Ethernet LAN Shortest ethernet frame time = 5.76uS?

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Subject Author Date
Shortest ethernet frame time = 5.76uS? James Harris 09-09-08
Posted by James Harris on September 9, 2008, 4:55 pm
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Could someone check my calculations? I'm looking for the shortest time
an Ethernet frame will occupy the wire. I make it about 5uS on 100Mbit
as follows.

Minimum frame size in bytes is 46 payload + 14 header + 4 FCS = 64.
Preamble (still needed on full duplex) of 8 octet times
Total of 72 octets * 8 bits = 576 bits.
Each bit time is 1/100,000,000 seconds. (IIRC the signalling rate is
125,000,000 bits with a 4B/5B encoding so we do get 100,000,000 data
bits in a second.)

Finally 576 * 1/100,000,000 = 5.76uS

On 10Mbit I make the figure ten times as much 57.6uS
On Gbit I think they kept the basic timing budget from 100Mbit so make
it's smallest-frame time about the same as for 100Mbit.

Corrections welcome. Does the above look about right?

--
James

Posted by Bob F. on September 9, 2008, 5:10 pm
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> Could someone check my calculations? I'm looking for the shortest time
> an Ethernet frame will occupy the wire. I make it about 5uS on 100Mbit
> as follows.
>
> Minimum frame size in bytes is 46 payload + 14 header + 4 FCS = 64.
> Preamble (still needed on full duplex) of 8 octet times
> Total of 72 octets * 8 bits = 576 bits.
> Each bit time is 1/100,000,000 seconds. (IIRC the signalling rate is
> 125,000,000 bits with a 4B/5B encoding so we do get 100,000,000 data
> bits in a second.)
>
> Finally 576 * 1/100,000,000 = 5.76uS
>
> On 10Mbit I make the figure ten times as much 57.6uS
> On Gbit I think they kept the basic timing budget from 100Mbit so make
> it's smallest-frame time about the same as for 100Mbit.
>
> Corrections welcome. Does the above look about right?
>
> --
> James


At one point on the cable or over the legal length of cable?

--
Regards, Bob F.


Posted by James Harris on September 9, 2008, 5:48 pm
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>
>
>
>
> > Could someone check my calculations? I'm looking for the shortest time
> > an Ethernet frame will occupy the wire. I make it about 5uS on 100Mbit
> > as follows.
>
> > Minimum frame size in bytes is 46 payload + 14 header + 4 FCS = 64.
> > Preamble (still needed on full duplex) of 8 octet times
> > Total of 72 octets * 8 bits = 576 bits.
> > Each bit time is 1/100,000,000 seconds. (IIRC the signalling rate is
> > 125,000,000 bits with a 4B/5B encoding so we do get 100,000,000 data
> > bits in a second.)
>
> > Finally 576 * 1/100,000,000 = 5.76uS
>
> > On 10Mbit I make the figure ten times as much 57.6uS
> > On Gbit I think they kept the basic timing budget from 100Mbit so make
> > it's smallest-frame time about the same as for 100Mbit.
>
> > Corrections welcome. Does the above look about right?
>
> > --
> > James
>
> At one point on the cable or over the legal length of cable?

Umm, well if the speed of signal propagation over copper cable is
about 0.7 times the speed of light (not sure of the exact figure but
that's about right, isn't it?) we have the speed as

0.7 * 300,000,000 metres per second = 210,000,000

If that's right then in 1uS the signal would propagate 210 metres so I
think the length of the cable becomes almost insignificant - up to a
standard length of 100m anyway. Maybe it would add 0.5uS to the total,
over that length.

Incidentally I think the old co-ax thinnet relied on reflected wave
signalling so signals had to go to a terminator and bounce back for
each bit - so the whole cable became energised for that bit long
before the next bit was put on the wire.

Just some more comments that people may be able to shoot down... Still
I don't mind being the clay pigeon. It would be good to get some of
these ideas confirmed ... or denied.

