The Nyquist theorem and the re-understanding of Shannon's theorem

the Nyquist theorem and the re-understanding of Shannon's theorem

@ (computer network)

About the code element, first review the concept:

http://blog.csdn.net/u011240016/article/details/53333682 Nyquist theorem

Under ideal low-pass conditions – no noise, limited bandwidth channels, and a limit code element rate of 2W baud.
W is the channel bandwidth, in Hz.

The number of discrete states for each code element is expressed in V, then:

Limit transfer rate = 2wlog2v 2wlog_2v

The emphasis is on Shannon's theorem. Shannon's theorem

Bandwidth constrained channel limit transmission rate with Gaussian white noise interference.

Limit transmission rate of the Channel = WLOG2 (1+s/n) wlog_2 (1+s/n)

Specifically, Shannon's theorem mainly revolves around the calculation of S/N.

Because a pit is set up here, the S/n used in the calculation often is a large number, and the smaller representation is given in the topic.

S: The average power of the channel transmission signal.
N: Gaussian white noise power inside the channel.

If there is a ratio of this word, is the direct to the S/N, otherwise db this unit, you need a conversion. This conversion is very important and will bring two different answers to the topic, one right, one error.

That

XdB = 10LOG10 (S/n) DB 10log_{10} (S/n) db

In addition, the Shannon theorem has the following points: the greater the signal-to-noise ratio, the higher the limit transmission rate of the channel a certain transmission bandwidth and a certain signal-to-noise ratio, the upper limit of the information transmission rate determines that as long as the transmission rate of the channel is lower than the limit transmission rate, there must be progress in the space, it The actual transfer rate is much lower than that obtained by Shannon theorem.

In addition, the usage scenarios of the two theorems are summarized in this way, without noise directly with the Nyquist.

There is noise, if the Nyquist can be calculated, that is, given the conditions of V, then also to calculate, and Shannon's theorem to do a comparison. The minimum value is taken.

The classic example is the following:

The binary signal is transmitted over a 4kHz channel with a signal-to-noise ratio of 127:1, and the maximum data rate can be achieved:

A. 28000b/s
B. 8000b/s
C. 4000b/s
D. can be infinitely large

Analysis: See the signal-to-noise ratio, immediately get started can be calculated:

Rate = 4KLOG2 (1+127) =4k⋅7=28kbps=28000b/s 4klog_2 (1+127) = 4k\cdot 7 = 28kbps = 28000b/s

Do not rush to choose, to want to binary signal means that the direct use of a code is 2 states, that is, V = 2 o'clock, you can calculate the Nyquist.

Rate = 2⋅4k⋅log22=8000b/s 2\cdot 4k \cdot log_22 = 8000b/s

That would be interesting. The Nyquist is smaller than the Shannon, and the two are smaller.

With the above accumulation, you can look at the following practical very simple question:

(2016.34) If the connection R2 and R3 link frequency bandwidth is 8Hz, the signal to noise ratio is 30dB, the actual transmission rate of the link is about 50% of the theoretical maximum, then the actual transmission rate of the link is about: C
A. 8kbps
B. 20kbps
C. 40kbps
D. 80kbps

Analysis: See DB, immediately think of to convert.

10LOG10 (S/n) =30 10log_{10} (S/n) = 30

→s/n=1000 \rightarrow S/n = 1000

Resulting channel limit rate = 8k⋅log21000=80kbps,210=1024 8k\cdot log_2{1000} = 80kbps, 2^{10} = 1024

Because this is actually the half, so the result is 40kbps.

If the direct surrogate is 30, get the limit rate is 40kbps, then take half, the result is 20kbps, feel like, is actually wrong.

This is also the biggest trap of the topic, the basic grasp of the solid, then the problem is very concise, can be calculated soon.