wimax advanced

Prepared by: Eng. Ahmed Zaaza

ATMATMtransporttransport

IPIPtransporttransport

Convergence Sub-layerConvergence Sub-layer

Common part Sub-layerCommon part Sub-layer

Privacy Sub-layerPrivacy Sub-layer

Physical Sub-layerPhysical Sub-layer





WIMAX layered architecture

WIMAX physical sub-layer

Data Source RandomizerReed-

SolomonEncoder

ConvolutionalEncoder

Sub-CarrierMapping IFFT

Interleaver

To air interface

Channel coding

WIMAX physical sub-layer1. Randomizer:


SolomonEncoder



Interleaver

To air interface

Channel coding



SolomonEncoder



Interleaver

To air interface

1. Randomizer:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Data input

Data output

• Randomizer is used to change the bit order to avoid long sequence of consecutive zeros & ones.



SolomonEncoder



Interleaver

To air interface

1. Randomizer:

• On the downlink sub-frame, Preambles are not randomised, the randomization begins from the information bits.

• At the start of the DL sub-frame, the randomizer is initiated with the sequence: 1 0 0 1 0 1 0 1 0 0 0 0 0 0 0



SolomonEncoder



Interleaver

To air interface

1. Randomizer:

• At the start of burst 2, the randomiser is initialised with the vector as shown.

UplinkUplinkSub-frameSub-frame

DownlinkSub-frame

PreamblePreamble FCHFCH MAPsMAPs UL PHY

Burst 1

UL PHY

Burst n

.…

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

Frame number

DIUC BSID1 1 1

WIMAX physical sub-layer2. Channel coding:


SolomonEncoder



Interleaver

To air interface

Channel coding

WIMAX physical sub-layer2. Channel coding:Why we must use coding before sending data over a channel?

1001110100011010

Data bits without coding

Error from channel

1001010101111010

Data bits with error

Are these bits have error or not ?Are these bits have error or not ?I don’t know . . . . I don’t know . . . .


Linear Block Code

Message block Message block UU of K-bits of K-bitsParity check bitsParity check bits

K - bits

Coded message V of N - bits

(N-K ) bits

• All possible messages = all possible code words = 2 ^ K

• How we could get the coded message V from the un-coded message U.


Linear Block Code• We must have G-matrix.

Parity matrix

P

Identity matrix

(Unity matrix)I

G =

N

K

K

N - K

• If we have a un-coded message U then we multiply it with G to get V

V = U*G

Contains only K code wordsContains only K code words

WIMAX physical sub-layer2. Channel coding: Linear Block Code

• Example:

1 1 0 1 0 0 0

0 1 1 0 1 0 0

1 1 1 0 0 1 0

1 0 1 0 0 0 1

G =

N

K

K

N - K

V = U*G

V = 0 0 0 1 1 0 1 (coded message to be send)

K = 4K = 4N-K = 3N-K = 3

if U = 1101

Check bits U


• At the receiver we use H-matrix:

• In our example:

Parity matrix

transposeP-1

Identity matrix

(Unity matrix)I

H =

N

K

N-K

N - K

1 0 0 1 0 1 1

0 1 0 1 1 1 0

0 0 1 0 1 1 1

H = N

K

N-K

N - K


• We send the constructed V:

U V=U*G channel with error R = V + e

• In our example:

V = 0001101R = 0001111 Syndrome (S) = R * HT

S = (0001111) * = 1 1 1

1 0 0

0 1 0

0 0 1

1 1 0

0 1 1

1 1 1

1 0 1

The sixth bit has an error

R = 0001111

U = 0001101

WIMAX physical sub-layer2. Channel coding: Cyclic Block Code

• This code called Cyclic because if we make right rotate to any code word, it gives another code word.

• It represents the un-coded message by a polynomial U(x) and the code by another polynomial g(x).

• Code dimensions can be written as follows: C (N,K)where: N is the length of the code word.

K is the length of the un-coded message. N-K is the number of added check bits.

• The greatest power of the code (generator) polynomial indicates the number of added check bits (N-K).


