Reliability of many valued electronic signals.

[mooncrater]

Homework Statement

In this book of mine , I saw this statement,"The digital-system designer is restricted to the use of binary signals because of the lower reliabilty of many-valued electronic circuits." . Why is so? How can we say that the many-valued electronic circuits are not reliable?

Homework Equations

~nil~

The Attempt at a Solution

A binary signal has 0s and 1s. A hexal signal has 0s, 1s, 2s, 3s, 4s, and 5s.
Therefore a hexal signal has more discrete input signals than a binary signal.
But the thing I don't understand is that why do they say the other signals are unreliable?

 

[axmls]

I believe it means "versus analog signals." For instance, any hex digit can be represented by 4 binary digits, so that's no problem. The issue is that noise interferes with analog signals, causing them to lose quality. Sure, digital signals have noise added to them too, but as long as the signal stays in the correct "zone", there is no loss in quality, because a 1 is a 1 is a 1 and likewise for a 0.

 

[mooncrater]

I believe it means "versus analog signals." For instance, any hex digit can be represented by 4 binary digits, so that's no problem. The issue is that noise interferes with analog signals, causing them to lose quality. Sure, digital signals have noise added to them too, but as long as the signal stays in the correct "zone", there is no loss in quality, because a 1 is a 1 is a 1 and likewise for a 0.

So, since there are many signals , noise (disturbances) would make nearby signals approximately the same, because of which we would not be able to differentiate anyone of those signals . Whereas , in the case of a binary signal , a 0 and 1 are too different , that noise would not affect them enough make them unrecognisable. Am I right?

 

[axmls]

Well, almost.

In the case of an analog signal, we may need some quite accurate values to produce a quality product. For instance, let's say we need the signal to equal 1.7 when x = 3.5. Then we had better hope no noise comes in and interferes with the signal right there, because it could easily change that value, and that messes with the quality.

For a digital (in this case, binary) signal, however, let's say a 1 is 5 volts and a 0 is 0 volts. Then what we do is, we set a range around 5 volts and a range around 0 volts, and declare anything in those ranges to be 1 or 0, respectively. So if we send out a signal that equals 5V at some point, and some noise gets in and changes it to 5.3V or 4.32V, then the receiver looks at it, decides it's closer to 5V, and declares it a 1, just like the transmitter intended. So we literally can get data that is as high-quality as the transmitter intended.

This is also why people who spend large amounts of money for "high quality" digital cables instead of a cheap 10$ cable have gotten scammed. With digital, you either have the correct value, or you don't. It is impossible to increase the quality. If the cable gets a 1 when it needs a 1 and a 0 when it needs a 0, then the quality cannot be improved.

Combine this with some clever error-correcting codes that detect when there's been a bit error, and digital transmission can be quite reliable.

 

[mooncrater]

Now its clear , thanks...:)