## Abstract

Coupling losses of rectangular waveguide resonators are discussed in this paper in terms of fourier analysis theorem. Compared to the traditional time-consuming method, the scheme presented in this paper will decrease the simulation time considerably. Under the conditions given in the paper, the EH 11-mode coupling coefficient is calculated numerically. The conclusions can be applied to higher-order mode.

## 1. Introduction

Many study on the coupling losses of waveguide have been done theoretically in detail by D.R.Hall et al[13]. Coupling losses is usually represented as the value of coupling coefficient, which in general is obtained using diffraction integral method in space-domain. As this approach requires calculation on double integral[4] which is much time-consuming, in order to reduce the simulation time, it can only handle the three special cases, viz., semiconfocal, half-concentric and plane-parallel geometries.

This article proposed a new method to analyze waveguide resonator. As will be seen later, the new method avoid calculating double integral using fourier analysis in frequency-domain. When apply to the three cases mentioned above, the approach obtains the same results as in space domain case, moreover, it gives more detailed analysis than the conventional one.

## 2. Theoretical analysis

Consider the following example shown in Fig.l. A spherical mirror with dimension of b ×b and radius of curvature R is put at a distance d from a rectangular waveguide of dimension 4 2a×2b, which coincides with the z-axis.

Assume that the reflectivity of the spherical mirror equals to l, the electric field is X-polarized(the results are qualitatively the same for Y-polarized case) and there is only EH 11 mode in the rectangular waveguide, then we get

Fig. 1. Schematic of the rectangular waveguide

$EH11(x,y,z=0)=(a×b)−12·cos(π×x2a)·cos(π×y2b)$

apply fourier analysis[5] to Eq. (1), yielding

$EH11(x,y,z=0)=∫−∞∞∫−∞∞A1(fX,fY)·exp[2jπ(fX·x+fY·y)]dfXdfY$

where

$A1(fX,fY)=∫−aa∫−bbEH11(x,y,z=0)·exp[−2jπ(fX·x+fY·y)]dxdy$
$=(a×b)−12·−π·cos(2πfX·a)a×[(2πfX)2−(π2a)2]·−π·cos(2πfY·b)b×[(2πfY)2−(π2b)2]$

Eq. (3) shows that the wave can be expressed as the sum of a series of plane waves, where α and β are determined by cos α=fx ·λ,cosβ=fy ·λ, and the amplitude is determined by A 1(fx , fy ). In face, when cos2α+cos2 β>1, the z component of plane wave attenuates exponentially, so the corresponding component is ignored in our discussion. Equivalently, we consider the spheric mirror as a lens of focal length equals to R/2, as shown in Fig.2, when the constant phase factor is ignored, the complex amplitude permeation ratio is given by

$t(x,y)=p(x,y)·exp[−jk2f0(x2+y2)]=p(x,y)·exp[−jkR(x2+y2)]$

Fig.2. Equavalent schematic of the waveguide and mirror

where aperture function

$p(x,y)={1−b′, k=2π/λ. According to the fourier analysis theorem, we denote the angular spectrum of U 1, U 2 by term A 1, A 2 respectively. When the Fresnel condition is satisfied, A 2 can be written in the form

$A2(fX,fY)=A1(fX,fY)·exp(jk·d)·exp[−jk·d·(λ2·fX2+λ2·fY22)]$

Based on the Fourier transform properties, we obtain the returned field distribution function U 3 at the waveguide entrance, viz.,

$U3=1jλ·d·exp(jk·d)·exp[jk2d·(x2+y2)]·F{U2·t·exp[jk2d·(x2+y2)]}∣fY=yλ·dfX=xλ·d$

where the Fourier transform is denoted as F{….},then

$U3=1jλ·d·exp(jk·d)·exp[jk2d·(x2+y2)]·{A2*F{p(x,y)}*F{exp[jC·(x2+y2)]}}∣fY·=yλ·dfX=xλ·d$

where*denotes convolution integral, the constant $C=k·(12d−1R)$. As the x,y components of U 3 are decomposible, we rewrite U 3 in the form of U 3 (x, y)=U 3 (xU 3 (y),and similarly

P(x, y)=P(xP(y);

A 1(fX ,fY )=A 1(fXA l(fY );A 2(fX ,fY )=A 2(fXA 2(fY );

EH 11 (x,y,z=0)=EH 11 (x,z=0)·EH 11 (Y,z=0)

