Analysis of Impedance in Conducting Materials
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ECE 6382
ECE 6382
David R. Jackson
Fall 2023
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A round wire made of conducting material is examined.
The wire has a conductivity of
.
Impedance of Wire
Impedance of Wire
2
We neglect the
z
variation of the fields inside the wire (
|
k
z
| << |
k
1
|
).
Inside the wire:
Impedance of Wire (cont.)
Impedance of Wire (cont.)
3
(The field must be finite on the
z
axis, no
variation.
)
N
o
t
e
:
This assumes that there are no sources
inside the wire.
(for some constant
A
)
Hence, we have
Impedance of Wire (cont.)
Impedance of Wire (cont.)
4
where
(skin depth of metal)
We can also write the field as
(
J
0
is an even function.)
Impedance of Wire (cont.)
Impedance of Wire (cont.)
5
Therefore, we can write
Recall:
Impedance of Wire (cont.)
Impedance of Wire (cont.)
6
The current flowing in the wire is
Hence
Impedance of Wire (cont.)
Impedance of Wire (cont.)
7
The impedance per unit length defined as:
Hence,
N
o
t
e
:
This assumes that the wire is fed
(excited) from the
outside
.
Impedance of Wire (cont.)
Impedance of Wire (cont.)
8
We have the following helpful
integration identity:
Hence
Impedance of Wire (cont.)
Impedance of Wire (cont.)
9
Hence, we have
where
Impedance of Wire (cont.)
Impedance of Wire (cont.)
10
At low frequency (
a
<<
):
where
At high frequency (
a
>>
):
(ECE 3318)
(ECE 6340)
Circular
Circular
Waveguide
Waveguide
TM
z
mode:
11
r
The waveguide is homogeneously filled,
so we have independent TE
z
and TM
z
modes.
N
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:
T
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p
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x
(
d
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s
)
.
Circular Waveguide (cont.)
Circular Waveguide (cont.)
(1)
variation
Choose
(uniqueness of solution)
12
(2) The field should be
finite
on the
z
axis (
= 0
)
is not allowed
Circular Waveguide (cont.)
Circular Waveguide (cont.)
13
(3) B.C.’s:
Circular Waveguide (cont.)
Circular Waveguide (cont.)
14
Hence
Circular Waveguide (cont.)
Circular Waveguide (cont.)
15
TM
np
mode:
Circular Waveguide (cont.)
Circular Waveguide (cont.)
16
Cutoff Frequency: TM
Cutoff Frequency: TM
z
z
17
(We assume a
lossless
dielectric for the cutoff discussion.)
Cutoff Frequency: TM
Cutoff Frequency: TM
z
z
(cont.)
(cont.)
Ordering of modes by cutoff frequency: TM
01
, TM
11
, TM
21
, TM
02
, …
x
np
values
18
TE
TE
z
z
Modes
Modes
In this case the boundary condition is different:
19
Set
TE
TE
z
z
Modes (cont.)
Modes (cont.)
Hence
20
At the boundary, the first term on the RHS is zero:
TE
TE
z
z
Modes (cont.)
Modes (cont.)
21
N
o
t
e
:
p
= 0
is not included
(see next slide).
TE
TE
z
z
Modes (cont.)
Modes (cont.)
p
=
0
(trivial soln.)
This generates other field
components that are zero; the
resulting field has only
H
z
and
violates the magnetic Gauss law.
22
Cutoff Frequency: TE
Cutoff Frequency: TE
z
z
23
(We assume a lossless dielectric for the cutoff discussion.)
TE
11
, TE
21
, TE
01
, TE
31
, ……..
x
´
np
values
Cutoff Frequency:TE
Cutoff Frequency:TE
z
z
24
TE
TE
11
11
Mode
Mode
TE
10
mode of
rectangular waveguide
TE
11
mode of
circular waveguide
The dominant mode of circular waveguide is the TE
11
mode.
The TE
11
mode can be thought of as an evolution of the TE
10
mode of
rectangular waveguide as the boundary changes shape.
25
(from Wikipedia)
Dielectric Rod
Dielectric Rod
Modes are
hybrid
* unless:
26
N
o
t
e
:
We can have
TE
0p
, TM
0p
modes
*This means that we need both
E
z
and
H
z
.
Unknown wavenumber:
where
(
k
z
is unknown)
<
a
Dielectric Rod (cont.)
Dielectric Rod (cont.)
27
Representation of fields inside the rod:
To see choice of sin/cos, examine the field components (for example
E
):
Dielectric Rod (cont.)
