Wide/Narrow Band Spectrograms Wide band (left) Combines harmonics Voiced speech vocal fold pulses (glottis air puffs) show as vertical lines Narrow band(right) Individual harmonics Narrow-band displays formants horizontally No vocal pulses shown Display parameters Generally log power (log(amplitude2) Frame shift: 1 ms typical Spectrogram for a vowel sound Spectrograms: vowel with varying pitch

Frame Positioning Pitch-synchronous Pitch-asynchronous Tradeoffs Centered around a pitch period Varied size frames Unvoiced sections assume fixed pitch period Challenge: Determine exact pitch period locations Pitch-asynchronous Fixed frames and shifts typically 25-30 ms frame width with a 10 ms frame shift Tradeoffs Too large: contains more than one phoneme Too small: cannot determine F0 or the harmonics

Source Filter Separation Source: F0 correlating to pitch and intonation Filter: The spectral envelope Three separation approaches: Filter bank, cepstral analysis, and linear prediction Importance: Spectrum and pitch need to be studied separately

Filter Bank Time Domain Frequency Domain Advantage Disadvantage Series of linear band pass filters Frequency Domain Window a frame (Ex: Hamming) Perform Fourier Transform Warp frequencies (Ex: Mel scale) Compute weighted sum of each bin Advantage simple and robust for finding spectral envelope Okay for ASR (unless language is tonal) Disadvantage Lose too much detail to find pitch. Peaks can fall between harmonics; not good for TTS

The Cepstrum Definition: c[n] = F -1{log(|F(x[n]|} Note: Sometimes the Cepstrum is taken on the square of the spectrum rather than on the log of the spectrum Treat the spectrum as a wave Formant frequency is slow Glottal pulses are fast Cepstrum separates the two Cepstral Terminology Cepstrum is Spectrum in reverse Quefrency instead of frequency Lifter instead of filter

Separating Source from Vocal Filter Excites particular fundamental frequencies The glottis source sometimes is noisy Filter The source is filtered resulting in vocal tract resonances Goal: Separate excitation frequencies from the filter Process Time domain convolves source with filter (u[n] * v[n]) Convolution multiplies in the frequency domain (UV) Log converts multiplication to a sum (log(UV) = log(U) + log(V)) The V (filter) varies slowly; the U (excitation) varies quickly. The inverse operation separates u[n] and v[n] into different quefrencies Observations There are no pitch excitations in unvoiced speech Cepstral analysis works well for speech recognition applications

Cepstrum Process Illustration Time Speech Signal Frequency Log Frequency Cepstrums of Excitation After FFT After log(FFT) After inverse FFT of log Spectral envelope on the left, F0 is one of the excitations

Cepstrum Samples Note: Band passing frequencies below 100 or greater than 900 can help

Cepstral Mean Normalization (CMN) For Automatic Speech Recognition For each window we perform a Cepstral analysis Mel scaled Quefrencies summed into 13 to 39 bins Each bin represents a Cepstral vector X = {x0, x1, …, xT-1} Compute the mean of each vector coefficient µk = 1/T ∑t=0,T-1xt where k is a vector coefficient Subtract uk from coefficient k of each vector X

Cepstral Evaluation The Cepstral process eliminates phase data. However, human perception largely, but not totally, ignores phase Use the lower quefrencies to study the vocal filter Use the peak to study pitch and glottis behavior Zeroing the pitch portion of the Cepstrum and transforming back to the frequency domain is an approach for speech recognition Disadvantage of Cepstrals: They are difficult to interpret using a visual plot

Time Domain Pitch Detection Recall the autocorrelation pitch detection algorithm Correlate a window of speech with a previous window Find the best match Problem: too many false peaks Peak and center clipping Algorithm to reduce false peaks clip the top/bottom of a signal Center the remainder around 0 Other alternatives Researchers propose many other pitch detection algorithms There are much debate as to which is the best

Epoch Detection Simply determining the pitch is not sufficient for synthesis Unit selection requires accurate anchors to be able to merge segments of speech Otherwise clicks and other artifacts will be heard Pitch-marking or epoch-detection attempt to accurately mark pitch points Mark peaks or troughs Mark Instant of glottal closure (large negative pulse) There are many algorithms proposed, but this remains an open research area

Linear Prediction Coding (LPC) Originally developed to compress (code) speech Although coding pertains to compression, the term LPC has much broader implications in NLP LPC is equivalent to the vocal tract model (Week 6) LPC is another computational method to Compute vocal tract reflection coefficients Compute vocal tract filter coefficients LPC is useful to separating source (glottis) from filter (vocal tract)

Linear Predictive Encoding (LPC) One approach: There are many others with better compression Pseudo Code WHILE not EOF READ sample n (s[n]) x = prediction() error = x – s[n] IF error too large to fit in compressed size WRITE special code WRITE s[n] ELSE WRITE error Concept Guess at the next value using a set of previous values Instead of outputting the actual data, output the error from the guess Less bits should be needed if the guess is good

