The Phonetic Analysis of Speech Corpora

1.1.

crplot(p=-pi/2)

crplot(k=15, p=pi/2)

cfun <- function(A, k, p, N)

{

n = 0:(N-1)

}
For example:

n = 0:19

par(new=T)

cr(1.5, 1.4, pi/3, 20, type="l")
1.5. The sinusoids cancel each other out.
o = cr(p=c(0, pi), values=T)

round(o, 4)

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2.

a = fapply(keng.dft.5[,700:9000], sum, power=T)

b = fapply(keng.dft.5[,2700:9000], sum, power=T)

d = a - b

boxplot(d ~ factor(keng.l), ylab="dB-difference")

Yes, this parameter separates the front and back allophones of /k/ quite well (Fig. 8.25).

plot(sib.mid, sib.l, fun=mean, power=T, xlab="Frequency (Hz)", ylab="Intensity (dB)")

boxplot(m[,1] ~ factor(sib.l), ylab="1^st spectral moment (Hz)")

dplot(f2geraus, f2geraus.l, average=T, normalise=T, xlab= "Normalized time", ylab="F2 (Hz)", leg="bottomright")

m = trapply(f2geraus, moments, simplify=T)

skew = m[,3]

boxplot(skew ~ factor(f2geraus.l), ylab="F2-skew")

temp2 = vowlax.l == "a" & vowlax.spkr == "68"

dat = rbind(dip.fdat[temp1,2], vowlax.fdat[temp2,2])

labs = c(dip.l[temp1], vowlax.l[temp2])

coeffs = trapply(dat, moments, simplify=T)

boxplot(coeffs[,3] ~ labs, ylab="F2-skew")

plot(stops10[,0:4000], stops10.lab, fun=mean, power=T, xlab="Frequency (Hz)", ylab="Intensity (dB)", leg="bottomleft")

m = fapply(stops10[,0:4000], moments, minval=T)

boxplot(sqrt(m[,2]) ~ stops10.lab, ylab="Square root of 2nd moment (Hz)")
There is some evidence that the spectral variance is less for [g] in the right panel of Fig. 8.29. However, the data for [g] is dispersed as shown by the large inter-quartile range (the extent of the rectangle) in comparison with [b, d]: this probably comes about because [g] bursts occur before both front and back vowels which, of course, induce a good deal of variation in initial [g].
7.1.

dplot(dat, lab, ylab="F2 (Hz)", norm=T, average=T, ylim=c(1600, 2100))

m = trapply(dat, dct, 2, simplify=T)

boxplot(m[,3] ~ factor(lab), ylab="k2")
8.1.

mid = dcut(timevow.dft, .5, prop=T)

plot(bark(mid[,0:3000]), timevow.l, fun=mean, power=T, xlab="Frequency (Bark)", ylab="Intensity (dB)")
Since the slope of the spectrum for [ʊ] in the left panel of Fig. 8.31 falls more steeply than for the other vowels, it should be distinguished from them on k₁. The spectra for [ɪ, a] have a ⋃-shape and ⋂-shape respectively whereas that of [ʊ] shows less evidence of a parabolic shape. Therefore, k₂ should certainly distinguish [ɪ] and [a] (they might have roughly the same value on k₂ but one will be positive and the other negative) and both [ɪ, a] should be distinguished on k₂ from the comparatively more flat [ʊ] spectra.
8.2

mid.dct = fapply(bark(mid[,0:3000]), dct, 2)

eplot(mid.dct[,2:3], timevow.l, centroid=T, xlab="Bark-scaled k1", ylab="Bark-scaled k2")
As Fig. 8.31 shows, the predictions from 8.1 are more or less supported.
Fig. 8.31 about here
8.3

# Logical vector for [ɪ]

temp = timevow.l == "I"
# n is an index of the first [ɪ]

n = 1:length(timevow.l)

n = n[temp][1]
# Bark spectrum for this [ɪ] at the segment's temporal midpoint

ispec = bark(dcut(timevow.dft[n,0:4000], .5, prop=T))

smooth = dct(ispec, 5, fit=T)

# smooth = fapply(ispec, dct, 5, fit=T)

both = rbind(ispec, smooth)

fs = 2 * max(trackfreq(ispec))

both = as.spectral(both, fs)
# A plot of the raw and overlaid smoothed spectrum

plot(both, c("raw", "smooth"), xlab="Frequency (Bark)", ylab="Intensity (dB)")

At various stages in this book, different classes of speech sounds have been compared with each other in terms of their acoustic or articulatory proximity. In this Chapter, the quantification of the similarity between speech sounds is extended by introducing some topics in classification. This will provide an additional set of tools for establishing both how effectively classes of speech sounds are separated from each other based on a number of parameters and for determining the likelihood that a given value within the parameter space belongs to one of the classes. One of the applications of this methodology in experimental phonetics is to quantify whether adding a new parameter contributes any further information to the separation between classes; another is to assess the extent to which there is a correspondence between acoustic and perceptual classification of speech data. More generally, using probability in experimental phonetics has become increasingly important in view of research advances in probabilistic linguistics (Bod et al, 2003) and developments in exemplar models of speech perception and production that are founded on a probabilistic treatment of speech signals. Probability theory has been used in recent years in forensic phonetics (Rose, 2002) as a way of expressing the likelihood of the evidence given a particular hypothesis.

