# Directional Couplers

This is a continuation from the previous tutorial - grating waveguide couplers.

Directional couplers are multiple-waveguide couplers used for codirectional coupling. They can be used in many different applications, including power splitters, optical switches, wavelength filters, and polarization selectors. We consider in this tutorial two-channel directional couplers, which consist of two parallel waveguides, as shown schematically in figure 4 below.

For simplicity, we consider only the case where each waveguide supports only its own fundamental mode. Coupling between the two single-mode waveguides in such a two-channel directional coupler is simply described by (57) and (58) in the two-mode coupling tutorial with

$\tag{30}2\delta=(\beta_b+\kappa_{bb})-(\beta_a+\kappa_{aa})$

In addition, $$c_{aa}=c_{bb}=1$$ for evaluation of the coupling coefficients using (47) [refer to the two-mode coupling tutorial] because the waves in both waveguides are forward propagating. In general, the two waveguides are not necessarily identical. Then the directional coupler is not symmetric, and $$\kappa_{ab}\ne\kappa_{ba}^*$$, as discussed in the two-mode coupling tutorial. If the two waveguides are identical, the directional coupler is symmetric. Then, $$\kappa_{ab}=\kappa_{ba}^*,\kappa_{aa}=\kappa_{bb},\text{and }\beta_a=\beta_b$$. In either case, the general solutions for codirectional coupling obtained in the two-mode coupling tutorial can be used directly if the coupling coefficients and the phase mismatch are known.

Coupling Coefficient

The coefficients involved in the coupling between two waveguides are given by (47) in the two-mode coupling tutorial. They are more complicated than those in the coupling between two modes in the same waveguide because of the existence of the overlap coefficient and the fact that $$\kappa_{ab}\ne\kappa_{ba}^*$$ in general.

As a result, many parameters have to be calculated in order to obtain the coupling coefficients $$\kappa_{ab}$$ and $$\kappa_{ba}$$ using (47). Here we consider a simple example, namely, the two-channel directional coupler with step-index waveguides on the same substrate shown in figure 4 above. We assume that the waveguides are planar slab waveguides for simplicity although practical direction couplers are often made of channel waveguides. We also consider only isotropic waveguides where TE and TM modes do not couple.

As shown in figure 4 above, the two waveguides have widths $$d_a$$ and $$d_b$$ and guiding layer refractive indices $$n_a$$ and $$n_b$$, respectively. They are separated by a distance $$s$$ between the two near edges of the guiding layers. The index of refraction of the substrate is $$n_2$$. When $$n_a=n_b=n_1$$ and $$d_a=d_b=d$$, the coupler is symmetric. Otherwise, it is asymmetric.

To calculate the relevant coefficients, we first identify the perturbation $$\Delta\epsilon$$ for each waveguide. For waveguide $$a$$, the susceptibility step of waveguide $$b$$ above the substrate is the perturbation, and vice versa. Therefore, we have

$\tag{31}\Delta\epsilon_a=\begin{cases}\epsilon_0(n_b^2-n_2^2),\;&s+d_a/2\lt x \lt s+d_a/2+d_b,\\0,\; &\text{otherwise}\end{cases}$

and

$\tag{32}\Delta\epsilon_b=\begin{cases}\epsilon_0(n_a^2-n_2^2),\;&-d_a/2\lt x \lt d_a/2,\\0,\; &\text{otherwise}\end{cases}$

where we have chosen the origin of the $$x$$ coordinate to be at the center of waveguide $$a$$. Using the field distributions and the characteristic parameters of planar waveguide modes obtained in the step-index planar waveguides tutorial, we can calculate the relevant coefficients using (41) and (42) in the coupled-mode theory tutorial.

For coupling between TE modes, the only nonzero component for the electric field is $$\hat{\mathcal{E}}_y$$ given by (55) in the step-index planar waveguides tutorial. Therefore, we have

