Computer Science Stack Exchange is a question and answer site for students, researchers and practitioners of computer science. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm reading “Speech segmentation without speech recognition” by Dong Wang, Lie Lu and Hong-Jiang Zhang. The algorithm I'm looking at is a V/C/P (Vowel/Consonant/Pause) classification algorithm on a digital speech signal. It is described as such:

  1. Audio data is segmented into 20ms-long non-overlapping frames, where features, including ZCR, Energy and Pitch, are extracted.

  2. Energy and pitch curve is smoothed.

  3. The Mean_En and Std_En of energy curve are calculated to coarsely estimate the background noise energy level, as:

    NoiseLevel = Mean_En - 0.75 Std_En. 

    Similarly the threshold of ZCR (ZCR_dyna) is defined as:

    ZCR_dyna = Mean_ZCR + 0.5 Std_ZCR
  4. Frames are classified as V/C/P coarsely by using the following rules, where FrameType is used to denote the type of each frame.

    If ZCR > ZCR_dyna then FrameType = Consonant 
    Else if Energy < NoiseLevel, then  FrameType = Pause 
    Else FrameType = Vowel   
  5. Update the NoiseLevel as the weighted average energy of the frames at each vowel boundary and the background segments.

  6. Re-classify the frames using algorithm in step 4 with the updated NoiseLevel. Pauses are merged by removing isolated short consonants. Vowel will be split at its energy valley if its duration is too long

I do not understand step #5. Like I don't know how to interpret the wording - is there another way to describe what they are doing? I get that we want to update the NoiseLevel variable and re-run step #4 for every frame, I just don't understand how exactly.

share|cite|improve this question
Maybe you can ask the authors for clarification? – mayank Feb 5 '13 at 17:48
Yeah I can try that, though I was hoping it would be obvious to someone else so I could get a quicker response =) – YoungMoney Feb 5 '13 at 17:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.