The introduction to this topic that is presented in this Chapter will begin with a brief overview of Bayes' theorem and with the classification of single-parameter data using a Gaussian distribution. This will be extended to two parameter classifications which will provide the tools for defining an ellipse and the relationship to principal components analysis. Some consideration will also be given to a support vector machine, a non-Gaussian technique for classification. At various stages in this Chapter, the problem of how to classify speech signals as a function of time will also be discussed.
9.1. Probability and Bayes theorem

The starting point for many techniques in probabilistic classification is Bayes' theorem which provides a way of relating evidence to a hypothesis. In the present context, it allows questions to be answered such as: given that there is a vowel with a high second formant frequency, what is the probability that the vowel is /i/ as opposed to /e/ or /a/? In this case, 'F2 is high' is the evidence and 'the vowel is /i/ as opposed to /e/ or /a/' is the hypothesis.

Consider now the following problem. In a very large labelled corpus of speech, syllables are labelled categorically depending on whether they were produced with modal voice or with creak and also on whether they were phrase-final or not. After labelling the corpus, it is established that 15% of all syllables are phrase-final. It is also found that creak occurs in 80% of these phrase-final syllables and in 30% of non-phrase final syllables. You are given a syllable from the corpus that was produced with creak but are not told anything about its phrase-label. The task is now to find an answer to the following question: what is the probability that the syllable is phrase-final (the hypothesis) given (the evidence) that it was produced with creak?

Since the large majority of phrase-final syllables were produced with creak while most of the non phrase-final syllables were not, it would seem that the probability that the creak token is also a phrase-final syllable is quite high. However, the probability must be computed not just by considering the proportion of phrase-final syllables that were produced with creak, but also according to both the proportion of phrase-final syllables in the corpus and the extent to which creak is found elsewhere (in non-phrase final syllables). Bayes' theorem allows these quantities to be related and the required probability to be calculated from the following formula:

• 
  p(E|H ) i.e., p(creak|final): the probability of there being creak, given that the syllable is phrase-final, is 0.8.
  

• 
  p(H), i.e., p(final): the probability of a syllable in the corpus being phrase final is 0.15.
  

• 
  p(E|¬H), i.e. p(creak|not-final): the probability that the syllable is produced with creak, given that the syllable is phrase-medial or phrase-initial, is 0.3.
  

• 
  p(¬H) = 1 - p(H), i.e., p(not-final): the probability of a syllable in the corpus being phrase-initial or medial (in non-phrase-final position) is 0.85.
  

The answer to the question, p(H|E), the probability that the observed token with creak is also phrase-final is given using (2) as follows:

0.68
Notice that the denominator is the same whichever of these two probabilities is calculated.

These calculations lead to the following conclusion. If you take a syllable from this corpus without knowing its phrase label, but observe that it was produced with creak, then it is more likely that this syllable was not phrase-final. The probability of the two hypotheses (that the syllable is phrase-final or non-phrase-final) can also be expressed as a likelihood ratio (LR) which is very often used in forensic phonetics (Rose, 2002). For this example:
LR = 0.68/0.32

LR

2.125

The above example can be used to introduce some terminology that will be used in later examples. Firstly, the quantities p(H) and p(¬H) are known as the prior probabilities: these are the probabilities that exist independently of any evidence. Thus, the (prior) probability that a syllable taken at random from the corpus is non phrase-final is 0.85, even without looking at the evidence (i.e., without establishing whether the syllable was produced with creak or not). The prior probabilities can have a major outcome on the posterior probabilities and therefore on classification. In all the examples in this Chapter, the prior probabilities will be based on the relative sizes of the classes (as in the above examples) and indeed this is the default of the algorithms for discriminant analysis that will be used later in this Chapter (these prior probabilities can be overridden, however and supplied by the user). p(E|H ) and p(E|¬H) are known as conditional probabilities. Finally the quantities that are to be calculated, p(H|E) and p(¬H|E), which involve assigning a label given some evidence, are known as the posterior probabilities. Thus for the preceding example, the posterior probability is given by:

prior = 0.15

conditional.neg = 0.3

prior.neg = 1 - prior

posterior = (conditional * prior) / (( conditional * prior) + (conditional.neg * prior.neg))

