Class Exercise 13 – Mode-locking
Question 1:
We analyzed in class active mode locking assuming several assumption (e.g. parabolic gain
function) and the following modulated absorption: 1 (t )  exp  m 1  cos mt   . Let’s
assume that the modulator is given by 2 (t )  exp  m 1  cos mt  / 2  cos 10mt  / 2  .
You may continue to assume all the other assumptions.
a) Derive the equation for the nth-harmonic in the frequency domain treatment for modelocking of this system.
b) Compare qualitatively (i.e. larger, smaller or indifference) the pulse duration and the
mode-locking threshold between the two modulators. Explain your answers.
Answer:
(a) Without modulation the field in the nth mode varies according to:
dEn g0  l
4 g0 n 2  2

En 
En
dt
T
T g2
The modulation will add an additional loss term:
ln 2 (t ) 
 m (1  cos(mt ) / 2  cos(10mt ) / 2)
En 
En 
T
T
  eimt  e imt ei10mt  e i10mt 
in t
  m 1 

 En e
T 
4
4

Assuming that the modulation frequency is m  2 T   we get:
ln 2 (t ) 



En   m En eint  m ein1t  ein1t En  m ein10t  ein10t En
T
T
4T
4T




We get that each mode is coupled to its nearest neighbors and to its 10th neighbors. Therefore,
the equation for the nth harmonic becomes:
dEn g0  l
4 g n 2  2


En  0 2 En  m En10  En1  4En  En1  En10
dt
T
T g
4T


(b) The spectrum will be larger, because of the energy that is transferred to farther modes,
and therefore the pulse duration will be smaller. The mode-locking threshold will be larger
(we will have to pump higher) since we increased the loss. Each mode now losses energy to
four other modes instead of two modes as in the original modulation.
:)‫ (לקוח ממבחן‬2 ‫שאלה‬
‫פתרון‪:‬‬