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Shannon derived the maximum datarate of a source coder's output that can be transmitted through a bandlimited additive white noise channel with no error.

In addition to the Noisy Channel Coding Theorem and its converse , Shannon also derived the capacity for a bandlimited (to W Hz) additive white noise channel. For this case, the signal set isunrestricted, even to the point that more than one bit can be transmitted each "bit interval." Instead of constrainingchannel code efficiency, the revised Noisy Channel Coding Theorem states that some error-correcting code exists suchthat as the block length increases, error-free transmission is possible if the source coder's datarate, B A R , is less than capacity.

C W 2 logbase --> 1 SNR  bits/s
This result sets the maximum datarate of the sourcecoder's output that can be transmitted through the bandlimited channel with no error.
The bandwidth restriction arises not so much from channel properties, but fromspectral regulation, especially for wireless channels.
Shannon's proof of his theorem was very clever, and did not indicate what this code might be; it has never been found.Codes such as the Hamming code work quite well in practice to keep error rates low, but they remain greater than zero. Untilthe "magic" code is found, more important in communication system design is the converse. It states that if your data rateexceeds capacity, errors will overwhelm you no matter what channel coding you use. For this reason, capacity calculationsare made to understand the fundamental limits on transmission rates.

The first definition of capacity applies only for binary symmetric channels, and represents the number ofbits/transmission. The second result states capacity more generally, having units of bits/second. How would youconvert the first definition's result into units of bits/second?

To convert to bits/second, we divide the capacity stated in bits/transmission by the bit interval duration T .

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The telephone channel has a bandwidth of 3 kHz and a signal-to-noise ratio exceeding 30 dB (at least they promise this much). The maximum data rate a modem can produce forthis wireline channel and hope that errors will not become rampant is the capacity.

C 3 3 2 logbase --> 1 10 3 29.901 kbps
Thus, the so-called 33 kbps modems operate right at the capacity limit.

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Note that the data rate allowed by the capacity can exceed the bandwidth when the signal-to-noise ratio exceeds 0 dB. Our results for BPSK and FSK indicated the bandwidth they requireexceeds 1 T . What kind of signal sets might be used to achieve capacity? Modem signal sets send more than one bit/transmissionusing a number, one of the most popular of which is multi-level signaling . Here, we can transmit several bits during one transmission interval by representingbit by some signal's amplitude. For example, two bits can be sent with a signal set comprised of a sinusoid with amplitudesof ± A and ± A 2 .

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can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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Source:  OpenStax, Fundamentals of electrical engineering i. OpenStax CNX. Aug 06, 2008 Download for free at http://legacy.cnx.org/content/col10040/1.9
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