What is the minimum data rate ?

Thread Starter

sama3505

Joined Jun 10, 2016
18
Include the all equations how to do this.
What is the minimum data rate needed if you wish to sample a 10V analogue signal with at least 20 mV precision and 512 samples per second?
 

Marley

Joined Apr 4, 2016
502
10V divided by 20mV = 500 steps minimum in the ADC.
500 steps could be represented by 2^9 bits (actually = 512). So a 9-bit binary value.
Transmitting 9 bits 512 times a second = 4608 bits per second.
This is an absolute minimum. In reality, there would have to be extra bits transmitted for timing and identification.
 

Thread Starter

sama3505

Joined Jun 10, 2016
18
so if we get 30 steps ,are we getting it as 2^5 =32 ,so its a 5 bit binary value ? right?
and did you multiply that 10v/20mv =0.5 x 100 as well ?
 

Marley

Joined Apr 4, 2016
502
Correct, 30 steps can be represented as a 5-bit binary value.
With 30 steps, the resolution will only be 10V / 30 = 0.333V (or 333mV).
So for more resolution you need more steps and more bits. To more accurately represent a changing analog value you need:
  • more steps = more bits
  • sample more often = more samples per second
So, more accuracy = greater data rate. This is one of the first rules of information theory.
 

Thread Starter

sama3505

Joined Jun 10, 2016
18
Correct, 30 steps can be represented as a 5-bit binary value.
With 30 steps, the resolution will only be 10V / 30 = 0.333V (or 333mV).
So for more resolution you need more steps and more bits. To more accurately represent a changing analog value you need:
  • more steps = more bits
  • sample more often = more samples per second
So, more accuracy = greater data rate. This is one of the first rules of information theory.
Thanks

can you have a look on this problem as well.
http://forum.allaboutcircuits.com/threads/how-to-find-the-torque-constant.124860/
 
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