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Dipole antenna Simulation by HFSS

Work frequency 1GHz, new project, add new design, edit add the following variablesAn antenna model is established, which is two metal cylinders. After editing a oscillator, the other half can use the image command to create the parameters as follows, the material is PECA rectangular sheet is formed between two cylinders, connecting two cylinders to feed the power.The excitation method for setting the rectangular face is the lumped port excitation. Fro

POJ 3020 Antenna Placement binary map matching

Click to open the link topic linkAntenna Placement Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7386 Accepted: 3651 DescriptionThe Global aerial Centre have been allotted the task of building the fifth generation of mobile phone nets in swed En. The most striking reason so they got the job, is their discovery of a new, highly noise resistant, antenna. It's called 4DAir, and

Antenna Placement (Hungarian algorithm, least path overlay)

Antenna Placement Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6991 Accepted: 3466 DescriptionThe Global aerial Centre have been allotted the task of building the fifth generation of mobile phone nets in swed En. The most striking reason so they got the job, is their discovery of a new, highly noise resistant, antenna. It's called

Bluetooth antenna artboard Attention point

Figure 1 is the cc2540 of the package fixed ground to facilitate heat dissipation, so there is no way to line.Figure 2 here to make as many holes as possible, the ground as large as possible, do not appear in the ANT1 to go to the place where there is no connection to a pieceThe white box is the antenna as far as possible after the connection line, so try to add a little bit, and the antenna and the ground

POJ 3020-antenna placement-binary map matching

Test Instructions: There are k points in a n*m matrix, and you must place the antennas over all the observation points. Each radar can only have two observation points on the antenna, and the two points must be adjacent. Calculates the minimum number of antennas.Practice: Connect all adjacent observation points and build a map. Run it once. The Hungarian algorithm calculates the maximum number of covers, divided by two, the number of antennas. Also ad

Current Situation and Future Trend of RFID-1.7 antenna shape and material of RFID tags

Antennas are rectangular, circular, and rod-shaped. Generally, the longer the communication distance is, the larger the size of the antenna. However, in actual use, when tags are attached to an item, even the RFID tags of the business card size are too large. In fact, the size and shape of the RFID tag in use need to be selected based on the communication distance. Currently, antenna materials include al

Poj 3020 antenna placement (Bipartite Graph Matching)

Label: style blog HTTP Io color OS AR for SP Poj 3020 antenna placement Question Link Question: Given a map, the place of '*' should be covered. Each time we can use a rectangle of 1X2 to cover the map, we can use at least a few to cover the map. Idea: bipartite graph matching to find the Maximum Independent Set, adjacent * edge, and then find the maximum independent set.Depending on the data range, you can also use the outline DP. Code: #in

"POJ 3020" antenna Placement

"POJ 3020" antenna PlacementThe maximal independent set problem of binary graphs' O ' means that the break point requires that all * connections be connected to one or two consecutive lines per route *The maximum match can satisfy only the longest path that can be formed by only two consecutive connections, and the points that are not connected need to be circled separately.i.e. N (total * number)-m (maximum match) +m (maximum match)/2(due to the fact

Antenna package the j2s program

Antenna package the j2s programAuthor: JeruMany people like to use ant to compile and package Java applications. They can easily generate war packages and run them in J2EE containers. I also saw an article on sun's website.How to use ant to package the j2_program: http://wireless.java.sun.com/midp/articles/ant/I tested it. It's not bad, but the disadvantage is that I need to write the manifest. mf and jad files in advance, so I am unwilling to

Antenna Package J2ME Program

Program | Package Antenna Package J2ME Program Author: Jeru Many people like to use ant to compile packaged Java applications, and it's easy to generate war packages running in the Java EE container, and I also saw an article on Sun's website tutorial on using ant to package J2ME programs: http://wireless.java.sun.com/midp/articles/ant/ Test a bit, but also good, but the drawback is that the need to write their own MANIFEST.MF and jad documents, so

Antenna Magus 5.5.0 Pro+compusoft.winner.v9.0a2+eplan Harness ProD Studio 2.5

Antenna Magus 5.5.0 pro\Compusoft.winner.v9.0a2_hao\EPLAN Harness ProD Studio 2.5\Ixray Ixforten 4000 version 4.9.8uu\Mintec.minesight.3d.v7.0.3\NI. Vision.v2015.SP1 1DVD NI Vision 2015\Pix4d.pix4dmapper.pro.v2.0.104.macosx\Techlog learning materials software training \Techlog2014.2\multiphase flow simulation software Schlumberger OLGA 2015.1\[CATIA. P2. V5r16 catia.cadam.drafting.v5r16sp1\338\CPFD. Barracuda.v17.0.0.rarHypermill8.4.zipMpcci v3.03 Lin

POJ-3020 Antenna Placement Minimum path overlay

The main problem: There are N cities, to establish a radio station in these n cities, each radio station can only cover 2 neighboring cities, ask at least how many wireless power stations need to buildProblem-Solving ideas: English topic good pit, looked at half a day.This problem is similar to POJ-2446 chessboard.All cities can be divided into two point sets, then the connection between the radio station to represent the coverage of the relationship.Because all cities have to be covered, so acc

Poj3020 antenna placement (Minimum side coverage)

Question: In a rectangle, there are n Cities '*'. Now all N cities need to cover wireless networks. If a base station is placed, it can cover at most two adjacent cities.How many base stations can be placed to ensure that all cities are wireless? Ideas: Number each city, create two-way edge, run Hungary, and then the number of cities-matching number/2 is the answer Assume that each city needs to build a base station, and then how many matches will reduce the number of base stations. Because the

Poj 3020 Antenna Placement Bipartite Graph Matching

Black and white coloring and then finding the maximum matching of the Bipartite Graph [cpp]# Include # Include # Include Using namespace std;Const int maxn = 40 + 9;Int n, m;Char a [maxn] [maxn];Struct{Int to, next;} E [maxn * 10*4];Int head [maxn *

Poj 3020 antenna placement, minimum path overwrite of the Bipartite Graph

Question: In a rectangle, there are n Cities '*'. Now all N cities need to cover wireless networks. If a base station is placed, it can cover at most two adjacent cities.How many base stations can be placed to ensure that all cities are wireless?

Poj 3020 antenna placement solution report

Question link: http://poj.org/problem? Id = 3020 First of all, ignore the one that may mislead others (maybe I have poor understanding, or I am misled by a little bit ). An H * w graph is provided. "*" indicates point of interest, and "O" ignores it.

Poj 3020 antenna placement Hungary Algorithm

There are n cities in a rectangular grid. Now all N cities need to be covered in wireless networks. For any base station, it can overwrite any adjacent two grids. So if we build a base station together, we can make all the cities covered and use the

Antenna placement poj3020

This is a classic binary match. After the grid is marked as black and white, the grid containing '*' is then divided into two groups based on black and white to create edges. The final result is the total number of '*'-the number of matches. #

Poj 3020 antenna placement (binary matching)

This question requires a more detailed understanding of binary matching! At the same time, you need to use a little bit of brains! Unlike hdu4185, the idea of creating a graph is that 4185 does not need to cover all vertices, but this question

Poj 3020 antenna placement

The basis of binary matching. You only need to find out a and B. Search for the four directions of each "*" and record their relationships; Because we read both a-> B and B-> A, we need to divide it by 2; #include#includechar str[41][11];bool mat[40

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