• Example:•C (7,4) N=7 , K=4 , m=3• Generator polynomial g(x) = X³ + X² + 1• Message polynomial U(x) = X² + X + 1 (mean 1 1 1

0 )Step (1):• Multiply U(x) * X^(m) = X³ * (X² + X + 1) = X + X⁴ + X³

Step (2):• Divide U(x) * X^(m) by g(x)

Step (3):• V(x) = U(x) * X^(m) + b(x) = X + X ⁴ + X³ + X

= 0 1 0 1 1 1 0

1 + X + X² + X⁴

5X + X⁴ + X³X³ + X + 1X² + X

X + X ³ + X²

5

5

X ⁴ + X²X ⁴ + X² + X

XReminder b(x)

5

Check bits U

• U(x) = R(x) + S(x)• R(x) = 0 1 1 1 1 1 0

= X + X ⁴ + X³ + X² + X + X²


In our example:• V(x) = X + X ⁴ + X³ + X

= 0 1 0 1 1 1 0• R(x) = 0 1 1 1 1 1 0

= X + X ⁴ + X³ + X² + X

5X + X⁴ + X³ + X² + XX³ + X + 1X² + X

X + X ³ + X²5

X ⁴ + XX ⁴ + X² + X

X²Reminder S(x)

5

channel with error V(x) R(x) = V (x)+ e(x)

5

Syndrome (S(x)) = R(x) mod g(x)

5

U(x) = X + X ⁴ + X³ + X = 0 1 0 1 1 1 05

• BCH codes is a type of codes used to encode block of K symbols each symbol consist of S-bits and adds some symbols as parity check each of s-bits.

• It uses a term called finite field (Galois Fields (GF)).• Because we transmit binary data so, we use GF(2^s). This GF

contain elements as follows: F={ 0,1,α, α ², α ³, α ⁴, . . . . . ., α^(2^m – 1) }

• Example: GF(2^3) = {0,1, α, α ², α ³, α⁴, α ,α ,α } contains 8 non zero symbols

Each element consist of 3 bits.

WIMAX physical sub-layer2. Channel coding: BCH code

S bitsS bits S bitsS bits S bitsS bits S bitsS bits. . . .

K-symbols

S bitsS bits S bitsS bits

N-K symbols

5 6 7

Equal 1


• So, each one element have a distribution according to a given Field Generator Polynomial ex: P(x) = X + X + 1

• The generator polynomial is generated from the roots & there conjugate of the Field Generator Polynomial.

• Example: P(x) = x³ + x + 1 S = 3

i α α² α α

0 1 0 0 11 α 0 1 02 α² 1 0 03 α³ = α + 1 0 1 1 4 α⁴ = α (α + 1 ) = α² + α 1 1

05 α = α (α² + α ) = α³ + α² = α² + α + 1 1 1

16 α = α (α² + α + 1) = α³ + α² + α = α² + 1 1 0

17 α = α (α² + 1) = α³ + α = α + 1 + α = 1 0 0

1 repeated

s

567

i 1 0


• BCH code has a dimension of C(N,K) of s-bits.

• The number of parity symbols = N – K = 2T

• The code can correct up to T error symbols

May ( T ) complete symbols have errors.

May ( T ) bits have errors each one in difference symbol.

WIMAX physical sub-layer1. Reed-Solomon Encoder:


SolomonEncoder



Interleaver

To air interface

Channel coding

WIMAX physical sub-layer2. Channel coding: Reed – Solomon

code• It is a part from the BCH code.

• Its dimension is: RS(255,239)

• N = 255 , K = 239 , S = 8• N-K = 16 = 2T• T = 8 it can correct 8 symbols.

• It uses GF( 2^8 ) i.e: each symbol consist of 8 bit.

• Field generator polynomial: p(x) = x8 + x4 + x3 + x2 + 1.• Code generator polynomial: g(x) = (x + m0) (x + m1) (x + m2)…(x +

m2T-1)

• Coding rate = 239/255 = 0.937

WIMAX physical sub-layer1.Convolutional

Encoder:


SolomonEncoder



Interleaver

To air interface

Channel coding

WIMAX physical sub-layer2. Channel coding: Convolutional

encoder• It is defined by three parameters n, k, mWhere:• n: number of output coded bits• k: number of input data bits enter the encoder simultaneously.• m: is the number of registers of the encoder menus one (m+1 =

registers ).

