# 代做 R | 统计代写 | math代写 – Part b

### Part b

Pocklington’s equation

subdivide the wire into equally spaced segments of length

expand the current with pulse functions

where

substitute (2) into equation (1)

where

multiply by weighting function and integrate over the length of the wire

which gives

let

so we have linear system equations

with plane wave excitation

`````` I ( z ) + k G ( z , z ) dz =
L /
``````
``````L /
[(
z^2
``````

#### ^22

``````)  ]   jEz ( z ) ( 1 )
``````

#### N = L / N

``````I ( z )= I u ( z )
n = 1
``````
``````N
n n (^2 )
``````
``````un ( z )={
``````

#### 0

``````zn  2  z  zn + 2
otherwise
``````
``````I K ( z , z ) dz =
n = 1
``````
``````N
n
zn /
``````
``````zn +/
Ez ( z ) ( 3 )
``````
``````K ( z , z )= + k G ( z , z )

``````
``````j
(
z^2
``````

#### )

``````wm ( z )=  ( z  zm ), 1  m  N
``````
``````I w ( z ) K ( z , z ) dzdz =
n = 1
``````
``````N
n
L /
``````
``````L /
m
zn /
``````
``````zn +/
w ( z ) E ( z ) dz
L /
``````
``````L /
m z (^4 )
``````
``````I K ( z , z ) dz =
n = 1
``````
``````N
n
zn /
``````
``````zn +/
m   Ez^ ( zm^ ) (^5 )
``````
``````Zmn = K ( z , z ) dz
zn /
``````
``````zn +/
m
``````

#### IN

`````` Ez^ ( z^1 )
Ez ( z 2 )
``````
``````
Ez ( zN )
``````

#### ( 7 )

``````Ez ( z )= E 0 sin ejkz cos  ( 8 )
``````

where is the angle of incidence.

with very thin wire

where

``````
``````
``````Zmn = K ( z , z ) dz
zn /
``````
``````zn +/
m
``````
``````= [( 1 + jkR )( 2  R  3 a )+( kaR )] dz
4
``````
``````j

zn /
``````
``````zn +/
R^5
``````
``````e  jkR 2 2 2
``````
``````R = a^2 +( zm  z )^2
``````