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 . 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
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 . 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
To calculate a numerical value, we put and . The for our model is