EE 5336/7336:

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EE 5336/7336
Integrated Photonics Chapter 4 Step-Index
Circular Waveguides

Marc P. Christensen Assistant Professor Electrical
Engineering Department Southern Methodist
University Dallas, TX 75275 mpc_at_engr.smu.edu (214)
768-1407
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Chapter 4

• Step-index Circular Waveguides
• a.k.a. step-index fiber
• Most prevalent geometry.
• Harder to intuit the modes.
• (at least I think so)
• Same approach
• Pick a co-ordinate system that makes sense
• Write down the wave equation in each region
• Use boundary equations to relate the regions
• Numerically solve any equations which cannot be
  solved analytically

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Fiber is everywhere!

• Why?
• No inherent performance advantage
• Cost!
• 100s of 1000s of kms each year
• When you think long haul telecom,
• Think fiber.
• When you think photonic integrated circuitry
  (i.e., signal processing and routing)
• Think rectangular or ridge waveguides.

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Geometry of a step index fiber
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Solutions for Modes

• Cylindrical Co-ordinate System
• Solve wave equation in this geometry
• 2-D
• Leads to 2 different mode numbers
• Coupling between differing field components.

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The Wave Eqn in Cylindrical Co-ordinates
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Field Coupling
r, f
z
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z-component Wave Eqn.
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Solving the Ez Wave Eqn
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Solving the Ez Wave Eqn
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Solving the Ez Wave Eqn
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Solving the Ez Wave Eqn
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Bessel Functions

• Just a function, no different than Sin or Cos
• Orthogonal to each other.
• Defined everywhere (all r)
• (check out Appendix B for some useful properties,
  well introduce them as needed here)

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Bessel Functions of 1st Kind - J
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Modified Bessel Functions of 2nd Kind - K
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Bessel Function Approximations

• At large values of the arguments

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Guiding Condition
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Solution for Ez and Hz
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Take a step back?
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Getting to r and f
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Inside the core (rlta)
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Outside the core (rgta)
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End of Session
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Boundary Conditions
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Boundary Conditions
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Characteristic Equation
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Relating the Coefficients
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Simplest Case is when n 0
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Bessel Identity Simplifies Characteristic Eqn
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Bessel Identity Simplifies Characteristic Eqn
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Mathematica Example

• File
• Chapter 04 Examples

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Hybrid Modes (n !0)

• A0 TE
• B0 TM
• AgtB Ez is larger than Hz so most
• transverse field is H,
• so we call it HE
• AgtB Hz is larger than Ex so most
• transverse field is E,
• so we call it EH

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Meridonal vs. Skew
r, f
z
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Too Ugly to Solve!
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Weakly Guiding Approximation

• ncore nclad

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Using that same old Bessel Identity
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Now we have
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Degeneracy

• Same Beta (same propagation speed)
• TE0m is degenerate with TM0m
• HEn1,m is degenerate with EHn-1,m

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Hypothesize it and see if it is True

• Linearly Polarized Mode

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Is it nearly Transverse E?
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Is it nearly Transverse E?
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Is it nearly Transverse E?
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Is it nearly Transverse E?
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Linearly Polarized Modes

• LP1,m sum of TE0,m or TM0,m and HE2,m
• LPn,m sum of HEn1,m and EHn-1,m
• LP0,m HE1,m

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Linearly Polarized Modes

• Useful, but not real!
• Useful because they tell how a polarized source
  will launch into a fiber.
• Not real because the weakly guiding approximation
  is just that (an approximation), Actual betas
  are nearly equal which means that over long
  distances this breaks down

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End of Session
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Normalized Frequency and Cutoff

• Determine if a given wavelength will have a given
  mode in the fiber.
• Recall We are looking for the intersections of
  curves. In 1-D we said that every time Tan(kh)
  went to Infinity we picked up another mode.
• This time the equation contains Jn1/Jn
• So now we pick up a new mode at every root of
  the denominator (Jn)

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1st Graph of Solutions
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List of Cutoffs

• TE0,m kagtmth root of J0(ka)
• HE1,m kagtmth root of J1(ka)
• EHn,m kagtmth root of Jn(ka)
• HEn,m

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Approximate Graph
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V and lc

• Vkmaxa
• All modes with ka less than V can exist within a
  fiber!
• We call the cutoff condition Vlt2.405 because
  TE0,1, TM0,1, and HE2,1 cease to exist there.
  Only HE1,1 will propagate.
• Alternatively we can define the wavelength lc
  above which V2.405 and the fiber is single mode.

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Example 1

• l1.3 mm, ncore1.5, nclad1.48
• What is the maximum radius of the fiber for which
  the fiber will be single mode?
• Va (2 p/1.3 mm)Sqrt1.52-1.4822.405
• a2.038 mm

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Example 2

• l1.3 mm, ncore1.5, nclad1.48
• If we increase the radius to a10 mm,
• What modes exist in the waveguide?
• V10 mm(2 p/1.3 mm)Sqrt1.52-1.482
• V11.799
• So now what?

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J0
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J1
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J2
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J3
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J4
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J5
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J6
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J7
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J8
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2J1ka J2
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2J2ka/2 J3
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2J3ka/3 J4
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2J4ka/4 J5
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2J5ka/5 J6
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2J6ka/6 J7
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2J7ka/7 J8
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2J8ka/8 J9
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2J9ka/9 J10
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The Fundamental Mode

• HE11 mode is the lowest mode (cutoff of zero!).
• It looks Gaussian!!!????

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Mode Field Diameter
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So Do I Always Need to Count Every Mode???

• Did I have to before?
• How did I avoid it?
• Can we do the same here?
• Well need to use the approximations for the
  Bessel Functions Then as good as the
  approximations are, our estimates of modes are

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Characteristic Eqn for LP modes
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Characteristic Eqn for LP modes
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Number of Modes
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Allowed Modes Triangle