# 6.17 Digital communication system properties

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Several properties of digital communication systems make them preferable to analog systems.

Results from the Receiver Error module reveals several properties about digital communication systems.

• As the received signal becomes increasingly noisy, whether due to increased distance from the transmitter (smaller  ) or to increased noise in the channel (larger ${N}_{0}$ ), the probability the receiver makes an error approaches $1/2$ . In such situations, the receiver performs only slightly better thanthe "receiver" that ignores what was transmitted and merely guesses what bit was transmitted. Consequently, it becomesalmost impossible to communicate information when digital channels become noisy.
• As the signal-to-noise ratio increases, performance gains--smaller probability of error ${p}_{e}$ -- can be easily obtained. At a signal-to-noise ratio of 12 dB, the probability the receiver makes an errorequals $10^{-8}$ . In words, one out of one hundred million bits will, on the average, be in error.
• Once the signal-to-noise ratio exceeds about 5 dB, the error probability decreases dramatically. Adding 1 dBimprovement in signal-to-noise ratio can result in a factor of 10 smaller ${p}_{e}$ .
• Signal set choice can make a significant difference in performance. All BPSK signal sets, baseband or modulated,yield the same performance for the same bit energy. The BPSK signal set does perform much better than the FSK signalset once the signal-to-noise ratio exceeds about 5 dB.

Derive the expression for the probability of error that would result if the FSK signal set were used.

The noise-free integrator output difference now equals $A^{2}T=\frac{{E}_{b}}{2}$ . The noise power remains the same as in the BPSK case, which from the probability of error equation yields ${p}_{e}=Q(\sqrt{\frac{^{2}{E}_{b}}{{N}_{0}}})$ .

The matched-filter receiver provides impressive performance once adequate signal-to-noise ratios occur. You might wonder whetheranother receiver might be better. The answer is that the matched-filter receiver is optimal: No other receiver can provide a smaller probability of error than the matchedfilter regardless of the SNR . Furthermore, no signal set can provide better performance than the BPSK signal set,where the signal representing a bit is the negative of the signal representing the other bit. The reason for this resultrests in the dependence of probability of error ${p}_{e}$ on the difference between the noise-free integrator outputs: For a given ${E}_{b}$ , no other signal set provides a greater difference.

How small should the error probability be? Out of $N$ transmitted bits, on the average $N{p}_{e}$ bits will be received in error. Do note the phrase "on the average" here: Errors occur randomly because of the noiseintroduced by the channel, and we can only predict the probability of occurrence. Since bits are transmitted at a rate $R$ , errors occur at an average frequency of $R{p}_{e}$ . Suppose the error probability is an impressively small number like $10^{-6}$ . Data on a computer network like Ethernet is transmitted at a rate $R=100\mathrm{Mbps}$ , which means that errors would occur roughly 100 per second. This error rate is very high, requiring a much smaller ${p}_{e}$ to achieve a more acceptable average occurrence rate for errors occurring. Because Ethernet is a wireline channel, which meansthe channel noise is small and the attenuation low, obtaining very small error probabilities is not difficult. We do have sometricks up our sleeves, however, that can essentially reduce theerror rate to zero without resorting to expending a large amount of energy at the transmitter. We needto understand digital channels and Shannon's Noisy Channel Coding Theorem .

#### Questions & Answers

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Joseph
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nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
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learn
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da
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Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
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narayan
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Bhagvanji
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Damian
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Professor
I think
Professor
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Rafiq
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Damian
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LITNING
scanning tunneling microscope
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Rafiq
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brayan
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Damian
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Renato
how did you get the value of 2000N.What calculations are needed to arrive at it
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