Linear Algebra Background N equations and P unknowns If N<P, ∞ number of potential solutions x + y = 5 Solutions are along the line y = 5-x If N=P, there is at most one unique solution Solution: x + y = 5 and x – y = 3, solution x=4, y=1 If N>P, there cannot even be one solution No solutions for: x+y = 4, x – y = 3, 2x + 7 = 7 The best we can do is find the closes fit

Least Squares: minimize error First Approach: Linear algebra – find orthogonal projections of vectors onto the best fit Second Approach: Calculus – Use derivative with zero slope to find best fit

Solving n equations and n unknowns Gaussian Elimination Complexity: O(n3) Successive Iteration Complexity varies Cholskey Decomposition More efficient, still O(n3) Levenson-Durbin Complexity: O(n2) Works for symmetric Toplitz matrices Definitions for any matrix, A Transpose (AT): Replace aij by aji for all i and j Symmetric: AT = A Positive Definite: No complex solutions Toplitz: Diagonals to the right all have equal values Lower/Upper triangular: No non zero values above/below diagonal

Symmetric Toeplitz Matrices Example Flipping rows and columns produces the same matrix Every diagonal to the right contains the same value

Levinson Durbin Algorithm Step 0 E0 = 1 [r0 Initial Value] Step 1 E1 = -3 [ (1-k12)E0] k1 = 2 [r1/E0] Step 2 E2 = -8/3 [ (1-k22)E1] k2 = 1/3 [(r2 – a11r1)/E1] Step 3 E3 = -5/2 [(1-k32)E2 k3 = 1/4 [(r3 – a21r2 – a22r1)/E2] Step 4 E4 = -12/5 [(1-k42) E3] k4 = 1/5 [r4 – a31r3 – a32r2 – a33r1)/E3] a11=2 [k1] a21=4/3 [a11-k2a11] a22=1/3[k2] a31=5/4 [a21-k3a22] a32=0 [a22-k3a21] a33=1/4 [k3] a41=6/5 [a31-k4a33] a42=0 [a32-k4a32] a43=0[a33-k4a31] a44=1/5[k4] Verify results by plugging a41, a42, a43, a44 back into the equations 6/5(1) + 0(2) + (0)3 + 1/5(4) = 2, 6/5(2) + 0(1) + 0(2) + 1/5(3) = 3 6/5(3) + 0(2) + 0(1) + 1/5(2) = 4, 6/5(4) + 0(3) + 0(2) + 1/5(1) = 5

Levinson-Durbin Pseudo Code E0 = r0 FOR step = 1 TO P kstep = ri FOR i = 1 TO step-1 THEN kstep -= ai-1,i * rstep-i kstep /= Estep-1 Estep = (1 – k2step)Estep-1 astep,step = kstep-1 For i = 1 TO step-1 THEN astep,i = astep-1,I – kstep*astep-1, step-i Note: ri are the row 1 matrix coefficients

Cholesky Decomposition Requirements: Symmetric (same matrix if flip rows and columns) Positive definite matrix (no complex solutions) Solution Factor matrix A into: A = LLT where L is lower triangular Perform forward substitution to solve: L(LT[ak]) = [bk] Use the resulting vector, [xi], in the above step to perform a backward substitution to solve for LT[ak] = [xi] Complexity Factoring step O(n3/3) Forward substitution: n2 Backward substitution: n2

Cholesky Factorization Result:

Cholesky Factorization Pseudo Code FOR k=1 TO n-1 lkk = a½kk FOR j = k+1 TO n ljk = ajk/ lkk FOR j = k+1 TO n FOR i = j TO n aij = aij – lik ljk lnn = ann Column index: k Row index: j Elements of matrix A: aij Elements of matrix L: l

Illustration: Linear Prediction {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16} Goal: Estimate yn using the three previous values yn ≈ a1 yn-1 + a2 yn-2 + a3 yn-3 Three ak coefficients, Frame size of 16 Thirteen equations and three unknowns Note: The equation is an IIR filter

LPC Basics Predict x[n] from x[n-1], … , x[n-P] Concept en = yn - ∑k=1,P ak yn-k en is the error between the projection and the actual value The goal is to find the coefficients that produce the smallest en value Concept Square the error take the partial derivative with respect to each ak Find the solution with zero derivative (the minimum). Result : P equations with P unknowns