Let $aλ=m$, $da=n$, $2dR=q$, $b′a=l$, $bλ=m′$, $db=n′$, $b′b=l′$, the Fresnel number in X direction can be written as $N1=a2λ·d=m/n$, the Fresnel number in Y direction is $N2=b2λ·d=m′/n′$. coupling coefficent is

$∣C11(N1,N2)∣2=∣C11(N1)∣2·∣C11(N2)∣2$

where

$|C11(N1)|2=|∫−aaU3(x)⋅EH11(x,z=0)dx|2;|C11(N2)|2=|∫−aaU3(y)⋅EH11(y,z=0)dy|2$

For simplicity, we only consider U 3 (x) in the following discussion, and for the sake of clarity, the constant phase factor is ignored as it has no effect on our conclusion, we can derive from Eq. (7)

$U3(x)=1λ·d·exp(jk2dx2)·{A2(fX)*F{p(x)}*F{exp(jCx2)}}∣fX=xλ·d$

1f we assume b >>a, viz., l>>1,then we have,

$A2(fX)*F{p(x)}*F{exp(jCx2)}≈A2(fX)*F{exp(jCx2)}$
$A2(fX)*F{p(x)}*F{exp(jCx2)}=4aπ·cos(2πa·fX)(4a·fX)2−1·exp(−jk·d·λ22·fX2)*F{exp(jCx2)}$
$U3(x)=4·exp(jk2dx2)πλ·da·{cos(2πa·fX)·exp(−jk·d·λ22·fX2)(4a·fX)2−1*F{exp(jCx2)}}∣fX=xλ·d$

Follows, we will discuss the formula (11) under two different conditions.

I For R=2d,as C=0 in this case,we obtain from Eq. (11)

$U3(x)=4πλ·da·cos(2πN1·xa)(4N1·xa)2−1$

and

$∣C11(N1)∣2=∣∫−aaU3(x)·EH11(x,z=0)dx∣2==∣∫018N1π·cos(2πN1·x)·cos(πx2)(4N1·x)2−1dx∣2$

further analysis will prove that $∣C11(N1)∣2∝1N1$ for large N 1.

II or R≠2d, with different range of Fresnel number N 1

(1) For N 1, satisfying $N1>25·∣2−q1−q∣$, we derive from Eq. (11)

$C11(N1)=∫0111−q·(sign(1−x1−q)+sign(1+x1−q))·cos(π(1−q)2·x)·exp(jπN1·q·x2q−1)·cos(πx2)dx$

where sign(..) denotes the signum function.

(2) For N 1, satisfying $N1<∣q−1∣20∣q∣$, we have

$C11(N1)=∫0532π2·(1−q)·1−q·exp(−jπ·x2N1·2−q1−q)·cos(2π·x1−q)·cos(2π·x)[16·x2−(1−q)2]·(16·x2−1)dx$

As shown in Eq.(8), coupling coefficent can be written as

$∣C11(N1,N2)∣2=∣C11(N1)∣2·∣C11(N2)∣2=∣C11(N)∣2$

where $N=N1·N2$ denotes the systemic Fresnel number. When simulated with adaptive Newton Cotes numerical integral method of variable step, the integral precision is 1×10-6. For square waveguide, a=b, viz., N=N 1=N 2. It is desirable to discuss our results in the form of different values of q.

A q=1,Numerical simulation result for Eqs. (13),(16) is shown in Fig.(3).1t is evident from the plot that the Fresnel number N should be small enough to achieve large coupling coefficient |C 11 (N)|2.

B q=2,as N 1 >0 satisfies condition $N1>25·∣2−q1−q∣$ in this case, numerical simulation result for Eqs. (14),(16) is shown in Fig.(4).Likewise, the Fresnel number N should be small enough to achieve large coupling coefficient in co-focal geometry case.

C q=0, as N<∞ satisfies condition $N1<∣q−1∣20∣q∣$, numerical simulation result for

Fig.3. EH 11 mode-coupling coefficient as a function of the Fresnel number N at q=1 for the square waveguide

Fig.4. EH 11 mode-coupling coefficient as a function of the Fresnel number N at q=2 for the square waveguide

Fig. 5. EH 11 mode-coupling coefficient as a function of the Fresnel number N at q=0 for the square waveguide

Eqs. (15),(16) is shown in Fig.(5).It is evident for the plane-parallel case that if the Fresnel number N is large enough then large coupling coefficient |C 11 (N)|2 will be obtained, and the distance d between the mirror and the waveguide could be smaller accordingly, which is desirable in the design of compact waveguide laser.

