Eq Question

Discussion in 'Mastering' started by shredfit, Jul 16, 2003.

  1. shredfit

    shredfit Guest

    It is my understanding that boosting frequencies with an EQ will cause phase problems. Cutting frequencies will not cause these problems.

    I don't understand the physics of this.

    Is this due to time delay's in the EQ's filter?

    Does this happen with all EQ types?

    Is the phenomemon just ostensible? or related to specific sound cycle lengths?

    Just trying to undertand this

  2. gilfo1

    gilfo1 Guest

    Boosting or cutting causes the same amount of phase shift, but I think one shift is in the opposite direction of the other.
  3. shredfit

    shredfit Guest

    Ok, I thought phase was referring to the time (ahead or behind) a particular waveform.

    So, if two 100Hz waveforms are not ahead or behind each other (with respect to time) the resulting amplitude will be doubled.

    However, If is takes 0.01 seconds for a 100Hz to complete a full cycle and you delay the other waveform signal by 0.01 seconds the resulting amplitude will be nothing... at least this is how I understand it.

    If you change a group of frequencies with an EQ will it effect the phase of certain logarithm multiples of that frequency?

    For understanding only:
    Say for example, I have an EQ with a Q value of infinity, so I cold pull out or boost ONLY one particular frequency...say 100Hz

    What would then happen to the phase of all other frequencies? Would they be changed? I realize that this is not practical in the real world but only a model for understanding

  4. Ethan Winer

    Ethan Winer Active Member


    > It is my understanding that boosting frequencies with an EQ will cause phase problems. <

    This is a common misconception. Yes, boosting and cutting with an EQ do both create phase shift, but that's never a problem.

    Here's a mini-article I wrote about this issue:


  5. falkon2

    falkon2 Well-Known Member

    Wow... excellent article!
  6. falkon2

    falkon2 Well-Known Member

    I've been thinking a lot about the mathematics behind equalizing, and through numerous sessions on the Best-Seat-In-The-House (you know what I mean... ;) ), is this possible in the digital domain? Have similiar ideas been thought of or implemented? Skip to avoid mindless blabber: ;)

    Say we have a sample.

    Every frame of the sample was averaged with it's adjacent frames (or fractions of values of the adjacent frames).

    Would this act the same way as a low-pass filter? signals half the frequency of the sampling rate would be reduced by the averaging, while DC component would remain virtually untouched.

    Different frequencies? Add the number of adjacent samples to average with - like averaging each sample with the six nearest samples (or fractions thereof) would affect lower frequencies than the first example.

    High pass instead of low pass? Source minus low pass.

    Am I way in over my head again with these trains of thought? I always assumed that equalizers behaved on some such principle, rather than phase shifts. (An artifact of being brought up during the digital age, I suppose... it never occured to me that these type of calculations are impossible in the analog realm ;) )
  7. Ethan Winer

    Ethan Winer Active Member


    > Every frame of the sample was averaged ... Would this act the same way as a low-pass filter? <

    Jump directly to the head of the class!

    This is exactly what happens with an integrator, which is a series resistor and then a capacitor to ground. It's the classic way to make a low-pass filter.

  8. falkon2

    falkon2 Well-Known Member

    Wait.. in real-time analog, how would one "sample" know the value of the next "sample" in line? (it hasn't even occured yet!)

    ...Oh, something occured to me right the moment I typed the question... knowing the values that occured before is already adequate for averaging purposes, so the values occuring after are ignored?
  9. Ethan Winer

    Ethan Winer Active Member


    Yes, a resistor / capacitor low-pass filter simply averages the voltages continually.


Share This Page