Dielectric Rod (cont.)
28
From the Appendix:
where
>
a
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Use
Dielectric Rod (cont.)
Dielectric Rod (cont.)
29
Representation of potentials outside the rod:
K
n
(
x
)
=
modified Bessel function of the second kind.
Useful identity:
Another useful identity:
Dielectric Rod (cont.)
Dielectric Rod (cont.)
30
The modified Bessel functions decay exponentially.
Dielectric Rod (cont.)
Dielectric Rod (cont.)
31
Hence, we choose the following forms in the air region (
>
a
)
:
Dielectric Rod (cont.)
Dielectric Rod (cont.)
32
Match
E
z
, H
z
, E
, H
at
= a
:
Dielectric Rod (cont.)
Dielectric Rod (cont.)
Example:
so
33
or
To have a non-trivial solution, we require that
Dielectric Rod (cont.)
Dielectric Rod (cont.)
34
This is a transcendental equation for the unknown
k
z
(for a given frequency
).
There will be an infinite number
of solutions (
p
= 1, 2,…
), for
each assumed value of
n
.
Dominant mode
(lowest cutoff frequency):
HE
11
(
f
c
=
0)
This is the mode that is used in fiber-optic guides (single-mode fiber).
Dielectric Rod (cont.)
Dielectric Rod (cont.)
35
Dielectric Rod (cont.)
Dielectric Rod (cont.)
36
Sketch of normalized wavenumber
At higher frequencies, the fields are more tightly bound to the rod.
HE
11
Scattering by Cylinder
Scattering by Cylinder
A TM
z
plane wave is incident on a PEC cylinder.
37
Top view of plane wave
Scattering by Cylinder (cont.)
Scattering by Cylinder (cont.)
From the plane-wave properties, we have
38
The total field is written as the sum of incident and scattered parts:
For
a
:
N
o
t
e
:
For any wave of the form
exp(-
jk
z
z
)
,
all field components can be put in terms
of
E
z
and
H
z
. This is why it is convenient to work with
E
z
.
Please see the Appendix.
Scattering by Cylinder (cont.)
Scattering by Cylinder (cont.)
where
We first put into cylindrical form using the Jacobi-Anger identity*:
Assume the following form for the scattered field:
39
*This was derived previously using the generating function.
Scattering by Cylinder (cont.)
Scattering by Cylinder (cont.)
At
40
Hence
This yields
or
Scattering by Cylinder (cont.)
Scattering by Cylinder (cont.)
We then have
41
The other components of the scattered field can be found
from the formulas in the Appendix.
and
Appendix
Appendix
42
For any wave of the form
exp(-
jk
z
z
)
,
all field components can be put in terms of
E
z
and
H
z
(derivation omitted).
Appendix (cont.)
Appendix (cont.)
43
These may be written more compactly as
where
Appendix (cont.)
Appendix (cont.)
44
This allows us to calculate the field components in terms of
E
z
and
H
z
in cylindrical coordinates.
In cylindrical coordinates we have
Appendix (cont.)
Appendix (cont.)
45
In cylindrical coordinates we then have
In these notes, the impedance of a round wire made of conducting material is examined in detail. The analysis includes the consideration of the wire's conductivity, current flow, and impedance per unit length. Various integration identities and equations are derived to understand the behavior of the wire under different conditions.
