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diff --git a/content/know/concept/step-index-fiber/index.pdc b/content/know/concept/step-index-fiber/index.pdc new file mode 100644 index 0000000..8847fff --- /dev/null +++ b/content/know/concept/step-index-fiber/index.pdc @@ -0,0 +1,427 @@ +--- +title: "Step-index fiber" +firstLetter: "S" +publishDate: 2022-02-11 +categories: +- Physics +- Optics +- Fiber optics + +date: 2022-01-31T19:29:33+01:00 +draft: false +markup: pandoc +--- + +# Step-index fiber + +As light propagates in the $z$-direction through an optical fiber, +the transverse profile $F(x,y)$ of the [electric field](/know/concept/electric-field/) +can be shown to obey the *Helmholtz equation* in 2D: + +$$\begin{aligned} + \nabla_{\!\perp}^2 F + (n^2 k^2 - \beta^2) F = 0 +\end{aligned}$$ + +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. +In [polar coordinates](/know/concept/cylindrical-polar-coordinates/) +$(r,\phi)$ this equation can be rewritten as follows: + +$$\begin{aligned} + \pdv[2]{F}{r} + \frac{1}{r} \pdv{F}{r} + \frac{1}{r^2} \pdv[2]{F}{\phi} + \mu F = 0 +\end{aligned}$$ + +Where we have defined $\mu \equiv n^2 k^2 \!-\! \beta^2$ for brevity. +From now on, we only consider choices of $\mu$ that do not depend on $\phi$ or $z$, +but may vary with $r$. + +This Helmholtz equation can be solved by *separation of variables*: +we assume that there exist two functions $R(r)$ and $\Phi(\phi)$ +such that $F(r,\phi) = R(r) \, \Phi(\phi)$. +Inserting this ansatz: + +$$\begin{aligned} + R'' \Phi + \frac{1}{r} R' \Phi + \frac{1}{r^2} R \Phi'' + \mu R \Phi = 0 +\end{aligned}$$ + +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 $\ell^2$: + +$$\begin{aligned} + r^2 \frac{R''}{R} + r \frac{R'}{R} + \mu r^2 + = -\frac{\Phi''}{\Phi} + = \ell^2 +\end{aligned}$$ + +This gives an eigenvalue problem for $\Phi$, +and the well-known *Bessel equation* for $R$: + +$$\begin{aligned} + \boxed{ + \Phi'' + \ell^2 \Phi = 0 + } + \qquad \qquad + \boxed{ + r^2 R'' + r R' + (\mu r^2 \!-\! \ell^2) R = 0 + } +\end{aligned}$$ + +We will return to $R$ later; we start with $\Phi$, because it has the +simplest equation. Since the angle $\phi$ is limited to $[0,2\pi]$, +$\Phi$ must be $2 \pi$-periodic, so: + +$$\begin{aligned} + \Phi(0) = \Phi(2\pi) + \qquad \qquad + \Phi'(0) = \Phi'(2\pi) +\end{aligned}$$ + +The above equation for $\Phi$ with these periodic boundary conditions +is a [Sturm-Liouville problem](/know/concept/sturm-liouville-theory/). +Consequently, there are infinitely many allowed values of $\ell^2$, +all real, and one of them is lowest, known as the *ground state*. + +To find the eigenvalues $\ell^2$ and their corresponding $\Phi$, +we in turn assume that $\ell^2 < 0$, $\ell^2 = 0$, or $\ell^2 > 0$, +and check if we can then arrive at a non-trivial $\Phi$ for each case. + +* For $\ell^2 < 0$, solutions have the form $\Phi(\phi) = A \sinh\!(\phi \ell) + B \cosh\!(\phi \ell)$, + where $A$ and $B$ are unknown linearity constants. + At least one of these constants must be nonzero for $\Phi$ to be non-trivial, + but the challenge is to satisfy the boundary conditions: + + $$\begin{alignedat}{3} + \Phi(0) &= \Phi(2 \pi) + \:\quad &&\implies \quad\:\: + 0 &&= A \sinh\!