Measurement and Remote Control – Electrical Engineering Industry

A code word

a n , a n  1 ... a 1 a 0

in there

a i =0 can be expressed as a polynomial in the variable x

and the coefficients are a i . .

For example, the codeword 1001101 can be written as a polynomial:

1. x 6  0. x 5  0. x 4  1. x 3  1. x 2  0. x 1  1. x 0  x 6  x 3  x 2  1

At this point, we can perform ordinary algebraic operations with that polynomial. As for addition, it must be done according to mod 2, which means:

x a x a  0

a . x a  0  x a

0  0  0

To build cyclic codes, people use irreducible polynomials (cannot be reduced

be ) as a generating polynomial to construct the codes. Encryption method :

To be the information carrier element of the codeword, we choose the codewords of the full codeword with length m. This codeword is called the original codeword, denoted G(x). To form the cyclic codeword F(x), we

receive codeword G(x) with x K , where K is the number of residual elements. That means we lengthen the codeword

G(x) produces K more elements. Then divide the polynomial G(x). x K gives the generating polynomial P(x), then

Take the remainder R(x) plus the polynomial G(x).

F ( x )  G ( x ). x K  R ( x ) .

F(x) will be divisible by the generating polynomial P(x)

x K , we will get the cyclic codeword:

According to this encoding, the mp elements with high exponents are the information carrying elements, while the remaining k elements with low exponents are the residual elements. Because the residual element and the message carrier element stand separately, the cyclic code is a type of delimiter code.

For example, let n=7

m=4 K=3

P ( x )  x 3  x 2  1 Let's encrypt message 1011 Solution:

Core

P ( x )  x 3  x 2  1  1101

G ( x )  x 3  x  1  1011

G ( x ). x K :

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G ( x ). x K  ( x 3 x  1). x 3 x 6 x 4 x 3  1011000

divide

G ( x ). xK

P ( x )

 x 6  x 4  x 3

x 3 x 2  1

x 3

x 2

x

x 3 x 2  1

 1011000  110  100

1101 1101

Residual: R ( x )  x 2  100

We have the cycle codeword:

F ( x )  G ( x ). x K  R ( x )

 x 6 x 4 x 3 x 2 1 { 0111 { 00

mgtin du

A simple method to find cyclic codewords is the matrix method.

In this method, we use a displacement generation matrix  P  x   . This matrix has m rows and n columns.

The leading edge is the polynomial G ( x ). xK

The rows after the exponent K decrease gradually to 0.

According to the example above, we can establish the displacement generation matrix as follows:

 G ( x ). x K 



 G ( x ). x



 P  x   m  n  L

K  1 



 G ( x ). x 1 







 G ( x ). x 0 

1	0	1	0	0
1	1	0	1	0
0	1	1	0	1
0	0	1	1	0

Maybe you are interested!

 1

P ( x ) 4  7 

 0

 0





0 a 1



a 2

0  a 3

a



 m  4



 n  7

Cyclic code words found = combination between rows

a 1 a 2 a 3 a 4

of matrix  P ( x )  .

According to the above example, we have m=4  so the number of code words available for the cyclic code is now: N  2 m  2 4  16 code words

Omitting the first zero codeword, we have 15 codewords left, which are:

Codeword 1: a 1 1101000

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a 2

2 :

0110100
3 : a 3	0011010
4 : a 4	0001101
5 : a 1  a 2	1011100
6 : a 1  a 3	1110010
7 : a 1  a 4	1100101
8 : a 2  a 3	0101110
9 : a 2  a 4	0111001
10 : a 3  a 4	0010111
11: a 1  a 2  a 3	1000110
12 : a 1  a 3  a 4	1111011
13 : a 1  a 2  a 4	1010001
14 : a 2  a 3  a 4	0100011

15 : a 1  a 2  a 3  a 4 1001011

From the above example we see that:

The codeword found in the above example is F(x)=1011100; found by adding the rows a 1 a 2

How to choose the generating polynomial P(x)?

The generating polynomial P(x) satisfies two conditions:

-The degree of P(x) is less than or equal to the number of residual elements K in it. That means: is the degree of the polynomial P(x).

-The number of 1's in P(x) is not less than the code distance d min

l K . With pussy

If there are multiple polynomials that satisfy the above conditions, the shortest polynomial should be chosen. The following table shows a number of non-reducible polynomials chosen as generating polynomials for cyclic codes:

Polynomials are not possible

regulation	Schedule	soy sauce	Duong
regulation	In system 2		In system 10
P ( x )  x  1		11		3
P ( x 2 )  x 2  x  1		0111		7
P ( x 3 )  x 3  x  1 P ( x 3 )  x 3  x 2  1 P ( x 4 )  x 4  x  1 P ( x 4 )  x 4  x 3  1		1011 1101 10011		11 13 19
P ( x 4 )  x 4  x 3  x 2	x  1	11001		25
		11111		31

Decoding method:

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The received codeword can be written as:

F'(x)=F(x)+E(x)

In which: F(x) is the transmitted codeword

E(x) is the wrong codeword in the received codeword.

At the receiver side, divide F'(x) by P(x). If the division has no remainder, then the received codeword is correct. If there is a remainder, the received codeword is incorrect. Residual analysis can determine which elements are incorrect. There are many ways to decode. Here's one way:

- Step 1 : calculate the remainder

R ( x )  F '( x ) . If R(x)=0  the codeword is correct.