posterior

0.32
The idea of training and testing is also fundamental to the above example: the training data is made up of a large number of syllables whose phrase-position and voice quality labels are known while training itself consists of establishing a quantifiable relationship between the two. Testing involves taking a syllable whose phrase-label is unknown and using the training data to make a probabilistic prediction about what the phrase-label is. If the syllable-token is taken from the same data used for training (ithe experimenter 'pretends' that the phrase label for the syllable to be tested is unknown), then this is an example of a closed test. An open test is if the token, or tokens, to be tested are taken from data that was not used for training. In general, an open test gives a much more reliable indication of the success with which the association between labels and parameters has been learned in the training data. All of the examples in this Chapter are of supervised learning because the training phase is based on prior knowledge (from a database) about how the parameters are associated with the labels. An example of unsupervised learning is kmeans-clustering that was briefly touched upon in Chapter 6: this algorithm divides up the clusters into separate groups or classes without any prior training stage.
9.2 Classification: continuous data

The previous example of classification was categorical because the aim was to classify a syllable in terms of its position in the phrase based on whether it was produced with creak or not. The evidence, then allows only two choices. There might be several choices (creak, modal, breathy) but this would still be an instance of a categorical classification because the evidence (the data to be tested) can only vary over a fixed number of categories. In experimental phonetics on the other hand, the evidence is much more likely to be continuous: for example, what is the probability, if you observe an unknown (unlabelled) vowel with F1 = 380 Hz, that the vowel is /ɪ/ as opposed to any other vowel category? The evidence is continuous because a parameter like F1 does not jump between discrete values but can take on an infinite number of values within a certain range. So this in turn means that the basis for establishing the training data is somewhat different. In the categorical example from the preceding section, the training consisted of establishing a priori the probability that a phrase-final syllable was produced with creak: this was determined by counting the number of times that syllables with creak occurred in phrase-final and non-phrase-final position. In the continuous case, the analogous probability that /ɪ/ could take on a value of 380 Hz needs to be determined not by counting but by fitting a probability model to continuous data. One of the most common, robust and mathematically tractable probability models is based on a Gaussian or normal distribution that was first derived by Abraham De Moivre (1667-1754) but which is more commonly associated with Karl Friedrich Gauss (1777-1855) and Pierre-Simon Laplace (1749-1827). Some properties of the normal distribution and its derivation from the binomial distribution are considered briefly in the next section.

Consider tossing an unbiased coin 20 times. What is the most likely number of times that the coin will come down Heads? Intuitively for an unbiased coin this is 10 and quantitatively is it given by μ = np, where μ is a theoretical quantity known as the population mean, n is the number of times that the coin is flipped, and p the probability of 'success', i.e., of the coin coming down Heads. Of course, in flipping a coin 20 times, the outcome is not always equal to the population mean of 10: sometimes there will be fewer and sometimes more, and very occasionally there may even be 20 Heads from 20 coin flips, even if the coin is unbiased. This variation comes about because of the randomness that is inherent in flipping a coin: it is random simply because, if the coin is completely unbiased, there is no way to tell a priori whether the coin is going to come down Heads or Tails.

The process of flipping the coin 20 times and seeing what the outcome is can be simulated as follows in R with the sample() function.
sample(c("H", "T"), 20, replace=T)

# (You are, of course, most likely to get a different output.)

"T" "H" "H" "H" "H" "H" "H" "H" "H" "H" "T" "T" "H" "T" "H" "T" "T" "T" "T" "T"
The above lines have been incorporated into the following function coin() with parameters n, the number of times the coin is tossed, and k, the number of trials, that is the number of times that this experiment is repeated. In each case, the number of Heads is summed.
coin<- function(n=20, k=50)

{

# n: the number of times a coin is flipped

result = NULL

for(j in 1:k){

singleexpt = sample(c("H", "T"), n, replace=T)

result = c(result, sum(singleexpt=="H"))

}

result

trials8

14 5 11 8 8 11 7 8

In this coin flipping experiment, there is, then, variation about the theoretical population mean, μ, in the number of Heads that are obtained per 20 coin flips and in Fig. 9.1, this variation is displayed in the form of histograms for 50, 500, and 5000 trials. The histograms were produced as follows:

k50 = coin(k=50); k500 = coin(k=500); k5000 = coin(k=5000)

hist(k50,col=col,xlim=xlim, xlab="",ylab ="Frequency count",main="k50")

hist(k500, col=col, xlim=xlim, , xlab = "", ylab = "", main="k500")