Fig.6. EH 11 mode-coupling coefficient as a function of the Fresnel number N at several values of q for the square waveguide

Fig.7. EH 11 mode-coupling coefficient as a function of the Fresnel number N at several values of q for the square waveguide

D As similar analysis in b, c cases, we obtain for different values of q a number of curves in which the coupling coefficient |C 11 (N)|2 is plotted as a function of the Fresnel number N as shown in Figs.(6),(7),(8),respectively. It turns out that the coupling coefficient |C 11 (N)|2 increases as the value of q decrease for large Fresnel number N.

Fig.8. EH 11 mode-coupling coefficient as a function of the Fresnel number N at q=0.001 for the square waveguide

Fig.9. EH 11 mode-coupling coefficient as a function of the Fresnel number N 1 and N 2 at q=0

The analysis mentioned above can apply to the broad waveguide, we will not discuss the case in details here, only the three-dimension distribution curves of coupling coefficient plotted as a function of parameters N 1, N 2 are shown in Figs.(9),(10),(11),respectively.

Fig.10. EH 11 mode-coupling coefficient as a function of the Fresnel number N 1 and N 2 at q=1

Fig. 11. EH 11 mode-coupling coefficient as a function of the Fresnel number N 1 and N 2 at q=2

## 3. Conclusion

For a given parameter q, one can use the fomula (13) or (14) to calculate |C 11(N 1)|2 and |C 11(N 2)|2 under different conditions $N1,2>25·∣2−q1−q∣$ or $N1,2<∣1−q∣20∣q∣$ chosen by the different values of Fresnel number N 1,2 in X or Y direction, respectively. As have been seen in the previous section, a compact design of the rectangular waveguide laser requires that we should use the plate-parallel geometry to achieve low coupling losses. The discussion above is for the EH 11 mode, higher-order mode cases will be handle in the same manner. As a comparison with the diffraction integral method in space-domain, the most benefit using fourier analysis in frequency-domain is that it avoid double integral calculation which in turn not only simplifies the numerical simulation and also improves the precision.

1. J.W. Goodman, Introduction to Fourier Optics, (Second Edition, McGraw-Hill, New York, 1996)

2. D.R. Hall and H.J. Baker, Laser Focus World. 10, 77 (1989)

3. C.A. Hill and D.R. Hall, “Coupling Loss theory of single-mode waveguide resonators,” Appl. Opt. 24, 1283–1290 (1985). [CrossRef]   [PubMed]

4. J.J. Degnan and D.R. Hall, “Finite-aperture waveguide-lasers resonators,” IEEE J.Quant.Electron. QE-9, 901–910 (1973). [CrossRef]

5. W. Xinbing, X. Qiyang, X. Minjie, and L. Zaiguang, “Coupling Losses and mode properties in planar waveguide resonators,” Opt.Commun. 131, 41–46 (1996). [CrossRef]

### References

• View by:

1. J.W. Goodman, Introduction to Fourier Optics, (Second Edition, McGraw-Hill, New York, 1996)
2. D.R. Hall and H.J. Baker, Laser Focus World. 10, 77 (1989)
3. C.A. Hill and D.R. Hall, “Coupling Loss theory of single-mode waveguide resonators,” Appl. Opt. 24, 1283–1290 (1985).
[Crossref] [PubMed]
4. J.J. Degnan and D.R. Hall, “Finite-aperture waveguide-lasers resonators,” IEEE J.Quant.Electron. QE-9, 901–910 (1973).
[Crossref]
5. W. Xinbing, X. Qiyang, X. Minjie, and L. Zaiguang, “Coupling Losses and mode properties in planar waveguide resonators,” Opt.Commun. 131, 41–46 (1996).
[Crossref]

#### 1996 (1)

W. Xinbing, X. Qiyang, X. Minjie, and L. Zaiguang, “Coupling Losses and mode properties in planar waveguide resonators,” Opt.Commun. 131, 41–46 (1996).
[Crossref]

#### 1989 (1)

D.R. Hall and H.J. Baker, Laser Focus World. 10, 77 (1989)

#### 1973 (1)

J.J. Degnan and D.R. Hall, “Finite-aperture waveguide-lasers resonators,” IEEE J.Quant.Electron. QE-9, 901–910 (1973).
[Crossref]

