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Quantization Error Voltage

  1. Aug 19, 2010 #1
    1. The problem statement, all variables and given/known data

    Q1
    A linear PCM system has an input signal 2cos6000PIt volt. Determine,
    (a) the minimum sampling rate required,
    (b) the number of bits per PCM codeword required for a signal to quantization noise ratio of at least 40 dB,
    (c) the maximum quantization error voltage,
    (d) the dynamic range in dB.


    (a) 6000 Hz (b) n = 7 (c) 15.63 mV (d) 42 dB




    3. The attempt at a solution

    a) By Nyquist theorom , for a analog signal to be accurate reproduced, it should be sampled at a rate of not less than 2 times the highest frequency.

    2cos6000PIt = 2cosPIft
    therefore highest f= 6000/2 = 3000Hz
    Sample f= 2 x 3000Hz = 6kHz

    b)SNq = 6n (in dB)

    Therefore 40dB <=6n (bigger or equals to)
    n= 7 bits

    c) How to Do


    d)Dynamic Range = 6n = 6*7 = 42 dB
     
  2. jcsd
  3. Aug 19, 2010 #2

    Zryn

    User Avatar
    Gold Member

    This quote says that given the highest frequency component of B Hz, 2B Hz is sufficient a sampling rate to prevent aliasing.

    Nyquist requires a sampling frequency greater than twice the highest frequency component. Thus given the highest frequency component of B Hz, you must have 2B + 1 Hz sampling rate to prevent aliasing.

    Lets say you have an analogue sin wave with an amplitude of x, and you split that wave into 2^n different voltage levels for your digital approximation. How many volts are there per digital voltage level? Now lets say that the analogue signal is right smack bang in the middle of one of these digital levels, what happens if you go up to the next level compared to down to the previous level? This is the choice that represents the quantisation error at its worst (or maximum).
     
  4. Aug 19, 2010 #3
    So Voltage peak = 2 V
    V peak to peak = 2*2 = 4V
    Num of quantization levels = 2^n = 2^7 = 128 levels
    error voltage = (4/128) /2 = 15.63mv (2 deci place) <-correct buT

    BUT why use number of quantization levels?shouldt we use number of steps than find step size. Than find the middle value of the step size?
    V peak to peak = 2*2 = 4V
    Number of steps = 2^n -1 = 2^7 -1= 127 levels
    error voltage = (4/127) /2 = 15.75mv (2 deci place) <-wrong ans though.
     
  5. Aug 19, 2010 #4

    Zryn

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    Gold Member

    A fair question!

    If you draw a picture using 1sin(t) and quantize this signal using 1bit, and then 2 bits, how does that work?

    There's a trick to where you place your quantization levels that makes all the difference! I drew a (very rough) picture to illustrate the problem which I believe is the basis of your confusion.
     

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