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Synchronisation Conditions for Two Semiconductor Lasers

Alistair Mees James Ong February 3, 2000

1 Abarbanel-Kennel-Illing laser system

Abarbanel, Kennel and Illing 1] model an experimental laser system using 1+ n I_ = C 1 + (IJ? 1) ? 1 I S " !# 1+ Jn n = J 1 ? 1 + (I ? 1) I ? n + max(?J ; min(A; h(I ))) _

N S P C N S P C

"

#

where

Write the system in the form I_ = f (I; n) n = g(I; n; I ) _

h(I ) = ? (I ? 1) ? B; I < 1 (I ? 1) ? B; I 1

and couple with a second system I_0 = f (I 0; n0) n0 = g(I 0; n0; I 00) _ via

I 00 = cI + (1 ? c)I 0 where c is the coupling constant, with c = 0 representing no coupling and c = 1 representing full coupling.

1

2 Lyapunov function

De ne ? = I0 ? I = n0 ? n ? = I0 ? I and set up a Lyapunov function in the form 1 V = 2 ?2 + 2 giving _ _ V = ?? +

2

+

Zt

t?

j?( )j2d

(1)

_ + (?2 ? ?2 )

3 Conditions arising from linearisation

_ ? = f (I 0; n0) ? f (I; n) _ = g(I 0; n0; I 00) ? g(I; n; I ) After linearisation, we have _ ? @I f (I; n)? + @nf (I; n) =

N S P C

8 " 2 # > < C 1+ Jn ?1 ? PI6 4 > S 1 + (I ? 1) S :

+

C N 2 S

_

@I g(I; n; I )? + @ng(I; n; I 2 + (1 ? c)@I g(I;3n; I )? ) 8 9 > 1+ n < = I (1 + J n) 7> P 6 J = ?J > 1 + (I ? 1) ? 4 2 5> ? C : 1 + (I ? 1) ; # " I ? 1 + (I ? 1) + 1 + (1 ? c) ?

N S N S P C P C N S P C

J 1 + (I ? 1)

P C

I

1+ 1 + (I ? 1)

N S P C

39 > 7= ? Jn 2 5> ;

where takes on the values 0, ? or . 2

Substituting this into (1), we get _ V =

N S P C

2 0 2 1+ n P ?2 6 C B 4 S @ 1 + (IJ? 1) ? 1 ? C I 6 4 "

+?

C N 2 S

1+ 1 + (I ? 1)

N S P C

31 3 7C 7 Jn 2 5A + 5

J 1 + (I ? 1)

N S

? 1 + (I ? 1)

J 1+

2

Jn

N S

P C

I

=

2 ?2 4?I

+ 4IC

?

"

P C

0 @1 ?

1 + (I ? 1) + 1 + (1 ? c)

P C

I

#

IP 1 + (I ? 1)

P C

13 A5

C

? ? ?2

1+ N n SJ

P S

1 ? 2I C 1 + (I ? 1) P

P C

!2

2 4

1+ N n ? S SJ ! + ? 1 + 1I ? 1) (

P C

C S

+

C N 2 S 2

I ? J 1+ J

?

"

0 N n @1 ? SJ

2 + 1 ? (1 ?4c) 1 + (I ? 1)

N S P C

I

1 + (I ? 1)

2

I

P

#

P C

13 A5

C 2

?

? ? (1 ? c) 2

:

Assuming that the coe cient of ?2 is negative, we can complete the square to eliminate the ? cross term. Rewrite the above equation as 2 _ = ?A?2 ? B ? ? C 2 ? ? ? (1 ? c) V 2 and complete the square to get 2 2 2 _ = ?C + ? B ? ?2 A ? B ? ? ? (1 ? c) : V 2C 4C 2 Thus we just want the following conditions to be satis ed: C > 0 (2) 3

We can simplify each term by substituting the following values: J = 0:6667 9 ?1 S = 1:458 10 s 9 ?1 N = 1:333 10 s 9 ?1 P = 2:4 10 s 11 ?1 C = 2:4 10 s . With these values, we get 1 50 2 A = 1:646I (1 + 0:609n) 1 + 0:01(I ? 1) ? I : ? 41I15 (1 + 0:609n) + 164:6 ? 1 B = ? 1 + 0:01(I ? 1) :01I 225:7I ? 0:6667 (1 + 0:609n) 1 ? 1 + 0001(I ? 1) : 2 2 c 0:914I : + 1 ? (1 ?4 ) C = 1 + 0:01(I ? 1)

3.1 Check of the condition

B2 A ? 4C > 0

(3)

This is trivially true for sensible values of I and small values of .

3.2 Check of the condition

B Now consider the condition A ? 4C2 > 0. Since C > 0, this is equivalent to 4AC ? B 2 > 0: Break up each A, B and C into smaller components: A = A1 ? B = B1 + B2 2 C = C 1 + C2

C >0

A ? BC > 0 4

2

4

The condition becomes 4(A1 ? )(C1 2 + C2 ) ? (B1 + B2)2 > 0 since > 0. Call the left hand side of the inequality f ( ; ). Thus we want to show that f ( ; ) > 0: Since f ( ; ) is a quadratic function in both and and we suspect that in each individually, we have a negative quadratic, consider choosing the critical point where @f = 0 to maximise the value of f with respect to . @ Alistair - FYI: I would suspect that this choice of would let me prove synchronisation over a large range of c, since it is the least restrictive. Rewriting f as a quadratic in , we get

f( ; ) =

2

2 2 4(A1 ? )C1 ? B1 ] + 4(A1 ? )C2 ? 2 B1 B2] ? B2 > 0:

Since > 0 we want to show that f~ > 0 where ? ? 2 B2)2 2 f~( ) = ?B2 ? (4(A1A ?)C)2C ?B1B 2) 4(4( 1 1 1 Now, rewrite component C1 considering its dependance on c: ~ C1 = C1(1 ? c)2 This gives a new form to f~:

2 f~( ) = ?B2 ?

If we choose at the critical point, which is 2 + B = ?2(A1 ? )CC ? B 21B2 ; 4(A1 ? ) 1 1 we get 2 A1 )C ? 2 2 f ( ; ) = ? B2 ? (4(4(4(? ? 2 )C ? B1B)2) A B2

1 1 1

(4(A1 ? )C2 ? 2B1B2)2 2 ~ 4(4(A1 ? )C1(1 ? c)2 ? B1 ) 2 ~ If 4(A1 ? )C1(1 ? c)2 ? B1 6= 0 then the problem f~ > 0 becomes equivalent to a simple quadratic which can be solved to give a range 1?

s

2 2 B1 B2 ? (2(A1 ? )C2 ? B1B2)2] < c 1 2 ~ 4B2 (?C1)(A1 ? )

5

where we note that we still have a free parameter . If we try to make the range as large as possible, all we need to do is try to make the part under the square root as large as possible. If we do this, we have to solve a cubic for :

2 2 0 = (A1 ? ) B1 B2 ? (2(A1 ? )C2 ? B1B2)2 ? 2 (2(A1 ? )C2 ? 2B1B2)(?2C2)] 2 2 + B1 B2 ? (2(A1 ? )C2 ? B1B2)2]

References

1] H. D. I. Abarbanel, M. B. Kennel, and L. Illing. Synchonization and communication with electro-optically modulated semiconductor lasers. Technical report, Institute for Nonlinear Science, University of California, San Diego, 1999.

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