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Presentation Transcript
ECE 6382 Fall 2023 David R. Jackson Notes 22 Applications of Bessel Functions Note:j is used in this set of notes instead of i. 1
Impedance of Wire A round wire made of conducting material is examined. ( ) , 0 0 ( ) , , 1 1 a = k 1 c j z = 1 1 c The wire has a conductivity of . We neglect the zvariation of the fields inside the wire (|kz| << |k1|). 2
Impedance of Wire (cont.) Inside the wire: ( 0 AJ = (The field must be finite on the z axis, no variation.) ) jk z z E k e z 1 ( ) J k sin( cos( ) 1 = jk z Recall: e z (for some constant A) ( ) ) Y k 1 = = 2 2 z k k k k k 1 1 1 1 1 c ( ) , 0 0 = 1 j ( ) , , 1 1 1 1 1 a j 1 1 1 = j z 1 ( ) = /4 j e 1 Note: ( ) = /4 j 2 e 1 This assumes that there are no sources inside the wire. 2 3
Impedance of Wire (cont.) Hence, we have ( ) , 0 0 ( ) , , 1 1 = /4 j 2 z E AJ e 0 a z where 2 Note: = (skin depth of metal) We can also write 1 ( ) = 1 E AJ j 0 z We can also write the field as 3 /4 3 /4 = = j j 2 2 z E AJ e AJ e (J0 is an even function.) 0 0 4
Impedance of Wire (cont.) 3 /4 = j 2 z E AJ e 0 ( ) , 0 0 ( ) , , 1 1 Recall: a ( ( ) ) ( ( ) ) ( ) ( ) x 3 /4 j Ber Re x J xe z 3 /4 j Bei Im J xe Therefore, we can write = + Ber 2 Bei 2 z E A j 0 0 5
Impedance of Wire (cont.) The current flowing in the wire is = I J dS z ( ) S , 0 0 2 a ( ) , , = d d J 1 1 z a 0 0 a = 2 J d z z 0 a = 2 E d z 0 a 3 /4 = j Hence 2 2 I A J e d 0 0 6
Impedance of Wire (cont.) The impedance per unit length defined as: ( ) I E a z Z ( ) l , 0 0 ( ) , , 1 1 Hence, a a 3 /4 j 2 J e 0 z = Z l a 3 /4 j 2 2 J e d 0 0 Note: This assumes that the wire is fed (excited) from the outside. 7
Impedance of Wire (cont.) We have the following helpful integration identity: ( ) , 0 0 ( ) , , 1 1 ( ) x xdx ( ) x = J xJ a 0 1 Hence z 2 2 a L 1 1 ( ) x xdx ( ) x L 3 /4 3 /4 3 /4 = = j j j 2 J e d e J e xJ 0 0 1 0 2 2 0 0 2 1 ( ) L 3 /4 = j e LJ 1 2 where 1 3 /4 j 2 x e ( ) L 3 /4 = j a e J a 1 3 /4 j 2 2 L e 1 3 /4 = j 2 dx d e 1 a 3 /4 3 /4 = j j 2 a e J e 1 2 2 1 3 /4 = j 2 xdx d e 8
Impedance of Wire (cont.) Hence, we have a 3 /4 j 2 J e 0 = Z l 1 a 3 /4 3 /4 j j 2 2 a e J e 1 2 0, 0 1, 1, a where a a a 3 /4 = + j z 2 Ber 2 Bei 2 J e j 0 0 0 a a a 3 /4 = + j 2 Ber 2 Bei 2 J e j 1 1 1 9
Impedance of Wire (cont.) At low frequency (a << ): 1 0, 0 lZ ( ) (ECE 3318) 2 a 1, 1, a At high frequency (a >> ): z Z Z s (ECE 6340) l 2 a where ( ) = + 1 Z R j s s 1 = = ( ) R 1 surface resistance of metal s 2 10
Circular Waveguide The waveguide is homogeneously filled, so we have independent TEz and TMz modes. TMz mode: ( z E = a ) , , z = k k r 0 r Note: The relative permittivity could be complex (due to loss). z ( ) J k sin( cos( ) = jk z e z ( ) ) Y k 2 = 2 2 z k k k 11
Circular Waveguide (cont.) [0,2 ] (1) variation + 2 , ) = ( , , ) = ( , z z (uniqueness of solution) n n cos( ) Choose ( ) J k n = jk z cos( ) n e z ( ) Y k n 12
Circular Waveguide (cont.) (2) The field should be finite on the z axis ( = 0) n Y k ( ) is not allowed = jk z cos( ) ( ) n J k e z n 2 = 2 2 z k k k 13
Circular Waveguide (cont.) (3) B.C. s: ( ) , , = 0 z E a z Hence = ( ) 0 n J k a 14
Circular Waveguide (cont.) = ( ) 0 n J k a ( ) x n J Sketch shown for n 0 nx 3 x nx nx 1 2 x = k a x np a = k np ( ) 0 = Note:xn0= 0 is not included since (for n > 0) (trivial solution). 0 n J 15
Circular Waveguide (cont.) TMnp mode: = = jk z cos( ) 0,1,2 E n J x e n z z n np a 1/2 2 x np a = = 2 1,2,3, k k p z 16