• The coding rate Rc = k/n (k is chosen to be always 1)


encoder• Example of convolutional coder of: R = ½ , k = 1 , n = 2 , m = 2

Input data bitsm

Output coded bits

1u

2u

First coded bit

Second coded bit

21,uu

(Branch word)


encoder• Example of convolutional coder of: R = ½ , k = 1 , n = 2 , m = 2

• Message m = (101)

0000 0011

1100 0011

11 00

11 00

1 11 1 1 01 0

0 00 0 1 01 0

UU11 UU22 UU11 UU22

UU11 UU22UU11 UU22

UU11

UU22

UU11

UU22

UU11

UU22

UU11

UU22

TT11 TT22

TT33 TT44


encoder

1100 000000 001 11 1 0 00 0UU11 UU22 UU11 UU22

UU11

UU22

UU11

UU22

TT55 TT66

n = 2, k = 1, m= 2,

L = 3 input bits 10 output bits

k = (101) U = (11 10 00 10 11)

EncoderEncoder


encoder• Polynomial representation:Polynomial representation:• We define n generator polynomials, one for each modulo-2 We define n generator polynomials, one for each modulo-2

adder. Each polynomial is of degree adder. Each polynomial is of degree mm or less and describes or less and describes the connection of the shift registers to the corresponding the connection of the shift registers to the corresponding modulo-2 adder.modulo-2 adder.

Example: for m = 2Example: for m = 2

The output sequence is found as follows:The output sequence is found as follows:

22)2(2

)2(1

)2(02

22)1(2

)1(1

)1(01

1..)(

1..)(

XXgXggX

XXXgXggX

g

g

)()( with interlaced )()()( 21 XXuXXuXV gg


encoder• Let us see the output from the equation:

g1(x) = 1 + X + X ²g2(x) = 1 + X ²

u(x) * g1(x) = (1 + X ²) (1 + X + X ²) = 1 + X + X³ + X⁴u(x) * g2(x) = (1 + X ²) (1 + X ²) = 1 + X⁴

u(x) * g1(x) = 1 + X + 0(X²) + X³ + X⁴ output from first branch U1

u(x) * g2(x) = 1 + 0(X) + 0(X²) + 0(X³) + X⁴ output from second branch U2

1 11 1UU11 UU22

1 01 0UU11 UU22

0 00 0UU11 UU22

1 01 0UU11 UU22

1 11 1UU11 UU22


encoder• Convolutional coding in wimax:

R = ½ , k = 1 , n = 2 , M = 6


encoder• Convolutional coding in wimax:

• OFDM symbol:

6 zero bits to initialize the encoder

• OFDMA symbol:

ConvolutionalConvolutionalEncoderEncoder

ConvolutionalConvolutionalEncoderEncoderData blockData block 6 bits6 bits Data blockData block 12 bits12 bits

Data blockData block

WIMAX physical sub-layer1. Interleaver:


SolomonEncoder



Interleaver

To air interface

Channel coding


•Convolutional codes are suitable for memory less channels with random error events.•Some errors have bursty nature:

Statistical dependence among successive error events (time-correlation) due to the channel memory.

Like errors in multipath fading channels in wireless communications.

A burst error of length 3 can not be A burst error of length 3 can not be corrected.corrected.

