O„zbekiston respublikasi oliy va o„rta maxsus ta‟lim vazirligi

 34   3

^{4 } 
24   2 4 

Uzatiladigan kodli kombinatsiya quyidagi ko‗rinishda bo‗ladi:

F  (x₁ , x₂ , x₃ , x₄ , x₁₂ , x₃₄ , x₁₃ , x₂₄ ).

Dekodlashtirishning iteratsion usuli quyidagilarni o‗z ichiga oladi.

1. 
   
   Birinchi bosqichda biz tekshiruvchi simvollarning gorizontal nobog‗liqligini hisoblaymiz:
   

_ H₁
H_{2  } x₂  x₁₂
x₁  x_{12 }

H  ^
H
^  ^
H x  x
x  x ^

 3 4   4
34 3
34 

Bu yerda  operatsiya quyidagi tarzda aniqlanadi:

A  B  ()  sign(A)  sign(B)  min( A, B ).

2. 
   
   Ikkinchi bosqichda tekshiruvchi simvollarning vertikal nobog‗liqligi hisoblanadi:
   

^V1
V2 
^(L3
 H₃ ) 
^x13
(L₄
 H ₄ ) 
^x24 ^

V  ^
V
^  ^
V (L

• 
  
  H )  x
  

(L 
H )  x ^

 3 4   1
1 13 2 2
24 

Birinchi iteratsiya natijasi quyidagi matritsa bo‗ladi:

X₁  L  H  V

3. 
   
   Uchinchi bosqichda yana tekshiruvchi simvollarning gorizontal nobog‗liqligini hisoblaymiz:
   
4. 
   

_(L₂

• 
  
  V₂ ) 
  

^x12
(L₁

• 
  
  V₁ ) 
  

^x12 ^

H  ^
(L

• 
  
  V )  x
  

(L  V )  x ^

 4 4
34 3 3
34 

Qayd etamizki, V matritsa elementlarini biz oldingi 2 bosqichdan olamiz.

5. 
   
   To‗rtinchi bosqichda tekshiruvchi simvollarning vertikal nobog‗liqligi hisoblanadi:
   

_V₁
V_{2 
}(L₃

• 
  
  H₃ ) 
  

^x13
(L₄

• 
  
  H₄ ) 
  

^x24 ^

V  ^

V

^  ^
V (L

• 
  
  H )  x
  

(L 
H )  x ^

 3 4   1
1 13 2 2
24 

Qayd etamizki, H matritsa elementlarini oldingi 3 bosqichdan olamiz. Ikkinchi iteratsiya natijasi quyidagi matritsa bo‗ladi:

X₂  L  H V

Boshqa barcha iteratsion algoritmlar kabi bizga davriylik so‗nggi iteratsiyalarda hisoblash natijalari qanchalik kuchli o‗zgarishini tekshirish

uchun talab qilinadi. Agar X ^k  X ^k1 hisoblash jarayoni to‗htatilsa
X  X ^k

bilan belgilanishi mumkin. Agar
X ^k miqdor X^k-1 dan kuchli farqlansa, 3-

4 bosqichlarni takrorlash zarur. Yakuniy bosqichda ―Yengil‖ qarorlardan
―Qat‘iy‖ javoblarga o‗tish zarur. Bularning barchasi uchun manfiy komponentli matritsa:
_ X₁ X _{2 }

X  ^
 ^X 3

^X 4 

a  ¹  ¹
_{1 }1 
sign( X₁ )
^{1  sign( X 2 )}


2 2sign(x)

yoki
a  ^
2 _^{1 }
sign( X ₃ )
1  sign( X

4 ⁾

Misol.

a  (1011)
xabarni kodlashtiring?

