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ECE 6341 Spring 2016 Prof. David R. Jackson ECE Dept. Notes 10 1 Dielectric Rod z a r , r This serves as a model for a fiber-optic guide. 2 Fiber Optic Guides Two types of fiber-optic guides: 1) Single-mode fiber This fiber carries a single mode (HE11). This requires the fiber diameter to be on the order of a wavelength. It has less loss, dispersion, and signal distortion. It is often used for long-distances (e.g., greater than 1 km). The diameter is typically 8 m. It requires more expensive electro-optic equipment. 2) Multi-mode fiber This fiber has a diameter that is large relative to a wavelength (e.g., 10 wavelengths). It operates on the principle of total internal reflection (critical-angle effect). It can handle more power than the single-mode fiber, but has more dispersion. The diameter is typically 50 m. It is often used for LANs, etc. 3 Single-mode Fiber It usually has a yellow jacket. 4 Multi-mode Fiber A 1.25 Gbit/s multi-mode fiber http://en.wikipedia.org/wiki/Multi-mode_fiber 5 Fiber-optic Cable Fibers (single-mode or multi-mode) may be bundled together into a “fiber optic cable” that has one or more fibers. Simplex: single fiber in a cable Duplex: two fibers in a cable Cable Types: (L to R): Zipcord, Distribution, Loose Tube, Breakout http://www.thefoa.org/tech/ref/basic/cable.html 6 Single-mode Fiber: Operation Fields decay away from the rod HE11 mode A single mode (HE11 mode) propagates on the dielectric rod. The mode is similar to the TM0 mode on a grounded slab. It has a zero cutoff frequency. 7 Multi-mode Fiber: Operation Higher index core region A multimode fiber can be explained using down an A laser bouncing acrylic rod, illustrating the geometrical optics and total internal reflectionThe of light internal reflection. in a multi-mode fiber “ray” of light is optical actually a superposition of many waveguide modes (hence the name “multimode”). http://en.wikipedia.org/wiki/Optical_fiber 8 Single-mode vs. Multi-mode Fibers Graded index 9 Dielectric Rod z 0 1 r , r a Modes are hybrid unless 0 For example, assume TMz: (n 0) Note: We can have TE0p, TM0p modes Az 1 jk z 1 E j z j H 10 Dielectric Rod (cont.) At = a : 1H 1 0 H 0 E 1 E 0 so 1 0 1 1 1 0 11 0 0 Hence, for n > 0 we have 11 0 0 (Not true in general!) 11 Dielectric Rod (cont.) Representation of potentials inside the rod: Az1 A J n k 1 sin n e jkz z Fz1 B J n k 1 cos n e where k21 k12 kz2 jk z z < a: (kz is unknown) 12 Dielectric Rod (cont.) To see choice of sin/cos, examine the field components (for example E): 1 Fz k z Az E The field E is assumed (arbitrarily) to vary as sin(n) for our choice of potentials. Fz cos n Az sin n We then have for the field components: E sin n H cos n E cos n H sin n Ez sin n H z cos n 13 Dielectric Rod (cont.) > a: Representation of potentials outside the rod: Use H n 2 (k 0 ) H n 2 ( j 0 ) where k 0 k k 2 0 2 1/2 z j 0 0 k z2 k02 Note: 0 is interpreted as a positive real number in order to have decay radially in the air region, for a bound (non-leaky) mode. 14 Dielectric Rod (cont.) Useful identity: Another useful identity: H n ( jx) 1 2 K n ( x) n 1 2 j H n1 ( jx) n 1 1 H n ( jx) Kn (x) = Modified Bessel function of the second kind. 15 Dielectric Rod (cont.) Modified Bessel function of the second kind 1 1 0.8 K1 x 0.6 K0( x) K1( x) 0.4 K0 x 0.2 3 3.69110 0 0 1 3 510 2 3 xx 4 5 5 16 Dielectric Rod (cont.) The modified Bessel function of the first kind grows exponentially, so we don’t want this one. I n x i J n ix n I0 I1 I2 x 17 Dielectric Rod (cont.) Hence, we choose Az 0 CKn ( 0 )sin(n ) e jkz z Fz 0 DKn ( 0 ) cos(n ) e jkz z 18 Dielectric Rod (cont.) Summary of potentials: Az1 A J n k 1 sin n e jkz z Fz1 B J n k 1 cos n e jkz z Az 0 CKn ( 0 )sin(n ) e jkz z Fz 0 DKn ( 0 ) cos(n ) e jk z z 19 Dielectric Rod (cont.) Match Ez , Hz , E , H at = a: M 11 M 21 M 31 M 41 M 12 M 22 M 32 M 13 M 23 M 33 M 42 M 43 Ez1 Ez 0 Example: 1 j11 M 14 A 0 M 24 B 0 M 34 C 0 M 44 D 0 2 2 Ez k Az j z 2 1 k12 kz2 AJ n k1a 1 j0 0 k 2 0 k z2 CK n 0 a 2 