what is the antima diameter required for a microwave signal of 6GHz ?
374573
The diameter is unimportant, it is the length that matters.\n\nA frequency is related to a wavelength by:\nThe speed of light = wavelength x frequency.\n\nThis wavelength is the length of the aerial (Simplisticly) there are a number of modifiers that affect the length too, but this is a basic start
580858
Spectrum analyzer.  See Berkeley Varitronics Systems BUMBLE BEE below has a handheld tool for wireless LAN frequencies of 2.4 GHz.  You can get other instruments for other frequencies -- some spectrum analyzers will frequency sweep.\n\nA very low cost solution, although not quantative, is to use a portable AM radio tuned away from a station and the volume turned up -- carry this around the room or use from your car's radio.\n\nSee the US Frequency Chart below.
925425
If you want to build it to a specific value, I recomend getting a copy of The Radio Amateurs Handbook from your local library.  It has info and formula for air core, ferrite core, and toroid core inductors.  You might try their website www.arrl.com
737159
A typical router can handle about 128 computer or other device connections.  \n\nI would not worry about attenna parameters, unless you are a total geek.   But then you would have done a search at google and learned already. \n\nIf one of your computers has a low signal strength, then look for solid obstructions.  Brick walls, chimneys and other solid stone or brick wil STOP the signal cold.  It needs straight Line Of Sight to work well.    Also, a cheap cordless phone can interfere with the signal.\n\nGood luck and Happy Computing!
434630
dont forget the formula I= R/U\nsort them and u will get the answer\n\n Current: 625mA (milli amperes)\nResistance: 19.2 ohms
341004
i think you are right, you only need one.\n\nw = h*c/wavelength-eV\n\nboth should give the same w, about 1.9eV, \n\nwavelentgh = h*c/w
698460
Assume microwaves travel at the speed of light...\n\nv = 300,000,000 m/s\nwavelength(lambda) = 16 cm = .16 m\n\nFrequency = v/lambda\n\nFreq. = 300,000,000/.16\n\nFreq. = 1.875E9 Hz... or 1.875 GHz
143763
a signal emmited in the band of the 2 GHz.
65252
High frequency radio waves. Around 1700 Mhz.
492234
4.0 mmols to around 7.0 mmols
1060977
you can buy a 6 foot extesnsion cable
483654
You usually have the point to make it shorter. so 1.6ghz would be 1160. 2.6 is 2600, 2.0 is 2000 etc
277573
Uh, no software will make a standard computer without a special radio reception card of some type sensitive to "radio waves".  If you are referring to the Wi-Fi specification, Blue tooth or some other specification, there 'could' be a group of API calls or a Java collection that could allow the specification to work.  Otherwise you are stuck with assembly language or C programs written to take control of a specialized piece of hardware.  All in all, this isn't a real good answer, simply because the question doesn't give enough detail of what you are using and wanting to achieve.  Re state it with more detail and maybe someone can give you a better answer.
445689
Radio shack has a book (about $20) that has every freq you could ever need in it , even your local #'s
880368
For the best compromise of cost vs performance I would go with a 1/2 wavelength inverted V.  Support the center feed point at 150-200 ft above ground.  This antenna will be broadly non-directional.  The feed point will be close to 50 Ohms Z. For additional gain and small broadside directivity.  You can lengthen the V to One wavelength.  This creates two half wavelengths in phase.  The feed point will be high-Z at a voltage loop requiring a matching network.  Hope this helps you out.
860088
A typical wirless card puts out about 50 mW (50 thousandths of a Watt) of RF radiation.  At one meter away, that power drops to about 20 pico Watts (or 20 billionths of a Watt).  Bottom line, there's very little power right next to the thing and there is essentially no RF power 36 inches away.  Relax.
898015
Nominal= 36ohms\nTolerance=5% \nMultiply nominal by .05 , then add that figure to nominal for Max,.subtract it from nominal for Min.,,and study your ohm's LAW!! ;-)=