Information Technology Reference
In-Depth Information
1
1
T
=
=
s
=
100
ns
6
bit
rate
10
×
10
Thus the maximum and minimum times to transmit a frame will be:
(
)
T
=
7
+
1
+
6
+
6
+
2
+
1500
+
4
+
12
×
8
×
100
ns
=
1
.
ms
max
(
)
T
=
7
+
1
+
6
+
6
+
2
+
46
+
4
+
12
×
8
×
100
ns
=
0
067
μs
min
It may be assumed that an electrical signal propagates at about half the speed of light
(
c=
3
10
8
m/s). Thus, the time for a bit to propagate a distance of 500 m is:
×
dist
500
T
=
=
=
3
33
µ
s
500
m
8
speed
1
5
×
10
by which time, the number of bits transmitted will be:
T
3
.
33
µ
s
500
m
Number
of
bits
transmitt
ed
=
=
=
33
.
33
T
100
ns
bit
Thus, if two nodes are separated by 500 m then it will take more than 33 bits to be transmitted
before a node can determine if there has been a collision of the line, as illustrated in Figure
26.4. If the propagation speed is less that this, it will take even longer. This shows the need
for the preamble and the requirement for a maximum segment length.
For a distance of 500m, there are approximately
33.33 bits transmitted before the sender can sense
a collision
Figure 26.4
Bits transmitted before a collision is detected
26.3.1 Ethernet II
The first standard for Ethernet was Ethernet I. Most currently available systems implement
either Ethernet II or IEEE 802.3 (although most networks are now defined as being IEEE
802.3 compliant). An Ethernet II frame is similar to the IEEE 802.3 frame; it consists of
eight bytes of preamble, six bytes of destination address, six bytes of source address, two
bytes of frame type, between 46 and 1500 bytes of data, and four bytes of the frame check
sequence field.
