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by: Thomas

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# 4nec2 simulation 1 ECE 4370

Thomas
AU

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antenna
COURSE
Antenna Engineering
PROF.
TYPE
Study Guide
PAGES
3
WORDS
KARMA
50 ?

## Popular in Electrical Engineering

This 3 page Study Guide was uploaded by Thomas on Sunday May 1, 2016. The Study Guide belongs to ECE 4370 at Auburn University taught by in Spring 2016. Since its upload, it has received 25 views. For similar materials see Antenna Engineering in Electrical Engineering at Auburn University.

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Date Created: 05/01/16
Problem 1:                           Figure 1: Dipole antenna current magnat 300 MHz Figure 2 Dipole antenna current magnitude  at 150 MHz This current look not as I would expect it to. Because the current magnitude is proportional to  cosβz, The plot should be a arc. Figure 3 Dipole antenna current magnitude  at 75 MHz The value of the current is 1.18+j3e­5A, 0.04+j2.71A, 8e­3+j5.58A for f=300MHz, 150 MHz, and  75MHz, and the value of the impedance is 72.1­j2e­3, 13.7­j893, 3.21­j2174 for f=300MHz, 150  MHz, and 75MHz. Problem 2: Figure 4: Dipole antenna current magnitude  at 600 MHz I redo the segmentation to 15, and from the Example1.out notepad, we could find the maximum  value of current on the dipole is 8.7730e­6 A. The locations are (0.0000 ,0.0004 ,0.0000) and  (0.0000, ­0.0004 ,0.0000). the input impedance is 3523+j218. Problem 3: “Auto Segmentation”, this setting used for optimizer and sweeper functions. Reduce for memory concerns, increase for accuracy.  The minimum value of SWR is 1.43 approximately, at frequency 299MHz. The wavelength at  8 6 this frequency is c/f=3*10 /299*10 =1.003m. The antenna­length­to­wavelength ratio at this  frequency is 0.02538/1.003=0.0253. When the radius of the dipole is a=0.01m. The minimum value of SWR is 1.42 approximately, at frequency 275MHz. The wavelength at this frequency is c/f=3*10 /275*10 =1.09m. The  antenna­length­to­wavelength ratio at this frequency is 0.02545/1.09=0.0233. When the radius of the dipole is a=0.05m. The minimum value of SWR is 1.59 approximately, at frequency 254MHz. The wavelength at this frequency is c/f=3*10 /254*10 =1.18m. The  antenna­length­to­wavelength ratio at this frequency is 0.02410/1.18=0.0204. The frequency of the lowest SWR is smaller and smaller as the radius of the dipole is bigger. Problem 4: Figure 5: Frequency sweep from 200 MHz to 1200MHz at 2 MHz steps From the data, the plot seems like periodicity.  Problem 5: Figure 6: 2D pattern plot with    90 Figure 7: 2D pattern plot with    90

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