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Berkeley COMPSCI 188 - Lecture 21: Speech / Viterbi

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CS 188: Artificial Intelligence Fall 2008AnnouncementsHidden Markov ModelsState Path TrellisViterbi AlgorithmExampleDigitizing SpeechSpeech in an HourSpectral AnalysisAdding 100 Hz + 1000 Hz WavesSpectrumPart of [ae] from “lab”Back to SpectraAcoustic Feature SequenceState SpaceHMMs for SpeechMarkov Process with BigramsDecodingEnd of Part II!CS 188: Artificial IntelligenceFall 2008Lecture 21: Speech / Viterbi11/13/2008Dan Klein – UC Berkeley1AnnouncementsP5 up, due 11/19W9 up, due 11/21 (note off-cycle date)Final contest: download and get started!Homework solution and review sessions coming2Hidden Markov ModelsAn HMM isInitial distribution:Transitions:Emissions:Query: most likely seq:X5X2E1X1X3X4E2E3E4E59State Path TrellisState trellis: graph of states and transitions over timeEach arc represents some transitionEach arc has weightEach path is a sequence of statesThe product of weights on a path is the seq’s probabilityCan think of the Forward (and now Viterbi) algorithms as computing sums of all paths (best paths) in this graphsunrainsunrainsunrainsunrain10Viterbi Algorithmsunrainsunrainsunrainsunrain12Example13Digitizing Speech14Speech in an HourSpeech input is an acoustic wave form s p ee ch l a bGraphs from Simon Arnfield’s web tutorial on speech, Sheffield:http://www.psyc.leeds.ac.uk/research/cogn/speech/tutorial/“l” to “a”transition:15Frequency gives pitch; amplitude gives volumesampling at ~8 kHz phone, ~16 kHz mic (kHz=1000 cycles/sec)Fourier transform of wave displayed as a spectrogramdarkness indicates energy at each frequency s p ee ch l a bfrequencyamplitudeSpectral Analysis16Adding 100 Hz + 1000 Hz WavesTime (s)0 0.05–0.96540.99017Spectrum1001000Frequency in HzAmplitudeFrequency components (100 and 1000 Hz) on x-axis18Part of [ae] from “lab”Note complex wave repeating nine times in figurePlus smaller waves which repeats 4 times for every large patternLarge wave has frequency of 250 Hz (9 times in .036 seconds)Small wave roughly 4 times this, or roughly 1000 HzTwo little tiny waves on top of peak of 1000 Hz waves19Back to SpectraSpectrum represents these freq componentsComputed by Fourier transform, algorithm which separates out each frequency component of wave. x-axis shows frequency, y-axis shows magnitude (in decibels, a log measure of amplitude) Peaks at 930 Hz, 1860 Hz, and 3020 Hz.20Acoustic Feature SequenceTime slices are translated into acoustic feature vectors (~39 real numbers per slice)These are the observations, now we need the hidden states Xfrequency……………………………………………..e12e13e14e15e16………..23State SpaceP(E|X) encodes which acoustic vectors are appropriate for each phoneme (each kind of sound)P(X|X’) encodes how sounds can be strung together We will have one state for each sound in each wordFrom some state x, can only:Stay in the same state (e.g. speaking slowly)Move to the next position in the wordAt the end of the word, move to the start of the next wordWe build a little state graph for each word and chain them together to form our state space X24HMMs for Speech25Markov Process with BigramsFigure from Huang et al page 61826DecodingWhile there are some practical issues, finding the words given the acoustics is an HMM inference problemWe want to know which state sequence x1:T is most likely given the evidence e1:T:From the sequence x, we can simply read off the words27End of Part II!Now we’re done with our unit on probabilistic reasoningLast part of class: machine


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Berkeley COMPSCI 188 - Lecture 21: Speech / Viterbi

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