#### Baker, H.J.

D.R. Hall and H.J. Baker, Laser Focus World. 10, 77 (1989)

#### Degnan, J.J.

J.J. Degnan and D.R. Hall, “Finite-aperture waveguide-lasers resonators,” IEEE J.Quant.Electron. QE-9, 901–910 (1973).
[Crossref]

#### Goodman, J.W.

J.W. Goodman, Introduction to Fourier Optics, (Second Edition, McGraw-Hill, New York, 1996)

#### Hall, D.R.

D.R. Hall and H.J. Baker, Laser Focus World. 10, 77 (1989)

J.J. Degnan and D.R. Hall, “Finite-aperture waveguide-lasers resonators,” IEEE J.Quant.Electron. QE-9, 901–910 (1973).
[Crossref]

#### Minjie, X.

W. Xinbing, X. Qiyang, X. Minjie, and L. Zaiguang, “Coupling Losses and mode properties in planar waveguide resonators,” Opt.Commun. 131, 41–46 (1996).
[Crossref]

#### Qiyang, X.

W. Xinbing, X. Qiyang, X. Minjie, and L. Zaiguang, “Coupling Losses and mode properties in planar waveguide resonators,” Opt.Commun. 131, 41–46 (1996).
[Crossref]

#### Xinbing, W.

W. Xinbing, X. Qiyang, X. Minjie, and L. Zaiguang, “Coupling Losses and mode properties in planar waveguide resonators,” Opt.Commun. 131, 41–46 (1996).
[Crossref]

#### Zaiguang, L.

W. Xinbing, X. Qiyang, X. Minjie, and L. Zaiguang, “Coupling Losses and mode properties in planar waveguide resonators,” Opt.Commun. 131, 41–46 (1996).
[Crossref]

#### IEEE J.Quant.Electron. (1)

J.J. Degnan and D.R. Hall, “Finite-aperture waveguide-lasers resonators,” IEEE J.Quant.Electron. QE-9, 901–910 (1973).
[Crossref]

#### Laser Focus World. (1)

D.R. Hall and H.J. Baker, Laser Focus World. 10, 77 (1989)

#### Opt.Commun. (1)

W. Xinbing, X. Qiyang, X. Minjie, and L. Zaiguang, “Coupling Losses and mode properties in planar waveguide resonators,” Opt.Commun. 131, 41–46 (1996).
[Crossref]

#### Other (1)

J.W. Goodman, Introduction to Fourier Optics, (Second Edition, McGraw-Hill, New York, 1996)

### Cited By

Optica participates in Crossref's Cited-By Linking service. Citing articles from Optica Publishing Group journals and other participating publishers are listed here.

### Figures (11)

Fig. 1. Schematic of the rectangular waveguide
Fig.2. Equavalent schematic of the waveguide and mirror
Fig.3. EH 11 mode-coupling coefficient as a function of the Fresnel number N at q =1 for the square waveguide
Fig.4. EH 11 mode-coupling coefficient as a function of the Fresnel number N at q=2 for the square waveguide
Fig. 5. EH 11 mode-coupling coefficient as a function of the Fresnel number N at q = 0 for the square waveguide
Fig.6. EH 11 mode-coupling coefficient as a function of the Fresnel number N at several values of q for the square waveguide
Fig.7. EH 11 mode-coupling coefficient as a function of the Fresnel number N at several values of q for the square waveguide
Fig.8. EH 11 mode-coupling coefficient as a function of the Fresnel number N at q =0.001 for the square waveguide
Fig.9. EH 11 mode-coupling coefficient as a function of the Fresnel number N 1 and N 2 at q = 0
Fig.10. EH 11 mode-coupling coefficient as a function of the Fresnel number N 1 and N 2 at q =1
Fig. 11. EH 11 mode-coupling coefficient as a function of the Fresnel number N 1 and N 2 at q =2

