This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

adc harmonic calculator

Intellectual 310 points

Replies: 7

Views: 1033

HI

I am a student

i downloaded the adc harmonic calculator from www.ti.com/.../adc-harmonic-calc

I wish to know the formula/reference to calculate the "min freq" and "max freq"  for each harmonics HD2, HD3, HD4, etc...

7 Replies

  • The signal min/max is simply calculated from the user input of center frequency -/+ BW/2.

    From there, the high order harmonics of min/max can be calculated in the normal way.  The trick is to get the "folding" handled correctly.  The Nyquist frequency is defined as Fs/2.  The aliased frequency of a signal will be:

    f-(N-1)*Fs/2   When signal falls in an odd Nyquist zone (N)

    Fs/2-[f-(N-1)*Fs/2]  When signal falls in an even Nyquist zone (N)

    For example, say Fs is 100 MHz.  With a 30 MHz signal the signal falls within 1st Nyquist zone so the frequency is simply 30 - 0*100/2 = 30.  A 60 MHz signal will fall within the second Nyquist zone so the tone will fall at 100/2 - (60-(2-1)*100/2 = 50 - (60-50) = 40 MHz.

    --RJH

  • In reply to RJ Hopper:

    Thanks RJ !!

    Thanks a ton for your reply.

    To put my confusion more clearly, lets consider the following case:

    (1) Signal BW = 25 MHz ; (2) Filter BW = 33 MHz ; (3) Signal center frequency = 296 MHz ; (4) Sampling freq = 250 MHz 

    The value of the min/max for fundamental tone : Center freq after aliasing = 296MHz - 250 MHz = 46MHz ; min freq = 46MHz - (25/2)MHz = 33.5MHz ; max freq =  46MHz + (25/2)MHz = 58.5MHz.  This OK.

    But the confusion begins with  the second harmonic , and so on .

    According to my calculations,

    The value of the min/max for HD2 tone : Center freq after aliasing = (296*2)MHz - (250*2) MHz = 92MHz ; min freq = 92MHz - (25/2)MHz = 79.5MHz ; max freq =  46MHz + (25/2)MHz = 104.5MHz.

    But the toolbox outputs : min freq = 125MHz ; max freq =  59MHz.

    This is confusing.  Also my basic doubt is why the BW of the harmonics change. As far as I understand, sampling does not change the BW.

    Please elaborate.

  • In reply to Pankaj Jha:

    Sorry for the typo error : Read it as

    The value of the min/max for HD2 tone : Center freq after aliasing = (296*2)MHz - (250*2) MHz = 92MHz ; min freq = 92MHz - (25/2)MHz = 79.5MHz ; max freq = 92MHz + (25/2)MHz = 104.5MHz.
  • In reply to Pankaj Jha:

    Sampling does not change the BW per se, but the high order harmonics will have higher BW than the original signal.  The second harmonic, for example, will have 2x the bandwidth; third harmonics will have 3x and so on.

    Take your example.  Signal [min,max] at 296 MHz +/- 25/2 MHz = [283.5,308.5] MHz corresponding to a BW of 25 MHz.  The HD2 [min,max] is [567,617] MHz corresponding to a BW of 50 MHz. So far this is straight analog and nothing to do with sampling.

    Now, once sampling is taken into account we want to see how the fundamental and harmonic signals alias down into the 1st Nyquist zone.  I think one error in your calculations is that you are treating the sampling frequency like a local oscillator of an analog mixer and mixing HD2 with second harmonic of sampling frequency.  This is not an accurate model.

    In your example the Nyquist zone is set at 250/2 = 125 MHz. 

    • 283.5 MHz resides in 3rd Nyquist zone
    • 308.5 MHz resides in 3rd Nyquist zone
    • 567 MHz resides in 5th Nyquist Zone
    • 617 MHz resides in 5th Nyquist zone

    From there I can utilize the equation I posted previously: f-(N-1)*Fs/2   When signal falls in an odd Nyquist zone (N)

    • 283.5 MHz falls at 33.5 MHz
    • 308.5 MHz falls at 58.5 MHz
    • 567 MHz falls at 67 MHz
    • 617 MHz falls at 117 MHz

    In this case the HD2 completely falls within one Nyquist zone so the BW is kept "intact".  Conceivably, the higher harmonics can straddle two different Nyquist zones and the signals would fold back on top of each other.

    --RJH

  • In reply to RJ Hopper:

    Hi RJH !!!!

    Thanks a ton for your reply. You really helped to correct my concepts on sampling.

    All things fall into place once we accept the fact that "the high order harmonics will have higher BW than the original signal. The second harmonic, for example, will have 2x the bandwidth; third harmonics will have 3x and so on."

    So, one last question. Can you please elaborate why " the high order harmonics will have higher BW than the original signal."

    Any image/ web-link/ book/ pdf/ paper explaining the same would be very helpful.
  • In reply to Pankaj Jha:

    I think the concept can be visualized through simple math.  Take a signal whose BW is 20MHz centered at 100 MHz.  Visualize this as a continuous series of discrete tones from 90 MHz to 110 MHz.  Now look at the second harmonics of the edge tones.  the lower edge lands at: 2x90 = 180 MHz.  The upper edge lands at 2x110 = 220 MHz.  The BW of the second harmonic is now 220 - 180 = 40 MHz or 2x the original 20 MHz signal.

    If you need further visual "proof", observe a modulated signal on a spectrum analyzer and then zone in on the center of the HD2 signal to observe its spread.

    -RJH

  • In reply to RJ Hopper:

    Thanks a ton for your reply. Now things are completely clear to me.
    You really save my life.

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.