hist(k5000, col=col, xlim=xlim, , xlab = "", ylab = "", main="k5000")
Fig. 9.1 about here
In all three cases, the most frequently observed count of the number of Heads is near μ = 10. But in addition, the number of Heads per 20 trials falls away from 10 at about the same rate, certainly for the histograms with 500 and 5000 trials in the middle and right panels of Fig 9.1. The rate at which the values fall away from the mean is governed by σ, the population standard deviation. In the case of flipping a coin and counting the number of Heads, σ is given by where n and p have the same interpretation as before and q = 0.5 = 1 - p is the probability of 'failure', i.e., of getting Tails. So the population standard deviation for this example is given by sqrt(20 * 0.5 * 0.5) which evaluates to 2.236068. The relationship between the sample standard deviation, s, and the population standard deviation σ is analogous to that between m and μ: the greater the number of trials, the closer s tends to σ. (For the data in Fig. 9.1, the sample standard-deviations can be evaluated with the sd() function: thus sd(k5000) should be the closest to sqrt(20 * 0.5 * 0.5) of the three).

When a coin is flipped 20 times and the number of Heads is counted as above, then evidently there can only be any one of 21 outcomes ranging in discrete jumps from 0 to 20 Heads. The probability of any one of these outcomes can be calculated from the binomial expansion given by the dbinom(x, size, prob) function in R. So the probability of getting 3 Heads if a coin is flipped four times is dbinom(3, 4, 0.5) and the probability of 8 Heads in 20 coin flips is dbinom(8, 20, 0.5). Notice that an instruction such as dbinom(7.5, 20, 0.5) is meaningless because this is to ask the probability of there being 7 ½ Heads in 20 flips (and appropriately there is a warning message that x is a non-integer). The binomial probabilities are, then, those of discrete outcomes. The continuous case of the binomial distribution is the Gaussian or normal distribution which is defined for all values between plus and minus infinity and is given in R by the dnorm(x, mean, sd) function. In this case, the three arguments are respectively the number of successful (Heads) coin flips, μ, and σ. Thus the probability of there being 8 Heads in 20 coin flips can be estimated from the normal distribution with:

0.1195934

hist(k50, freq=F, main="50 trials", xlab="", col=col)

curve(dnorm(x, 20*.5, sqrt(20*.5*.5)), 0, 20, add=T, lwd=2)
# These are the probabilities of getting 0, 1, ..20 Heads from the binomial distribution

binprobs = dbinom(0:20, 20, .5)

points(0:20, binprobs)

hist(k500, freq=F, main="500 trials", xlab="Number of Heads in 20 coin flips", col=col)

curve(dnorm(x, 20*.5, sqrt(20*.5*.5)), 0, 20, add=T, lwd=2)

points(0:20, binprobs)

hist(k5000, freq=F, main="5000 trials", xlab="", col=col)

curve(dnorm(x, 20*.5, sqrt(20*.5*.5)), 0, 20, add=T, lwd=2)

points(0:20, binprobs)

Fig. 9.2 about here

Some of the other important properties of the normal distribution are:

• 
  the population mean is at the centre of the distribution and has the highest probability.
  
• 
  the tails of the distribution continue at the left and right edge to plus or minus infinity.
  
• 
  the total area under the normal distribution is equal to 1.
  
• 
  the probability that a value is either less or greater than the mean is 0.5.
  

The R function pnorm(q, mean, sd) calculates cumulative probabilities, i.e., the area under the normal curve from minus infinity to a given quantile, q. Informally, this function returns the probability that a sample drawn from the normal distribution is less than that sample. Thus, pnorm(20, 25, 5)returns the probability of drawing a sample of 20 or less from a normal distribution with mean 25 and standard deviation 5. To calculate the probabilities within a range of values requires, therefore, subtraction:

0.8399948

15.20018 34.79982

Without the second and third arguments, the various functions for computing values from the normal distribution return so-called z-scores or the values from the standard normal distribution with μ = 0 and σ = 1. This default setting of qnorm() can be used to work out the range of values that fall within a certain number of standard deviations from the mean. Thus without the second and third arguments, qnorm(c(0.025, 0.975)) returns the number of standard deviations for samples falling within 95% of the mean of any normal distribution, i.e. the well-known value of ±1.96 from the mean (rounded to two decimal places). Thus the previously calculated range can also be obtained as follows:

25 - 1.959964 * 5

15.20018
# Upper range

25 + 1.959964 * 5

34.79982
An example of the use of dnorm() and qnorm() over the same data is given in Fig. 9. 3 which was produced with the following commands:
xlim = c(10, 40); ylim = c(0, 0.08)

curve(dnorm(x, 25, 5), 5, 45, xlim=xlim, ylim=ylim, ylab="", xlab="")

region95 = qnorm(c(0.025, 0.975), 25, 5)

values = seq(region95[1], region95[2], length=2000)

values.den = dnorm(values, 25, 5)

par(new=T)

plot(values, values.den, type="h", col="slategray", xlim=xlim, ylim=ylim, ylab="Probability density", xlab="Values")
Fig. 9.3 about here