Assuming you are talking about Ethernet or FDDI TP-PMD ...
A minor correction; if you used a long loopback cable, it should be
only 50 meters long, the round-trip length is 100 meters, which is
the max. length.
The electrical characteristics of a 100 meter cable are very
frequency dependent. Any simple network to simulate just the DC
loss characteristics, or the loss at one frequency, will not
correctly represent the situation at all frequencies.
This is especially true for 100BASE-TX, or FDDI, where the received
signal after 100 meters does not look anything like the transmitted
signal. The low frequency components are there, but the high
frequencies are severely attenuated.
There is an example of a "Twisted-pair model" provided as Figure
14-7 in the ISO/IEC 8802-3 spec. This will give you an idea of the
complexity required to model the cable, in the slower 10 Mbit case.
For 100BASE-T4, see Figure 23-23 in IEEE 802.3u, which applies to
CAT 3 cable and 100BASE-T4's bit rate. I don't know of a similar
model for 100BASE-TX or TP-PMD, which use 125 Mbit signaling.
If all you want to do is simple go/no-go testing and don't mind
some inaccuracy, a simple network similar to these or even your own
might suffice. (By the way, in your very simple circuit, you should
make R1=R2, and then choose R3 to make the impedance 100 ohms
looking into either end with the other end terminated.)
But sometimes there is no substitute for using a real cable.
Regards,
Andy
> L1 (50 cm )
> >--R1------------------+
> |
> <---------------R2 +---+
> L2 (50 cm) |
> R3
> |
> ----
> / / / (gnd)
>
> R1 : about 8.55 ohms to works as DC resistor for 100 m cable.
>
> Because of Characteristic imped. Z0 for cable is 100 ohms
> R2 // R3 = 100 ohms (to avoid reflection).
>
> R2/(R2+R3) = attenuation of 100 m cable.
NOTE: These formulas aren't right.