Let us use a block interleaver 3X3Let us use a block interleaver 3X3

A1 A2 A3 B1 B2 B3 C1 C2 C32 errors

A1 A2 A3 B1 B2 B3 C1 C2 C3

Interleaver

A1 B1 C1 A2 B2 C2 A3 B3 C3

A1 B1 C1 A2 B2 C2 A3 B3 C3

Deinterleaver

A1 A2 A3 B1 B2 B3 C1 C2 C31 error 1 error 1 error


WIMAX physical sub-layer1. Mapping:


SolomonEncoder



Interleaver

To air interface

Channel coding

• The The bit rate bit rate defines the rate at which information is passed. defines the rate at which information is passed. • The The baud baud (or (or signallingsignalling)) rate rate defines the number of symbols per defines the number of symbols per

second. Each symbol represents second. Each symbol represents nn bits, and has bits, and has MM signal states, signal states, where where M = 2M = 2nn. This is called . This is called M-ary signallingM-ary signalling..


Amplitude Shift Keying (ASKAmplitude Shift Keying (ASK))

Pulse shaping can be employed to remove spectral Pulse shaping can be employed to remove spectral spreading.spreading.

ASK demonstrates poor performance, as it is heavily ASK demonstrates poor performance, as it is heavily affected by noise and interference.affected by noise and interference.

BasebandData

ASK modulatedsignal

A cos ct A cos ct0 0


Frequency Shift Keying (FSK)Frequency Shift Keying (FSK)

Bandwidth occupancy of FSK is dependant on the spacing of the two Bandwidth occupancy of FSK is dependant on the spacing of the two symbols. A frequency spacing of 0.5 times the symbol period is symbols. A frequency spacing of 0.5 times the symbol period is typically used.typically used.

BasebandData

FSK modulatedsignal

f1 f1f0 f0

where f0 = A cos(c-)t and f1 = A cos(c+)t


Phase Shift Keying (PSK)Phase Shift Keying (PSK)

Binary Phase Shift Keying (BPSK) demonstrates better Binary Phase Shift Keying (BPSK) demonstrates better performance than ASK and FSK.performance than ASK and FSK.

PSK can be expanded to a M-ary scheme, employing multiple PSK can be expanded to a M-ary scheme, employing multiple phases and amplitudes as different states.phases and amplitudes as different states.

BasebandData

Binary PSK modulatedsignal

s1 s1s0 s0

where s0 = -A cos ct and s1 = A cos ct


Modulation - QPSKModulation - QPSK

Quadrature Phase Shift Keying is effectively two independent Quadrature Phase Shift Keying is effectively two independent BPSK systems (BPSK systems (I and QI and Q), and therefore exhibits the same ), and therefore exhibits the same performance but performance but twicetwice the bandwidth efficiency. the bandwidth efficiency.

Cos Wc t

900

Odd Data

Even Data

(NRZ)

(NRZ)

QPSK

Q-Channel

I-Channel

Q

(-1,-1)

(-1,1)

I

(1,1)

(1,-1)Wc = Carrier Frequency, I = In phase channel, Q = Quadrature channel


Multi-level (M-ary) Phase and Amplitude ModulationMulti-level (M-ary) Phase and Amplitude Modulation

Amplitude and phase shift keying can be combined to transmit Amplitude and phase shift keying can be combined to transmit several bits per symbol (in this case M=4). These modulation several bits per symbol (in this case M=4). These modulation schemes are often refered to as schemes are often refered to as linearlinear, as they require linear , as they require linear amplification.amplification.16QAM has the largest distance between points, but requires 16QAM has the largest distance between points, but requires very linear amplification. 16PSK has less stringent linearity very linear amplification. 16PSK has less stringent linearity requirements, but has less spacing between constellation points, requirements, but has less spacing between constellation points, and is therefore more affected by noise.and is therefore more affected by noise.M-ary schemes are more bandwidth efficient, but more M-ary schemes are more bandwidth efficient, but more susceptible to noise.susceptible to noise.16 PSK 16 APSK16 QAM


MappingMapping

The encoded and interleaved binary serial input data shall be divided into groups of NcpC (1, 2, 4, or 6) bits and converted into complex numbers representing BPSK, QPSK, 16-QAM, or 64-QAM constellation points. The conversion shall be performed according to Gray-coded constellation mappings


• d = (I + jQ) × KMODd = (I + jQ) × KMOD• The output values, d, are formed by multiplying the The output values, d, are formed by multiplying the

resulting (I+jQ) value by a normalization factor resulting (I+jQ) value by a normalization factor KMOD.KMOD.

