Communication Systems Engineering Episode 2 Part 10 pptx
Communication Systems Engineering Episode 2 Part 1 pptx
... (TO), go to 2
RETRANSMISSION BECAUSE OF ERRORS FOR GO
BACK 4 ARQ
0
3
4
t
1
SN
RN
0 1 1
1
1
2
x
3 4
3
Window
Node A
Node B
(0,3)
(1,4) (2, 5)
2 1 2
1
Packets
0
1
2 3
delivered ...
2
3
2 2
1
Packets
0
1
2
3
4
delivered
• When an error occurs in the reverse direction the ACK may still arrive in
time. This is the case here where the packet from B to A...
Communication Systems Engineering Episode 2 Part 2 pptx
...
= E[(n-E[n]) ] = E[n ]-E[n] = T
22
22 22
λ
λλ
σλ
Eytan Modiano
Slide 19
Example (fast food restaurant)
• Customers arrive at a fast food restaurant at a rate of 100 per hour
and take 30 seconds ... we can tolerate
Circuits P
B
20 1%
15 8%
7 30%
Eytan Modiano
Slide 22
Delay formulas
• M/G/1
• M/M/1
• M/D/1
DX=+
−
λµ
µλ
/
DX=+
−
λµ
µλ
/
( )2
DX
X
=+
−
λ
λµ
2
21(/)
Delay components...
Dictionary of Engineering Episode 2 Part 10 pptx
... holes in soft to medium-hard rocks.
{ ra
¨
k bit } 120 Њ conical diamond with rounded point, 1/16,
1/8, 1/4, or 1 /2 inch (1.5875, 3.175, 6.35, 12. 7
rockbolt
[
ENG
]
A bar, usually constructed of
steel, ... processing and commu-
ploys a very long (22 feet or 7 meters) alternating-
nication. In manufacturing, the scheduling func-
tion coordinates the flow of parts and products current arc...
Communication Systems Engineering Episode 1 Part 9 pptx
... Modiano
Slide 22
Example (syndrome decoding)
• Simple (4 ,2) code
Data codeword
00 0000
01 0101
10 1 010
11 1111
Standard array 0000 0101 101 0 1111 Syndrome
100 0 1101 0 010 0111 10
0100 0001 1 110 1011 01
1100 ... code, n = 6, k = 3, R = 1 /2
000 →→
→→
000000 100 →→
→→
100 101
001 →→
→→
0 0101 1 101 →→
→→
101 110
010 →→
→→
0101 11 110 →→
→→
1100 10
011 →→
→→
01...
Communication Systems Engineering Episode 1 Part 10 ppsx
...
Example:
100 1
Send T = 1101 0101 1
1101 0101 1
Receive T’ = 1101 0101 1
(no errors)
No way of knowing how many
errors occurred or which bits are
In error
100 1
0100 0
100 1
000 1101
100 1
0100 1
100 1 ... 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 = 100 1
M = 1101 01 =>...
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 ... duration packets)
– P(success) = e
-g
x e
-g
= e
-2g
– Throughput (success rate) = ge
-2g
– Max throughput at g = 1 /2, Throughput = 1/2e ~ 0.18
– Stabilization issues are simila...
Communication Systems Engineering Episode 2 Part 4 pps
...
Eytan Modiano
Slide 10
Comparison of MAC protocols
Load (packets/second)
Delay in 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 ... Example (Ethernet)
– Transmission rate = 10 Mbps
– Packet length = 100 0 bits, D
Tp
= 10
-4
sec
– Cable distance = 1 mile,
τ
= 5x10
-6
sec
–
β
= 5x10 -2 and E = 80%
•...
Communication Systems Engineering Episode 2 Part 5 ppsx
...
• Start with D3=1 and D2 =100
1
– After one iteration node 2 receives D3=1 and
D2 = min [1+1, 100 ] = 2
• In practice, link lengths occasionally change
3 2
1
1
1
100
– Suppose link between ...
1
2
3
4
N = {1 ,2, 3,4}
A = {(1 ,2) , (2, 1),(1,4),
(4 ,2) , (4,3),(3 ,2) }
• Directed walk: (4 ,2, 1,4,3 ,2)
• Directed path: (4 ,2, 1)
• Directed cycle: (4 ,2, 1,4)
• Dat...
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