Communication Systems Engineering Episode 3 Part 2 ppsx
Communication Systems Engineering Episode 1 Part 2 ppsx
... H(X
1
) + H(X
2
|X
1
) + H(X
3
|X
2
,X
1
) + …+ H(X
n
|X
n-1
…X
1
)
2. H(X,Y) = H(X) + H(Y|X) = H(Y) + H(X|Y)
3. If X
1
, , X
n
are independent then:
H(X
1
, , X
n
) = H(X
1
) + H(X
2
) + …+H(X
n
) ...
Log
(
) ≤
∑
P
i
Log(M) = Log(M)
i=1
P
i
i=1
Eytan Modiano
Slide 6
1
16 .36 : Communication Systems Engineering
Lecture 2: Entropy
Eytan Modiano
Eytan Modiano
S...
Communication Systems Engineering Episode 1 Part 10 ppsx
... check is a modulo 2 sum of some of the data bits
Example:
c
1
= x
1
+ x
2
+ x
3
c
2
= x
2
+ x
3
+ x
4
c
3
= x
1
+ x
2
+ x
4
Example
r = 3, G = 1001
M = 110101 => M2
r
= 110101000 ... of 1 ,2, or 3 errors (d > 3)
2) All bursts of errors of r or fewer bits
3) Random large numbers of errors with prob. 1 -2
-r
• Standard DLC's use a CRC with r=16 with...
Communication Systems Engineering Episode 3 Part 4 pdf
... MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Aeronautics and Astronautics
16 .36 : Comm. Sys. Engineering Date Issued: April 17
Problem Set No. 8 Date Due: April 29
Problem 1
Suppose a cyclic ... any errors occur?
C) Find the generator matrix for the (7 ,3) block code based on the above generator. (note
that a (7 ,3) code has 3 information bits and 4 check bits).
D) Using...
Handbook of Mechanical Engineering Calculations ar Episode 3 Part 2 ppsx
... zh)
2
ϭ [(6) (2. 82 ϫ 10
Ϫ
9
) (21 00)/
14.7] [3 / (2
ϫ 0.0015)]
2
ϭ 2. 41.
Also,
⌬
T
ϭ 6
Na
2
␣
(RT)
0.5
/p
a
h
3
ϭ (6) (2. 82 ϫ 10
Ϫ
9
)(8)(0.0087)
2
(0. 63) (1 . 32 2 ϫ
10
8
)
2
/[(14.7)(0.0015)
3
] ... Operation 22 .21
Journal Bearing Operation Analysis
22 .22
Bearing Type Selection of a Known
Load 22 . 23
Shaft Bearing Length and Heat
Generation 22 .28
Rol...
Communication Systems Engineering Episode 1 Part 3 docx
... ||< 1/ 2
t / /
Π() =
12 | t |= 12
0 otherwise
12
Π
∞
2
π
/
F[(t)]
=
∫
−∞
Π(t)e
− jft
dt
=
∫
− 12
e
− j 2
π
ft
dt
/
e
−
π
−
e
π
π
f
= =
Sin()
=
Sinc f
−
jf
π
f
2
π
() ...
− j 2
π
f
o
t
xt() cos(
2
π
f
o
t) =
()
2
π
()
2
(
x tHence,( ) cos(
2
π
f
o
t) ⇔
Xf − f
o
) + X( f + f
o
)
2
• Example: x(t)= sinc(t), F[sinc(t)] =
Π
(f)...
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