• The normalization factor, KMOD, depends on the The normalization factor, KMOD, depends on the base modulation modebase modulation mode

Modulation KMOD

BPSK 1

QPSK 1/ 16-QAM 1/64-QAM 1/


Encoding tables for different modulation schemes

Input b0 I-out Q-out

0 -1 0

1 1 0

Input (b1)

Q-out

0 -1

1 1

Input (b0 ) I-out

0 -1

1 1

BPSK encoding table

QPSK encoding table


16 - QAM encoding table

Input (b0b1)

I-out

00 -3

01 -1

11 1

10 3

Input (b0b1)

Q-out

00 -3

01 -1

11 1

10 3



64-QAM encoding table

Input (b0b1b2) I-out

000 -7

001 -5

011 -3

010 -1

110 1

111 3

101 5

100 7

Input (b3b4b5)

Q-out

000 -7

001 -5

011 -3

010 -1

110 1

111 3

101 5

100 7



Constellation of different modulation schemesConstellation of different modulation schemes


WIMAX physical sub-layer1. IFFT :


SolomonEncoder



Interleaver

To air interface

Channel coding

90


• Let us introduce the meaning of the orthogonality.

0 π/2 π 3π/2 2π

1

0.5

0.5

1

Sin (f)Cos (f)Sin(f) * Cos(f) = 0


• We can obtain orthogonal signals by choosing their frequencies have integer multiple of the basic frequency.

• sin sin αα , sin 2 , sin 2αα , sin3 , sin3αα are orthogonal signals are orthogonal signals

0 π/2 π

1

0.5

0.5

1

Sin 3α Sin α Sin 2α 1

0.5

0.5

1

sin 2sin 2αα * sin3 * sin3αα = 0 = 0 sin sin αα * sin2 * sin2αα = 0= 0

0 π/2 π

+ + ++++-- -- -- -- -- --


t

f

A

f1

f2 = n*f1

f3 = m*f1

f1 f2 f3

f

• Un-modulated orthogonal frequencies.

A


t

f

A

f1

f2 = n*f1

f3 = m*f1

f1 f2 f3

f

• modulated orthogonal frequencies.

A


• The solution for multicarrier using several oscillator.

S/PS/P

t

Oscillator array

S(t)

OFDM symbol in time domain

• Disadvantages:• It requires multiple oscillator for all subcarriers ex.: 256, 512, . . . • It requires very sharp filters (rectangular) to select each carrier.


• The solution for multicarrier using inverse fast forrier transform.

S/PS/P

t

S(t)


Samples

1

0

)/2(

).()(N

n

Nnkj

nxKX

1

0

)/2(

).()/1()(N

k

Nnkj

kxNnX

• DFT (FFT):

• IDFT (IFFT):


• The solution for multicarrier using inverse fast forrier transform.

S/PS/P

t

S(t)


Samples

1

0

1

0

)/2cos().()/2sin().()(N

n

N

n

NnkjnbjNnkjnaKX

])/2cos().()/2sin().()[/1()(1

0

1

0

N

k

N

k

NnkjkbNnkjkaNnX

• DFT (FFT):

• IDFT (IFFT):


• OFDM symbol in frequency domain:

t

S(t)


f


• OFDM symbol in frequency domain:

f

Lower

Guard

band

Upper

Guard

band

Datasubcarrier

s

Pilotsubcarrier

s

DC

∆f = 1/Tb

WIMAX physical sub-layer1. Cyclic prefix :

• OFDM & Inter-Symbol interference:

• We need to eliminate the corruptedPeriod of the next symbol.

• So, we adds a Guard period between Each two successive symbols.

• Solution:

t

T (symbol)

T (symbol)

CorruptedSymbol Symbol Symbol . . . . .

Guard Period

SymbolSymbol

SymbolSymbol

SymbolSymbol

WIMAX physical sub-layer1. Cyclic prefix :

• We solve the adjacent carrier interference butThe ISI still exist.

