Image Processing Reference
In-Depth Information
data rate, thus the time to transmit bit of data over the network is
T
bit
=
µs for Mb/s CAN and
ns for Mb/s Ethernet.
The data rate of a network must be considered together with the packet size and overhead. Data
are not just sent across the network one bit at a time. Instead, data are encapsulated into packets,
with headers specifying (for example) the source and destination addresses of the packet, and often
a checksum for detecting the transmission errors. All industrial networks have a minimum packet
size; for example the minimum packet size is bits for CAN and bytes for Ethernet. A min-
imum “interframe time” between two packets is required between subsequent messages to ensure
that each packet can be distinguished individually; this time is specified by the network protocol and
is included herein as part of the frame.
The transmission time for a message on the network can be computed from the network's data
rate, the message size, and the distance between two nodes. As most of these quantities can be com-
puted exactly (or approximated closely), transmission time is considered a deterministic parameter
in a network system. The transmission time can be written as the sum of the frame time and the
propagation time,
T
tx
=
T
frame
+
T
prop
where
T
frame
isthetimerequiredtosendthepacketacrossthenetwork
T
prop
is the time for a message to propagate between any two devices
m
As the typical transmission speed in a communication medium is
is the propagation time
T
prop
is negligible on a small scale. In the worst case, the propagation delays from one end to the other
of the network cable for two typical control networks are
T
prop
×
/
. µs for m Ethernet,
∗
and
=
T
prop
µs for m CAN. The propagation delay is not easily characterized because the distance
between the source and destination nodes is not constant among different transmissions, but typically
itislessthanµs(ifthedevicesarelessthanmapart).Somenetworks(e.g.,Ethernet)arenota
single trunk but have multiple links connected by hubs, switches, and/or routers that receive, store,
and forward packets from one link to another. he delays associated with these interconnections can
dominate propagation delays in a complex network, and must also be considered when determining
the transmission delays [].
The frame time,
T
frame
, depends on the size of the data, the overhead, any padding, and the bit time.
Let
N
data
be the size of the data in terms of bytes,
N
ovhd
be the number of bytes used as overhead,
N
pad
bethenumberofbytesusedtopadtheremainingpartoftheframetomeettheminimumframesize
requirement, and
N
stuff
be the number of bytes used in a stuffing mechanism (on some protocols).
†
The frame time can then be expressed by the following equation:
=
T
frame
=[
N
data
+
N
ovhd
+
N
pad
+
N
stuff
]×
×
T
bit
(.)
In Ref. [], these values are explicitly described for Ethernet, ControlNet, and DeviceNet protocols.
The effective bandwidth of a control network will depend not only on the physical bandwidth, but
also on the efficiency of encoding the data into packets (how much overhead is needed in terms of
addressing and padding), how efficiently the network operates in terms of (long or short) interframe
times, and whether network time is wasted due to message collisions. For example, to send one bit
of data over a kb/s CAN network, a bit message is needed, requiring µs. To send the same
one bit of data over Mb/s Ethernet, an byte message is needed ( byte frame size plus bytes
for interframe separation), requiring a . µs
T
frame
. hus, even though the raw network speed is
∗
Because Ethernet uses Manchester biphase encoding, two bits are transmitted on the network for every bit of data.
†
The bit-stuffing mechanism in DeviceNet is as follows: if more than bits in a row are “,” then a “” is added and vice
versa. Ethernet uses Manchester biphase encoding, and, therefore, does not require bit stuffing.
