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.net barcode reader library Highfrequency dipole and other doublet antennas in Software
150 Highfrequency dipole and other doublet antennas Data Matrix ECC200 Recognizer In None Using Barcode Control SDK for Software Control to generate, create, read, scan barcode image in Software applications. Paint ECC200 In None Using Barcode maker for Software Control to generate, create Data Matrix image in Software applications. 35 : 1 Reading DataMatrix In None Using Barcode decoder for Software Control to read, scan read, scan image in Software applications. Painting ECC200 In C#.NET Using Barcode creation for .NET Control to generate, create ECC200 image in VS .NET applications. VSWR
Encoding Data Matrix In Visual Studio .NET Using Barcode maker for ASP.NET Control to generate, create Data Matrix ECC200 image in ASP.NET applications. Data Matrix 2d Barcode Creator In .NET Framework Using Barcode generation for .NET Control to generate, create Data Matrix image in .NET framework applications. 15 : 1 B D 1:1 Lower band edge C Design frequency FREQUENCY 66 VSWR versus frequency for several cases E Upper band edge Encode DataMatrix In VB.NET Using Barcode creator for .NET framework Control to generate, create Data Matrix ECC200 image in .NET framework applications. Barcode Generator In None Using Barcode generator for Software Control to generate, create bar code image in Software applications. the band Curve E represents an antenna that is resonant outside the upper limit of the band, so this antenna is too short, and must be lengthened Because this situation is possible, the antenna elements are typically made longer than needed when they are first cut How much to cut That depends on two factors: how far from the desired frequency the resonant point is found, and which band is being used The latter requirement comes from the fact that the frequency per unit length varies from one band to another Look at an example of how to calculate this figure The procedure is simple: 1 2 3 4 Calculate the length required for the upper end of the band Calculate the length required for the bottom end of the band Calculate the difference in lengths for the upper and lower ends of the band Calculate the width of the band in kilohertz by taking the difference between the upper frequency limit and the lower frequency limit 5 Divide the length difference by the frequency difference; the result is in kilohertz per unit length Example 61 Calculate the frequency change per unit of length for 80 and for 15 m Solution For 80 m (35 to 40 MHz): Make ANSI/AIM Code 128 In None Using Barcode encoder for Software Control to generate, create Code 128 Code Set A image in Software applications. UCC.EAN  128 Generation In None Using Barcode printer for Software Control to generate, create EAN128 image in Software applications. The dipole radiation pattern 151 468 = 117 ft 4 MHz
Make Code 3 Of 9 In None Using Barcode creator for Software Control to generate, create Code39 image in Software applications. Paint European Article Number 13 In None Using Barcode printer for Software Control to generate, create UPC  13 image in Software applications. 1 Lft = 2 Lft =
Print Identcode In None Using Barcode maker for Software Control to generate, create Identcode image in Software applications. Bar Code Reader In Java Using Barcode scanner for Java Control to read, scan read, scan image in Java applications. 468 = 1337 ft 35 MHz 3 Difference in length: 1337 ft 117 ft = 167 ft 4 Frequency difference: 4000 kHz 3500 kHz = 500 kHz frequency 500 kHz 5 Calculate : = 30 kHz/ft unit length 167 ft For 15 m (210 2145 MHz): 1 Lft = 2 Lft = 468 = 2182 ft 2145 468 = 2229 ft 21 Code 128 Code Set A Decoder In Visual Studio .NET Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET framework applications. Bar Code Maker In None Using Barcode encoder for Word Control to generate, create barcode image in Office Word applications. 3 Difference in length: 2229 ft 2182 ft = 047 ft 4 Convert to inches: 047 ft 12 in/ft = 564 in 5 Frequency difference: 21,450 kHz 21,000 kHz = 450 kHz frequency 450 kHz 6 Calculate : = 80 kHz/in unit length 564 in At 80 m, the frequency change per foot is small, but at 15 m small changes can result in very large frequency shifts You can calculate approximately how much to add (or subtract) from an antenna under construction from this kind of calculation If, for example, you design an antenna for the socalled net frequency on 15 m (21,390 kHz), and find the actual resonant point is 21,150 kHz, the frequency shift required is 21,390 21,150, or 240 kHz To determine how much to add or subtract (as a first guess): 1 The factor for 15 m is 80 kHz/in, which is the same as saying 1 in/80 kHz 2 The required frequency shift is 240 kHz 3 Therefore: Length change = 240 kHz = 3 in 1 in 80 kHz [612] [613] Bar Code Recognizer In Java Using Barcode recognizer for Java Control to read, scan read, scan image in Java applications. Make Data Matrix In ObjectiveC Using Barcode generation for iPad Control to generate, create ECC200 image in iPad applications. Each side of the antenna must be changed by half of the length calculated above, or 15 in Because the first resonant frequency is less than the desired frequency, the length should be shortened 15 in Once the length is correct, as proven by the VSWR curve, the connections at the center insulator are soldered and made permanent, and the antenna rehoisted to the operating level Impedance matching The difference between resonance and impedance matching is seen in the value of the VSWR minimum While the minimum indicates the resonant point, its value is a measure of the relationship between the feedpoint GS1  12 Recognizer In Java Using Barcode scanner for Java Control to read, scan read, scan image in Java applications. Scan Barcode In .NET Framework Using Barcode Control SDK for ASP.NET Control to generate, create, read, scan barcode image in ASP.NET applications. 152 Highfrequency dipole and other doublet antennas impedance of the antenna and the characteristic impedance of the transmission line Earlier in this chapter you learned that 1 Zo > Rr: VSWR = 2 Zo < Rr: VSWR = where Zo is the coaxialcable characteristic impedance Rr is the radiation resistance of the antenna Although knowledge of the VSWR will not show which situation is true, you can know that there is a high probability that one of them is true, and you can experiment to find which is the case Of course, if the VSWR is less than about 15:1 or 2:1, then forget about it the improvement is not generally worth the expense and cost When coupled to a transmitter that is equipped with the tunable output network (most tubetype transmitters or final amplifiers), then it can accommodate a relatively wide range of reflected antenna impedances But modern solidstate final amplifiers tend to be a little more picky about the load impedance For these transmitters, a coaxtocoax antenna tuning unit (ATU) is needed Rr Zo [615] Zo Rr [614]