### Equations (19)

$EH 11 ( x , y , z = 0 ) = ( a × b ) − 1 2 · cos ( π × x 2 a ) · cos ( π × y 2 b )$
$EH 11 ( x , y , z = 0 ) = ∫ − ∞ ∞ ∫ − ∞ ∞ A 1 ( f X , f Y ) · exp [ 2 jπ ( f X · x + f Y · y ) ] df X df Y$
$A 1 ( f X , f Y ) = ∫ − a a ∫ − b b EH 11 ( x , y , z = 0 ) · exp [ − 2 jπ ( f X · x + f Y · y ) ] dx dy$
$= ( a × b ) − 1 2 · − π · cos ( 2 πf X · a ) a × [ ( 2 πf X ) 2 − ( π 2 a ) 2 ] · − π · cos ( 2 πf Y · b ) b × [ ( 2 πf Y ) 2 − ( π 2 b ) 2 ]$
$t ( x , y ) = p ( x , y ) · exp [ − j k 2 f 0 ( x 2 + y 2 ) ] = p ( x , y ) · exp [ − j k R ( x 2 + y 2 ) ]$
$A 2 ( f X , f Y ) = A 1 ( f X , f Y ) · exp ( jk · d ) · exp [ − jk · d · ( λ 2 · f X 2 + λ 2 · f Y 2 2 ) ]$
$U 3 = 1 jλ · d · exp ( jk · d ) · exp [ j k 2 d · ( x 2 + y 2 ) ] · F { U 2 · t · exp [ j k 2 d · ( x 2 + y 2 ) ] } ∣ f Y = y λ · d f X = x λ · d$
$U 3 = 1 jλ · d · exp ( jk · d ) · exp [ j k 2 d · ( x 2 + y 2 ) ] · { A 2 * F { p ( x , y ) } * F { exp [ jC · ( x 2 + y 2 ) ] } } ∣ f Y · = y λ · d f X = x λ · d$
$∣ C 11 ( N 1 , N 2 ) ∣ 2 = ∣ C 11 ( N 1 ) ∣ 2 · ∣ C 11 ( N 2 ) ∣ 2$
$| C 11 ( N 1 ) | 2 = | ∫ − a a U 3 ( x ) ⋅ E H 11 ( x , z = 0 ) d x | 2 ; | C 11 ( N 2 ) | 2 = | ∫ − a a U 3 ( y ) ⋅ E H 11 ( y , z = 0 ) d y | 2$
$U 3 ( x ) = 1 λ · d · exp ( j k 2 d x 2 ) · { A 2 ( f X ) * F { p ( x ) } * F { exp ( jCx 2 ) } } ∣ f X = x λ · d$
$A 2 ( f X ) * F { p ( x ) } * F { exp ( j Cx 2 ) } ≈ A 2 ( f X ) * F { exp ( j Cx 2 ) }$
$A 2 ( f X ) * F { p ( x ) } * F { exp ( j Cx 2 ) } = 4 a π · cos ( 2 πa · f X ) ( 4 a · f X ) 2 − 1 · exp ( − j k · d · λ 2 2 · f X 2 ) * F { exp ( j Cx 2 ) }$
$U 3 ( x ) = 4 · exp ( j k 2 d x 2 ) π λ · d a · { cos ( 2 π a · f X ) · exp ( − j k · d · λ 2 2 · f X 2 ) ( 4 a · f X ) 2 − 1 * F { exp ( j C x 2 ) } } ∣ f X = x λ · d$
$U 3 ( x ) = 4 π λ · d a · cos ( 2 π N 1 · x a ) ( 4 N 1 · x a ) 2 − 1$
$∣ C 11 ( N 1 ) ∣ 2 = ∣ ∫ − a a U 3 ( x ) · EH 11 ( x , z = 0 ) dx ∣ 2 == ∣ ∫ 0 1 8 N 1 π · cos ( 2 π N 1 · x ) · cos ( πx 2 ) ( 4 N 1 · x ) 2 − 1 dx ∣ 2$
$C 11 ( N 1 ) = ∫ 0 1 1 1 − q · ( sign ( 1 − x 1 − q ) + sign ( 1 + x 1 − q ) ) · cos ( π ( 1 − q ) 2 · x ) · exp ( j π N 1 · q · x 2 q − 1 ) · cos ( πx 2 ) dx$
$C 11 ( N 1 ) = ∫ 0 5 32 π 2 · ( 1 − q ) · 1 − q · exp ( − j π · x 2 N 1 · 2 − q 1 − q ) · cos ( 2 π · x 1 − q ) · cos ( 2 π · x ) [ 16 · x 2 − ( 1 − q ) 2 ] · ( 16 · x 2 − 1 ) dx$
$∣ C 11 ( N 1 , N 2 ) ∣ 2 = ∣ C 11 ( N 1 ) ∣ 2 · ∣ C 11 ( N 2 ) ∣ 2 = ∣ C 11 ( N ) ∣ 2$