as a general rule of thumb, about 5 times the highest frequency for quality reproduction, although one can sometimes get away with a minimum of 2 times.
7.5KHz
Study the optical transmission property of glass, to learn more. also: http://en.wikipedia.org/wiki/Optical_fiber You might wish to refine the question to differentiate between optical bandwidth and data bandwidth. If a laser can be modulated and the beam directed through an optical fiber, then the data or the analog bandwidth is about equal to the modulation bandwidth capability of the laser in Hz per second.
That is what that piece of equipment samples at. Others can sample at billions of samples per sec. It all depends on the A/D sample rate that was picked.
In data communication computer use periodic analog signal because it Need Less Bandwidth. So, by using Periodic Analog Signals it is easy to select the Medium through Which data is travelled otherwise it is Much Difficult (aprox. Impossible ) .
Digital data requires less bandwidth than analog data (e.g. new digital television broadcast compared to old analog broadcast), so more data can fit in the same frequency (e.g. DSL compared to analog modems). This is the primary advantage, although it is also generally stated that data is easier to recover and more tolerant towards interference.
Bandwidth is defined as a frequency span - the difference between a high frequency and a lower frequency. For the low end voice it depends if its male bass bariton tenor or female alto and soprano. A bass voice goes down to 100 Hz. The harmonics go up to 15 kHz. So the bandwidth for voices is arround 15 kHz.
The theoretical minimum sampling rate required in order to avoid frequency aliasing is 3.4 KHz.
Advantage of Analog Communication: 1)Transmission bandwidth required is less. 2)No need of synchronization.
PCM technique is used to convert analog voice signals into digital. In PCM the analog frequency is first sampled and then converted into binary bits. Each samples are taken as 8bits long. Basic communication theory requires that a minimum sampling rate of twice the frequency of the signal to be sampled will result in an accurate representation of the original signal.Human voice can have max 4000hz frequency, therefore sampling rate should be 8000 samples/sec.Which implies required bit rate for transmitting voice is 8000*8 = 64000 bits/sec = 64kbps.
6 Megacycles in analog. This includes picture, sound and guard bands.
it could be either one, depending on application. need more information to answer.
ssb modulation scheme
It must be an analog filter. The high frequencies must be attenuated before any digital sampling takes place, to ensure that these high frequencies are not sampled as false signals. See the related link.
Study the optical transmission property of glass, to learn more. also: http://en.wikipedia.org/wiki/Optical_fiber You might wish to refine the question to differentiate between optical bandwidth and data bandwidth. If a laser can be modulated and the beam directed through an optical fiber, then the data or the analog bandwidth is about equal to the modulation bandwidth capability of the laser in Hz per second.
Measurement of the capacity of a communications signal. For digital signals, the bandwidth is the data speed or rate, measured in bits per second (bps). For analog signals, it is the difference between the highest and lowest frequency components, measured in hertz (cycles per second). For example, a modem with a bandwidth of 56 kilobits per second (Kbps) can transmit a maximum of about 56,000 bits of digital data in one second. The human voice, which produces analog sound waves, has a typical bandwidth of three kilohertz between the highest and lowest frequency sounds it can generate.
That is what that piece of equipment samples at. Others can sample at billions of samples per sec. It all depends on the A/D sample rate that was picked.
In data communication computer use periodic analog signal because it Need Less Bandwidth. So, by using Periodic Analog Signals it is easy to select the Medium through Which data is travelled otherwise it is Much Difficult (aprox. Impossible ) .
Error resulting from trying to represent a continuous analog signal with discrete, stepped digital data. The problem arises when the analog value being sampled falls between two digital "steps." When this happens, the analog value must be represented by the nearest digital value, resulting in a very slight error. In other words, the difference between the continuous analog waveform, and the stair-stepped digital representation is quantization error.