\tag{33}\begin{align}\tilde{\kappa}_{aa}&=\omega\displaystyle\int\limits_{-\infty}^{\infty}\Delta\epsilon_a|\hat{\mathcal{E}}_{a,y}|^2\text{d}x\\&=\frac{1}{\beta_ad_a^E}\cdot\frac{h_b^2+\gamma_b^2}{h_a^2+\gamma_a^2}\cdot\frac{h_a^2}{2\gamma_a}(1-\text{e}^{-2\gamma_a d_b})\text{e}^{-2\gamma_a s}\end{align}

and

\tag{34}\begin{align}\tilde{\kappa}_{ab}&=\omega\displaystyle\int\limits_{-\infty}^{\infty}\Delta\epsilon_b\hat{\mathcal{E}}_{a,y}^*\hat{\mathcal{E}}_{b,y}\text{d}x\\&=\frac{1}{\sqrt{\beta_a\beta_bd_a^Ed_b^E}}\sqrt{\frac{h_a^2+\gamma_a^2}{h_b^2+\gamma_b^2}}\cdot\frac{h_ah_b}{h_a^2+\gamma_b^2}[(\gamma_a+\gamma_b)+(\gamma_a-\gamma_b)\text{e}^{-\gamma_bd_a}]\text{e}^{-\gamma_bs}\end{align}

where $$h_a$$ and $$h_b$$ are the parameter $$h_1$$, $$\gamma_a$$ and $$\gamma_b$$ are the parameter $$\gamma_2$$, and $$d_a^E$$ and $$d_b^E$$ are the effective waveguide thickness for the TE modes in waveguide $$a$$ and waveguide $$b$$, respectively.

The coefficient $$\tilde{\kappa}_{bb}$$ can be obtained by simply interchanging the indices $$a$$ and $$b$$ in (33), while $$\tilde{\kappa}_{ba}$$ can be obtained by interchanging the indices $$a$$ and $$b$$ in (34). It can be seen that $$\tilde{\kappa}_{aa}\ne\tilde{\kappa}_{bb}$$, and $$\tilde{\kappa}_{ab}\ne\tilde{\kappa}_{ba}$$, in general, as expected.

By following a procedure that reduces (41) to (44) for TE modes [refer to the wave equations for optical waveguides tutorial], it can be shown that, for TE modes, the overlap coefficient defined by (41) in the coupled-mode theory tutorial can be reduced to

\tag{35}\begin{align}c_{ab}&=c_{ba}^*=\frac{\beta_a+\beta_b}{\omega\mu_0}\displaystyle\int\limits_{-\infty}^{\infty}\hat{\mathcal{E}}_{a,y}^*\hat{\mathcal{E}}_{b,y}\text{d}x\\&=\frac{\beta_a+\beta_b}{\sqrt{\beta_a\beta_bd_a^Ed_b^E}}\left[\sqrt{\frac{h_a^2+\gamma_a^2}{h_b^2+\gamma_b^2}}\frac{h_ah_b}{h_a^2+\gamma_b^2}\left(\frac{1}{\gamma_a-\gamma_b}+\frac{\text{e}^{-\gamma_bd_a}}{\gamma_a+\gamma_b}\right)\text{e}^{-\gamma_bs}\right.\\&\qquad\qquad\qquad\left.+\sqrt{\frac{h_b^2+\gamma_b^2}{h_a^2+\gamma_a^2}}\frac{h_bh_a}{h_b^2+\gamma_a^2}\left(\frac{1}{\gamma_b-\gamma_a}+\frac{\text{e}^{-\gamma_ad_b}}{\gamma_b+\gamma_a}\right)\text{e}^{-\gamma_as}\right]\end{align}

Similar, but somewhat more complicated, formulas can be obtained for coupling between TM modes.

We now consider the simple case of a symmetric directional coupler. In this case, $$n_a=n_b=n_1$$, and $$d_a=d_b=d$$. Therefore, we have $$\beta_a=\beta_b\equiv\beta,h_a=h_b=h_1\equiv h$$, and $$\gamma_a=\gamma_b=\gamma_2\equiv\gamma$$, and the coefficients are much simplified. For coupling between TE modes of the same order, we have

$\tag{36}\tilde{\kappa}_{aa}=\tilde{\kappa}_{bb}=\frac{1}{\beta d_E}\cdot\frac{h^2}{2\gamma}(1-\text{e}^{-2\gamma d})\text{e}^{-2\gamma s}$

$\tag{37}\tilde{\kappa}\equiv\tilde{\kappa}_{ab}=\tilde{\kappa}_{ba}^*=\frac{2}{\beta d_E}\cdot\frac{h^2\gamma}{h^2+\gamma^2}\text{e}^{-\gamma s}$

and

$\tag{38}c\equiv c_{ab}=c_{ba}^*=\frac{2}{d_E}\frac{h^2}{h^2+\gamma^2}\left(s+\frac{\text{e}^{-\gamma d}}{\gamma}\right)\text{e}^{-\gamma s}$

Similar simplified formulas can be obtained for coupling between TM modes. The coupling coefficient used in the coupled-mode equation is

$\tag{39}\kappa=\frac{\tilde{\kappa}-c^*\tilde{\kappa}_{aa}}{1-|c|^2}$

Because now

$\tag{40}\beta_a+\kappa_{aa}=\beta_b+\kappa_{bb}$

we have $$\delta=0$$, and the coupling is always phase matched, as expected.

