Computer Engineering

Paper details:

Course : Optical Computer System

ELG5103 Optical Communications Systems Assignment 2 Waves in a Slab Waveguide
Hint: The electric field distribution varies depending on the confinement factor
and the dispersion relation. To find the electric field distribution inside the core
and the cladding you will need to find K & y which will involve solving the
dispersion relation for a particular value of V – this may be done with a few lines
of Matlab. See supplementary material.

3) To keep the notation light-weight it may be assumed that one is only concerned with
TE-like modes with transverse amplitude profiles adequately described by scalar
valued eigenfunctions (pp associated to an eigenvalue 3,, corresponding to the
propagation constant of the mode i.e.

up (x, z) = (pp (x) exp(i,6’pz)
The (pp are orthonormal with respect to an inner product:

(‘pp”pq) = 6m
The inner product may take different forms but it is always linear in the first argument
and conjugate linear in the second argument and for purpose of this problem may be
taken1 as:

(em wq) – I epwqu – 6pc:

The set of eigenfunctions is complete.
Suppose light is launched into a waveguide corresponding a total forward going field
1? , exploit the orthonormality of the modes relative to the inner product to find the
coefficients of the expansion of the field ‘1’ as linear combination of the modes. What

I The electric 5,, and magnetic fields of propagating modes of a dielectric waveguide satisfy the orthogonality condition:

fi(£,,xfl;+5; xup).ds= (SW

No longitudinal component takes part in this integral and so one may considerthe transverse components only.

In the special case of TE & TM modes, the TE mode is always orthogonal to a TM mode and vice versa. It is therefore only necessary

to consider the orthogonality within the set of TE and TM modes:

For TE one notes that only the 3 component of H x contributes and:

as),
HI. O( E = IKE),
For TM one notes that only the 3 component of E , contributes and:
1 as 1
5 -” = ‘ -E
x cc 722(x) dz ll? 122(x) y
Hence with suitable normalization one finds:
(‘PpKM : [(pp<pq’dx = 5m TE
and
OS 1
(ewq) = I W em dx = 5w W
ELG5103 Optical Communications Systems Assignment 2 Waves in a Slab Waveguide
Hint: The electric field distribution varies depending on the confinement factor
and the dispersion relation. To find the electric field distribution inside the core
and the cladding you will need to find K & y which will involve solving the
dispersion relation for a particular value of V – this may be done with a few lines
of Matlab. See supplementary material.

3) To keep the notation light-weight it may be assumed that one is only concerned with
TE-like modes with transverse amplitude profiles adequately described by scalar
valued eigenfunctions (pp associated to an eigenvalue 3,, corresponding to the
propagation constant of the mode i.e.

up (x, z) = (pp (x) exp(i,6’pz)
The (pp are orthonormal with respect to an inner product:

(‘pp”pq) = 6m
The inner product may take different forms but it is always linear in the first argument
and conjugate linear in the second argument and for purpose of this problem may be
taken1 as:

(em wq) – I epwqu – 6pc:

The set of eigenfunctions is complete.
Suppose light is launched into a waveguide corresponding a total forward going field
1? , exploit the orthonormality of the modes relative to the inner product to find the
coefficients of the expansion of the field ‘1’ as linear combination of the modes. What

I The electric 5,, and magnetic fields of propagating modes of a dielectric waveguide satisfy the orthogonality condition:

fi(£,,xfl;+5; xup).ds= (SW

No longitudinal component takes part in this integral and so one may considerthe transverse components only.

In the special case of TE & TM modes, the TE mode is always orthogonal to a TM mode and vice versa. It is therefore only necessary

to consider the orthogonality within the set of TE and TM modes:

For TE one notes that only the 3 component of H x contributes and:

as),
HI. O( E = IKE),
For TM one notes that only the 3 component of E , contributes and:
1 as 1
5 -” = ‘ -E
x cc 722(x) dz ll? 122(x) y
Hence with suitable normalization one finds:
(‘PpKM : [(pp<pq’dx = 5m TE
and
OS 1
(ewq) = I W em dx = 5w W