T (symbol)

Symbol Symbol Symbol

Guard Period

SymbolSymbol

SymbolSymbol

SymbolSymbol

t

T(s)T(g)

T(g)

T(b)

Copy final part of the symbol


S/PS/P D/AD/A

CPCP3.5 GHz3.5 GHzSamplin

g Frequen

cy(fs)• Fs = n * BW

• n: sampling factor – depends on the used BW – Possible values are: 8/7 , 86/75 , 144/125 , 316/275 , 57/50.


• OFDM symbol duration:

f

LowerGuar

dband

Upper

Guard

band

Datasubcarriers

Pilotsubcarriers

DC

∆f = 1/Tb

• OFDM symbol duration = Useful symbol time + guard time (CP)= (1/one sub-carrier spacing) + G * Useful symbol time= (1/∆f) ( 1 + G)= (1/(fs/Nfft)) ( 1 + G)= (1/(n*BW/Nfft)) ( 1 + G)


• OFDM symbol Data rate:

f

LowerGuar

dband

Upper

Guard

band

Datasubcarriers

Pilotsubcarriers

DC

∆f = 1/Tb

• Data rate= number of un-coded bites per OFDM symbol/ OFDM symbol duration

= (data sub-carriers * bits per FFT point * coding rate ) / OFDM symbol duration.


• OFDM & OFDMA PHY layer:

f

LowerGuardband

UpperGuardband

Datasubcarriers

Pilotsubcarriers

DC

∆f = 1/Tb

•OFDM (fixed WiMAX) PHY layer use only 256 sub-carrier so, the sub-carrier spacing is variable with the selected BW.

•OFDMA (mobile WiMAX) PHY layer use only 256, 512, 1024, 2048 sub-carriers so, the sub-carrier spacing is fixed with the selected BW.





Privacy Sub-layerPrivacy Sub-layer

Physical Sub-layerPhysical Sub-layer





WIMAX layered architecture

WIMAX Privacy Sub-layer

The fundamental services we need:

WIMAX Privacy Sub-layerAttacks Types:

User A User B

Attacker• Modification

User A User B

Attacker

• Attack availability

User A User B

Attacker• Attack Entity

Authentication


Confidentiality using Encryption:

• Types of Encryption:

1.Symmetric Key Encryption: used in traffic encryption.2.Asymmetric Key Encryption: used in key encryption.



Symmetric Key Encryption



• It is based on the idea that the Transmitter have the secret key of the Receiver .

MS 1SC: 2008

MS 2SC: 2009

Data encrypted with key: 2009Data encrypted with key: 2008



• The simplest example of the encryption is the XOR function.

0 1 1 0 1 1 0

1 0 1 1 0 1 1 1 1 0 1 1 0 1Plain

text

KeyCipher text 0 1 1 0 1 1 0

Key

1 0 1 1 0 1 1 Received Plain text

• But the XOR have some critical disadvantages:• Known text attack.• Chosen text attack.• Flipping attack.

• Note: both Plaintext & Key must be the same in size.



• So, we must develop another method doesn’t depend on the XOR directly.

• We use Simple DES technique (Data Encryption Standard).

KeyKeyschedulerschedulerKey 10-

bit EncryptionEncryption

Plain text 8-bit

Cipher text 8-bit

Key1 8-bit

Key2 8-bit

Simple DES (Data Encryption Standard)



KeyKeyschedulerschedulerKey

10-bitEncryptionEncryption

Plain text 8-bit

Cipher text 8-bit

Key1 8-bit

Key2 8-bit3 5 2 7 4 10 1 9 8 63 5 2 7 4 10 1 9 8 6

LS - 1LS - 1 LS - 1LS - 1

LS - 2LS - 2 LS - 2LS - 2

Compression Permutation

Compression Permutation

Key 1

Key 2

Key 10-bit



KeyKeyschedulerschedulerKey

10-bitEncryptionEncryption

Plain text 8-bit

Cipher text 8-bit

Key1 8-bit

Key2 8-bit

2 6 3 1 4 8 5 72 6 3 1 4 8 5 7

F1F1

SwitchSwitch

F2F2

Key 1

Key 2

Plain text 8-bit

P-1P-1

Cipher text 8-bit



EncryptionPlain text 8-bit

Cipher text 8-bit

Key 10-bit

• We want to calculate the encryption power.• 2^10 / 2^8 = 2^2 (Keys per Cipher)• This means that each one plain text can have 4(2^2) different Cipher

text.


