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Electromagnetics Assignment: Wave Propagation & Field Calculations

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Module 1 Assignment
Problem 1:
An electric field propagating in a lossless non-magnetic media is characterized by
𝐸(𝑦, 𝑑) = 100.0 π‘π‘œπ‘ (4πœ‹ × 106 𝑑 − 0.1257𝑦) π‘Žπ‘§
𝑉
π‘š
Find:
A. The wave amplitude, frequency, propagation velocity, wavelength, and relative
permittivity of the media.
B. Write the expression for H(y, t), including the direction (unit vector) and units.
Solution:
𝑉
A. Using the general form 𝐸(𝑦, 𝑑) = 𝐸0 cos(πœ”π‘‘ − 𝛽𝑦) π‘Žπ‘§ π‘š
𝑉
i.
Wave amplitude = 𝐸0 = 100 π‘š
ii.
Frequency = 𝑓 = 2πœ‹ =
πœ”
4πœ‹×106
2πœ‹
2πœ‹
2πœ‹
= 2 𝑀𝐻𝑧
iii.
Wavelength = πœ† = 𝛽 = 0.1257 = 49.986π‘š
iv.
Propagation velocity = 𝑣𝑝 = πœ† βˆ™ 𝑓 = 49.986π‘š βˆ™ 2 βˆ™ 106 𝑠 = 99.97 βˆ™ 106 𝑠
v.
Relative permittivity = 𝑣𝑝 =
1
𝐸̅
Μ… (𝑦, 𝑑) = =
B. 𝐻
πœ‚
100.0
πœ‚
𝑐2
𝑐
(3βˆ™108 )
√ π‘Ÿ π‘Ÿ
𝑝
π‘π‘œπ‘ (4πœ‹ × 106 𝑑 − 0.1257𝑦) π‘Žπ‘₯ π‘š =
√πœ‚0
100.0 βˆ™ √3
π‘π‘œπ‘ (4πœ‹ × 106 𝑑 − 0.1257𝑦) π‘Žπ‘₯
𝐴
=
π‘š
𝐴
=
π‘š
√377
𝐴
𝐻(𝑦, 𝑑) = 0.796 π‘π‘œπ‘ (4πœ‹ × 106 𝑑 − 0.1257𝑦) π‘Žπ‘₯
π‘š
π‘π‘œπ‘ (4πœ‹ × 106 𝑑 − 0.1257𝑦) π‘Žπ‘₯
Problem 2:
A magnetic field propagating in free space is given by
𝐻(𝑧, 𝑑) = 20. 𝑠𝑖𝑛(πœ‹ × 108 𝑑 + 𝛽𝑧) π‘Žπ‘₯
Find f, λ.
Solution:
2
, πœ‡π‘Ÿ = 1, πœ–π‘Ÿ = 𝑣2 = (99.97βˆ™106 )2 = 9.005
πœ‡ πœ–
𝐴
100.0 βˆ™ √πœ–
π‘š
𝐴
π‘š
πœ”
πœ‹×108
𝑐
2πœ‹
3βˆ™108
i.
Frequency = 𝑓 = 2πœ‹ =
ii.
Wavelength = πœ† = 𝑓 = 50βˆ™106 = 6π‘š
= 50 𝑀𝐻𝑧
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