Figure 6: Left: Changes in the amplitude A vs. time. Right: Changes in the number of the carriers n vs. time.
A desirable feature of a laser is the constant amplitude, shown in
Figure [6]. Right after the laser turns on, the amplitude
varies for a while and then gets stabilized to a constant. We call the
frequency before the laser gets stabilized the relaxation oscillation
frequency, which we are going to calculate in this section.
The reason we want to obtain the value of the relaxation oscillation
frequency is that physically, we easily get resonance when we add a
frequency similar to that of the system; in resonance we can obtain
large responses.
When we add the term for the TM injection, we
will get various different behaviors from the SRE system,
depending on the injection strength and the detuning between the
injection and the free running laser. When their detuning has a value
similar to the relaxation oscillation frequency of the laser, we
often observe interesting non-linear behavior.
To calculate the relaxation oscillation frequency( 
,
hereafter), it is convenient to use the laser equation in terms of
amplitude.
where 
. 
and 
have damped
oscillations as seen in Figure [6]. The
frequency changes along with time. The relaxation oscillation
frequency is the frequency seen when the system relaxes close to its
stable state. So we will find a steady state solution and add a small
perturbation in order to calculate the relaxation oscillation
frequency. First, when 
, we have 
. We can
also find the steady state solutions other than this trivial one by
letting 
and 
. Substituting
, the equations can be written as
From Eqns. (31) and (30) , we obtain the following solution.
For ease of computation, that is, in order to transform this complicated system into a simple 2nd order ODE, we changed the variables into non-dimensional quantities.
We can also write the steady state solutions in terms of 
and
.
where 
and
. We differentiate 
and 
with respect to and obtain the following differential
equation.
Let 
and 
be small perturbation terms and add them to 
and 
respectively.
Substituting Eqns. (34),(35) into Eqns. (32),(33), the equations are rewritten as
In the equations, since we assume p,q are small, 
is so small
that it can be neglected. Substituting in, we obtain
Then we differentiate Eqn. (37) with respect to and
substitute p from Eqn. (37) again to obtain the 2nd order differential
equation of with respect to .
The equation above is dimensionless. But the value of has
dimension. So We write the equation in terms of t again
to obtain an ODE in terms of t.
The solution for the above ODE is
where is
To calculate a numerical value, we put 
and 
. The for our
model is
