Communication Systems Engineering Episode 1 Part 4 docx
Dictionary of Engineering Episode 1 Part 4 docx
... kalиənиdər}
ture of 1 gram of water from 14 .5Њ to 15 .5ЊCat
calibrating tank
[
ENG
]
A tank having known ca-
a constant pressure of 1 standard atmosphere;
pacity used to check the volumetric accuracy of
equal ... escape of air from the
ər}
working chamber of a pneumatic caisson.
board-foot
[
ENG
]
Unit of volume in measuring
{ blo
¯
au
˙
t}
lumber; equals 14 4 cubic...
Engineering Analysis with Ansys Software Episode 1 Part 4 docx
... and displayed on the ANSYS Graphics” window.
Ch02-H6875.tex 24 /11 /2006 17 : 2 page 49
2 .4 Solution stage 49
A
Figure 2 . 14 Type of analysis definition.
Select the type of analysis that is appropriate ... beam on the ANSYS
Graphics window as shown in Figure 3.7.
Ch02-H6875.tex 24 /11 /2006 17 : 2 page 46
46 Chapter 2 Overview of ANSYS structure and visual capabilities...
Communication Systems Engineering Episode 1 Part 1 pps
... Lecture
4-Feb L1
6-Feb L2
11 -Feb L3
13 -Feb L4
18 -Feb
20-Feb L5
25-Feb L6
27-Feb L7
4-Mar L8
6-Mar L9
11 -Mar L10
13 -Mar L 11
18 -Mar L12
20-Mar L13
25-Mar
27-Mar
1- Apr L14
Eytan Modiano ...
8-Apr L16
10 -Apr L17
15 -Apr L18
17 -Apr L19
22-Apr
24-Apr L20
29-Apr L 21
1- May L22
6-May L23
8-May L24
13 -May L25
15 -May L26
5 /19 - 5/23
Topic Reading
Packet...
Communication Systems Engineering Episode 1 Part 2 ppsx
... experiment
X = x
1
with probability p
– X = x
2
with probability (1- p)
– H(X) = pLog
2
(1/ p) + (1- p)Log
2
(1/ (1- p)) = H
b
(p)
– H(X) is maximized with p =1/ 2, H
b
(1/ 2) = 1
Not surprising ... Rules for entropy
1. Chain rule:
H(X
1
, , X
n
) = 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|...
Communication Systems Engineering Episode 1 Part 3 docx
...
Slide 6
jt jf
Rectangle pulse
t
1 ||< 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 ... F
1
[ X
δ
( f )T
s
Π( )]
2W
∞
t
() =
∑
xnT
s
)Sinc( − n)xt (
n
=−∞
T
s
Eytan Modiano
Slide 11
Properties of the Fourier transform
ã Linearity
x1(t) <=>...
Communication Systems Engineering Episode 1 Part 4 docx
... E.g., N =4,
D = 0 .11 75, H(x) = 1. 911
– Recall: uniform quantizer, D= 0 .11 88, H(x) = 1. 9 04 (slight improvement)
0.9 816
1. 51
0 .45 28
-0.9 816
-0 .45 28
Eytan Modiano
Slide 12
- 1. 51
EX ... 3
fx
Uniform quantizer example
ã N =4, X~N(0 ,1)
x
() =
2
πσ
1
e
− x
2
/ 2
σ
2
,
σ
2
= 1
ã From table 6.2,
=0.9957, D=0 .11 88, H(Q)= 1. 9 04
Notice th...
Communication Systems Engineering Episode 1 Part 5 ppt
...
0. 25
a
3
0. 25
a
4
0 .1
a
5
0 .1
0.3
0. 25
0. 25
0.2
0.3
0. 25
0. 45
+
+
+
0 .55
0. 45
+
1. 0
1
0
0
1
0
1
0
1
Letter Codeword
a
1
11
a
2
10
a
3
01
a
4
0 01
a
5
000
n bits symbol
HX p
p
Shannon Fanocodes ... holds!)
n
1
n
2
n
3
n
4
n
5
Eytan Modiano
Slide 11
Huffman code example
A = {a
1
,a
2
,a
3
, a
4
, a
5
} and p = {0.3, 0. 25, 0. 25,...