Confidentiality using Encryption:• If we want to increase the encryption system power then we increase

the Key length.• But this will be very difficult in hardware because this require change

in all the system.• So, we use multi-encryption systems.

EncryptionPlain text 8-bit

Cipher text 8-bit

Key1 10-bit

Encryption Encryption

Key2 10-bit

Key3 10-bit


Confidentiality using Encryption:• Multi-encryption modes:

EPlain text 8-bit

Cipher text 8-bit

Key1

E E

Key2

Key3

EEE mode

EPlain text 8-bit

Cipher text 8-bit

Key1

D E

Key2

Key3

EDE mode


Confidentiality using Encryption:• Modes of operation of block cipher:

• There are many modes of operation for the encryption block.1. ECB (Electronic Code Block)2. CBC (Cipher Block Chaining)3. CFB (Cipher Feed Back)4. OFB (Output Feed Back)5. CTR (Counter mode)

• The question is:How will the block operate in multi Plain text input?

Encryption

Plain text

Cipher text

Key


Confidentiality using Encryption:ECB (Electronic Code

Block)

Encryption

Plain text 1

Cipher text 1

Key Encryption

Plain text 2

Cipher text 2

Key Encryption

Plain text 3

Cipher text 3

Key

Features:•If blocks received out of order, the system still run.•No error propagation.•Can’t sense any of the attacks (deletion, insertion, exchanging, substitution).• Suitable to one block ciphering.


Confidentiality using Encryption:CBC (Cipher Block

Chaining)

Features:•Ci = EK (Pi + Ci-1)•Error propagation occur.•Very sense to any of these attacks (deletion, insertion, exchanging, substitution).• Suitable to multi-block ciphering.

Encryption

Plain text 1

Cipher text 1

Key Encryption

Plain text 2

Cipher text 2

Key Encryption

Plain text 3

Cipher text 3

Key

IV

WIMAX Privacy Sub-layerConfidentiality using Encryption:

CBC (Cipher Block Chaining)

Decryption

Plain text 1

Cipher text 1

Key Decryption

Plain text 2

Cipher text 2

Key Decryption

Plain text 3

Cipher text 3

Key

If Ci is corrupted then the error propagates to Pi & Pi+1

IV


WIMAX Traffic encryption algorithms

• WIMAX uses 3-DES algorithm in CBC mode EDE which:• Key = 56 bits.• Plain text & Cipher text = 64 bits.

Encryption

Plain text 1

Cipher text 1

Key Encryption

Plain text 2

Cipher text 2

Key Encryption

Plain text 3

Cipher text 3

Key

IV


WIMAX Traffic encryption algorithms

DES - CBCDES - CBC User messageUser message StuffingStuffingn * 64

7 - bit7 - bit PP 7 - bit7 - bit PP 7 - bit7 - bit PP. . . . .

64 bit

IVIV

XORXOR CBC - IVCBC - IVPHY – Synch fieldPHY – Synch field

Encrypted messageEncrypted message



Asymmetric Key Encryption



• It is based on the idea that the Transmitter doesn’t have the secret key of the Receiver .

What is the solution?

?MS 1 MS 2



• We must know the MOD operation which mean the reminder of the division operation.

• 7 mod 5 = 2 7/5 = 1 & 2 reminder

• (6+3) mod 5 = 4• (x mod n) + (y mod n) mod n = (x + y) mod

n


Confidentiality using Encryption:• Additive inverse:• Y + (-Y) = 0

• -y mod n = ?• The we must get the value of –y in the modulo system.

• Example:

5 – 7 mod 10 = 5 + (-7 + 10) mod 10 = 5 + 3 mod 10 = 8


Confidentiality using Encryption:• Multiplicative inverse:• Y * (1/Y) = 1

• y/x mod n = ?• The we must get the value of (1/X) in the modulo system.

