From 1d700ab734aa9b6711eb31796beb25cb7659d8e0 Mon Sep 17 00:00:00 2001 From: Prefetch Date: Tue, 20 Dec 2022 20:11:25 +0100 Subject: More improvements to knowledge base --- source/know/concept/step-index-fiber/index.md | 157 ++++++++++++++------- .../step-index-fiber/transcendental-full.png | Bin 122545 -> 109224 bytes .../step-index-fiber/transcendental-half.avif | Bin 21001 -> 19600 bytes .../step-index-fiber/transcendental-half.jpg | Bin 95385 -> 84184 bytes .../step-index-fiber/transcendental-half.png | Bin 88438 -> 90521 bytes .../step-index-fiber/transcendental-half.webp | Bin 48626 -> 43374 bytes 6 files changed, 107 insertions(+), 50 deletions(-) (limited to 'source/know/concept/step-index-fiber') diff --git a/source/know/concept/step-index-fiber/index.md b/source/know/concept/step-index-fiber/index.md index c0c95d1..210a339 100644 --- a/source/know/concept/step-index-fiber/index.md +++ b/source/know/concept/step-index-fiber/index.md @@ -14,17 +14,20 @@ the transverse profile $$F(x,y)$$ of the [electric field](/know/concept/electric can be shown to obey the *Helmholtz equation* in 2D: $$\begin{aligned} - \nabla_{\!\perp}^2 F + (n^2 k^2 - \beta^2) F = 0 + \nabla_{\!\perp}^2 F + (n^2 k^2 - \beta^2) F + = 0 \end{aligned}$$ With $$n$$ being the position-dependent refractive index, -$$k$$ the vacuum wavenumber $$\omega / c$$, -and $$\beta$$ the mode's propagation constant, to be determined later. +$$k = \omega / c$$ the vacuum wavenumber, +and $$\beta$$ the mode's propagation constant (i.e. wavenumber), +to be found later. In [polar coordinates](/know/concept/cylindrical-polar-coordinates/) -$$(r,\phi)$$ this equation can be rewritten as follows: +$$(r,\phi)$$ this can be rewritten as follows: $$\begin{aligned} - \pdvn{2}{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdvn{2}{F}{\phi} + \mu F = 0 + \pdvn{2}{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdvn{2}{F}{\phi} + \mu F + = 0 \end{aligned}$$ Where we have defined $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ for brevity. @@ -37,14 +40,16 @@ such that $$F(r,\phi) = R(r) \, \Phi(\phi)$$. Inserting this ansatz: $$\begin{aligned} - R'' \Phi + \frac{1}{r} R' \Phi + \frac{1}{r^2} R \Phi'' + \mu R \Phi = 0 + R'' \Phi + \frac{1}{r} R' \Phi + \frac{1}{r^2} R \Phi'' + \mu R \Phi + = 0 \end{aligned}$$ We rearrange this such that each side only depends on one variable, by dividing by $$R\Phi$$ (ignoring the fact that it may be zero), and multiplying by $$r^2$$. Since this equation should hold for *all* values of $$r$$ and $$\phi$$, -this means that both sides must equal a constant $$\ell^2$$: +this means that both sides must equal a constant, +which we call $$\ell^2$$: $$\begin{aligned} r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2 @@ -57,11 +62,13 @@ and the well-known *Bessel equation* for $$R$$: $$\begin{aligned} \boxed{ - \Phi'' + \ell^2 \Phi = 0 + \Phi'' + \ell^2 \Phi + = 0 } \qquad \qquad \boxed{ - r^2 R'' + r R' + (\mu r^2 \!