Finding the Best LPC Estimate One linear prediction equation: en = yn - ∑k=1,P ak yn-k Over a whole frame we have n equations and k unknowns Sum en over the entire frame: E = ∑n=0,N-1(yn - ∑k=1,P ak yn-k) Square the total error: E2 = ∑n=0,N-1 (yn - ∑k=1,P ak yn-k)2 Take partial derivative with respect to each aj; generates P equations (Ej) Like a regular derivative treating only aj as a variable 2Ej = 2(∑n=0,N-1 (yn - ∑k=1,P akyn-k)yn-j) Calculus Chain Rule: if y = y(u(x)) then dy/dx = dy/du * du/dx Set each Ej to zero (zero derivative) to find the minimum P errors for j = 1 to P then 0 = ∑n=0,N-1 (yn - ∑k=1,P akyn-k)yn-j (j indicates the equation) Rearrange terms: for each j of the P equations, ∑n=0,N-1 ynyn-j=∑n=0,N-1∑k=1,Pakyn-kyn-j=∑k=1,P∑n=1,Nakyn-kyn-j =∑k=1,Pakφ(j,k)=φ(j,0) Result: P equations and P unknowns where φ(j,k) = ∑n=0,N-1 yn-kyn-j

Covariance Method Now we have P equations and P unknowns Result from previous slide (equation j): ∑n=0,N-1ynyn-j = ∑k=1,P∑n=0,N-1akyn-kyn-j A more concise notation when φ(j,k) = ∑n=0,N-1 yn-kyn-j is φ(j,0)=∑k=1,Pakφ(j,k) Now we have P equations and P unknowns Because φ(j,k) = φ(k,j), the matrix is symmetric Solution requires O(n3) iterations (ex: Cholskey’s decomposition) Why covariance? It’s not probabilistic, but the matrix looks similar

Covariance Example Recall: φ(j,k) = ∑n=start,start+N-1 yn-kyn-j Where equation j is: φ(j,0) = ∑k=1,Pakφ(j,k) Signal: {…, 3, 2, -1, -3, -5, -2, 0, 1, 2, 4, 3, 1, 0, -1, -2, -4, -1, 0, 3, 1, 0, …} Frame: {-5, -2, 0, 1, 2, 4, 3, 1}, Number of coefficients: 3 φ(1,1) = -3*-3 +-5*-5 + -2*-2 + 0*0 + 1*1 + 2*2 + 4*4 + 3*3 = 68 φ(2,1) = -1*-3 +-3*-5 + -5*-2 + -2*0 + 0*1 + 1*2 + 2*4 + 4*3 = 50 φ(3,1) = 2*-3 +-1*-5 + -3*-2 + -5*0 + -2*1 + 0*2 + 1*4 + 2*3 = 13 φ(1,2) = -3*-1 +-5*-3 + -2*-5 + 0*-2 + 1*0 + 2*1 + 4*2 + 3*4 = 50 φ(2,2) = -1*-1 +-3*-3 + -5*-5 + -2*-2 + 0*0 + 1*1 + 2*2 + 4*4 = 60 φ(3,2) = 2*-1 +-1*-3 + -3*-5 + -5*-2 + -2*0 + 0*1 + 1*2 + 2*4 = 36 φ(1,3) = -3*2 +-5*-1 + -2*-3 + 0*-5 + 1*-2 + 2*0 + 4*1 + 3*2 = 13 φ(2,3) = -1*2 +-3*-1 + -5*-3 + -2*-5 + 0*-2 + 1*0 + 2*1 + 4*2 = 36 φ(3,3) = 2*2 +-1*-1 + -3*-3 + -5*-5 + -2*-2 + 0*0 + 1*1 + 2*2 = 48 φ(1,0) = -3*-5 +-5*-2 + -2*0 + 0*1 + 1*2 + 2*4 + 4*3 + 3*1 = 50 φ(2,0) = -1*-5 +-3*-2 + -5*0 + -2*1 + 0*2 + 1*4 + 2*3 + 4*1 = 23 φ(3,0) = 2*-5 +-1*-2 + -3*0 + -5*1 + -2*2 + 0*4 + 1*3 + 2*1 = -12

Auto-Correlation Method Assume: all values of the signal outside of 0<j<N-1 is zero Correlate from -∞ to ∞ (most values are 0) The LPC formula for φ becomes: φ(j,k)=∑n=0,N-1-(j-k) ynyn+(j-k)=R(j-k) The Matrix is now in the Toplitz format The Levinson Durbin algorithm applies Implementation complexity: O(n2)

Auto Correlation Example Recall: φ(j,k)=∑n=0,N-1-(j-k) ynyn+(j-k)=R(j-k) Where equation j is: R(j) = ∑k=1,P R(j-k)ak Notation: j is the row, k is the column Signal: {…, 3, 2, -1, -3, -5, -2, 0, 1, 2, 4, 3, 1, 0, -1, -2, -4, -1, 0, 3, 1, 0, …} Frame: {-5, -2, 0, 1, 2, 4, 3, 1}, Number of coefficients: 3 R(0) = -5*-5 + -2*-2 + 0*0 + 1*1 + 2*2 + 4*4 + 3*3 + 1*1 = 60 R(1) = -5*-2 + -2*0 + 0*1 + 1*2 + 2*4 + 4*3 + 3*1 = 35 R(2) = -5*0 + -2*1 + 0*2 + 1*4 + 2*3 + 4*1 = 12 R(3) = -5*1 + -2*2 + 0*4 + 1*3 + 2*1 = -4