(2 \pi \ell) + B \big( \cosh\!(2 \pi \ell) - 1 \big) + \\ + \Phi'(0) &= \Phi'(2 \pi) + \: \quad &&\implies \quad \:\: + 0 &&= A \ell \big( \cosh\!(2 \pi \ell) - 1 \big) + B \ell \sinh\!(2 \pi \ell) + \end{alignedat}$$ + + This only has non-trivial solutions + if the determinant of the system matrix is zero: + + $$\begin{aligned} + 0 + &= \mathrm{det} + \begin{bmatrix} + \sinh\!(2 \pi \ell) & \cosh\!(2 \pi \ell) - 1 \\ + \cosh\!(2 \pi \ell) - 1 & \sinh\!(2 \pi \ell) + \end{bmatrix} + = 2 \big( \cosh\!(2 \pi \ell) - 1 \big) + \end{aligned}$$ + + This can only be zero if $\ell = 0$, + which contradicts the premise that $\ell^2 < 0$, + so we conclude that $\ell^2$ cannot be negative, + because no non-trivial solutions exist here. + +* For $\ell^2 = 0$, the solution is $\Phi(\phi) = A \phi + B$. + Putting this in the boundary conditions: + + $$\begin{alignedat}{3} + \Phi(0) &= \Phi(2 \pi) + \qquad &&\implies \qquad + A &&= 0 + \\ + \Phi'(0) &= \Phi'(2 \pi) + \qquad &&\implies \qquad + B &&= B + \end{alignedat}$$ + + $B$ can be nonzero, so this a valid solution. + We conclude that $\ell^2 = 0$ is the ground state. + +* For $\ell^2 > 0$, all solutions have the form + $\Phi(\phi) = A \sin\!(\phi \ell) + B \cos\!(\phi \ell)$, therefore: + + $$\begin{alignedat}{3} + \Phi(0) &= \Phi(2 \pi) + \quad &&\implies \quad + 0 &&= A \sin\!(2 \pi \ell) + B \big(\cos\!(2\pi \ell) - 1\big) + \\ + \Phi'(0) &= \Phi'(2 \pi) + \quad &&\implies \quad + 0 &&= A \big(\cos\!(2 \pi \ell) - 1\big) - B \sin\!(2 \pi \ell) + \end{alignedat}$$ + + This system only has nontrivial solutions + if the determinant of its matrix is zero: + + $$\begin{aligned} + 0 + &= \mathrm{det} + \begin{bmatrix} + \sin\!(2 \pi \ell) & \cos\!(2 \pi \ell) - 1 \\ + \cos\!(2 \pi \ell) - 1 & -\sin\!(2 \pi \ell) + \end{bmatrix} + = 2 \big(\cos\!(2 \pi \ell) - 1\big) + \end{aligned}$$ + + Meaning that $\ell$ must be an integer. + We revisit the boundary conditions and indeed see: + + $$\begin{alignedat}{3} + 0 &= A \sin\!(2 \pi \ell) + B \big(\cos\!(2 \pi \ell) - 1\big) + \qquad &&\implies \qquad + 0 &&= 0 + \\ + 0 &= A \big(\cos\!(2 \pi \ell) - 1\big) - B \sin\!(2 \pi \ell) + \qquad &&\implies \qquad + 0 &&= 0 + \end{alignedat}$$ + + So $A$ and $B$ are *both* unconstrained, + and each integer $\ell$ is a doubly-degenerate eigenvalue. + The two linearly independent solutions, + $\sin\!(\phi \ell)$ and $\cos\!(\phi \ell)$, + represent the polarization of light in the mode. + For simplicity, we assume that all light is in a single polarization, + so only $\cos\!(\phi \ell)$ will be considered from now on. + +By combining our result for $\ell^2 = 0$ and $\ell^2 > 0$, +we get the following for $\ell = 0, 1, 2, ...$: + +$$\begin{aligned} + \boxed{ + \Phi_\ell(\phi) = A \cos(\phi \ell) + } +\end{aligned}$$ + +Here, $\ell$ is called the **primary mode index**. +We exclude $\ell < 0$ because $\cos\!(x) \propto \cos\!(-x)$ +and $\sin\!(x) \propto \sin\!