Cutoff Frequency: TMz (We assume a lossless dielectric for the cutoff discussion.) = 2 2 2 zk k k x np a = = zk = k k 0 x np a = 2 f c 2 x np a = c 2 TM c , k f f = TM f x = k c np z 2 a 2 x np a = 2 TM c , j j k f f r 17
Cutoff Frequency: TMz (cont.) xnp values 0 1 2 3 4 5 p \ n 1 2.405 3.832 5.136 6.380 7.588 8.771 2 5.520 7.016 8.417 9.761 11.065 12.339 3 8.654 10.173 11.620 13.015 14.372 4 11.792 13.324 14.796 Ordering of modes by cutoff frequency: TM01, TM11, TM21, TM02, 18
TEz Modes ( ) = , , H z z = jk z cos( ) ( ) n J k e z n In this case the boundary condition is different: ( ) , , 0 a z 19
TEz Modes (cont.) Set ( ) , , = 0 E a z = H j E H 1 H = E z j z At the boundary, the first term on the RHS is zero: ( ) , , = 0 H a z Hence n J k a = ( ) 0 20
TEz Modes (cont.) n J k a = ( ) 0 Recall: 1 n ( ) = n ( ) ~ J x , 0 0,1,2,.... x x n n 2 ! n ( ) x n J Sketch shown for n 1 nx 3 x nx nx 1 2 Note: p = 0 is not included (see next slide). np = k a x x np a = = 1,2,3,..... k p 21
TEz Modes (cont.) np = = jk z cos( ) 1,2, n J x e p z n a ( ) np = = = If 0, 0 p n 1 p x but cannot be zero for ( ) 0 np = = n 0 J x J 0 (trivial soln.) n n a p =0 ( ) np = = 00 1 J x J n = 0 0 a This generates other field components that are zero; the resulting field has only Hz and violates the magnetic Gauss law. = = jk z jkz e e z k = 0 22
Cutoff Frequency: TEz (We assume a lossless dielectric for the cutoff discussion.) = 2 2 2 zk k k x zk = np a 0 = = k k x np a = 2 f c 2 x c np a = 2 TE c , k f f np = TE f x = k c z 2 a 2 x np a = 2 TE c , j j k f f r 23
Cutoff Frequency:TEz x np values 0 1 2 3 4 5 p \ n 1 3.832 1.841 3.054 4.201 5.317 5.416 2 7.016 5.331 6.706 8.015 9.282 10.520 3 10.173 8.536 9.969 11.346 12.682 13.987 4 13.324 11.706 13.170 TE11, TE21, TE01, TE31, .. 24
TE11 Mode The dominant mode of circular waveguide is the TE11 mode. Electric field Magnetic field Image:TE10.gif Image:TE11.gif (from Wikipedia) TE11 mode of circular waveguide TE10 mode of rectangular waveguide The TE11 mode can be thought of as an evolution of the TE10 mode of rectangular waveguide as the boundary changes shape. 25
Dielectric Rod z 0 1 Unknown wavenumber: a , r r k k k 0 1 z = n 1 r Modes are hybrid* unless: = Note: We can have TE0p, TM0p modes = 0 ( 0) n *This means that we need both Ez and Hz. 26
Dielectric Rod (cont.) Representation of fields inside the rod: ( ( ) ) ( ( ) ) = jk z sin E AJ k n e z 1 1 z n <a = jk z cos H BJ k n e z 1 1 z n where 2 = 2 2 z k k k (kz is unknown) 1 1 27
Dielectric Rod (cont.) To see choice of sin/cos, examine the field components (for example E ): From the Appendix: 1 j H jk E = E z z z 2 2 z 2 2 z k k k k 28
Dielectric Rod (cont.) >a Representation of potentials outside the rod: Use ( ) 2 n ( ) 2 n = ( ) ( ) H k H j 0 0 where ( ) 1/2 = = 2 0 2 z k k k j 0 0 = 2 2 0 zk k 0 Note: 0 is interpreted as a positive real number in order to have decay radially in the air region. 29
Dielectric Rod (cont.) Useful identity: ( ) ( ) 2 n ( ) 1 n + 1 n = + ( ) 1 ( ) H jx H jx Another useful identity: 2 ( ) 1 n + = ( 1) n ( ) ( ) H jx j K x n Kn (x) = modified Bessel function of the second kind. 30
Dielectric Rod (cont.) The modified Bessel functions decay exponentially. 1 1 0.8 ( ) K x 1 0.6 K0 x ( ) K1 x ( ) 0.4 ( ) x K 0 0.2 3 3.691 10 0 0 1 2 3 4 5 3 x x 5 5 10 31
Dielectric Rod (cont.) Hence, we choose the following forms in the air region ( > a): = jk z ( )sin( ) E CK n e z 0 0 z n = jk z ( )cos( ) H DK n e z 0 0 z n = 2 2 0 zk k 0 32