0 k z2 K n 0 a so M 11 1 j11 k 2 1 k z2 J n k 1a M 13 1 j0 0 k M 12 M 14 0 20 Dielectric Rod (cont.) M 11 M 21 M 31 M 41 M 12 M 22 M 32 M 13 M 23 M 33 M 42 M 43 M 14 A 0 M 24 B 0 M 34 C 0 M 44 D 0 To have a non-trivial solution, we require that det M (k z , ) 0 kz = unknown (for a given frequency ) 21 Dielectric Rod (cont.) Cutoff frequency: Note: This is an open structure, so cutoff means the boundary between proper and improper behavior (kz = k0). k z k0 c 0 0 Set Then det M (c 0 0 , c ) 0 The unknown is now c. 22 Dielectric Rod (cont.) Dominant mode (lowest cutoff frequency): HE11 (fc = 0) E Note: The notation HE means that the mode is hybrid, and has both Ez and Hz, although Hz is stronger. (For an EH mode, Ez would be stronger.) The field shape is somewhat similar to the TE11 circular waveguide mode. The physical properties of the fields are similar to those of the TM0 surface wave on a slab (For example, at low frequency the field is loosely bound to the rod.) 23 Dielectric Rod (cont.) When will the next mode be at cutoff? This determines the upper frequency limit for the single-mode fiber. The next mode (i.e., with the next lowest cutoff frequency) is the TM01 mode. (Recall: For n = 0, the modes are TMz and TEz.) k z k0 Cutoff: k k 0 0 (in air region) 2 E j z 1 Az 2 1 2 Ez k k2 2 j z j 1 The tangential field in the air region at the boundary becomes zero at the cutoff frequency: the rod acts like a circular waveguide with a PEC wall. 24 Dielectric Rod (cont.) TM01 mode at cutoff: k 1a x01 2.405 k12 k02 a 2.405 or k0 a n12 1 2.405 Hence, to have only the HE11 mode, we have the following frequency restriction: k0 a 2.405 n 1 2 1 n1 = index of refraction of rod 25 Impedance of Wire A round wire made of conducting material is examined. Goal: Determine the impedance per unit length (in the z direction). a z The wire has a conductivity of . We neglect the z variation of the fields inside the wire. (The wire is short compared with a wavelength.) 26 Impedance of Wire (cont.) Inside the wire: Ez AJ 0 k (The field must be finite on the z axis.) k c 1 j j a j z e j /4 2 2 e j /4 27 Impedance of Wire (cont.) Hence, we have j /4 Ez AJ 0 2 e a where 2 z (skin depth) We can also write the field as j 3 /4 j 3 /4 Ez AJ 0 2 e AJ 0 2 e 28 Impedance of Wire (cont.) j 3 /4 Ez AJ 0 2 e Definition of Kelvin functions: Im J xe Ber x Re J xe j 3 /4 Bei x j 3 /4 a z Therefore, we can write Ez A Ber0 2 jBei0 2 29 Impedance of Wire (cont.) The current flowing in the wire is I J z dS S 2 a J z d d Ez AJ 0 2 e j 3 /4 0 0 a 2 J z d a 0 a 2 Ez d z 0 I 2 A J 0 2 e j 3 /4 d 0 a Hence 30 Impedance of Wire (cont.) The internal impedance per unit length is defined as: Ez a Zl I a Hence, AJ 0 2 e j 3 /4 Zl a 2 A J 0 2 e j 3 /4 d 0 a z Note: The internal impedance accounts for the internal stored energy and power dissipation. 31 Impedance of Wire (cont.) We also have the following helpful integration identity: J x xdx xJ x 0 1 a z Hence a 0 j 3 /4 j 3 /4 J0 2 e e d 2 xJ1 x 0 L LJ1 L where L 2 a e j 3 /4 2 L J x xdx 0 0 Use : x 2 j 3 /4 e 2 j 3 /4 dx d e 32 Impedance of Wire (cont.) Hence, we have a j 3 /4 J0 2 e Zl 2 a j 3 /4 a j 3 /4 j 3 /4 2 e 2 e J 2 e 1 2 where a a a a J 0 e j 3 /4 Ber0 jBei0 a a a J 1 e j 3 /4 Ber1 jBei1 z 33 Impedance of Wire (cont.) At low frequency (a << ): Zl 1 0 r j a 2 8 a At high frequency (a >> ): Zs Zl 2 a z where Z s Rs 1 j Rs 2 1 (surface resistance of metal) 34 Impedance of Wire (cont.) Transmission Line Model An extra impedance per unit length Zi is added to the TL model in order to account for the internal impedance of the conductors. Note: The form of Zi will depend on the shape of the conductors: we have only analyzed the case of a round wire. L0 Dz R Dz Zi Dz C Dz G Dz The external inductance per unit length L0 is calculated assuming perfectly conducting wires. 35 Impedance of Wire (cont.) Transmission Line Model For a twin lead (two wires running parallel) we have: Zi 2Z l We assume that the wires are far enough apart so that the current is uniform inside each one. i + - v z 36