Communication Systems Engineering Episode 2 Part 7 doc
Handbook of Corrosion Engineering Episode 2 Part 7 docx
... commonly
076 51 62_ Ch09_Roberge 9/1/99 6:11 Page 820
therefore usually clad with aluminum alloys containing about 1% zinc,
such as 70 72 , or aluminum-zinc-magnesium alloys such as 70 08 and
70 11, which ... dry) and 1B (exterior, normally dry) and is particularly
076 51 62_ Ch09_Roberge 9/1/99 6:11 Page 821
Protective Coatings 79 7
TABLE 9.4 Description of the Main Advanced Te...
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 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 2 Part 1 pptx
... Back 7 ARQ
SN
0 3 4 5
t
1
6
RN
0 1 2
3
5
Window (0,6)
(1, 7) (5 ,11 ) (2, 8) (3,9)
Node A
Node B
2
0 5
Packets
delivered
0
1
2
3
4
5
ã Note that packet RN -1 must be accepted at B before ... developed for X .25 networks
ã They all use bit oriented framing with flag = 011 111 10
ã They all use a 16 -bit CRC for error detection
ã They all use Go Back N ARQ wi...
Communication Systems Engineering Episode 2 Part 2 pptx
... that we can tolerate
Circuits P
B
20 1%
15 8%
7 30%
Eytan Modiano
Slide 22
Delay formulas
ã M/G/1
ã M/M/1
ã M/D/1
DX=+
à
à
/
DX=+
à
à
/
( )2
DX
X
=+
à
2
21(/)
Delay components:
Service (transmission) ... arrivals in T
It can be shown that,
E[n] = T
E[n ] = T + ( T)
= E[(n-E[n]) ] = E[n ]-E[n] = T
22
22 22
λ
λλ
σλ
Eytan Modiano
Slide 19
Example (fast food restaurant)
ã Custo...
Communication Systems Engineering Episode 2 Part 3 pps
... group transmits
(1 ,2, 3, 4)
(1 ,2, 3)
4
success
collision
(2, 3)
collision
idle
collision
(2, 3)
success
success
Notice that after the idle slot,
collision between (2, 3) was
sure to happen ... the first success is 2, so the expected
number of slots to transmit 2 packets is 3 slots
Throughput over the 3 slots = 2/ 3
– In practice above algorithm cannot real...
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 2 Part 5 ppsx
...
1
2
3
4
N = {1 ,2, 3,4}
A = {(1 ,2) , (2, 1),(1,4),
(4 ,2) , (4,3),(3 ,2) }
ã Directed walk: (4 ,2, 1,4,3 ,2)
ã Directed path: (4 ,2, 1)
ã Directed cycle: (4 ,2, 1,4)
ã Data networks are best represented ... n2, ,nk) in which each adjacent node
pair is an arc.
ã A path is a walk with no repeated nodes.
1
2
4
3
1
2
4
3
Walk (1 ,2, 3,4 ,2) Path (1 ,2, 3,4)
Eytan Modian...
Communication Systems Engineering Episode 2 Part 6 pps
...
CONGESTION CONTROL
BER
EFFICIENCY
0
0.1
0 .2
0.3
0.4
0.5
0 .6
0.7
0.8
0.9
1
1.00E-07 1.00E- 06 1.00E-05 1.00E-04 1.00E-03
1,544
64 KBPS
16 KBPS
2. 4 KBPS
KBPS
ã TCP assumes dropped packets ... addresses)
– Allocate a block of contiguous addresses
E.g., 1 92. 4. 16. 1 - 1 92. 4. 32. 155
Bundles 16 class C addresses
The first 20 bits of the address field are the...
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