along with these how can a power line data transmission is enabled with these higher frequencies. can any one suggest me right from the basic to develop my own circuit regarding this.\n\nthanks in advance
121180
High frequency signals can be coupled onto a high voltage power line through the use of capacitors.  The capacitor works as a high-pass filter, keeping the high voltage from entering back into the signal generation circuit. \n\n\nCurrently, broadband over powerlines (BPL) is in development, and is actually being tested in a few communities.  The intent is to be able to deliver high speed internet access to power customers over their electrical power lines.
224080
That's because its supposed to be secret! Something like the frequency and much more. We aren't permitted to braodcast anything above permitted frequency, as a part of general public. (Frequency emission includes, usage of home-RF devices) Please make sure to get information from the military if you are making a geniune effort, other than, putting the stake at risk!
173980
yes you can. For real low freq. you can actually do a DC measurement. Just slow the sweep rate way down and you will see it.
536943
You will need an adapter.\nhttp://en.wikipedia.org/wiki/List_of_countries_with_mains_power_plugs,_voltages_and_frequencies
499675
Well, you could build a *relatively* simple circuit that simply cut the frequency in half, but then again, what kind of sensor are you dealing with?\n\nWhy do you want to tune it to a lower frequency? It is not simple to just shift a frequency, there are many things to take into account; input impedance, output impedance (both may be frequency dependent), signal amplitude, and required voltage swing.\nI would only attempt this if it was not a vital sensor I was messing with.
89377
This site has a wealth of information. I often use it as a starting point when looking for a new type of circuit:\nhttp://www.discovercircuits.com/list.htm
423137
Would help tremendously to know what frequency your transmitter works on....!\nBut have a look at these guys, they might have what you are looking for\nhttp://www.rentron.com/RF-Remote.htm
54960
You can always put a modulated signal over the 60Hz (or 50Hz if you are in other parts of the world) carrier.  There used to be a stereo system which used this concept to remote speakers around the house.
332109
It is impossible to give the simple answer for which you seem to be searching.\n\nThere are many frequencies which could be used to cover that distance.\n\nI recommend you read about radio communication in general and about h.f., v.h.f. and u.h.f. communications in particular.
388087
Your transmitter will work, your receiver won't.\nbut you could bounce the signal onto a simple solar cell\nconnected to a set of headphones and you should be\nable to receive a weak signal...\nYour receiver won't work because light won't move the paper and the grains of sand.....
83812
Go here for an automatic HJT analysis:\nhttp://hjt.networktechs.com/
256301
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.
214633
You've actually got several concepts and questions merged together.\n\nWhen you ask "how does the aerial setup the waves from only the ac signal?"  I assume you're asking how any DC components of the signal get filtered out?\n\nA transverse electromagnetic wave involves both a VARYING electric field and a VARYING magnetic field, appearing at right angles to each other and to the direction of travel of the wave.\n\nAny DC component has no varying component, therefore, no wave is generated.  Although DC flowing through a wire creates electric and magnetic fields, no wave is generated because there is no variance in the fields.  \n\nThere are other electronics concepts that require the changing electromagnetic fields of AC to work.  A transformer uses AC, not DC.  A capacitor will block DC signals, but allow AC signals to pass.\n\nAlso, as you indicated, an efficient antenna has a matched impedance to the signal source in order to maximize the amount of engergy propagated by the antenna in the desired direction(s).
66455
you can only run so many things on soo many frequencies, if everything was wireless we would all have brain cancer from the emi and everything would interfere, just like when your cell phone goes off and it makes your computer screen twitch
29612
Hmmm... good question... exactly what are you trying to do?\n\nMaximum current would depend on the gauge, type, and quality of wire used to make the cable.  Not all cables are made the same... some are higher quality and provide better transmission distances while others are generic, cheap, and won't handle the same load.\n\nIf you're an electronics person... it's kinda like comparing copper to aluminum... so the answer would be... it really depends on your cable...\n\nLater!
319718
well go to control panel-power settings and adjust it from there ... make sure you dont have the settings too sensitive ... enjoy ..
3837
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