Yechish.
a  (x₁ ,
x₂ ,
x₃ ,
x₄ )  (1011)
ekanligini hisobga olib matritsa

shaklida X_k ketma-ketligini taqdim qilamiz:

 ^x1
x₂  ^{1 0}

L  ^
x
^  ^{ },
x

 3 4 
_1 1_

Bunda gorizontal va vertikal tekshiriladigan belgilar quyidagi tarzda tuziladi:

 ^x12   ^x1
 x_{2  }1  0_
¹

^{ }  ^
x x
^  ^
 x 1 
^  ^{ }
1 0

 34   3
4  
  

 ^x₁₃   ^x₁
 x_{3  }1  1_
⁰

^{ }  ^
x x
^  ^
 x 0 
^  ^{ }

 24   2
4  
¹ ¹

Uzatiladigan kodli kombinatsiya quyidagi shaklda bo‗ladi:

F  (x₁ ,
x₂ ,
x₃ ,
x₄ ,
x₁₂ ,
x₃₄ ,
x₁₃ ,
x₂₄ )  (10111001 )

Kodlashtirish jarayonini quyidagi tarzda ifodalaymiz (4.11-rasm):

Axborot ketma-ketligi. Kodli ketma-ketlik.

4.11-rasm. Kodlashtirish jarayoni

F  (0.4,
0.5,
0.8,
0.9,
0.7,
0.3,
0.3,
0.5).

Misol. Qabul qilingan kodli kombinatsiyani dekodlashtiring?

F  (0.4,
0.5,
0.8,
0.9,
0.7,
0.3,
0.3,
0.5).

Yechish. Qabul qilinadigan axborot ketma-ketligini ajratamiz:

F  (x₁ ,
x₂ ,
x₃ ,
x₄ ,
x₁₂ ,
x₃₄ ,
x₁₃ ,
x₂₄ )  (0.4,
0.5,
0.8,
0.9,
0.7,
0.3,
0.3,
0.5).

axborot matritsasi

 ^x1

x_{2  }0.4

0.5_

L  ^
x
^  ^
x 0.8
^,
0.9

 3 4   
gorizontal (x₁₂, x₃₄) va vertikal (x₁₃, x₂₄) tekshiruvchi simvollar

_ x_{12  }0.7_
 x_{13  }0.3_

^{ }  ^{ },
x 0.3
^{ }  ^{ }
x 0.5

 34    ^{ 24   }

1. 
   
   Birinchi bosqichda tekshruvchi simvollarning gorizontal nobog‗liqligini hisoblaymiz:
   

_(x₂
 x₁₂ ) 
^x12 ^x1
 x_{12 
}0.5 
0.7

4. 
   
   
   

7. 
   
   _
   

H  ^
(x
 x ) x
x  x
^  ^
0.9 
0.3

7. 
   
   
   

^ .
0.3

 4 34
34 3
34   

A  B  (1)  sign(A)  sign(B)  min( A, B ) _{ni hisobga olgan holda} quyidagilarni olamiz:
x₂  x₁₂  0.5  0.7  (1)  sign(0.5)  sign(0.7) min( 0.5, 0.7 )  0.5, x₁  x₁₂  0.4  0.7  (1)  sign(0.4)  sign(0.7)  min( 0.4, 0.7 )  4.5, x₄  x₃₄  0.9  0.3  (1)  sign(0.9)  sign(0.3)  min( 0.9, 0.3)  0.3, x₃  x₃₄  0.8  0.3  (1)  sign(0.8)  sign(0.3)  min( 0.8, 0.3)  0.3,
va

_ H₁
H_{2  } 0.5
 0.4_

H  ^{ }  ^{ }.
H H  0.3  0.3
 3 4   

2. 
   
   Ikkinchi bosqichda tekshiruvchi simvollarning vertikal nobog‗liqligini hisoblaymiz:
   

^(L3
 H₃ )
^x13
(L₄
 H₄ ) 
^x24 ^
_(0.8 
0.3)
 0.3
(0.9 
0.3)
 0.5_

V  ^
(L
 H )  x
(L 
H ) 
^  ^
x (0.4
 0.5)
 0.3
(0.5 
0.4)
^.
 0.5

 1
yoki
1 13 2
2
^V1
24  
V_{2  }  0.3

 0.5_

V  ^
V
^  ^
V
0.1
 0.1^

 3 4   
Birinchi iteratsiya natijasi quyidagi matritsa bo‗ladi:

_0.4

4. 
   