Dielectric Rod (cont.) Match Ez , Hz , E , H at = a: M M M M M M M M M M M M M M M M 0 0 0 0 A B C D 11 12 13 14 21 22 23 24 = 31 32 33 34 41 42 43 44 ( ) ( ) = = E E AJ k a CK a Example: 1 0 1 0 z z n n ( ) ( ) ( ) or ( ) 0 ( ) 0 + + + = 0 AJ k a B C K a D 1 0 n n so ( ) ( ) = = = = , , 0 M J k a M K a M M 11 1 13 0 12 14 n n Recall: ( ( ) )sin( ( ) = jk z sin E AJ k n e z 1 1 z n = = 2 2 2 z 2 z 2 0 , k k k k k = jk z ) E CK n e 1 1 0 z 0 0 z n 33
Dielectric Rod (cont.) M M M M M M M M M M M M M M M M 0 0 0 0 A B C D 11 12 13 14 21 22 23 24 = 31 32 33 34 41 42 43 44 To have a non-trivial solution, we require that There will be an infinite number of solutions (p= 1, 2, ), for each assumed value of n. ( M k = det , ) 0 z This is a transcendental equation for the unknown kz (for a given frequency ). 34
Dielectric Rod (cont.) Dominant mode (lowest cutoff frequency): HE11(fc = 0) E This is the mode that is used in fiber-optic guides (single-mode fiber). 35
Dielectric Rod (cont.) Sketch of normalized wavenumber / zk k HE11 0 r 1.0 f ( ) 2 = = 2 z 2 0 / 1 k k k k k 0 0 0 z At higher frequencies, the fields are more tightly bound to the rod. 36
Scattering by Cylinder A TMz plane wave is incident on a PEC cylinder. y z ( , ) a k TMz x i Top view of plane wave H i x i E = = cos sin k k k k i e + = ( ) j k x k z 0 x i yH H x z 0 y 0 z i 37
Scattering by Cylinder (cont.) From the plane-wave properties, we have + = ( ) j k x k z i z 0cos E H e x z 0 y i The total field is written as the sum of incident and scattered parts: For a: = + i z s z E E E z 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. 38
Scattering by Cylinder (cont.) i We first put into cylindrical form using the Jacobi-Anger identity*: z E + 1 j = jk z i z jn cos ( ) E H e J k e z 0 0 y i n n = n Recall: = = = 2 0 2 z where 0cos k k k k k ( ) ( ) jkx jn n = x i e j J k e n = n Let k k k x Assume the following form for the scattered field: + 1 j ( ) 2 n = jk z s z jn cos ( ) E H e a H k e z 0 0 y i n n = n *This was derived previously using the generating function. 39
Scattering by Cylinder (cont.) ( ) = , , = a 0 E a z At z Hence ( ) ( ) , , = , , s i E a z E a z z z This yields ( ) ( )( n ) 2 = J k a a H k a n n or ( ) J k a n = a ( )( n ) n 2 H k a 40
Scattering by Cylinder (cont.) We then have ( ( ) ) J k a k a + 1 j ( ) 2 n n = jk z s jn cos ( ) E H e H k e z ( ) 2 n 0 0 y i z n H = n and H = s 0 (TM ) z z The other components of the scattered field can be found from the formulas in the Appendix. 41
Appendix For any wave of the form exp(-jkz z), all field components can be put in terms of Ez and Hz (derivation omitted). H y H x jk E x E y j = E z z z x 2 2 z 2 2 z k k k k jk j = E z z z y 2 2 z 2 2 z k k k k j E y E x jk H x H y = c H z z z x 2 2 z 2 2 z k k k k j jk = c H z z z y 2 2 z 2 2 z k k k k 42
Appendix (cont.) These may be written more compactly as j jk ( ) ( ) = E z H E z k z z t t t 2 2 z 2 2 z k k k j jk ( ) ( ) = H z E H z k z z t t t 2 2 z 2 2 z k k k where + x y t x y 43
Appendix (cont.) In cylindrical coordinates we have 1 = + t This allows us to calculate the field components in terms of Ez and Hz in cylindrical coordinates. 44
Appendix (cont.) In cylindrical coordinates we then have 1 j H jk E = E z z z 2 2 z 2 2 z k k k k 1 j H jk E = E z z z 2 2 z 2 2 z k k k k 1 j E jk H = H z z z 2 2 z 2 2 z k k k k 1 j E jk H = H z z z 2 2 z 2 2 z k k k k 45
