x(t)x[n]Quant level
000
001
010
011
100
101
110
111
x=011100100101101101111111111…
Quan8sa8on distor8on
19
Distor8on: Average distor8on for uniform input:
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“InfoTheory” — 2014/10/30 — 10:08 — page 278 — #286 ii
ii
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278 11. Rate Distortion
xQ(0) D2 x < xQ(M 1) + D
2 the limits should be
y =
8
>
<
>
:
m, xQ(m) D2 x < xQ(m) D
2 , 1 m M 20, x < xQ(0) + D
2M 1, x xQ(M 1) D
2
(11.61) eq:RD:Quant:y-v1
From (eq:RD:Quant:x_Q=11.59) this can equivalently be written as
y =
8
>
<
>
:
m, (2m M)D2 x < (2m M)D
2 + D, 1 m M 20, x < (2 M)D
2M 1, x (M 2)D
2(11.62) eq:RD:Quant:y-v2
The output values from the quantiser can be represented by a finite length bi-nary vector. The price for representing a real value with a finite levels is anerror introduced in the signal. In Figure
fig:RD:LinQuantDist11.8 this error, defined as the dif-
ference x xQ, is shown in the upper plot, and the corresponding distortionIDX:quantisation distortion,
d(x, xQ) = (x xQ)2 in the lower plot.
x
x xQM2 D
M2 D
x
d(x, xQ)
M2 D M
2 D
Figure 11.8: The quantisation error, x xQ, and distortion, d(x, xQ) = (x xQ)
2, for the linear quantisation function in Figurefig:RD:LinQuant11.7. fig:RD:LinQuantDist
An estimate of the distortion introduced can be made by considering a uni-formly distributed input signal, X U(M D
2 , M D2 ). Then all quantisation
levels will have uniformly distributed input with f (x) = 1D , and deriving the
average distortion can be made with normalised xQ = 0,
E
(X XQ(m))2|Y = m
=Z D/2
D/2x2 1
Ddx =
D2
12(11.63)
From the uniform assumption we have P(Y = m) = 1M and hence
E
(X XQ)2 =
M1
Âm=0
1M
D2
12=
D2
12(11.64)
d(x, xQ ) = (x − xQ )2
E X − XQ( )2⎡⎣
⎤⎦ = x2 1
Δdx =
−Δ/2
Δ/2
∫Δ2
12
Quan8za8on
Delta modula+on • Represent
change in amplitude with 1 bit • 1: +1 • 0: −1
• Must be faster sampling
20
n5 10 15
s(n)
1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 1
Examples Telephony Fmax= 4 kHz Fs= 8 kHz (samples per sec) 8 bit/sample => 64 kb/s
CD Fmax= 20 kHz Fs= 44.1 kHz (samples per sec) 16 bit/sample => 705.6 kb/s 2 channels (stereo) => 1.4 Mb/s
21
From bits to signals
Principles of digital communica8ons
22
Internet
Digital data
Analog signal
On-‐off keying
• Send one bit during Tb seconds and use two signal levels, “on” and “off”, for 1 and 0.
Ex.
23
t
s(t)
T 2T 3T
x=10010010101111100
A
0
Non-‐return to zero (NRZ)
• Send one bit during Tb seconds and use two signal levels, +A and -‐A, for 0 and 1.
Ex.
24
t
s(t)
T 2T 3T
x=10010010101111100
A
-A
Mathema8cal descrip8on
With g(t)=A, 0<t<T, the signals can be described as • On-‐off • NRZ
25
s(t) = ang(t − nT )n∑
an = xn
an = (−1)xn
Two signal alterna8ves • s0(t)=0 and s1(t)=g(t)
• s0(t)=g(t) and s1(t)=-‐g(t)
Manchester coding
• To get a zero passing in each signal 8me, split the pulse shape g(t) in two parts and use +/-‐ as amplitude.
Ex.
26
T
g(t)
t
A
T/2
-A
t
s(t)
T/2 T 3T/2
x=10010010101111100
A
-A
2T 3T 4T 5T 6T 7T
Mul8 level modula8on
• To transmit k bits in one signal alterna8ve of dura8on Ts, use M=2k levels.