Example 4

A symmetric directional coupler for $$\lambda=1$$ μm wavelength, figure 5 below, is made by placing two strip-loaded waveguides next to one another with a separation of $$s=1$$ μm. Find the coupling length for the TM00 mode of the individual waveguide.

Because the two waveguides are coupled to each other in the $$y$$ direction, we have to examine their characteristics in that direction. In the $$y$$ direction, the characteristics of TM00 for the strip waveguide are those of the TE0 mode of the effective vertical waveguide in the effective index method, as discussed in the example in the channel waveguides tutorial. Therefore, the results in (36) - (38) above for symmetric directional coupling between TE modes apply to this problem by using the mode parameters for TM00 listed in the table of example in the channel waveguides tutorial: $$\beta$$ = 10.8120 μm-1, $$h$$ = 0.435 μm-1, and $$\gamma$$ = 0.832 μm-1.

In applying these formulas, however, we have to replace $$d$$ with $$w$$ = 5 μm and $$\text{d}_\text{E}$$ with $$w_n$$ = 7.4 μm because those are the parameters of the two identical waveguides in the $$y$$ direction. Using these parameters and $$s$$ = 1 μm, we find that $$\tilde{\kappa}_{aa}=\tilde{\kappa}_{bb}$$ = 2.7 x 10-4 μm-1, $$\tilde{\kappa}=\tilde{\kappa}_{ab}=\tilde{\kappa}_{ba}^*$$ = 1.943 x 10-3 μm-1, and $$c=c_{ab}=c_{ba}^*$$ = 2.572 x 10-2.

From (39) above, we find the coupling coefficient to be $$\kappa$$ = 1.937 x 10-3 μm-1. Therefore, the coupling length of this directional coupler is

$l_c=\frac{\pi}{2|\kappa|}=811\,\mu m$

Supermodes

The variation of the field amplitudes in the two coupled waveguides of a directional coupler as a function of propagation distance is given by (62) and (63) in the two-mode coupling tutorial, which are the solutions for codirectionally coupled modes with initial conditions $$\tilde{A}(0)\ne0$$ and $$\tilde{B}(0)=0$$.

The complete field profile across the directional coupler can be obtained as a combination of the two mode fields. Substituting (62) and (63) into (52) [all three equations from the two-mode coupling tutorial] for the mode expansion coefficients $$A(z)$$ and $$B(z)$$, we have the total field in the directional coupler:

\tag{41}\begin{aligned}\mathbf{E}(\mathbf{r})&=\tilde{A}(0)\left[\hat{\boldsymbol{\mathcal{E}}}_a\left(\cos\beta_cz-\frac{\text{i}\delta}{\beta_c}\sin\beta_cz\right)\text{e}^{\text{i}(\beta_a+\kappa_{aa}+\delta)z}+\hat{\boldsymbol{\mathcal{E}}}_b\frac{\text{i}\kappa_{ba}}{\beta_c}\sin\beta_cz\text{e}^{\text{i}(\beta_b+\kappa_{bb}-\delta)z}\right]\\&=\tilde{A}(0)\left[\frac{(\beta_c-\delta)\hat{\boldsymbol{\mathcal{E}}}_a+\kappa_{ba}\hat{\boldsymbol{\mathcal{E}}}_b}{2\beta_c}\text{e}^{\text{i}(\bar{\beta}+\beta_c)z}+\frac{(\beta_c+\delta)\hat{\boldsymbol{\mathcal{E}}}_a-\kappa_{ba}\hat{\boldsymbol{\mathcal{E}}}_b}{2\beta_c}\text{e}^{\text{i}(\bar{\beta}-\beta_c)z}\right]\\&=\tilde{A}(0)[\boldsymbol{\mathcal{E}}_1(x,y)\text{e}^{\text{i}\beta_1z}+\boldsymbol{\mathcal{E}}_2(x,y)\text{e}^{\text{i}\beta_2z}]\end{aligned}