(-x)$, +and because $A$ is free to choose thanks to linearity. + +Let us now revisit the Bessel equation for the radial function $R(r)$, +which should be continuous and differentiable throughout the fiber: + +$$\begin{aligned} + r^2 R'' + r R' + \mu r^2 R - \ell^2 R = 0 +\end{aligned}$$ + +To continue, we need to specify the refractive index $n(r)$, contained in $\mu(r)$. +We choose a **step-index fiber**, +whose cross-section consists of a **core** with radius $a$, +surrounded by a **cladding** that extends to infinity $r \to \infty$. +In the core $r < a$, the index $n$ is a constant $n_i$, +while in the cladding $r > a$ it is another constant $n_o$. + +Since $\mu$ is different in the core and cladding, +we will get different solutions $R_i$ and $R_o$ there, +so we must demand that the field is continuous at the boundary $r = a$: + +$$\begin{aligned} + R_i(a) = R_o(a) + \qquad \qquad + R_i'(a) = R_o'(a) +\end{aligned}$$ + +Furthermore, for a physically plausible solution, +we require that $R_i$ is finite +and that $R_o$ decays monotonically to zero when $r \to \infty$. +These constraints will turn out to restrict $\mu$. + +Introducing a new coordinate $\rho \equiv r \sqrt{|\mu|}$ +gives the Bessel equation's standard form, +which has well-known solutions called *Bessel functions*, shown below. +Let $\pm$ be the sign of $\mu$: + +$$\begin{aligned} + \begin{cases} + \displaystyle + 0 = \rho^2 \pdv[2]{R}{\rho} + \rho \pdv{R}{\rho} \pm \rho^2 R - \ell^2 R + & \mathrm{for}\; \mu \neq 0 + \\ + \displaystyle + 0 = r^2 \pdv[2]{R}{r} + r \pdv{R}{r} - \ell^2 R + & \mathrm{for}\; \mu = 0 + \end{cases} +\end{aligned}$$ + +<a href="bessel.jpg"> +<img src="bessel.jpg" style="width:90%;display:block;margin:auto;"> +</a> + +Looking at these solutions with our constraints for $R_o$ in mind, +we see that for $\mu > 0$ none of the solutions decay +*monotonically* to zero, so we must have $\mu \le 0$ in the cladding. +Of the remaining candidates, $\ln\!(r)$, $r^\ell$ and $I_\ell(\rho)$ do not decay at all, +leading to the following $R_o$: + +$$\begin{aligned} + R_{o,\ell}(r) = + \begin{cases} + r^{-\ell} + & \mathrm{for}\; \mu = 0 \;\mathrm{and}\; \ell = 1,2,3,... + \\ + K_\ell(\rho) = K_\ell(r \sqrt{-\mu}) + & \mathrm{for}\; \mu < 0 \;\mathrm{and}\; \ell = 0,1,2,... + \end{cases} +\end{aligned}$$ + +Next, for $R_i$, we see that when $\mu < 0$ all solutions are invalid +since they diverge at $r = 0$, +and so do $\ln\!(r)$, $r^{-\ell}$ and $Y_\ell(\rho)$. +Of the remaining candidates, $r^0$ and $r^\ell$ have a non-negative slope +at the boundary $r = a$, so they can never be continuous with $R_o'$. +This leaves $J_\ell(\rho)$ for $\mu > 0$: + +$$\begin{aligned} + R_{i,\ell}(r) = + J_\ell(\rho) = J_\ell(r \sqrt{\mu}) + \qquad \mathrm{for}\; \mu > 0 \;\mathrm{and}\; \ell = 0,1,2,... +\end{aligned}$$ + +Putting this all together, we now know what the full solution for $F$ should look like: + +$$\begin{aligned} + F_\ell(r, \phi) + = R_\ell(r) \, \Phi_\ell(\phi) + = + \begin{cases} + A_\ell \: R_{i,\ell}(r) \, \cos\!(\phi \ell) + & \mathrm{for}\; r \le a + \\ + B_\ell \: R_{o,\ell}(r) \, \cos\!(\phi l) + & \mathrm{for}\; r \ge a + \end{cases} +\end{aligned}$$ + +Where $A_\ell$ and $B_\ell$ are constants to be chosen +based on the light's intensity, and to satisfy the continuity condition at $r = a$. + +We found that $\mu \le 0$ in the cladding and $\mu > 0$ in the core. +Since $\mu \equiv n^2 k^2 \!