Ex. Two bits per signal
27
t
s(t)
T 2T 3T
x=10 01 00 10 10 11 11 10 00
A
-A
3A
-3A
01:
10:
00:
11:
Pulses in frequency
28 0 2 4 6 8 10−50
−45
−40
−35
−30
−25
−20
−15
−10
−5
0
Normalised frequency
|G(f)
|2 [dB]
RecManTri
Bandwidth
• The bandwidth is the posi8ve frequency interval occupied by the main part of the signal power • Main lobe • WX% bandwidth contains X% of power in signal
Typical X: • 90% • 99% • 99.9%
29
Pulse Wlobe W90% W99%
Rect 2/T 1.7/T 20.6/T
Man 4/T ≈ 6/T ≈ 60/T
Tri 4/T 1.7/T 2.6/T
TRC 4/T 1.9/T 2.82/T
HCS 3/T 1.6/T 2.36/T
Some system parameters • With signal 8me Ts the signal rate is Rs=1/Ts. The
unit of the signal rate is oUen Hz. • If k bits are transmiqed each signal alterna8ve, the
8me per bit is Tb=Ts/k. The bit rate, or data rate, is Rb=1/Tb. The unit is b/s or bps. Rb=kRs
• Bandwidth efficiency Rb/W, in bps/Hz • If P signal power, then • Es = TsP is energy per signal alterna8ve • Eb = TbP = Es/k energy per bit
30
Modula8on in frequency
• Baseband signal centered around f=0. • Pass-‐band signal centered around f=f0, f0>>W. • ShiU a BB signal up in frequency to PB by
mul8plying with cos(2 pi f0 t). • ShiU a PB signal down to BB by mul8plying with
cos(2 pi f0 t) followed by low-‐pass filtering.
31
Modula8on in frequency
32
=
# F
s(t)
t
3A
1A
1A
3A
cos(2f0t)
t
sf0(t)
t
3A
1A
1A
3A
=S(f)
f
12 ((f + f0) + (f f0))
ff0 f0
12 (S(f + f0) + S(f f0))
ff0 f0
ASK (Amplitude ShiU Keying)
Use on-‐off keying at frequency f0.
33
t
s(t)
T 2T 3T
x=10010010101111100
A
0
BPSK (Binary Phase ShiU Keying)
Use NRZ at frequency f0, but view informa8on in phase
34
s(t) = (−1)xn g(t − nT )cos(2π f0t) = g(t − nT )cos(2π f0t + xnπ )n∑
n∑
t
s(t)
T 2T 3T
x=10010010101111100
A
-A
QPSK and M-‐PSK
Quadrature PSK • 2 bits/signal gives 4 different phase levels,
35
s(t) = g(t − nT )cos 2π f0t + xnπ2
⎛⎝⎜
⎞⎠⎟n
∑xn ∈0,1,2,3
M-‐PSK • k bits/signal gives M=2k
different phase levels,
s(t) = g(t − nT )cos 2π f0t +ϕn( )n∑
ϕn =xnM2π
xn ∈0,1,...,M −1
Signal space
Since and cos and sin behaves orthogonal, view PSK signals as points on a circle.