where

$$$\tag{42}\bar{\beta}=\frac{(\beta_a+\kappa_{aa})+(\beta_b+\kappa_{bb})}{2}$$$

$$$\tag{43}\boldsymbol{\mathcal{E}}_1=\frac{(\beta_c-\delta)\hat{\boldsymbol{\mathcal{E}}}_a+\kappa_{ba}\hat{\boldsymbol{\mathcal{E}}}_b}{2\beta_c},\qquad\qquad\boldsymbol{\mathcal{E}}_2=\frac{(\beta_c+\delta)\hat{\boldsymbol{\mathcal{E}}}_a-\kappa_{ba}\hat{\boldsymbol{\mathcal{E}}}_b}{2\beta_c}$$$

and

$$$\tag{44}\beta_1=\bar{\beta}+\beta_c,\qquad\qquad\beta_2=\bar{\beta}-\beta_c$$$

We see in (41) above that the total field in the coupler is a linear combination of two independent field patterns  $$\boldsymbol{\mathcal{E}}_1(x,y)$$ and $$\boldsymbol{\mathcal{E}}_2(x,y)$$ propagating with different propagation constants $$\beta_1$$ and $$\beta_2$$, respectively. They are the normal modes of the composite two-waveguide structure of the directional coupler. Such modes are known as the supermodes of the structure. Note that $$\boldsymbol{\mathcal{E}}_1$$ and $$\boldsymbol{\mathcal{E}}_2$$ given in (43) above is not normalized.

The characteristics of the supermodes clearly depend on the parameters $$\delta,\kappa_{ab},$$ and $$\kappa_{ba}$$. We first consider an asymmetric directional coupler with nonidentical waveguides for which $$\delta\ne0$$. When $$\delta^2\gg\kappa_{ab}\kappa_{ba}$$ and $$\delta\gt0$$, we have $$\beta_c\rightarrow\delta$$. As a result, $$\beta_1\rightarrow\beta_b+\kappa_{bb}\rightarrow\beta_b$$ and $$\beta_2\rightarrow\beta_a+\kappa_{aa}\rightarrow\beta_a$$. Therefore, the supermodes of the composite structure of the directional coupler are just those of the individual waveguides when phase mismatch is large. This situation is expected because in the limit that $$\delta^2\gg\kappa_{ab}\kappa_{ba}$$, the two waveguides are effectively decoupled, and a wave propagating in either one of them is not to be affected by the existence of the other. This can be see from the fact that in this limit, (43) above reduces to

$$$\tag{45}\boldsymbol{\mathcal{E}}_1\rightarrow\frac{\kappa_{ba}}{2\delta}\hat{\boldsymbol{\mathcal{E}}}_b\approx0,\qquad\boldsymbol{\mathcal{E}}_2\rightarrow\hat{\boldsymbol{\mathcal{E}}}_a-\frac{\kappa_{ba}}{2\delta}\hat{\boldsymbol{\mathcal{E}}}_b\approx\hat{\boldsymbol{\mathcal{E}}}_a$$$

Therefore, the total field $$\mathbf{E}(\mathbf{r})$$ in (41) consists of approximately only the normal mode field of waveguide $$a$$ propagating with $$\beta_a$$:

$$$\tag{46}\mathbf{E}(\mathbf{r})\approx\tilde{A}(0)\boldsymbol{\mathcal{E}}_2\text{e}^{\text{i}\beta_2z}\approx\tilde{A}(0)\hat{\boldsymbol{\mathcal{E}}}_a\text{e}^{\text{i}\beta_az}$$$

This is the result of the fact that we have assumed the initial excitation of only waveguide $$a$$. No power is coupled to waveguide $$b$$ throughout the length of the structure because of the large phase mismatch. If we assumed the initial excitation of waveguide $$b$$ only, we would have $$\boldsymbol{\mathcal{E}}_1\rightarrow\hat{\boldsymbol{\mathcal{E}}}_b$$ and $$\boldsymbol{\mathcal{E}}_2\rightarrow0$$. Then, the wave would propagate as the normal mode of waveguide $$b$$ as if waveguide $$a$$ did not exist. In the above discussion, we have assumed that $$\beta_b\gt\beta_a$$ so that $$\delta\gt0$$. In the case of $$\beta_a\gt\beta_b$$ so that $$\delta\lt0$$, the conclusion is the same, but with the asymptotic connection of $$\boldsymbol{\mathcal{E}}_1$$ and $$\boldsymbol{\mathcal{E}}_2$$ to $$\hat{\boldsymbol{\mathcal{E}}}_b$$ and $$\hat{\boldsymbol{\mathcal{E}}}_a$$ and that of $$\beta_1$$ and $$\beta_2$$ to $$\beta_b$$ and $$\beta_a$$ simply interchanged.

