What's the top frequency you can operate with on a ham radio?
489817
Some where around 300 ghz. Go to www.hamlink/freq.list.html
197001
a Scanner\n\nMost can be programmed for a several different frequencies which you then scan. Or you can set it for scan all frequencies between two.\n\nThe Federal govt. FCC has dictated frequency bands for certain functions. Scanners cover the non-broadcast radio bands between 30 and 950 MHz.
262987
I live in Los Angeles and we have a station at 97.1 FM
177735
Its not a ham radio, but can be if you want to pay to have it re-crystaled and aligned. Its a business band radio, but was sometimes used by fire or EMS.\n\nThe Motorola HT440 was either VHF or UHF depending on model number. VHF is 136-174 Mhz, UHF had two models, 403-430 Mhz, or 440-512 Mhz. \n\nIt is a 4-channel crystal radio, so converting to ham would require new crystals and re-aligning.\n\nModel H##HMB#124A is the VHF \nModel H##HMU#124A is the UHF
522363
you can use a fm modulator
445689
Radio shack has a book (about $20) that has every freq you could ever need in it , even your local #'s
505284
Sure, I have a scanner and the website I use to find frequencys is http://www.radioreference.com/modules.php?name=RR&stid=17 . \n\nHope that Helps :)
548231
Find or build a jammer for frequencys above 820 and below 900mhz. Set the jammer to where the calls are made. Have a antenna wire run throughout the area. Good luck.
238014
89.3  <<FM  \n89.5  <<FM
834597
No \nThere are no optimal frequencies for a transformer.
1012888
none at 91.4.. got a joy fm in florida at 91.5....\n\n\nhttp://www.thejoyfm.com/Home.aspx?pageID=3
216733
Here's a link that'll help you best\nhttp://radio.about.com/library/weekly/aa013003c.htm
572837
RTFM\nRead the fricken Manual.
208893
http://radioreference.com/modules.php?name=RR&stid=36
298935
TRANSMITTER ,RECEIVER, ANTENNA COAX SWITCH, COAXIAL CABLE FOR ANTENNA, ANTENNA,  ANTENNA STRUCTURE. 2 EACH. MAKE A SYSTEM.\n\nOR\n\nTRANSCEIVER , COAX , ANTENNA, ANTENNA SUPPORT! 2 EACH. MAKE A SYSTEM.\n\n\nTRANSCEIVER COULD BE FM, OR AM.  AIRCRAFT USE AM FOR PLANE TO PLANE AND PLANE TO GROUND.  POLICE , FIRE, UTILITIES AND MOST LAND MOBILE USE NARROW BAND FM.  CB IS AM MODULATION.  THE  NEW CB IN HIGHBAND IS 900 MHZ FM THIS IS WHAT YOU CAN GET AT WALMART IN THE LITTLE WALKI-TALKIES! WITH A 6 MILE RANGE!\n\nA SYSTEM IS SYMPLY A SENDING STATION AND A RECEIVING STATION.  IT CAN BE VERY COMPLEX OR AS SIMPLE AS A SET OF WALKIE TALKIES.  A SYSTEM IS A FUNCTIONAL DEFINITON FOR SOMETHING THAT WORKS!\n\nA SYSTEM COULD REFER TO A SYSTEM THAT YOU USE AROUND YOUR HOME TO KEEP IN COMMUNICATION WITH YOUR FAMILY!  OR IT COULD BE A SYSTEM THAT A BUSINESS USES TO STAY IN TOUCH WITH ITS EMPLOYEES!  A SYSTEM IS JUST A FUNCTIONAL DEFINITION OF A OPERATIONAL "SYSTEM". IT IS THAT SIMPLE!\n\nHOPE THAT CLEARS THAT UP FOR YOUR!\n\n\nDO ME A FAVOR AND VOTE TO BREAK THE TIE ON YOUR QUESTION ON RADAR AS A DISTANCE FINDER IF YOU HAVENT CHOSEN A ANSWER YET SO THAT ONE OF US WILL GET OUR POINTS !\n\n\nTHANK YOU!!!
10500
I think the previous answer is geared towards channel allocation and not bit rate vs bandwidth.  I will try to clarify the issue.\n\nFirst, you must understand how FM works and how a digital signal converts to FSK (frequency shift keying).  Therefore I will use a simple example.\n\nEach bit takes up some frequency.  For example, the bit '1' could take up 5 Hz carrier frequency on the bandwidth spectrum and '0' could take up 6 Hz on the bandwidth spectrum.  Therfore, there is a seperation of 1 Hz between bits.  \n\nIf I send a '1' followed by a '0' then on a meter I should see 5 Hz freqency and then a 6 Hz frequency back to back.  We could speed this up to infinity and therefore have these 2 frequencies transmitting all data at infinite speeds.\n\nIn the real world however, a '1' may fall around the 5 Hz frequency but not exactly on it.  It may be about 5 Hz.  Why around 5 Hz and not exactly on it everytime?  Because in the real world we have noise in transmission lines.  Noise can distort the frequency causing it to be a little less than 5 Hz or a little more than 5 Hz.  Imagine if the '1' bit is transmitted but on the receiving end it ends up receieved as 5.5 Hz.  What happens now?  It's not quite 5 Hz and it's not quite 6 Hz.  Noise caused it to move 0.5 Hz away from 5 Hz.  It is in the middle of 5 Hz and 6 Hz.\n\nNow the receiving end has to decide if this is '1' or '0'.  What are the odds the receiving end gets it right?  By slowing down the speed the 1s and 0s that are transmitted , you can hold steady the 5Hz or 6 Hz for that bit.  The faster you go, the less accurate the bit will fall exactly where it should on the frequency.\n\nSo, what are some possible solutions?  Well, you can slow the frequency down or make the frequency seperation larger.  For example, make a '1' be 5 Hz and make '0' be 10 Hz.  Now you have 5 Hz seperation between bits instead of 1 Hz seperation.  Now if the frequency of 5.5 Hz is received, the receiving end can determine that this should have been 5 Hz.  It's highly unlikely it was originally transmitted at 10 Hz.  Therefore, the receiving end detects a '1'.  \n\nBut what have you lost by doing this increase in frequency seperation?  You have taken up more bandwidth.  Every cable has only so much bandwidth and if you have several channels on a cable all transmitting at the same time, you need to limit the amount of bandwidth each channel takes up or you will run out.\n\nSo, therefore, since most companies don't want to backdown on speed to the consumer (you), then they upgrade their technology (i.e. cables) to higher bandwidth cables such as fiber optic cables.\n\nI hope this helped.  Others, please feel free to expand on this topic.  I've only discussed FSK modulation.  Other techniques such as PSK, ASK, QPSK, bit stuffing, NRZI, CRC, and FEC codes can improve bit rate over channels.\n\nYou can read more about bandwidth limits.  Shannon and Hartley came up with a theorem which still holds today and is used quite extensively.  I even know someone who worked with Shannon at Bell Labs back in the day.  Shannon's theorem is C = BW*log2(1+S/N) where C is the channel capacity in bits per second,  BW is bandwidth, S is the signal power and N is the noise power.\n\nWF\nTelecom Engineer
665774
ELF --> Extremely Low Frequency radio waves