Explain frequency division multiplexing.?
81045
Frequency Division Multiplexing refers to sharing the wide frequency spectrum/band into multiple narrow channels or bins to send data on each one of the bins as if they had the entire narrow bandwidth to themselves. This is the basis for DSL, Wireless, Digital TV, etc.\n\nRef: http://en.wikipedia.org/wiki/OFDM
800734
Speed of waves = wavelength X frequency of waves.
254300
http://en.wikipedia.org/wiki/Time-division_multiplexing
252949
It's basically one way to encapsulate data.  If you send a 1 MB transmission it is chopped into smaller bits so that it can be sent efficiently.  One way to do this is frame relay, another way is ATM.
789468
Here's how an LTRT works:\n\nThe Lt/Rt downmix sums the surround channels, attenuates them 3 dB (i.e. multiply by 0.707) and adds them out-of-phase to the left channel and in-phase to the right channel\n\n The formula for Lt/Rt:\n\nLt = L + (0.707*C) - (0.707*Ls) - (0.707*Rs)\nRt = R + (0.707*C) + (0.707*Ls) + (0.707*Rs)
126863
The RF waves emanating from an antenna are generated by the movement of electrical charges in the antenna. Electromagnetic waves can be characterized by a wavelength and a frequency. The wavelength is the distance covered by one complete cycle of the electromagnetic wave, while the frequency is the number of electromagnetic waves passing a given point in one second. The frequency of an RF signal is usually expressed in terms of a unit called the "hertz" (abbreviated "Hz"). One Hz equals one cycle per second. One megahertz ("MHz") equals one million cycles per second.
283860
Frequency = speed of light/wavelength.\n\nMake sure you keep the units straight when performing the calculations or it may not make sense.\n\nSo for your problem its (3 x 10^17 )/200 = 1.5 x 10^15 Hz
14165
This is a high speed internet connection, usually through cable, dsl or wifi.\n\n"A transmission medium capable of supporting a wide range of frequencies, typically from audio up to video frequencies. It can carry multiple signals by dividing the total capacity of the medium into multiple, independent bandwidth channels, where each channel operates only on a specific range of frequencies."
445689
Radio shack has a book (about $20) that has every freq you could ever need in it , even your local #'s
38648
multiplexor\nIn communications, a device that merges several low-speed signals into one high-speed transmission and vice versa.\n\nmul·ti·plex \n\nRelating to, having, or consisting of multiple elements or parts: “the whole complex and multiplex detail of the noble science of dinner” (Thomas Love Peacock).\nRelating to or being a system of simultaneous communication of two or more messages on the same wire or radio channel.
715101
The station generates the signal.  The carrier wave has to be at the specific frequency for which they are licensed, and can not deviate from that any more than standard modulation is allowed.  They are also licensed for how much broadcast power they use.
85883
Fiber Optic and Satellite
1098183
The first part of the sound is the destination number that is being dialled.\n\nThis follows the DTMF which is Dual Tone Multi-Frequency. The signal generated by a DTMF encoder is a direct algebraic summation, in real time, of the amplitudes of two sine (cosine) waves of different frequencies. i.e. pressing '1' will send a tone made by adding 1209 Hz and 697 Hz to the other end of the line.\n\nThe subsequent sound is the protocol for handshaking, acknowledgement, and authentication to establish the link. Once a valid connection had been made, the sound will stop.
1160585
4 Hertz = 4 times per second
1010952
Following site has very simplified way of explaining this. You might find it useful. \n\nPFC decoded ... http://www.dansdata.com/gz028.htm
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
479254
A modulator is used to impress the information from a single source upon a radio wave or other system carrier, such as the process that happens when you play a VCR through channel three of your TV. The audio and video program are modulated onto a high frequency radio wave which can be received by your TV (demodulated). A radio wave can easily pass through space unlike the baseband video and audio signals.\n\n   A multiplexer may be thought of as a time sharing system in which several program sources are alternately routed to an output. An example of this would be in security systems where the pictures from several cameras are combined to be seen on one monitor screen.\n\n   Modulation has a 'one to one' relationship, as well as does its inverse, 'demodulation'. Multiplexing has a 'many to one' relationship. Its inverse, demultiplexing, has a 'one to many' relationship useful for extracting the various original signals.\n\nWK