Телекоммуникационные системы
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71
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4.12 –
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72
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C2=log2 4=2
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l =256.
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4.2
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n = 2000–4000;
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2 / 12 |
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(Uc) ( 4.13), -
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73
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(Uc)
Uc
4.13 –
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1. ( 4.14).
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4.14 –
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( 4.15) y=f(x),
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1) –
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ln A |
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1 ln A |
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0 x 1 , A
1 x 1 A
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– , ; U – .
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=255.
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N
N-1
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4.15 –
2. .
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4.16.
75
Nc |
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1/2 |
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32 |
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16 |
32 64 123 |
256 |
512 |
1024 |
2048 |
4.16 – =87,6/13
8 -
. : 0,1,2..7. Nc 16
N.
.
,
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=87,6. 0 1
0.
16, 13 =87,6/13.
=87,6/13 -
U N. U N
4.2.1, 0.
4.1
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XYZ |
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32 |
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32 |
76
2048 0,
128 128 ( 256).
=87,6/13
P X Y Z A B C D
– (1 – ; 0 – -
),
XYZ – N ,
D – .
, , 00110101 -
Nc=3 N=5.
Sâõ U ýò3 5 3 (64 5 4) 0 84 0 .
3 = 4, 00110101
85 0, 86 0 87 0. 0 3 0.
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N .
2
2 0.
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:
1)( 1);
2)XYZ ( 2–4), -
U N;
3)( 5– 8), U N ,
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3. .
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n=12. -
12- .
8- .
4.2.
77
4.2 -
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12- |
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XYZ |
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0 0 0 0 0 0 0 A B C D |
P |
0 0 0 |
A B C D |
1 |
P 0 0 0 0 0 0 1 A B C D |
P |
0 0 1 |
A B C D |
2 |
P 0 0 0 0 0 1 A B C D x |
P |
0 1 0 |
A B C D |
3 |
P 0 0 0 0 1 A B C D x x |
P |
0 1 1 |
A B C D |
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P 0 0 0 1 A B C D x x x |
P |
1 0 0 |
A B C D |
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P 0 0 1 A B C D x x x x |
P |
1 0 1 |
A B C D |
6 |
P 0 1 A B C D x x x x x |
P |
1 1 0 |
A B C D |
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P 1 A B C D x x x x x x |
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1 1 1 |
A B C D |
:
1)« », , ;
2)«P», , ;
3)XYZ, -
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4)ABCD
XYZ;
5), « », -
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P XYZ ABCD ( XYZ). U N,
N . -
ABCD .
, -
N. -
: U = (U ÝÒN ( À 8 Â 4 Ñ 2 D 0.5) N ),
A,B,C,D 0 1, UN N -
. -
:
U = +934 0 . : 101110100110
(+1 29+1 28+1 27+1 25+1 22+1 21)
4.2.2 N=6 -
110, 1110 , -
.
P XYZ ABCD 1 1 1 0 1 1 01
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U = +512 0+(1 8+1 4+0 2+1 1+0.5) 32 0 = (512+432) 0 = 944 0
+10 0,
6=32.
4.3
,
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, , I,
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( ),
– , ( ).
NRZ RZ :
NRZ – , .
( 4.17).
t
4.17 – NRZ
RZ – , . -
( 4.18).
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4.18 – RZ
RZ
NRZ, .
, , ( 4.20)
, NRZ RZ ( 4.19).
79
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4.19 - ,
NRZ RZ .
S1
S2
4.20 - , -
NRZ(S1) RZ(S2)
.
, NRZ , . NRZ , RZ .
-
( ), . . -
NRZ RZ:
- NRZ ( 4.21).
t
4.21 – NRZ
,
, . -
, , ,
.
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- RZ ( 4.22).
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4.22– RZ
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(1 0)
(1, 0 –1), . . -
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4 ( |
– binary), ( – ternary), |
( -
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, 33=27
24=16 -
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4.3 - 4 |
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