-\! \ell^2) R = 0 + r^2 R'' + r R' + (\mu r^2 \!-\! \ell^2) R + = 0 } \end{aligned}$$ @@ -70,9 +77,11 @@ simplest equation. Since the angle $$\phi$$ is limited to $$[0,2\pi]$$, $$\Phi$$ must be $$2 \pi$$-periodic, so: $$\begin{aligned} - \Phi(0) = \Phi(2\pi) + \Phi(0) + = \Phi(2\pi) \qquad \qquad - \Phi'(0) = \Phi'(2\pi) + \Phi'(0) + = \Phi'(2\pi) \end{aligned}$$ The above equation for $$\Phi$$ with these periodic boundary conditions @@ -90,13 +99,17 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case. but the challenge is to satisfy the boundary conditions: $$\begin{alignedat}{3} - \Phi(0) &= \Phi(2 \pi) + \Phi(0) + &= \Phi(2 \pi) \:\quad &&\implies \quad\:\: - 0 &&= A \sinh(2 \pi \ell) + B \big( \cosh(2 \pi \ell) - 1 \big) + 0 + &&= A \sinh(2 \pi \ell) + B \big( \cosh(2 \pi \ell) - 1 \big) \\ - \Phi'(0) &= \Phi'(2 \pi) + \Phi'(0) + &= \Phi'(2 \pi) \: \quad &&\implies \quad \:\: - 0 &&= A \ell \big( \cosh(2 \pi \ell) - 1 \big) + B \ell \sinh(2 \pi \ell) + 0 + &&= A \ell \big( \cosh(2 \pi \ell) - 1 \big) + B \ell \sinh(2 \pi \ell) \end{alignedat}$$ This only has non-trivial solutions @@ -121,13 +134,17 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case. Putting this in the boundary conditions: $$\begin{alignedat}{3} - \Phi(0) &= \Phi(2 \pi) + \Phi(0) + &= \Phi(2 \pi) \qquad &&\implies \qquad - A &&= 0 + A + &&= 0 \\ - \Phi'(0) &= \Phi'(2 \pi) + \Phi'(0) + &= \Phi'(2 \pi) \qquad &&\implies \qquad - B &&= B + B + &&= B \end{alignedat}$$ $$B$$ can be nonzero, so this a valid solution. @@ -137,13 +154,17 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case. $$\Phi(\phi) = A \sin(\phi \ell) + B \cos(\phi \ell)$$, therefore: $$\begin{alignedat}{3} - \Phi(0) &= \Phi(2 \pi) + \Phi(0) + &= \Phi(2 \pi) \quad &&\implies \quad - 0 &&= A \sin(2 \pi \ell) + B \big(\cos(2\pi \ell) - 1\big) + 0 + &&= A \sin(2 \pi \ell) + B \big(\cos(2\pi \ell) - 1\big) \\ - \Phi'(0) &= \Phi'(2 \pi) + \Phi'(0) + &= \Phi'(2 \pi) \quad &&\implies \quad - 0 &&= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell) + 0 + &&= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell) \end{alignedat}$$ This system only has nontrivial solutions @@ -163,13 +184,17 @@ and check if we can then arrive at a non-trivial $$\Phi$$ for each case. We revisit the boundary conditions and indeed see: $$\begin{alignedat}{3} - 0 &= A \sin(2 \pi \ell) + B \big(\cos(2 \pi \ell) - 1\big) + 0 + &= A \sin(2 \pi \ell) + B \big(\cos(2 \pi \ell) - 1\big) \qquad &&\implies \qquad - 0 &&= 0 + 0 + &&= 0 \\ - 0 &= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell) + 0 + &= A \big(\cos(2 \pi \ell) - 1\big) - B \sin(2 \pi \ell) \qquad &&\implies \qquad - 0 &&= 0 + 0 + &&= 0 \end{alignedat}$$ So $$A$$ and $$B$$ are *both* unconstrained, @@ -185,7 +210,8 @@ we get the following for $$\ell = 0, 1, 2, ...