P ( x )

R(x)  0 codeword is wrong; Now continue to step 2.

- Step 2 : calculate the residual weight (calculate the number of 1's in R(x)).

If we call W: number of p elements 1 in R(x).

If W S , where S is the error correction level of the code; Then we add the received codeword with the remainder and we get the correct codeword.

W  S

then we continue to step 3.

- Step 3 : translate the received codeword 1 step forward (1 element), then divide by P(x) to find the remainder R(x). That translation process continues until W  is reachedS , then proceed to add the shifted codeword with the newly found remainder. Then to get the correct codeword, we have to translate back a number of steps equal to the number of steps translated before

there.

For example: know P(x)=1101; The code corrects 1 error (S=1). The received codeword is 1111100. Check if the codeword is correct or not and correct it if it is wrong.

-Step 1:

F ( x )  1111100  1001  111

P ( x )

1101

The remainder R(x) is 111 with W=3 S = 1, so we translate the codeword 1 p forward to get 0111110.

-Step 2: divide

0111110  100  1010

1101 1101

The remainder R(x) is 1010 with w=2  S

0011111

So we move the codeword forward by 1 more element, we get

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-Step 3: divide 0011111  10 101

1101 1101

The remainder R(x) is 101 with W=2 S so we move the codeword forward by one more element, ta

gets 1001111.

-Step 4: divide 1001111  110 

1101

The remainder R(x) is 1 with W=1=S So stop translating.

-Step 5: add 1001111+1= 1010000

-Step 6: translate back 3 elements.

We have the added codeword 1010000  0000101

Pay in 3 steps

-Step 7: compare 2 codewords: Original codeword: 1111100 Corrected codeword: 0000101

1234567

So the 1st, 2nd, 3rd, 4th, 7th p elements are wrong

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CHAPTER 7: COMMUNICATION CHANNEL

7.1 Overhead lines :

The communication channel is the connection between the transmitter and receiver of the communication system. In remote control, electric and electromagnetic channels are often used.

The basic requirements for a communication channel are to work reliably, with interference not exceeding the permissible value and with large bandwidth.

One type of transmission channel is an overhead line, which consists of wires and cables. Conductors include steel wire and copper wire

- Bandwidth of steel wire: 30 KHz

-Bandwidth of copper wire: 150 KHz

The disadvantage of this type is that it is affected by the environment.

The basic parameters of the conductor are: resistance R, inductance L, capacitance C, conductivity G, wave resistance Z.

The formula to calculate those parameters is:

Resistor:

R t  R o  1   t 

cu  0.0039

 steel  0.0046

R o : resistance at 0 C

, : coefficient n degrees.

The resistance also depends on the frequency due to the surface effect.

Inductance:

The inductance of a 2-wire wire is determined as:

L   4.ln a  K .   4

a: distance of 2 fibers (cm)



r

outside

 .10



r: fiber radius (cm)

: relative magnetic permeability

 cu  1

 steel  140

K: coefficient accounting for the influence of face effect

Capacitance of 2-wire wire:

.10  6



36.ln a

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Capacitance of 1 wire:

.10  6



C 2 hours

18.ln

Where: : dielectric constant

2  1

h: distance from ground to wire. a: distance between 2 fibers.

r: fiber radius.

Total wave impedance of the circuit:

R j  L G j  C

Z S G: conductivity.

When transmitting with frequency f 10 KHz, if R L and G C then we have: Z S 

If copper wire: Z S = 600  900 

When transmitting energy on the line, people need to pay attention to the total wave impedance Z S . Because when satisfied:

Z S = Z load

Then the total input impedance:

Z comes in

U in Z .

I come in

At this time, the transmission line reaches its highest energy level, giving us the highest transmission efficiency. Otherwise, there will be a wave reflection phenomenon: waves at the end of the line will continue to reach the beginning of the line and create noise.

An important parameter of the line is the propagation coefficient  .

 R j  L  G j  C 

j 

In which: : attenuation coefficient.

: phase shift coefficient of pressure and current.

 characterizes the conditions for spreading electromagnetic energy on the line. for 1 km of line is determined by the expression:

 ln v 1  ln I 1  1 ln P 1 .

v 2 I 2 2 P 2

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The unit of is nepe ( N P )

If a line has  1 N P means that at the end of the line the voltage and current decrease

go e=2, 718 times and the power reduces e 2  7.39 times.

Also measured in decibels (db):

 10 log P 1

P 2

 20 log I 1

I 2

 20 log v 1 db

v 2

The wire communication channel is large, making the bandwidth narrow.

For cables: cables have a larger bandwidth due to being smaller. For symmetrical cables, the bandwidth is 12 550KHz. For coaxial cable bandwidth up to 8850KHz. To overcome the phenomenon of decline, on the transmission line, every 250km, a signal amplification station is placed to restore and raise the signal to near its original value.

7.2 Power supply line :

Advantages: -save installation costs

-The line has a sturdy structure

-The direction of the line coincides with the direction of transmitting measurement information.

Disadvantage: requires separate equipment to modulate high-frequency signals transmitted on the line

electricity.

Resonance 1

Filter

Resonance 2

Filter

The diagram of remote control signal transmission along the power supply line is as follows:

Station BA 1