Communication Systems Engineering Episode 1 Part 6 potx
... coding achieves 1 bit difference between adjacent levels
ã Example M= 8 (can be generalized)
A
1
000
A
2
0 01
A
3
011
A
4
010
A
5
11 0
A
6
11 1
A
7
10 1
Eytan Modiano
A
8
10 0
Slide ... PAM
Eg
x y
,4 − QAM ⇒ A
m
, A
m
∈
{
+ / − 1
}
x y
16 − QAM ⇒ A
m
, A
m
∈
{
+ / − 1, + / − 3
}
Eytan Modiano
Slide 16
-3 -1 1 3
-3
-1
1
3
Bandpass signals...
Communication Systems Engineering Episode 1 Part 7 pps
... points
10
-1
P
e
10
-5
12
14
antipodal
orthogonal
3dB
E
b
/N
0
(dB)
Eytan Modiano
Slide 25
Signal Detection
ã After matched filtering we receive r = S
m
+ n
S
m
{S
1
, S
M
} ... Notes on Q(x)
Q(0) = 1/ 2
– Q(-x) = 1- Q(x)
– Q(
∞
) = 0, Q(-
∞
) =1
– If X is N(m,
σ
2
) Then P(X>x) = Q((x-m)/
)
ã Example: Pe = P[r<0|S1 was sent)
f
|
(| s1) ~ N( E
b...
Communication Systems Engineering Episode 1 Part 8 pdf
... attenuation
P
R1
= P
T
/L, P
T2
= P
R1
A, P
R2
= P
T2
/L, …
P
N1
= P
N
, P
N2
= P
N1
A/L + P
N
, …
Let A = L => P
RK
= P
T
/L, P
NK
= KP
N
P
RK
/P
NK
= P
T
/LKP
N
= 1/ K (P
R1
/P
N1
)
Received ... (E
b
/N
0
)
d
= 10 dB
ã PAM modulation
P
b
= Q( E N2
0 b
/ )
ã Repeater: Received (E
b
/N
0
)
u/d
= 1/ 2 (E
b
/N
0
)
u
= 10 dB - 3dB = 7dB
– => Pb = 5x10
-4
from...
Communication Systems Engineering Episode 1 Part 9 pptx
... 0000 010 1 10 10 11 11 Syndrome
10 00 11 01 0 010 011 1 10
010 0 00 01 111 0 10 11 01
110 0 10 01 011 0 0 011 11
Suppose 011 1 is received, S = 10 , co-set leader = 10 00
Decode: C = 011 1 + 10 00 = 11 11
G
HH
T
=
=
=
10 10
010 1
10 10
010 1
10
01
10
01
... = 6, k = 3, R = 1/ 2
000 →→
→→
000000 10 0 →→
→→
10...
Communication Systems Engineering Episode 1 Part 10 ppsx
...
10 01
Send T = 11 010 1 011
11 010 1 011
Receive T’ = 11 010 1 011
(no errors)
No way of knowing how many
errors occurred or which bits are
In error
10 01
010 00
10 01
00 011 01
10 01
010 01
10 01 ...
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 = 10 01
M = 11 010 1 => M2
r
= 11 010 100 0...
Communication Systems Engineering Episode 2 Part 4 pps
... seconds
0
2
4
6
8
10
12
14
16
18
20
0 0 .2 0 .4 0.6 0.8
ALOHA
SCHEMES
TDMA
(10 USERS)
Perfect
Scheduling
(M/M/1)
Reservation
with 20 %
overhead
packet length 24 00 bits
transmission ... for mixed voice and data
Slot 1 2 3 4 5 6
frame 1
frame 2
frame 3
frame 4
frame 5
15 3 20 2
15 7 3 9
7 9
7 9
18 7 3 15 9 6
18
idle
idle
idle
idle
2...
Communication Systems Engineering Episode 3 Part 4 pdf
... 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 the generator matrix from ... 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 ... code?
Probl...
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