Significance of admittance ``sqrt(y_odd * y_even)'':
For a coupled symmetric (homogeneous/inhomog. medium) transmission line terminated with its
characteristic admittance matrix, the input admittance matrix (which will be
the characteristic admittance matrix) is given as:
port 1 port 2
o-----o----^^^^^^^^^----o----o line 1 ---> inft.
| | o----------------------
| -y_even/2 | o----------------------
^ ^ line 2 ----> inft
y_even ^ ^ y_even
- - gnd - - gnd
1) Now looking at this circuit :
If we excite the given structure with
the source admittance y_t_odd (y_t_odd=y_odd,
a non-mode converting admittance ) connected at port 1
and port 2 and also make sure that the excitation is always differential (note: only possible
in the case of symmetric line, in all other asymmetric structures such an excitation is
not possible) the lines will be matched at the input port and the input admittance
seen at these input ports will be y_odd.
2) In case the lines are not excited differitially (e.g. digital
buses), then the optimum (also non-mode converting admittance)
admittance for a approximate match is given by sqrt(y_even y_odd).
NOTE: a) In case (1) only one channel but in case (2) two channels are present
for the data transmission.
b) In case (1) reference is with respect to the other line but in case 2 the reference
is with respect to the ground. Thereby in case (1) the effect of ground noise
on the data transmission is reduced.
On Wed, 11 Nov 1998, fabrizio zanella wrote:
> To calculate the differential impedance of a transmission line pair, I've
> always used
> 1) Zdiff = 2 * Zoo, where Zoo is the odd mode impedance of the pair. This
> formula is used by the pc board manufacturers.
> I was speaking to someone last week who has 30+ years in the industry and
> said this formula is wrong, that the real differential impedance should be
> calculated as
> 2) Zdiff = square root of (Zoo * Zee), where Zee is the even mode
> If I compare 1) to 2), I get quite different values of differential
> impedance. Does anyone have comments on this?
> Thanks, Fabrizio Zanella
> EMC corporation
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