36
cos(2π f0t + π2 ) = sin(2π f0t)
cos
sin
s0(t)
s1(t)
s2(t)
s3(t)
cos
sin
s0(t)
s2(t)
s4(t)
s6(t)
s1(t)s3(t)
s5(t) s7(t)
QPSK 8-‐PSK
M-‐PAM (Pulse Amplitude Modula8on)
• Use amplitude for informa8on carryer. • M=2k amplidudes
where
37
s(t) = ang(t − nT )cos(2π f0t)n∑ an = −M +1+ 2xn
g(t)cos1 3 5 7-1-3-5-7
s0(t) s1(t) s2(t) s3(t) s4(t) s5(t) s6(t) s7(t)
M-‐QAM (Quadrature amplitude Modula8on)
Use that cos and sin are orthogonal to combine two orthogonal PAM constella8ons
38
g(t)cos
g(t)sin
g(t)cos
g(t)sin
OFDM Orthogonal Frequency Division Mul8plexing
N QAM signals combined in an orthogonal way • Used in e.g. ADSL, VDSL, WiFi, DVB-‐C&T&H, LTE, LTE-‐A
39
f
W
Transmission media
Guided media • Twisted pair copper
cables • Coax cable • Fibre op8c cable
Unguided media • Radio • Microwave • Infra red
40
Fibre op8c
• Transmission is done by light in a glass core (very thin)
• Total reflec8on from core to cladding
• Bundle several fibres in one cable
• Not disturbed by radio signals
41
Op8cal network architekture
Point to point • Two nodes are connected by one dedicated fibre
Point to mul8-‐point • One one is connected to several end nodes • PON (Passive Op8cal Network)
2.5/1 Gbps
• Used in • Core network
• FqH 42
Twisted pair copper cables
Two conducters twisted around each other • Twis8ng decreases disturbances (and emission) • Used for • Telephony grid (CAT3) • Ethernet (CAT5, CAT6 and some8mes CAT 7)
43
Coax cable
One conductor surrounded by a shield • Used for • Antenna signals • Measurement instrumenta8ons
44
Radio structures
Single antenna system
MIMO (Mul8ple In Mul8ple Out)
45
Transmission impairments
When a signal is transmiqed on a link, it will deteriorate due to transmission impairment: • Aqenua8on, loss of energy
• Electromagne8c emission
• Distor8on, modifica8on of signal shape • Mul8-‐path propaga8on
• Noise, various noise sources
46
Signal-‐to-‐noise ra8o (SNR) = Average signal power Average noise power
Noise disturbances
Thermal noise (Johnson-‐Nyquist) • Generated by current in
a conductor • -‐174 dBm/Hz
(3.98*10-‐18 mW/Hz) • Rela8vely white Impulse noise • OUen user generated
Intermodula8on noise • Different systems disturb
each other Cross-‐talk • User in the same system
disturb each other Background noise • Misc disturbances
(e.g. Telephony cable -‐140 dBm/Hz)
47
Addi8ve white noise
A commonly used model for noise is the AWGN channel (Addi8ve White Gaussian Noise) • White noise with PSD(f) = N0/2 is added to the
signal • AUer LP filter and sampling the added noise
samples are Gaussian distributed
48
zn ∼ N 0, N0 / 2( )
Gaussian channel
A Gaussian channel is a sta8s8cal transmission model with input variable X of total power E[X2]<P and output variable Y=X+Z, where Z is Normal distributed with zero mean and variance N.
49
X
Z N0,pN
Y = X + Z
1
A glimpse of Informa8on Theory
• The channel represents all impairments during transmission. • The informa8on about X by observing Y is given by the mutual informa8on
50
Channel X Y
I(X;Y ) = f (x, y)log2f (x, y)f (x) f (y)
dxdyR2∫
Maximise for E[X2]=P over all distribu8ons to get the channel capacity For the Gaussian channel
C = maxf (x ),E[X2 ]=P
I(X;Y )
C = 12log2 1+
PN
⎛⎝⎜
⎞⎠⎟
= 12log2 1+ SNR( )
Shannon capacity
If s(t) is a band limited signal with bandwidth W and power P is transmiqed over an AWGN channel, the maximum achievable data rate [b/s] is given by
51
C =W log2 1+P
N0W⎛⎝⎜
⎞⎠⎟
=W log2 1+ SNR( )
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“InfoTheory” — 2014/11/5 — 10:13 — page 235 — #243 ii
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9.3. Fundamental Shannon limit 235
Consider a band limited Gaussian channel with bandwidth W and noise levelN0. If the transmitted power constraint is P, then the capacity is given by
C = W log
1 +P
N0W
[b/s] (9.52) Eq:Gauss:Capacity
The capacity maximises the throughput for a fixed signal power when usinga certain bandwidth. As the bandwidth increases the noise power will alsoincrease, whereas the signal power will remain, meaning that the capacity in-crease will diminish with increasing bandwidth. In Figure
fig:Gauss:GbandlimWvsC9.6 the capacity as a
function of the bandwidth is shown. As the bandwidth increases to infinity thefollowing limit value will be reached.
