Page 314 - Fiber Optic Communications Fund

Optical Amplifiers 295

Substituting Eq. (6.283) in Eq. (6.282), we find

̃ u(f)= triang (f∕f )rect(f∕(2f ))f . (6.288)
o
e
o
The product in Eq. (6.280) may be rewritten as a convolution in the frequency domain at f = 0as
̃
B 2 oe = ∫ ̃ u(f)H (−f)df, (6.289)
e
( ) ( )
f f
̃
̃ u(f)H (−f)= triang (f∕f )rect rect f , (6.290)
o
o
e
2f 2f
e e
( )
f
= triang (f∕f )rect f . (6.291)
o
o
2f e
First let us consider the case f < f . The integral in Eq. (6.289) is equal to the area of the shaded region shown
e o
in Fig. 6.32.
B 2 =(2f − f )f . (6.292)
oe o e e
2
If f ≥ f , B equals the area of the triangle,
e o oe
2
B 2 = f . (6.293)
oe o
Substituting Eqs. (6.292) and (6.293) in Eq. (6.271), we find
2 2
2 = R (2f − f )f if f ≤ f , (6.294)
sp−sp ASE o e e e o
2 2
= R ASE o 2 otherwise. (6.295)
f
f o

f o triang (f/f o )

f o – f e

–f o f e f o f o

Figure 6.32 Calculation of the variance of ASE–ASE beat noise for the case of ideal optical and electrical filters.