-\! \beta^2$ by definition, +this discovery places a constraint on the propagation constant $\beta$: + +$$\begin{aligned} + n_i^2 k^2 > \beta^2 \ge n_o^2 k^2 +\end{aligned}$$ + +Therefore, $n_i > n_o$ in a step-index fiber, +and there is only a limited range of allowed $\beta$-values; +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, +assuming $n_i$ and $n_o$ do not depend on $k$: + +$$\begin{aligned} + \xi_i(k) + \equiv \sqrt{ n_i^2 k^2 - \beta^2(k) } + \qquad \qquad + \xi_o(k) + \equiv \sqrt{ \beta^2(k) - n_o^2 k^2 } +\end{aligned}$$ + +It is important to note that the sum of their squares is constant with respect to $\beta$: + +$$\begin{aligned} + \xi_i^2 + \xi_o^2 = (\mathrm{NA})^2 k^2 +\end{aligned}$$ + +Where $\mathrm{NA}$ is the so-called **numerical aperture**, +often mentioned in papers and datasheets as one of a fiber's key parameters. +It is defined as: + +$$\begin{aligned} + \boxed{ + \mathrm{NA} + \equiv \sqrt{n_i^2 - n_o^2} + } +\end{aligned}$$ + +From this, we define a new fiber parameter: the $V$-**number**, +which is extremely useful: + +$$\begin{aligned} + \boxed{ + V + \equiv a \sqrt{\xi_i^2 + \xi_o^2} + = a k \: \mathrm{NA} + } +\end{aligned}$$ + +Now, the allowed values of $\beta$ are found +by fulfilling the boundary conditions (for $\mu \neq 0$): + +$$\begin{aligned} + A_\ell J_\ell(a \xi_i) + &= B_\ell K_\ell(a \xi_o) + \\ + A_\ell \xi_i J_\ell'(a \xi_i) + &= B_\ell \xi_o K_\ell'(a \xi_o) +\end{aligned}$$ + +To remove $A_\ell$ and $B_\ell$, +we divide the latter equation by the former, +meanwhile defining $X \equiv a \xi_i$ and $Y \equiv a \xi_o$ +for convenience, such that $X^2 + Y^2 = V^2$: + +$$\begin{aligned} + X \frac{J_\ell'(X)}{J_\ell(X)} = Y \frac{K_\ell'(Y)}{K_\ell(Y)} +\end{aligned}$$ + +We can turn this result into something a bit nicer +by using the following identities: + +$$\begin{aligned} + J_\ell'(x) = -J_{\ell+1}(x) + \ell \frac{J_\ell(x)}{x} + \qquad \quad + K_\ell'(x) = -K_{\ell+1}(x) + \ell \frac{K_\ell(x)}{x} +\end{aligned}$$ + +With this, the transcendental equation for $\beta$ +takes this convenient form: + +$$\begin{aligned} + \boxed{ + X \frac{J_{\ell+1}(X)}{J_\ell(X)} = Y \frac{K_{\ell+1}(Y)}{K_\ell(Y)} + } +\end{aligned}$$ + +All $\beta$ that satisfy this indicate the existence +of a **linearly polarized** mode. +These modes are called $\mathrm{LP}_{\ell m}$, +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 $\ell$. + +An example graphical solution of the transcendental equation +is illustrated below for a fiber with $V = 5$, +where red and blue respectively denote the left and right-hand side: + +<a href="modes.jpg"> +<img src="modes.jpg" style="width:90%;display:block;margin:auto;"> +</a> + +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, +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 $V_{02} = V_{21} = 3.832$. + +All differential equations have been linear, +so a linear combination of these solutions is also valid. +Therefore, the fiber modes represent independent "channels" of light. +However, in practice, they can interact nonlinearly, +and light can scatter between them, and between polarizations. + + + +## References +1. O. Bang, + *Applied mathematics for physicists: lecture notes*, 2019, + unpublished. +2. B.E.A. Saleh, M.C. Teich, + *Fundamentals of photonics*, 1st edition, 1991, + Wiley. |