--
James

Posted by Bob F. on September 9, 2008, 6:18 pm
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>>
>>
>>
>>
>> > Could someone check my calculations? I'm looking for the shortest time
>> > an Ethernet frame will occupy the wire. I make it about 5uS on 100Mbit
>> > as follows.
>>
>> > Minimum frame size in bytes is 46 payload + 14 header + 4 FCS = 64.
>> > Preamble (still needed on full duplex) of 8 octet times
>> > Total of 72 octets * 8 bits = 576 bits.
>> > Each bit time is 1/100,000,000 seconds. (IIRC the signalling rate is
>> > 125,000,000 bits with a 4B/5B encoding so we do get 100,000,000 data
>> > bits in a second.)
>>
>> > Finally 576 * 1/100,000,000 = 5.76uS
>>
>> > On 10Mbit I make the figure ten times as much 57.6uS
>> > On Gbit I think they kept the basic timing budget from 100Mbit so make
>> > it's smallest-frame time about the same as for 100Mbit.
>>
>> > Corrections welcome. Does the above look about right?
>>
>> > --
>> > James
>>
>> At one point on the cable or over the legal length of cable?
>
> Umm, well if the speed of signal propagation over copper cable is
> about 0.7 times the speed of light (not sure of the exact figure but
> that's about right, isn't it?) we have the speed as
>
> 0.7 * 300,000,000 metres per second = 210,000,000
>
> If that's right then in 1uS the signal would propagate 210 metres so I
> think the length of the cable becomes almost insignificant - up to a
> standard length of 100m anyway. Maybe it would add 0.5uS to the total,
> over that length.
>
> Incidentally I think the old co-ax thinnet relied on reflected wave
> signalling so signals had to go to a terminator and bounce back for
> each bit - so the whole cable became energised for that bit long
> before the next bit was put on the wire.
>
> Just some more comments that people may be able to shoot down... Still
> I don't mind being the clay pigeon. It would be good to get some of
> these ideas confirmed ... or denied.
>
> --
> James


"insignificant" depends on what's behind the question. Can you tell us what
you are trying to do, or what's concerning you so the question can be
answered more appropritely.

--
Regards, Bob F.


Posted by James Harris on September 9, 2008, 7:16 pm
Please log in for more thread options


>
>
>
>
>
>
> >> > Could someone check my calculations? I'm looking for the shortest time
> >> > an Ethernet frame will occupy the wire. I make it about 5uS on 100Mbit
> >> > as follows.
>
> >> > Minimum frame size in bytes is 46 payload + 14 header + 4 FCS = 64.
> >> > Preamble (still needed on full duplex) of 8 octet times
> >> > Total of 72 octets * 8 bits = 576 bits.
> >> > Each bit time is 1/100,000,000 seconds. (IIRC the signalling rate is
> >> > 125,000,000 bits with a 4B/5B encoding so we do get 100,000,000 data
> >> > bits in a second.)
>
> >> > Finally 576 * 1/100,000,000 = 5.76uS
>
> >> > On 10Mbit I make the figure ten times as much 57.6uS
> >> > On Gbit I think they kept the basic timing budget from 100Mbit so make
> >> > it's smallest-frame time about the same as for 100Mbit.
>
> >> > Corrections welcome. Does the above look about right?
>
> >> > --
> >> > James
>
> >> At one point on the cable or over the legal length of cable?
>
> > Umm, well if the speed of signal propagation over copper cable is
> > about 0.7 times the speed of light (not sure of the exact figure but
> > that's about right, isn't it?) we have the speed as
>
> > 0.7 * 300,000,000 metres per second = 210,000,000
>
> > If that's right then in 1uS the signal would propagate 210 metres so I
> > think the length of the cable becomes almost insignificant - up to a
> > standard length of 100m anyway. Maybe it would add 0.5uS to the total,
> > over that length.
>
> > Incidentally I think the old co-ax thinnet relied on reflected wave
> > signalling so signals had to go to a terminator and bounce back for
> > each bit - so the whole cable became energised for that bit long
> > before the next bit was put on the wire.
>
> > Just some more comments that people may be able to shoot down... Still
> > I don't mind being the clay pigeon. It would be good to get some of
> > these ideas confirmed ... or denied.
>
> > --
> > James
>
> "insignificant" depends on what's behind the question. Can you tell us what
> you are trying to do, or what's concerning you so the question can be
> answered more appropritely.

I mean the signal propagation delay (absolute maximum 0.5uS) appears
to be "almost insignificant" compared with the minimum frame time of
5.76uS. There's no application relevant. I just wanted to check my
figures. Of course the resulting thread may be useful for someone to
refer back to.

--
James

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