1. Bessel Function Examples
ECE 6382
Fall 2017
David R. Jackson
Notes 21
Bessel Function Examples
Notes are from D. R. Wilton, Dept. of ECE
Note: j is used in this set of notes instead of i.

2. Impedance of Wire
A round wire made of conducting material is examined. a
1, 1,  z
0, 0
The wire has a conductivity of .
We neglect the z variation of the fields inside the wire (|kz| << |k1|).

3. Impedance of Wire (cont.)
Inside the wire:
(The field must be finite on the z axis, no  variation.) a
1, 1,  z
0, 0
Note:
This assumes that the wire is fed (excited) from the outside.

4. Impedance of Wire (cont.)
Hence, we have a
1, 1,  z
0, 0
where
(skin depth)
We can also write the field as

5. Impedance of Wire (cont.)
1, 1,  z
0, 0
Recall:
Therefore, we can write

6. Impedance of Wire (cont.)
The current flowing in the wire is a
1, 1,  z
0, 0
Hence

7. Impedance of Wire (cont.)
The impedance per unit length defined as: a
1, 1,  z
0, 0
Hence,
Note:
This assumes that the wire is fed (excited) from the outside.

8. Impedance of Wire (cont.)
We have the following helpful
integration identity: a
1, 1,  z
0, 0
Hence
where

9. Impedance of Wire (cont.)
Hence, we have a
1, 1,  z
0, 0
where

10. Impedance of Wire (cont.)
At low frequency (a << ): a
1, 1,  z
0, 0
At high frequency (a >> ):
where

11. Circular Waveguide TMz mode: The waveguide is homogeneously filled,
so we have independent TEz and TMz modes. a z
TMz mode: r

12. Circular Waveguide (cont.)
(1)  variation
(uniqueness of solution)
Choose

13. Circular Waveguide (cont.)
(2) The field should be finite on the z axis
is not allowed

14. Circular Waveguide (cont.)
(3) B.C.’s:
Hence

15. Circular Waveguide (cont.)
xn1
xn2
xn3 x
Jn(x)
Plot shown for n  0
Note: xn0 is not included since (trivial solution).

16. Circular Waveguide (cont.)
TMnp mode:

17. Cutoff Frequency: TMz

18. Cutoff Frequency: TMz (cont.)
xnp values
p \ n 1 2 3 4 5
2.405
3.832
5.136
6.380
7.588
8.771
5.520
7.016
8.417
9.761
11.065
12.339
8.654
10.173
11.620
13.015
14.372
11.792
13.324
14.796
TM01, TM11, TM21, TM02, …

19. TEz Modes
In this case we have

20. TEz Modes (cont.)
Set
At he boundary, the first term on the RHS is zero:
Hence

21. p = 0 is not included (see next slide).
TEz Modes (cont.)
x'n1
x'n2
x'n3 x
Jn' (x)
Plot shown for n  1
Note:
p = 0 is not included (see next slide).

22. TEz Modes (cont.) p = 0 (trivial soln.)
This generates other field components that are zero; the resulting field that only has Hz violates the magnetic Gauss law.

23. Cutoff Frequency: TEz

24. Cutoff Frequency:TEz x´np values TE11, TE21, TE01, TE31, …….. p \ n 11 2 3 4 5
3.832
1.841
3.054
4.201
5.317
5.416
7.016
5.331
6.706
8.015
9.282
10.520
10.173
8.536
9.969
11.346
12.682
13.987
13.324
11.706
13.170
TE11, TE21, TE01, TE31, ……..

25. rectangular waveguide
TE11 Mode
The dominant mode of circular waveguide is the TE11 mode.
Electric field
Magnetic field
(from Wikipedia)
TE10 mode of
rectangular waveguide
TE11 mode of
circular waveguide
The TE11 mode can be thought of as an evolution of the TE10 mode of rectangular waveguide as the boundary changes shape.