   _{ } 0.5
   

 0.4_{ } 0.3
 0.5_{ } 0.4  0.4_

X₁  L  H V  ^
  ^
  ^
  ^{ }

_0.8

7. 
   
   _{ }  0.3
   

 0.3_{ } 0.1
 0.1_{ } 0.6 0.5 _

3. 
   
   Uchinchi bosqichda biz tekshiruvchi simvollarning nobog‗liqligini yana hisoblaymiz:
   

^(L2
 V₂ ) 
^x12
(L₁
 V₁ ) 
^x12 ^
_(0.5 
0.5)
 0.7
(0.4
 0.3) 

7. 
   
   _
   

H  ^
(L

• 
  
  V )  x
  

(L 
V ) 
^  ^
x (0.9 
0.1)
 0.3
(0.8 
0.1)  ^.

 4 4
34 3 3
34  

3. 
   

Qayd etamizki, V matritsa elementlarini oldingi ikki bosqichdan olamiz. Natijada quyidagi ko‗rinish olinadi:

_ 0 
0.7
0.1 
0.7_{ } 0
 0.1_

H  ^

7. 
   
   
   

0.3

7. 
   
   
   

^  ^
0.3  0.3
 0.3^.

   

4. 
   
   To‗rtinchi bosqichda V tekshiruvchi simvollar vertikal nobog‗liqligini hisoblaymiz:
   

^V1
V2 
^(L3
 H₃ ) 
^x13
(L₄
 H ₄ ) 
^x24 ^

V  ^
V
^  ^
V (L
 H )  x
(L 
H )  x ^

 3 4   1 1
13 2 2
24 

_(0.8 
0.3)  0.3
(0.9  0.3)
 0.5_{ }0.5 
0.3

4. 
   
   
   

0.5_

V  ^
(0.4 
0) 
0.3
(0.5 
0.1)
^  ^
 0.5 0.4 
0.3

3. 
   

^,

3. 
   


yoki
_ 0.3


^{V  } 0.3

  

 0.5_


 0.4^

Ikkinchi iteratsiya natijasi bo‗lgan matritsa quyidagi ko‗rinishga ega:

_0.4
0.5_{ } 0
 0.1_{ } 0.3
 0.5_{ } 0.1  0.1_

X ₂  L  H V  ^
  ^
  ^
  ^{ }

_0.8
0.9_{ } 0.3
 0.3_{ } 0.3
 0.4_{ }0.2 0.2 _

O‗quvchi 5 va 6 bosqichlarni hisoblashda qanday natijalarni olishini mustaqil ravishda ishlashi mumkin:

_ 0  0.1_
^{ 0.3
 0.5}
 ^{0.1
 0.1}











^{H  }  0.3  0.3^{,
V  } 0.3
 0.4^{,
X 3  }0.2

0.2 _

Modomiki

X  X ²  X ³
bo‗lsa, kelgusi iteratsiyalarni biz to‗htatamiz.

Oxirgi bosqichda ―Yengil‖ yechimlardan ―Qat‘iy‖ javoblarga o‗tamiz. X matritsa barcha manfiy komponentlari uchun qabul qilinadigan signal - 0, barcha ijobiylar uchun –1 miqdor bo‗ladi:

a 
yoki
^{1 }

1


2 _^{1 }
sign( X₁ )
sign( X ₃ )
^{1  sign( X )}

2
^ 

1  sign( X ₄ )_{
1 }1 

2 _1 
sign(0.1)
sign(0.2)
^{1  sign(0.1)}

1  sign(0.2) _

_{1 }1  1
1  1_{ 1 } 2  0_
^{1 0}

a  ^
  ^
  ^{ }

2 _1  1

1  1_
2 _ 2  2_
1 1_

Shunday qilib qayta kodlashtirilgan axborot ketma-ketligi quyidagi shaklda bo‗ladi:

a  (1011).

Ishlarda ko‗rsatilishicha, ―Yengil‖ yechimli qarordan foydalanish signal 8-16 darajalarda kvantlanganda optimal hisoblanadi.