Characterization of waveguide loss using distributed Bragg reflectors

Abstract

We propose and demonstrate a method to characterize waveguide loss using the spectral response of combinations of distributed Bragg reflectors. The method is independent of coupling efficiency and waveguide dispersion and does not require the introduction of bending loss into the device.

Appendix: Derivations

1.1 Coefficients of transmission and reflection of a DBR

We shall take as our starting point Eqs. (1, 3) of the paper. To uncouple the equations, first differentiate them:

$$\frac{{{\text{d}}^{2} A_{F} }}{{{\text{d}}z^{2} }} = - i \cdot \kappa \cdot \frac{{{\text{d}}A_{B} }}{{{\text{d}}z}}\exp \left( {i \cdot \Updelta \beta \cdot z} \right) + \kappa \cdot \Updelta \beta \cdot A_{B} \exp \left( {i \cdot \Updelta \beta \cdot z} \right) - \frac{{\alpha_{\text{DBR}} }}{2}\frac{{{\text{d}}A_{F} }}{{{\text{d}}z}} $$

(23)

$$\frac{{{\text{d}}^{2} A_{B} }}{{{\text{d}}z^{2} }} = i \cdot \kappa^{ * } \cdot \frac{{{\text{d}}A_{F} }}{{{\text{d}}z}}\exp \left( { - i \cdot \Updelta \beta \cdot z} \right) + \kappa^{ * } \cdot \Updelta \beta \cdot A_{F} \exp \left( { - i \cdot \Updelta \beta \cdot z} \right) + \frac{{\alpha_{\text{DBR}} }}{2}\frac{{{\text{d}}A_{B} }}{{{\text{d}}z}} $$

(24)

Complete the decoupling by substitution from Eqs. (1, 3) in order to obtain separate differential equations for the forward and backward propagating field amplitudes:

$$\frac{{{\text{d}}^{2} A_{F} }}{{{\text{d}}z^{2} }} = \left( {\kappa^{ * } \kappa + i \cdot \Updelta \beta \frac{{\alpha_{\text{DBR}} }}{2} + \frac{{\alpha_{\text{DBR}}^{2} }}{4}} \right)A_{F} + i \cdot \Updelta \beta \frac{{{\text{d}}A_{F} }}{{{\text{d}}z}} $$

(25)

$$\frac{{{\text{d}}^{2} A_{B} }}{{{\text{d}}z^{2} }} = \left( {\kappa^{ * } \kappa + i \cdot \Updelta \beta \frac{{\alpha_{\text{DBR}} }}{2} + \frac{{\alpha_{\text{DBR}}^{2} }}{4}} \right)A_{B} - i \cdot \Updelta \beta \frac{{{\text{d}}A_{B} }}{{{\text{d}}z}} $$

(26)

The general solution of Eqs. (25, 26) is Eqs. (4–6) of the paper.

The boundary conditions that we will apply to determine the undetermined coefficients C ₁, C ₂, D ₁, and D ₂ in Eqs. (4–6) are A _B (L _DBR) = 0 and A _F (0) = a nonzero constant. This corresponds to the physical arrangement in which only a forward propagating field is incident upon the DBR. It is possible to eliminate two of the coefficients by substituting the boundary conditions into Eqs. (4, 5). The general solutions may then be rewritten as:

$$A_{F} = 2 \cdot C_{1} \cdot \exp \left( {\frac{i \cdot \Updelta \beta \cdot z}{2}} \right)\sinh \left( {s \cdot z} \right) + A_{F} \left( 0 \right) \cdot \exp \left( {\frac{i \cdot \Updelta \beta \cdot z}{2}} \right)\exp \left( { - s \cdot z} \right) $$

(27)

$$A_{B} = 2 \cdot D_{1} \cdot \exp \left( {s \cdot L_{\text{DBR}} } \right) \cdot \exp \left( {\frac{ - i \cdot \Updelta \beta \cdot z}{2}} \right)\sinh \left[ {s\left( {z - L_{\text{DBR}} } \right)} \right]. $$

(28)

Substituting Eqs. (27, 28) into Eqs. (1, 2) and again employing the boundary conditions, the remaining coefficients may be determined:

$$C_{1} = \frac{{\left( {s - \frac{{i \cdot \Updelta \beta + \alpha_{\text{DBR}} }}{2}} \right)A_{F} \left( 0 \right)\exp \left( { - s \cdot L_{\text{DBR}} } \right)}}{{\left( {i \cdot \Updelta \beta + \alpha_{\text{DBR}} } \right)\sinh \left( {s \cdot L_{\text{DBR}} } \right) + 2 \cdot s \cdot \cosh \left( {s \cdot L_{\text{DBR}} } \right)}} $$

(29)

$$D_{1} = \frac{{ - i \cdot \kappa^{ * } \cdot A_{1} \left( 0 \right) \cdot \exp \left( { - s \cdot L_{\text{DBR}} } \right)}}{{\left( {i \cdot \Updelta \beta + \alpha_{\text{DBR}} } \right)\sinh \left( { - s \cdot L_{\text{DBR}} } \right) - 2 \cdot s \cdot \cosh \left( { - s \cdot L_{\text{DBR}} } \right)}}. $$