26. Scattering by Cylinder
A TMz plane wave is incident on a PEC cylinder. x y
Top view x z a
TMz qi

27. Scattering by Cylinder (cont.)
From the plane-wave properties, we have
The total field is written as the sum of incident and scattered parts:
For   a:
Note:
For any wave of the form exp(-jkzz), all field components can be put in terms of Ez and Hz. This is why it is convenient to work with Ez.
Please see the Appendix.

28. Scattering by Cylinder (cont.)
We first put into cylindrical form using the Jacobi-Anger identity*:
where
Assume the following form for the scattered field:
*This was derived previously using the generating function.

29. Scattering by Cylinder (cont.)
Hence
This yields or

30. Scattering by Cylinder (cont.)
We then have
and
The other components of the scattered field can be found
from the formulas in the Appendix.

31. Current Line Source z I (z) = I0 TMz : y Conditions: x 1)2) 3) 4)
Allowed angles:
Symmetry:
Hence
Radiation condition:
Symmetry:

32. Current Line Source (cont.)
Our goal is to solve for the constant A: C z I0
Choose a small circular path:

33. Current Line Source (cont.)
From Ampere’s law and Stokes’ theorem: C z I0
Examine the last term (displacement current):
where

34. Current Line Source (cont.)
Hence so
Therefore
Now use

35. Current Line Source (cont.)
Hence or so
Thus

36. Scattering From a Line Currenty z x a
We use the addition theorem to translate the Hankel function to the z axis.

37. Scattering From a Line Current (cont.)
The addition theorem tells us:
We use the first form, since the cylinder at  = a is inside the circle on which the line source resides (radius 0).

38. Scattering From a Line Current (cont.)
Incident field:
Assume a form for the scattered field:

39. Scattering From a Line Current (cont.)
Boundary Conditions ( = a):
Hence or

40. Scattering From a Line Current (cont.)
Final result: y z x a
Scattered field

41. *This means that we need both Ez and Hz.
Dielectric Rod a z 1
Unknown wavenumber:
Modes are hybrid* unless:
Note:
We can have
TE0p, TM0p modes
*This means that we need both Ez and Hz.

42. Dielectric Rod (cont.)  < a
Representation of potentials inside the rod:
where
(kz is unknown)

43. Dielectric Rod (cont.)
To see choice of sin/cos, examine the field components (for example E):
From the Appendix:

44. Dielectric Rod (cont.)  > a
Representation of potentials outside the rod:
Use
where
Note: 0 is interpreted as a positive real number in order to have decay radially in the air region.

45. Kn (x) = modified Bessel function of the second kind.
Dielectric Rod (cont.)
Useful identity:
Another useful identity:
Kn (x) = modified Bessel function of the second kind.

46. Dielectric Rod (cont.)
The modified Bessel functions decay exponentially.

47. Dielectric Rod (cont.)
Hence, we choose the following forms in the air region ( > a):

48. Dielectric Rod (cont.) Match Ez , Hz , Ef , Hf at  = a: Example: orso

49. kz = unknown (for a given frequency )
Dielectric Rod (cont.)
To have a non-trivial solution, we require that
kz = unknown (for a given frequency )

50. Dielectric Rod (cont.)
Dominant mode (lowest cutoff frequency): HE (fc = 0)
This is the mode that is used in fiber-optic guides (single-mode fiber).

51. At higher frequencies, the fields are more tightly bound to the rod.
Dielectric Rod (cont.)
Sketch of normalized wavenumber
At higher frequencies, the fields are more tightly bound to the rod.

52. Appendix For any wave of the form exp(-jkz z),
all field components can be put in terms of Ez and Hz.

53. Appendix (cont.)
These may be written more compactly as
where

54. Appendix (cont.) In cylindrical coordinates we have
This allows us to calculate the field components in terms of Ez and Hz in cylindrical coordinates.

55. Appendix (cont.)
In cylindrical coordinates we have