• Example:

7/6 mod 11 = 7 * 2 mod 11 6 * ? = 1 mod 11 = 14 mod 11 = 3 6 * 2 = 12 mod

11 = 1• Example:

5/4 mod 8 = No answer Because the GCD (Greatest common Divisor) of (4,8) not equal 1


Confidentiality using Encryption:• Prime numbers:

• X mod n where (n) is a prime number

• Then all numbers below this value have a GCD = 1 with this prime number.

• Example:

• 3/5 mod 7 = (3 * 3) mod 7 = 2

WIMAX Privacy Sub-layerConfidentiality using Encryption:• Euler Totient Ф(n):

X mod n• Example:• At n = 6

• Field = { 0, 1, 2, 3, 4, 5 } numbers less than 6 {1, 5} numbers have GCD = 1 with

6

• Then, Euler Totient Ф(6) = 2

• Example:• At n = 7 Prime number

• Field = { 0, 1, 2, 3, 4, 5, 6 } numbers less than 6 { 0, 1, 2, 3, 4, 5, 6 } numbers have GCD = 1 with

7

• Then, Euler Totient Ф(7) = 6 = (n – 1)

WIMAX Privacy Sub-layerConfidentiality using Encryption:• Euler Totient Ф(n):• Note: Ф(p) * Ф(q) = (p – 1)(q – 1) at p, q are prime

numbers.

• General Formula:

a mod n = 1Example:At n = 3 , a = 2

Ф(n) = Ф(n) = n – 1 = 3 – 1 = 2.

a = 2 ^ 2 = 4

Then, 4 mod 3 = 1

Ф(n)

Ф(n)


MS 1 MS 2

Public Key (1) Public Key (2)

Secrete Key (1)

Secrete Key (2)

• Each user have Secrete Key & Public Key.• It uses the Secrete key decrypt the data sent to it encrypted with its

Public key


MS 1 MS 2

Public Key (1)Public Key (2)

Secrete Key (1)

Secrete Key (2)

1. Data encrypted with Public Key (2)

2. Data decrypted with Secrete Key (2)

• Data Encryption/decryption sequence.


• Key generation process.

1. Pick two large prime numbers P & Q (each of 512 bits).

2. Calculate N = P*Q (RSA public modulus – 1024 bits).

3. Calculate Euler Totion Ф(N) = (P – 1)(Q - 1).4. Choose (e) – Public exponent - at which GCD(e,

Ф(N) ) = 1.5. Find (d) – Secrete exponent - at which d*e = 1 mod

N.

• Then:• Public key = (e , N).• Secrete key = (d , N).

• Lets have an example . . . .


• If there is a message x , where x < N

Encryption:• Then C – encrypted message – equals:

C = x mod N

Decryption:x = C mod N

= (x mod N) mod N

= (x mod N) mod N e * d = 1 mod N

= x mod N = x x < N

e

d

de

ed

WIMAX Privacy Sub-layerAuthentication using hash message:

• Hash function is a on way function.• Hash function features:1. If x is a message ,then hash message (Y) = H(x).2. Very low collision, i.e: low Prop. to find one hash message to

different X messages.

Encryption

Plain text 1

Key Encryption

Plain text 2

Key Encryption

Plain text 3

Hash message

Key

IV

WIMAX Privacy Sub-layerAuthentication using hash message:

• WIMAX makes HMAC (Hash Message Authentication Code) using SHA-1 Algorithm.

XORXOR

XORXORMAKMAK 00000000

opad

ipad

messagemessagemessagemessageXOR outputXOR output

SHASHA

SHA outputSHA outputXOR outputXOR output

SHASHA

messagemessage

HASHHASH+

Ipad = 36 , 36 , . . ,36Opad = 5C , 5C , . . .

5C

wimax advanced

Documents

x x x x x5x x

x x x x x x2

x x xstep

x x55x

x xxreminder sx5channel

v x ex5syndrome sx

channel coding

g channel