C• = limW!•
W log
1 +P/N0
W
= limW!•
log
1 +P/N0
W
W
= log eP/N0 =P/N0ln 2
W
C
Figure 9.6: Capacity as a function of the bandwidth W for a fixed total signalpower P. fig:Gauss:GbandlimWvsC
Denoting the achieved bit rate as Rb, it is required that this is not more than thecapacity, C• > Rb. Further, assume that the signalling time is Ts and that ineach signal k information bits are transmitted. Then the average power is theenergy per time instance, which can be derived according to
P =EsTs
=EbkTs
=EbTb
(9.53) Gauss:Limit:PT
where Es is the average energy per transmitted symbol, Eb the average energyper information bit and Tb = Ts/k the average transmission time for each infor-mation bit. The variable Eb is very important since it can be compared between
Fundamental Shannon limit
Leng the bandwidth approach infinity gives
The achieved bit rate is Rb=Ts/k. Then Hence, communica8on is only possible if
Approaching the limit means leng the computa8onal complexity go to infinity.
52
C∞ = limW→∞
W log 1+ P / N0
W⎛⎝⎜
⎞⎠⎟ =
P / N0
ln2 C∞
Rb= Eb / N0
ln2>1
Eb
N0
> ln2 = 0.69 = −1.59dB
Data communication
Two nodes that transmit data on a physical link cannot do this simultaneously on the same frequencies with the same coding scheme.
53
Data flow concepts
54
Duplexing
Duplex communica8on can be achieved by • TDD (Time Division Duplexing)
Each direc8on has its 8me slots to transmit in. • FDD (Frequency Division Duplexing)
Each direc8on has its frequency slots to transmit in
55
Multiplexing of links
Also, physical links need to be shared. This is called mulOplexing, where one physical link is divided into several channels.
56
Multiplexing techniques
5 basic types of mul8plexing techniques: • Space-‐Division mul8plexing (SDM) • Time-‐Division Mul8plexing (TDM) • Frequency-‐Division Mul8plexing (FDM) • Wavelength-‐Division Mul8plexing (WDM) • Code-‐Division Mul8ple Access (CDMA)
57
Space-Division Multiplexing (SDM)
SDM is used in cables. Each channel uses one line (op8cal fibre or twisted pair).
58
Time-Division Multiplexing (TDM)
In TDM, each channel occupies a por8on of the 8me in the link.
59
Frequency-Division Multiplexing (FDM)
FDM is an analog mul8plexing technique where each link has its own frequency band
Each channel uses a unique carrier frequency.
Ch 1 Ch 2
fCh N...
60
Wavelength-Division Multiplexing (WDM)
In an op8cal fibre different wavelengths can be combined.
61
Synchronous TDM
If a channel has nothing to send, its 8me slots will be empty!
62
Example: Empty slots
63
Statistical Time-Division Multiplexing
In Sta8s8cal TDM, the channels have no reserved 8me slots. Instead, slots are dynamically allocated. Data about the des8na8on is added to each slot. Sta8s8cal TDM usually has beqer performance than Synchronous TDM when not all channels transmit data all the 8me.
64
TDM comparison
65
CDMA (Code-Division Multiple Access)
CDMA, or Spread Spectrum, is a mul8ple access technique for wireless links. The original signal is changed in a spreading process.
66
Spread Spectrum techniques
• Frequency Hopping Spread Spectrum (FHSS) • A source uses many carrier frequencies. One
frequency is used at a 8me, but the frequencies are changed very oUen (e.g. 1000 8mes a second).
• Direct Sequence Spread Spectrum (DSSS) • Each data bit is replaced with n bits (called chips)
using a unique spreading code. The chip sequences for all users are orthogonal.
67
Comparison with FDM
68
DSSS
The original signal is combined with the spreading code.
69
Mod seq 1
Mod seq 2
Mod seq M
U 1
U 2
U N
US 1
US 2
US M
All added
Mod seq M
U N
Mod seq 2
U 2
Mod seq 1
U 1
DSSS example
70