In a strong coupling situation, where $$\kappa_{ab}\kappa_{ba}\gt\delta^2$$, we find from (44) and (42) that

$$$\tag{47}\beta_1\gt\beta_b+\kappa_{bb}\gt\beta_a+\kappa_{aa}\gt\beta_2$$$

if $$\delta\gt0$$. As a result, both $$\boldsymbol{\mathcal{E}}_1$$ and $$\boldsymbol{\mathcal{E}}_2$$ have significant contributions from both $$\hat{\boldsymbol{\mathcal{E}}}_a$$ and $$\hat{\boldsymbol{\mathcal{E}}}_b$$. Then, the supermodes are linear combinations of individual waveguide modes. At the input location $$z$$ = 0, the two supermodes are in phase, and

$$$\tag{48}\mathbf{E}(x,y,0)=\tilde{A}(0)(\boldsymbol{\mathcal{E}}_1+\boldsymbol{\mathcal{E}}_2)=\tilde{A}(0)\hat{\boldsymbol{\mathcal{E}}}_a$$$

as expected from the fact that only waveguide $$a$$ is initially excited. The maximum power transfer occurs when the two supermodes have a $$\pi$$ phase shift. This takes place at a distance of

$$$\tag{49}l_c=\frac{\pi}{\beta_1-\beta_2}=\frac{\pi}{2\beta_c}$$$

which is exactly the coupling length given in (67) obtained from solution of the coupled-mode equations [refer to the two-mode coupling tutorial]. At this distance, we find that the total field is

$$$\tag{50}\mathbf{E}(x,y,l_c)=\tilde{A}(0)(\boldsymbol{\mathcal{E}}_1-\boldsymbol{\mathcal{E}}_1)\text{e}^{\text{i}\beta_1l_c}=\tilde{A}(0)\frac{-\delta\hat{\boldsymbol{\mathcal{E}}}_a+\kappa_{ba}\hat{\boldsymbol{\mathcal{E}}}_b}{\beta_c}\text{e}^{\text{i}\beta_1l_c}$$$

because $$\beta_2l_c=\beta_1l_c-\pi$$. As expected, the power transfer to waveguide $$b$$ is not complete because $$\delta\ne0$$. This scenario is illustrated in figure 6(a) below.

We now consider the case of a symmetric coupler where the two waveguides are identical. Then, $$\beta_a=\beta_b$$, $$\kappa_{aa}=\kappa_{bb}$$, and $$\kappa_{ab}=\kappa_{ba}^*\equiv\kappa$$, where $$\kappa_{ab}$$ is real and positive as can be seen from (37). In addition, $$\delta$$ = 0, $$\beta_c$$ = $$\kappa$$, and $$\bar{\beta}=\beta_a+\kappa_{aa}=\beta_b+\kappa_{bb}$$. Therefore, the supermodes become the even and odd modes

$$$\tag{51}\boldsymbol{\mathcal{E}}_1=\frac{\hat{\boldsymbol{\mathcal{E}}}_a+\hat{\boldsymbol{\mathcal{E}}}_b}{2}\equiv\boldsymbol{\mathcal{E}}_\text{even},\qquad\boldsymbol{\mathcal{E}}_2=\frac{\hat{\boldsymbol{\mathcal{E}}}_a-\hat{\boldsymbol{\mathcal{E}}}_b}{2}\equiv\boldsymbol{\mathcal{E}}_\text{odd}$$$

with the following propagation constants:

$$$\tag{52}\beta_1=\bar{\beta}+\kappa\equiv\beta_\text{even},\qquad\beta_2=\bar{\beta}-\kappa\equiv\beta_\text{odd}$$$

Note again that $$\boldsymbol{\mathcal{E}}_\text{even}$$ and $$\boldsymbol{\mathcal{E}}_\text{odd}$$ given in (51) are not normalized. The total field in the coupler following initial input to waveguide $$a$$ only is then

$$$\tag{53}\mathbf{E}(\mathbf{r})=\tilde{A}(0)[\boldsymbol{\mathcal{E}}_\text{even}(x,y)\text{e}^{\text{i}\beta_\text{even}z}+\boldsymbol{\mathcal{E}}_\text{odd}(x,y)\text{e}^{\text{i}\beta_\text{odd}z}]$$$

It can be seen that the coupling length is now given by

$$$\tag{54}l_c^{PM}=\frac{\pi}{\beta_\text{even}-\beta_\text{odd}}=\frac{\pi}{2\kappa}$$$

which is consistent with that given in (86) [refer to the two-mode coupling tutorial] for phase-matched codirectional coupling.

