Crosstalk noise onto a floating line
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An active line-1 is terminated by resistances at both ends, and a passive line-2 is floating. The voltage on active line is the same in the case of a single line:
where
The voltage on passive line is proportional to that on active line:
where ξ is the crosstalk coefficient:
An example waveform for a rising step input v01(t) with rise time tr=τ/2:
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We will calculate the crosstalk voltage onto a floating line.
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We start from the general solutions in even and odd mode:
Transforming the variables of even and odd mode to those of each line:
we obtain the general solutions for each line:
where we have assumed that the propagation speeds are equal in even and odd mode.
For convenience, expressing (3),(4) in terms of Z0 and parameter ξ
,we have
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We are now ready to calculate the crosstalk noise onto a floating line.
Boundary conditions are:
then Laplace transformed boundary conditions are
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First, inserting the general solutions (1),(3') into the boundary conditions of line-1:
Eliminating the pair of Va2,Vb2:
After some manipulation we obtain:
where τ is the propagation time from x=0 to x=l, and ΓS,ΓT are reflection coefficients at source and termination end.
This result is the same in the case of a single line. (→Multiple reflection in a terminated line)
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Next, inserting the general solutions (2),(4') into the boundary conditions of line-2:
Eliminating the pair of Vb1,Vb2 or Va1,Va2:
We find that the voltage on passive line-2 is proportional to that on active line-1.
This parameter ξ indicates the ratio of amplitude between passive and active line, so it is called crosstalk coefficient.