(30)

The coefficients of reflection and transmission may be determined by substituting Eqs. (29, 30) into Eqs. (27, 28):

$$r_{F} = \frac{{A_{B} \left( 0 \right)}}{{A_{F} \left( 0 \right)}} = \frac{{ - i \cdot 2 \cdot \kappa^{ * } \cdot \sinh \left( { - s \cdot L_{\text{DBR}} } \right)}}{{\left( {i \cdot \Updelta \beta + \alpha_{\text{DBR}} } \right)\sinh \left( { - s \cdot L_{\text{DBR}} } \right) - 2 \cdot s \cdot \cosh \left( { - s \cdot L_{\text{DBR}} } \right)}} $$

(31)

$$t_{F} = \frac{{A_{F} \left( L \right)}}{{A_{F} \left( 0 \right)}} = \frac{{2 \cdot s \cdot \exp \left( {i\frac{\Updelta \beta }{2} \cdot L_{\text{DBR}} } \right)}}{{\left( {i \cdot \Updelta \beta + \alpha_{\text{DBR}} } \right)\sinh \left( {s \cdot L_{\text{DBR}} } \right) + 2 \cdot s \cdot \cosh \left( {s \cdot L_{\text{DBR}} } \right)}} $$

(32)

where r _F is the coefficient of reflection and t _F is the coefficient of transmission. It is trivial to manipulate Eqs. (31, 32) into Eqs. (7, 8).

1.2 First-order approximation of null point coefficient of reflection

To arrive at the null point reflectance to first order in α _DBR, begin by considering the parameter s described by Eq. (6) at the null point Δβ _null described by Eq. (10). Neglecting the term that is second order in α _DBR, the product s·L _DBR at such a point becomes:

$$s \cdot L_{\text{DBR}} \approx \sqrt { - \left( {n \cdot \pi } \right)^{2} + i \cdot \Updelta \beta_{\text{null}} \frac{{\alpha_{\text{DBR}} }}{2} \cdot L_{\text{DBR}}^{2} } . $$

(33)

Next, take the Taylor series of the radical in Eq. (33) and keep the terms to first order in α _DBR:

$$s \cdot L_{\text{DBR}} \approx n \cdot i\pi \left( {1 - \frac{{i \cdot \Updelta \beta_{\text{null}} \cdot \alpha_{\text{DBR}} \cdot L_{\text{DBR}}^{2} }}{{4\left( {n \cdot \pi } \right)^{2} }}} \right) = n \cdot i\pi + \frac{{\Updelta \beta_{\text{null}} \cdot \alpha_{\text{DBR}} \cdot L_{\text{DBR}}^{2} }}{4 \cdot n \cdot \pi }. $$

(34)

Next use the approximation in Eq. (34) to simplify the term in the denominator of coefficient of reflectance Eq. (7) containing the hyperbolic tangent:

$$\frac{{s \cdot L_{\text{DBR}} }}{{\tanh \left( {s \cdot L_{\text{DBR}} } \right)}} \approx \frac{n \cdot i\pi }{{\tanh \left( {n \cdot i\pi + \frac{{\Updelta \beta_{\text{null}} \cdot \alpha_{\text{DBR}} \cdot L_{\text{DBR}}^{2} }}{4 \cdot n \cdot \pi }} \right)}}. $$

(35)

Taking the hyperbolic tangent to first order and substituting in Eq. (10), Eq. (35) becomes:

$$\frac{{s \cdot L_{\text{DBR}} }}{{\tanh \left( {s \cdot L_{\text{DBR}} } \right)}} \approx \frac{n \cdot i\pi }{{\frac{{\Updelta \beta_{\text{null}} \cdot \alpha_{\text{DBR}} \cdot L_{\text{DBR}}^{2} }}{4 \cdot n \cdot \pi }}} = \frac{{ \pm 2 \cdot i \cdot n^{2} \cdot \pi^{2} }}{{\alpha_{\text{DBR}} \cdot L_{\text{DBR}} \sqrt {\kappa^{ * } \kappa \cdot L_{\text{DBR}}^{2} + \left( {n \cdot \pi } \right)^{2} } }}. $$

(36)

Substituting Eqs. (36) and (10) into Eq. (6) and keeping only terms of the first order in α _DBR results in the expression for reflectance:

$$r_{F} = \frac{{ \mp \left( {\alpha_{\text{DBR}} \cdot L_{\text{DBR}} } \right)\left( {\kappa^{ * } \cdot L_{\text{DBR}} } \right)\sqrt {\kappa^{ * } \kappa \cdot L_{\text{DBR}}^{2} + \left( {n \cdot \pi } \right)^{2} } }}{{2 \cdot n^{2} \cdot \pi^{2} }}. $$

(37)

It is trivial to verify that taking the magnitude of Eq. (37) results in Eq. (11).