Complete power transfer between the two waveguides is now accomplished at the coupling length because of perfect phase matching. Figure 6(b) above illustrates the evolution of supermode fields in this situation. It can be seen from (51) and figure 6(b) that the even supermode has a symmetric field pattern while the odd supermode has an antisymmetric field pattern. Because $$\kappa$$ is positive real, we also have $$\beta_\text{even}\gt\beta_\text{odd}$$.

For two waveguides of given structural parameters and index profiles, the coupling coefficient depends solely on the proximity between them, as can be seen from equations (33)-(38). As the spacing between the two waveguides is reduced, the coupling becomes stronger and, according to equation (52), the disparity between the propagation constants of the two supermodes is increased, resulting in a shorter coupling length.

Example 5

Find the propagation constants for the even and odd supermodes of the directional coupler described in example 4 above. What is the coupling length found from the beat length of these two supermodes?

Because $$c_{aa}=c_{bb}=1$$ for a directional coupler, the self-coupling coefficients can be found using (47) from the two-mode coupling tutorial and the parameters found in example 4 above as

$\kappa_{aa}=\kappa_{bb}=\frac{\tilde{\kappa}_{aa}-c\tilde{\kappa}}{1-|c|^2}=2.2\times10^{-4}\;\mu m^{-1}$

Because $$\beta_a=\beta_b=10.8120\;\mu m^{-1}$$ and $$\kappa_{aa}=\kappa_{bb}$$, we find that $$\bar{\beta}=\beta_a+\kappa_{aa}=\beta_b+\kappa_{bb}=10.812\;22\;\mu m^{-1}$$. Therefore, the propagation constants for the even and odd supermodes are, respectively

$\beta_\text{even}=\bar{\beta}+\kappa=10.814\;157\;\mu m^{-1}\text{ and }\beta_\text{odd}=\bar{\beta}-\kappa=10.810\;063\;\mu m^{-1}$

where $$\kappa=1.937\times10^{-3}\;\mu m^{-1}$$ found in example 4 above. The beat length is

$l_c^{PM}=\frac{\pi}{\beta_\text{even}-\beta_\text{odd}}=\frac{\pi}{2\kappa}=811\;\mu m$

which is exactly the coupling length found in example 4 above, as expected.

Asymmetric Directional Couplers

In a dual-channel directional coupler, the phase mismatch $$\delta$$ is determined by the symmetry between the two individual waveguides. In an asymmetric directional coupler, the two waveguides are not identical. In this case, $$\delta\ne0$$ in general. Nevertheless, it is possible to make $$\delta=0$$ for two nonidentical waveguides at a particular optical frequency, but not at all frequencies, by compensating for the difference in their thicknesses with a proper difference in their index profiles. Because of the different dispersion characteristics of the two nonidentical waveguides, $$\delta\ne0$$ at frequencies away from the phase-matching frequency. Therefore, such a coupler can be used as a frequency filter similar to the function of a DBR, but for copropagating waves rather than for contrapropagating waves as in the case of a DBR.

Similarly to the discussions for the DBR, the frequency selectivity of a phase-matched asymmetric directional coupler can be illustrated by considering the dispersion characteristics shown in figure 7 below.

The waveguides in the coupler are designed so that $$\beta_b+\kappa_{bb}=\beta_a+\kappa_{aa}$$ at a desired optical frequency $$\omega_0$$ located at the crossing of the two dispersion curves in figure 7. Therefore, $$\delta(\omega_0)=0$$ for perfect phase matching is accomplished at $$\omega_0$$. The two modes $$\hat{\boldsymbol{\mathcal{E}}}_a$$ and $$\hat{\boldsymbol{\mathcal{E}}}_b$$ are relatively well coupled within the range $$|\delta|\le|\kappa|$$, where $$|\kappa|=\sqrt{\kappa_{ab}\kappa_{ba}}$$ for coupling in the asymmetric directional coupler.

