CS计算机代考程序代写 ER Mid 2 Problem 1

Mid 2 Problem 1
In a spherical coordinate system, a ring (red) with a current I=I0 generates fields as
H(r,θ,φ)=jka2I0cos(θ)[1+ 1]e−jkr r 2r2 jkr
θˆ φˆ
rˆ
Hφ (r,θ,φ)=0
H (r,θ,φ)=−(ka)2I0sin(θ)[1+ 1 − 1 ]e E (r,θ,φ)=η(ka)2I0sin(θ)[1+ 1 ]e−jkr
− jkr θ 4r jkr (kr)2
φ
Er (r,θ,φ)=0
4r jkr
Eθ (r,θ,φ)=0
The ring radius a=20mm. The ring is in the free space. Current I0 =1A.
The frequency=2.5GHz. And phase constant k = 2π . λ
At every point in the 3-D space, there are two magnetic field vectors and one electrical field vector.
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1

(1)
At every point in the space, there are two magnetic field vectors and one electrical field vector.
Plot the fields
Hr (x, y, z) at a plane at Hθ (x,y,z)
Eφ (x,y,z)
zˆ
rˆ
z = 2cm, y=0 −1m ≤ x ≤ 1m
|Hr |, |Hθ |or|Eφ |
θˆ xˆ
xˆ
xˆ JCCHIAO©2020
φˆ
yˆ
2

(2)
Plot the fields in a 2-D map at a plane at |Hr ||Hθ |or|Eφ |
z=2cm,−1m≤x≤1m, −1m≤y≤1m,
zˆ
rˆ
yˆ
Then convert the magnitudes of fields to normalized values.
xˆ
Plot a “heat map” with red being 1 (maximum) and blue
θˆ
being almost zero (minimum).
yˆ
0.5
0 -0.5 -1.0
-1.0
Heat map example:
1.0
1.0
0.5
0.0
xˆ JCCHIAO©2020
φˆ
-0.5
0
0.5
1.0
3

Now plot the “heat map” with red being 0dB (maximum) and blue being -30dB for the three field component magnitudes.
Heat map example:
1.0 0.5 0 -0.5
-1.0 -1.0
0dB
-15dB
-30dB
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4
-0.5
0
0.5
1.0