0.01
0.001
RF frequency f/f 0e
a)
c)
b)
20
0
–20
–40
–60
–80
K
L , K
L · K
EF
magnitude (dB)
K
OA · K
OF’
magnitude (dB)
0.1
1
10
K L
A 0 = 1.5
10
K EF
0.01
0.001
100
100
200
1
2 3
100
0
–100
–200
50
0
–50
0.1
Optical frequency v/v 0L (u.a)
1 1.1
3
2
1
10
0.1
1 1.1
10
50
0
–50
–100
Phase
ϕ
L +
ϕ
EF ,
ϕ
EA (degree)
Phase
ϕ
OA ,
ϕ
OF (degree)
0.1
1
10
A 0 = 1.5
10
0.01
0.001
20
0
–20
–40
–60
–80
K
L , K
L · K
EF
magnitude (dB)
0.1
1
10
K EA
A 0 = 1.5
10
K L · K EF
ϕ EF
ϕ L
RF frequency f/f 0e (u.a)
0.01
0.001
200
100
0
–100
–200
Phase
ϕ
L +
ϕ
EF ,
ϕ
EA (degree)
0.1
1
10
A 0 = 1.5
10
ϕ L + ϕ EF
ϕ EA
ϕ OA
ϕ OF
Fig. 3.17 The example of the graphical solution of APB equation for OEO DM. (a) The module K L
and the argument φ L of QWLD for two excess values A 0 ¼ J 0L /J 0Lth (A 0 ¼ 1.5; 10) of the pumping
(bias) current J 0L above the threshold value J 0Lth , and also the module K EF and the argument φ EF .
(b) The module K L Á K EF and the argument φ L + φ EF for A 0 ¼ 1.5; 10. The module K EA and the
argument φ EA . The normalized frequency f/f 0e , which is 1, corresponds to the natural frequency of
the RF filter. (c) The example of graphical solution of APB equations for the laser in OEO. The
module and the argument of the gain for QWLD with the resonator Q-factor of 100 at different
pumping current. The module and the argument of the optical channel of the laser. Normalized
frequency ν/ν 0L , which is 1, corresponds to the natural frequency of the optical filter
3.4 Equations of Amplitude and Phase Balance for the Laser and for OEO
117