It can be seen from figure 7 that within this phase-matched range, mixing between $$\hat{\boldsymbol{\mathcal{E}}}_a$$ and $$\hat{\boldsymbol{\mathcal{E}}}_b$$ results in supermodes whose propagation constants $$\beta_1$$ and $$\beta_2$$ are different from either $$\beta_a$$ or $$\beta_b$$.

Outside this range, the modes in the two waveguides are effectively decoupled, and the supermodes effectively reduce to the individual modes. It can be seen from figure 7 that in the frequency range where $$\beta_b+\kappa_{bb}\gt\beta_a+\kappa_{aa}$$ for $$\delta\gt0$$, $$\beta_1\rightarrow\beta_b$$ and $$\beta_2\rightarrow\beta_a$$ at large phase mismatch, whereas in the range where $$\beta_b+\kappa_{bb}\lt\beta_a+\kappa_{aa}$$ for $$\delta\lt0$$, $$\beta_1\rightarrow\beta_a$$ and $$\beta_2\rightarrow\beta_b$$ at large phase mismatch. This observation is consistent with that discussed above regarding the asymptotic behavior of the supermodes at large phase mismatches.

The frequency bandwidth of the coupler can be found by considering the frequency dependence of $$\delta$$. We have

$$$\tag{55}\begin{split}\delta(\omega)&=\frac{[\beta_b(\omega)+\kappa_{bb}]-[\beta_a(\omega)+\kappa_{aa}]}{2}\\&=\frac{[\beta_b(\omega_0)+\kappa_{bb}]-[\beta_a(\omega_0)+\kappa_{aa}]}{2}+\frac{1}{2}\left(\frac{\text{d}\beta_b}{\text{d}\omega}-\frac{\text{d}\beta_a}{\text{d}\omega}\right)(\omega-\omega_0)+\cdots\\&\approx\frac{1}{2}\left(\frac{\text{d}\beta_b}{\text{d}\omega}-\frac{\text{d}\beta_a}{\text{d}\omega}\right)(\omega-\omega_0)\end{split}$$$

where the frequency dependence of $$\kappa_{aa}$$ and $$\kappa_{bb}$$ in the Taylor series expansion is ignored. It can be included if necessary.

At the phase-matching point, the length of the coupler needed for complete transfer of power from waveguide $$a$$ to waveguide $$b$$ is one of the odd multiples of $$l_c^{PM}$$:

$$$\tag{56}l=(2n+1)l_c^{PM},\qquad\;n=0,1,2,\cdots$$$

If the length is chosen to be one given in (56), we have $$\eta=1$$ at $$\delta=0$$. Then, $$\delta_{1/2}$$ for $$\eta=1/2$$ can be found from the root of the following equation:

$$$\tag{57}2\sin^2\left(|\kappa|l\sqrt{|\delta/\kappa|^2+1}\right)=|\delta/\kappa|^2+1$$$

where $$l$$ is one of those given in (56). The FWHM frequency bandwidth is then given by

$$$\tag{58}\Delta\omega=4\left|\frac{\delta_{1/2}}{\text{d}\beta_b/\text{d}\omega-\text{d}\beta_a/\text{d}\omega}\right|$$$

Rather than solving (57), the value of $$|\delta_{1/2}|$$ can also be found by reading the value of $$|\delta/\kappa|$$ for $$\eta=1/2$$ on the curve in figure 7(b) [in the two-mode coupling tutorial] corresponding to a given length of the coupler.

It is seen that for a fixed value of $$|\kappa|$$, the frequency bandwidth is narrower for a length corresponding to a higher multiple of $$l_c^{PM}$$. For example, a coupler with $$l=3l_c^{PM}$$ has a narrower bandwidth than one with $$l=l_c^{PM}$$.

By taking $$|\text{d}\beta_b/\text{d}\omega-\text{d}\beta_a/\text{d}\omega|=\Delta N_\beta/c$$, where $$\Delta N_\beta$$ is the effective group index difference between the two waveguide modes at the coupling wavelength, the bandwidth given in (58) is approximately bounded within the range:

$$$\tag{59}3.2\frac{|\kappa|c}{\Delta N_\beta}\ge\Delta\omega\gt0$$$

where the equals sign for the upper limit of the bandwidth applies only when $$l=l_c^{PM}$$ for $$|\kappa|l=\pi/2$$.