$$: $$\begin{aligned} \boxed{ - \Phi_\ell(\phi) = A \cos(\phi \ell) + \Phi_\ell(\phi) + = A \cos(\phi \ell) } \end{aligned}$$ @@ -198,7 +224,8 @@ Let us now revisit the Bessel equation for the radial function $$R(r)$$, which should be continuous and differentiable throughout the fiber: $$\begin{aligned} - r^2 R'' + r R' + \mu r^2 R - \ell^2 R = 0 + r^2 R'' + r R' + \mu r^2 R - \ell^2 R + = 0 \end{aligned}$$ To continue, we need to specify the refractive index $$n(r)$$, contained in $$\mu(r)$$. @@ -231,11 +258,13 @@ Let $$\pm$$ be the sign of $$\mu$$: $$\begin{aligned} \begin{cases} \displaystyle - 0 = \rho^2 \pdvn{2}{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R + 0 + = \rho^2 \pdvn{2}{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R & \mathrm{for}\; \mu \neq 0 \\ \displaystyle - 0 = r^2 \pdvn{2}{R}{r} + r \pdv{R}{r} - \ell^2 R + 0 + = r^2 \pdvn{2}{R}{r} + r \pdv{R}{r} - \ell^2 R & \mathrm{for}\; \mu = 0 \end{cases} \end{aligned}$$ @@ -249,7 +278,8 @@ Of the remaining candidates, $$\ln(r)$$, $$r^\ell$$ and $$I_\ell(\rho)$$ do not leading to the following $$R_o$$: $$\begin{aligned} - R_{o,\ell}(r) = + R_{o,\ell}(r) + = \begin{cases} r^{-\ell} & \mathrm{for}\; \mu = 0 \;\mathrm{and}\; \ell = 1,2,3,... @@ -267,8 +297,9 @@ at the boundary $$r = a$$, so they can never be continuous with $$R_o'$$. This leaves $$J_\ell(\rho)$$ for $$\mu > 0$$: $$\begin{aligned} - R_{i,\ell}(r) = - J_\ell(\rho) = J_\ell(r \sqrt{\mu}) + R_{i,\ell}(r) + = J_\ell(\rho) + = J_\ell(r \sqrt{\mu}) \qquad \mathrm{for}\; \mu > 0 \;\mathrm{and}\; \ell = 0,1,2,... \end{aligned}$$ @@ -295,7 +326,8 @@ Since $$\mu \equiv n^2 k^2 \!-\! \beta^2$$ by definition, this discovery places a constraint on the propagation constant $$\beta$$: $$\begin{aligned} - n_i^2 k^2 > \beta^2 \ge n_o^2 k^2 + n_i^2 k^2 > \beta^2 + \ge n_o^2 k^2 \end{aligned}$$ Therefore, $$n_i > n_o$$ in a step-index fiber, @@ -304,7 +336,7 @@ the fiber is not able to guide the light outside this range. However, not all $$\beta$$ in this range are created equal for all $$k$$. To investigate further, let us define the quantities -$$\xi_\mathrm{core}$$ and $$\xi_\mathrm{clad}$$ like so, +$$\xi_i$$ and $$\xi_o$$ like so, assuming $$n_i$$ and $$n_o$$ do not depend on $$k$$: $$\begin{aligned} @@ -318,7 +350,8 @@ $$\begin{aligned} It is important to note that the sum of their squares is constant with respect to $$\beta$$: $$\begin{aligned} - \xi_i^2 + \xi_o^2 = (\mathrm{NA})^2 k^2 + \xi_i^2 + \xi_o^2 + = (\mathrm{NA})^2 k^2 \end{aligned}$$ Where $$\mathrm{NA}$$ is the so-called **numerical aperture**, @@ -333,7 +366,7 @@ $$\begin{aligned} \end{aligned}$$ From this, we define a new fiber parameter: the $$V$$-**number**, -which is extremely useful: +which is very useful: $$\begin{aligned} \boxed{ @@ -360,16 +393,19 @@ meanwhile defining $$X \equiv a \xi_i$$ and $$Y \equiv a \xi_o$$ for convenience, such that $$X^2 + Y^2 = V^2$$: $$\begin{aligned} - X \frac{J_\ell'(X)}{J_\ell(X)} = Y \frac{K_\ell'(Y)}{K_\ell(Y)} + X \frac{J_\ell'(X)}{J_\ell(X)} + = Y \frac{K_\ell'(Y)}{K_\ell(Y)} \end{aligned}$$ We can turn this result into something a bit nicer by using the following identities: $$\begin{aligned} - J_\ell'(x) = -J_{\ell+1}(x) + \ell \frac{J_\ell(x)}{x} - \qquad \quad - K_\ell'(x) = -K_{\ell+1}(x) + \ell \frac{K_\ell(x)}{x} + J_\ell'(x) + = - \ell \frac{J_\ell(x)}{x} + J_{\ell-1}(x) + \qquad \qquad + K_\ell'(x) + = - \ell \frac{K_\ell(x)}{x} - K_{\ell-1}(x) \end{aligned}$$ With this, the transcendental equation for $$\beta$$ @@ -377,13 +413,14 @@ takes this convenient form: $$\begin{aligned} \boxed{ - X \frac{J_{\ell+1}(X)}{J_\ell(X)} = Y \frac{K_{\ell+1}(Y)}{K_\ell(Y)} + X \frac{J_{\ell-1}(X)}{J_\ell(X)} + = - Y \frac{K_{\ell-1}(Y)}{K_\ell(Y)} } \end{aligned}$$ All $$\beta$$ that satisfy this indicate the existence -of a **linearly polarized** mode. -These modes are called $$\mathrm{LP}_{\ell m}$$, +of a **linearly polarized mode**, +each labelled $$\mathrm{LP}_{\ell m}$$, where $$\ell$$ is the primary (azimuthal) mode index, and $$m$$ the secondary (radial) mode index, which is needed because multiple $$\beta$$ may exist for a single $$\ell$$. @@ -394,12 +431,32 @@ where red and blue respectively denote the left and right-hand side: {% include image.html file="transcendental-full.png" width="100%" alt="Graphical solution of transcendental equation" %} -This shows that each $$\mathrm{LP}_{\ell m}$$ has an associated cut-off $$V_{\ell m}$$, -so that if $$V > V_{\ell m}$$ then $$\mathrm{LP}_{lm}$$ exists, +For the ground state the light is well-confined in the core, +but for higher modes it increasingly leaks into the cladding, +thereby reducing the wavenumber $$\beta$$ (because $$n_i > n_o$$) +until the fiber can no longer guide the light $$\beta < n_o k$$, +and the modes thus stop existing. +Therefore, there is a mode cutoff when $$\beta = n_o k$$ +or equivalently when $$\xi_o = 0$$. +With this is mind, consider a slightly rearranged version +of the above transcendental equation: + +$$\begin{aligned} + X J_{\ell-1}(X) K_\ell(Y) + = - Y K_{\ell-1}(Y) J_\ell(X) +\end{aligned}$$ + +Because $$Y \equiv a \xi_o$$ and $$X^2 = V^2 - Y^2$$, +if $$\xi_o = 0$$ then $$Y = 0$$ and $$X = V$$, +and the right-hand side is zero; +for it to be satisfiable (i.e. for the mode to exist), +the other side should also vanish. +Therefore, all $$\mathrm{LP}_{\ell m}$$ have cutoffs $$V_{\ell m}$$ +equal to the $$m$$th roots of $$J_{\ell-1}(V_{\ell m}) = 0$$, +so if $$V > V_{\ell m}$$ then $$\mathrm{LP}_{\ell m}$$ exists, as long as $$\beta$$ stays in the allowed range. -The cut-offs of the secondary modes for a given $$\ell$$ -are found as the $$m$$th roots of $$J_{\ell-1}(V_{\ell m}) = 0$$. -In the above figure, they are $$V_{01} = 0$$, $$V_{11} = 2.405$$, and $$V_{02} = V_{21} = 3.832$$. +In the above figure, they are $$V_{01} = 0$$, +$$V_{11} = 2.405$$, and $$V_{02} = V_{21} = 3.832$$. 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