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Edwin Jan Klein - Universiteit Twente

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Appendix B. Crosstalk Derivation<br />

The adjacent channel crosstalk is the relative power fraction of a signal that is present<br />

on a channel adjacent to the channel on which this signal is transmitted and is defined<br />

as:<br />

⎛ PDrop<br />

( λ = λc<br />

+ ∆λcs<br />

) ⎞<br />

CT = 10log⎜<br />

⎟<br />

⎜<br />

⎟<br />

(B.1)<br />

⎝ PDrop<br />

( λ = λc<br />

) ⎠<br />

where λc is the wavelength of the channel for which the resonator is in full resonance<br />

and ∆λcs is the separation between the channels.<br />

Then with φ1 the roundtrip phase of a resonator in resonance:<br />

( 2π<br />

)<br />

ϕ 1 =<br />

λ<br />

c<br />

2<br />

. R.<br />

Neff<br />

and φ2 the roundtrip phase of the adjacent channel:<br />

2<br />

( 2π<br />

)<br />

ϕ2<br />

= . R.<br />

Neff<br />

λ + ∆λ<br />

Combined with the drop response of the resonator:<br />

P<br />

Drop<br />

c<br />

cs<br />

H<br />

=<br />

1+<br />

F . sin<br />

The expression for the crosstalk then becomes:<br />

C<br />

197<br />

(<br />

2 ϕ<br />

⎛ H ⎞<br />

⎜<br />

2 ⎟<br />

2<br />

⎜ 1+<br />

F<br />

⎟ ⎛<br />

C . sin ( ϕ 2 / 2)<br />

1+<br />

FC<br />

. sin ( ϕ1<br />

/<br />

CT = 10log<br />

= 10log<br />

⎜<br />

⎜<br />

⎟<br />

2<br />

H<br />

⎝1<br />

+ FC<br />

. sin 2<br />

⎜<br />

2 ⎟<br />

⎝ 1+<br />

FC<br />

. sin ( ϕ1<br />

/ 2)<br />

⎠<br />

/<br />

2)<br />

2)<br />

( ϕ / 2)<br />

The phase φ2 is equal to m.2π because λc is the resonance frequency. In addition FC<br />

can be written as:<br />

F C<br />

1<br />

= 2<br />

sin ( π / 2F<br />

)<br />

Combining (B.5) with (B.6) then gives the expression:<br />

⎛<br />

⎞<br />

⎜<br />

⎟<br />

2<br />

⎜<br />

1<br />

⎟ ⎛ sin ( π / 2F<br />

) ⎞<br />

CT = 10log<br />

= 10log<br />

⎜<br />

⎟<br />

⎜<br />

⎟<br />

2<br />

2<br />

1<br />

2<br />

⎝ sin ( π / 2F<br />

) + sin ( ϕ / 2)<br />

⎜1<br />

+<br />

. sin ( ϕ / 2)<br />

2<br />

⎟<br />

⎠<br />

2<br />

2<br />

⎝ sin ( π / 2F<br />

)<br />

⎠<br />

Then, since<br />

⎞<br />

⎟<br />

⎠<br />

(B.2)<br />

(B.3)<br />

(B.4)<br />

(B.5)<br />

(B.6)<br />

(B.7)

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