Symmetric Directional Couplers

In an ideal symmetric directional coupler, the two waveguides are identical, and the modes are always phase matched. the coupling efficiency is then simply that given by equation (85) in the two-mode coupling tutorial. For a desired coupling efficiency $$\eta_{PM}$$, the length of the coupler has to be:

$$$\tag{60}l=\frac{1}{\kappa}\left(n\pi\pm\sin^{-1}\sqrt{\eta_{PM}}\right)=2\left(n\pm\frac{1}{\pi}\sin^{-1}\sqrt{\eta_{PM}}\right)l_c^{PM}$$$

For complete transfer of power, the length of the coupler has to be exactly one of the odd multiples of $$l_c^{PM}$$ given in (56), as discussed above.

It is interesting to see that for a 50% coupling efficiency, we need $$l=(n+1/2)l_c^{PM}$$, where $$n=0,1,2,\cdots$$, and the shortest coupler length needed is exactly $$l_c^{PM}/2$$, as can be seen in figure 4(b) in the two-mode coupling tutorial. Launching optical power into one waveguide of such a coupler at its input end results in equal division of power between the two waveguides at the output end. Thus, the device functions as a 3-dB coupler or as a 50:50 power divider.

Any desired coupling efficiency can be obtained by properly choosing the length of the coupler for a given value of $$\kappa$$ or by choosing a proper value of $$\kappa$$ through the design of the coupler for a given length.

In many applications, it is often necessary to vary the coupling efficiency in the operation of a device for certain purposes. In some applications, this objective can be accomplished by altering the effective length of the coupler or the spacing between the waveguides. This approach is possible if the coupler is not integrated. An example is a directional coupler made of two closely placed optical fibers whose interaction length and spacing can be adjusted.

If the coupler has an integrated structure, it is certainly not easy to vary the length of the coupler at will, nor is it convenient to vary the spacing between the waveguides. In this situation, changes in the coupling efficiency of a coupler can be accomplished either by altering the value of $$\kappa$$ or by varying the propagation constants $$\beta_a$$ and $$\beta_b$$ in the two waveguides by different amounts to introduce a finite phase mismatch $$\delta$$ in the coupler. In fact, a change in $$\kappa$$ will necessarily result in changes in the propagation constants, and vice versa. Nevertheless, changing the value of $$\kappa$$, and thus the values of $$\beta_a$$ and $$\beta_b$$, does not necessarily result in a phase mismatch although creating a phase mismatch in an originally symmetric coupler will certainly cause changes in the coupling coefficients, as indicated by equation (44) in the coupled-mode theory tutorial. In practical devices, these changes can be created through the electro-optic effect, thus being controllable with an externally applied voltage, through nonlinear optical effects, thus being controllable with the optical power in the waveguides, or through any other effects that cause changes in the refractive index of the medium of the coupler.

The ability to vary the coupling efficiency of a directional coupler through a control signal results in many useful applications. An important example is an optical switch that functions between the cross state, in which power is completely transferred from the input channel to the other channel at the output, and the parallel state, in which power is completely passed through the input channel at the output without any transfer to the other channel. These states are illustrated in figure 8 below. The parallel state is also called the bar state

Another interesting example is the TE-TM polarization splitter illustrated in figure 9 below. It is possible to create a difference between the coupling coefficients, $$\kappa_{EE}$$ and $$\kappa_{MM}$$, of the TE and TM modes, respectively, even though the coupler has a symmetric structure.

This can be accomplished by fabricating the coupler in a birefringent medium such as LiNbO3 or a nonbirefringent electro-optic material such as GaAs and by applying a voltage to adjust properly the different refractive indices seen by the TE and TM fields. For a coupler of length $$l$$, polarization splitting as shown in figure 9, where TE polarization is in the parallel state while TM polarization is in the cross state, is achieved when

$$$\tag{61}l=\frac{n\pi}{\kappa_{EE}}=\frac{(2n\pm1)\pi}{2\kappa_{MM}}$$$

for an integer $$n$$. This is possible if there is a difference between the coupling coefficients of the two different polarizations of the amount

$$$\tag{62}\Delta\kappa\equiv|\kappa_{MM}-\kappa_{EE}|=\frac{\kappa_{EE}}{2n}$$$

For polarization splitting resulting in TE polarization in the cross state and TM polarization in the parallel state, we need $$\Delta\kappa=\kappa_{MM}/2n$$ and $$l=n\pi/\kappa_{MM}$$ instead.

The next part continues with the Surface Input and Output Couplers tutorial