CS485 Sylabus

### Sampling rate and accuracy --- the Nyquist rate

• Sampling rate and accuracy of the reconstructed analog signal

• Re-constructed signal and the original signal using a low sampling rate:

• Re-constructed signal and the original signal using a higher sampling rate:

• Conclussion:

 The faster (more frequent) you sample, the more accurate the approximation

• \$64,000 question:

 Should we sample as fast as possible ????

No !!!! The reason is cost !!

• Transmission rate

• Transmission rate:

 Transmission rate = the amount of data (in bits) transmitted per time unit (sec)

• Fact:

 We should try to keep the transmission rate as low as possible (More data sent = higher cost)

• Sampling rate and transmission rate

• Data rate at a low sampling rate:

Conclusion:

 We only need to sent 5 values when we sample at the lower sampling rate

• Data rate at a higher sampling rate:

Conclusion:

 We must sent 9 values when we sample at the higher sampling rate !!!!

• Therefore:

 The faster (more frequent) you sample, the higher the (transmission) data rate

• What is the best sampling rate ???

• High sampling rate:

 Good accuracy High transmission data rate (costly)

• Low sampling rate:

 Poor accuracy Low transmission data rate (cost-efficient)

• Goal: (what we want)

 At which data rate can we achieve good enough accuracy ????

• The Nyquist Sampling Rate

• The Nyquist (sampling) rate:

• Nyquist (sampling) rate = the minimum sampling rate required to avoid aliasing when sampling a continuous signal.

See: Wikipedia

• Aliasing:

 an effect that causes different signals to become indistinguishable when sampled.

See: Wikipedia

• In plain English:

 Nyquist rate = the lowest possible sampling rate that permits an accurate reconstruction of a input signal using samples

• Disclaimer:

 We will not derive the Nyquist rate :-)... because it requires a lot of background knowledge and is beyond the scope of this course...

• I will only state the result from Signal Processing:

 ``` Nyquist Rate = 2 × B where: B = highest frequency found in the (analog) input signal ```