LPC Transfer Function Predict the values of the next sample Ŝ[n] = ∑ k=1,p ak s[n−k] The error signal (e[n]), is the LPC residual e[n]=s[n]− ŝ[n] = s[n]− ∑ k=1,p ak s[n−k] Perform a Z-transform of both sides E(z)=S(z)− ∑k=1,pak S(z)z−k Factor S(z) E(z) = S(z)[ 1−∑k=1,p ak z−k ]=S(z)A(z) Compute the transfer function: S(z) = E(z)/A(z) Conclusion: LPC provides us with an all pole filter

LPC Coding and Synthesis Models Coding Model Synthesis Model Conclusion The LPC all-pole model can code and synthesizes speech

The LPC Model Possible solutions The LPC estimate An all-pole IR filter: yn = Gxn - ∑k=1,N ak yn The Gxn residual attempts to model the glottal source LPC estimates the separation of source from filter Challenges (Problems in synthesis) The residual does not accurately model the source (glottis) The filter does not model radiation from the lips The model does not account for nasal resonances Possible solutions Additional poles can increase the accuracy to a point 1 pole pair for each 1k of sampling rate 2 more pairs can better estimate the source and lips Introduce zeroes into the model More robust analysis of the glottal source and lip radiation

The LPC Spectrum Perform a LPC analysis Find the poles Plot the spectrum around the z-Plane unit circle What do we find concerning the LPC spectrum? Adding poles better matches speech up to about 22 for a 16k sampling rate The peaks tend to be overly sharp (“spiky”) because small radius changes greatly alters pole skirt widths

PARCOR Definition: [PAR]tial auto [COR]relation coefficients Review LPC coefficients are: a1, a2, … aP PARCOR coefficients are: k1, k2, … kP It is easy to compute PARCOR from LPC and visa versa Review Rectangular tubes have reflection coefficients rk = (Ak+1 – Ak)/(Ak+1 + Ak) With algebra the ratio of areas between tubes are: Ak/Ak+1 = (1-rk)/(1+rk) Importance LPC is equivalent to the tube model of the vocal tract Log[Ak+1/Ak] = log[(1-ki)/(1+ki)] We can adjust the LPC parameters based on PARCOR

Relationship to Tube Model Given PARCOR, compute LPC Given LPC, compute PARCOR FOR j = 1 TO P THEN xP,j = aj kp = xP,P FOR i = P TO 2 STEP -1 FOR j = 1 TO i-1 xi-1,j = (xi,j + kixi,i-j)/(1-ki2) ki-1 = xi-1,i-1 FOR i = 1 TO P xi,i = ki if (i>1) then FOR j = 1 TO i-1 xi,j = xi-1,j – kixi-1,i-j FOR j=1 TO P THEN aj = xP,j Notes: ki are PARCOR coefficients ai are LPC coefficients xi,j is a temporary work array

Line Spectrum Pairs Overview Characteristics Advantages Filter with an additional coefficient Uses the equations on the right The New filter models: A completely closed glottis Completely open lips Characteristics Spectrum shown as lines because of infinite amplitudes of formants Forces zeros and poles to be interspersed on the unit circle Advantages Easier to estimate formants Less sensitive to quantization errors

LPC and the Source Signal Experiments show Glottis requires both zeros and poles It requires less poles than the vocal function LPC combines the glottal and vocal tract poles If U(z) = I(z)G(z) U(z) = source function I(z) = Impulse sequence G(z) Glottal filter Transfer function Goal: separate glottal poles from the LPC predictor ∏k=0,Mbkz-k U(Z) = I (z) 1 - ∏j=0,Lajz-j

Closed Phase Analysis Find Instant of glottal closure Epoch detection algorithm Divide signal closed phase (glottis does not affect LPC predictors) open phases (glottis has significant impact) Strategy Compute the G filter over a number of pitch periods Perform an inverse filter to obtain the glottal signal

Open Phase Analysis Problem: It is not easy to find the instance of glottal closure Goal: add extra poles to the model Advantages Human hearing is more sensitive to peaks than to valleys The tube model and LPC are all-pole systems Disadvantages: Relationships between the poles and the formants becomes obscure Extra poles can approximate a zero, but not perfectly How can extra poles approximate zeros For example if x,y ≠ -1, then consider the following derivation 1-x = 1/(1+y) 1 = (1-x)(1+y) = 1 +y –x –xy = 1 + y(1-x) – x Therefore: y = x/(1-x)