M. Hasegawa-Johnson. DRAFT COPY.

Transcription

1 Lecture Notes in Speech Production, Speech Coding, and Speech Recognition Mark Hasegawa-Johnson University of Illinois at Urbana-Champaign February 7, 000

2 M. Hasegawa-Johnson. DRAFT COPY.

3 Chapter Linear Predictive Coding. All-Pole Model of the Speech Transfer Function The speech spectrum S(z) is produced by passing an excitation spectrum, U(z), through an all-pole transfer function H(z) = G=A(z): S(z) =H(z)U(z) = A(z) If we write the corresponding difference equation, we get Where the input signal u(n) is u(n) = ρ N 0 P s(n) = r f(n rn 0) Uncorrelated Gaussian Random Noise G U(z) (.) A(z) a k z k (.) a k s(n k)+gu(n) (.) Voiced Speech Unvoiced Speech A(z) can be written in terms of either the direct-form coefcients a k or the roots r i :. Normal Equations A(z) a k z k py i= (.) ( r i z ) (.5) Suppose we want to pick an A(z) which explains" as much as possible of S(z). If we force A(z) to explain" as much as possible, then the remaining excitation signal, e(n) = Gu(n) will be as simple" as possible. We can do this by choosing A(z) in order to minimize the energy of e(n): X X E = e (m) = G u (m) (.) m= m= Plugging in equation., we get X E = m= ψ s(m) a k s(m k)! (.7) 5

4 M. Hasegawa-Johnson. DRAFT COPY. Solving the equation k = 0 yields the generalized normal equations: X m= s(m)s(m i) = X a k m= s(m k)s(m i); i =;;p (.8) The generalized normal equations contain innite sums, which makes it uncomputable. In order to make it computable, we need to truncate it somehow... Autocorrelation Method Solving the Normal Equations In the autocorrelation method, we window the signal: Then minimize the following criterion: E n = m= s n (m) w(m)s(n + m) (.9) X ψ s n (m) Solving the n =@a k = 0 yields the autocorrelation normal equations: where This can also be written in matrix form: μr p = R() R() R(p) 7 5 ; R μ = R n (i) = R n (i) X m= a k s n (m k)! (.0) a k R n (ji kj) (.) s n (m)s n (jm ij) (.) μr p = μ Rμa (.) R(0) R() R(p ) R() R(0) R(p ) R(p ) R(p ) R(0) 7 5 ; μa = a a p 5 (.) Since the matrix μ R is Toeplitz" (all elements on a given diagonal are the same), it can be inverted efciently (O(p )) using the Levinson-Durbin recursion. ffl Advantage of autocorrelation method: the normal equations can be solved using a computationally efcient algorithm. ffl Disadvantage of autocorrelation method: if x(n) is actually a steady-state signal, autocorrelation will not nd the true spectrum. Levinson-Durbin Recursion The normal equations for autocorrelation can be solved by inverting a Toeplitz matrix: μr p = R() R() R(p) 7 5 ; R μ = μr = μ Rμa ) μa = μ R μr (.5) R(0) R() R(p ) R() R(0) R(p ) R(p ) R(p ) R(0) 7 5 ; μa = a a p 5 (.)

5 Lecture Notes in Speech..., DRAFT COPY. 7 This inversion can be done efciently using the following recursive algorithm, called the Levinson-Durbin recursion:" Finding the Minimum Energy The minimum energy is E min = E (0) = R(0) (.7) k i = R(i) P i j= a(i ) j R(i j) E (i ) ;» i» p (.8) a (i) i = k i (.9) a (i) j = a (i ) j k i a (i ) i j ;» j» i (.0) E (i) = ( k i )E (i ) (.) ψ s(m) a k s(m k)! = R n (0) Combined Equation for Energy and Coefcients (.) a k R n (k) (.) The equations for LPC coefcients and minimum error can be written together in matrix form: fl = E min ; R p =.. Covariance Method Solving the Normal Equations fl = R p a (.) R(0) R() R(p) R() R(0) R(p ) R(p) R(p ) R(0) 7 5 ; a = Instead of windowing the signal, we use the original signal s n (m), but window the error: E n = n+n X m=n ψ s(m) Solving the n =@a k = 0 yields the autocorrelation normal equations: where This can also be written as μfl = Φ(; 0) Φ(; 0) Φ(p; 0) Φ n (i; k) Φ n (i; 0) = 7 5 ; μφ= n+n X m=n a a p 7 5 (.5) a k s(m k)! (.) a k Φ n (i; k) (.7) s(m i)s(m k) (.8) μfl = μφμa (.9) Φ(; ) Φ(; ) Φ(;p) Φ(; ) Φ(; ) Φ(;p) Φ(p; ) Φ(p; ) Φ(p; p) 7 5 (.0)

6 8 M. Hasegawa-Johnson. DRAFT COPY. The matrix μφ is symmetric, but not Toeplitz, so inverting it requires O(p ) operations (using an algorithm called the Cholesky decomposition). ffl Advantage: if x(n) is steady-state, covariance can nd the true spectrum. ffl Disadvantage: solving the normal equations requires an O(p )matrixinversion operation. Finding the Minimum Energy The minimum error energy is: E min =Φ n (0; 0) a k Φ n (0;k) (.) Combined Equation for Energy and Coefcients The equations for LPC coefcients and minimum error can be written together in matrix form: Φ p =.. Choosing the LPC Order fl = Φ p a (.) Φ(0; 0) Φ(0; ) Φ(0; p) Φ(; 0) Φ(; ) Φ(; p) Φ(p; 0) Φ(p; ) Φ(p; p) 7 5 (.) The LPC order needs to be large enough to represent each formant using a complex-conjugate pole pair. There also need to be an extra or poles to represent spectral tilt. With everything together, we have: p ß (Number of Formants) + ( to ) (.) The number of formants in the spectrum is the Nyquist rate (F s =), divided by the average spacing between neighboring formants: F s = Number of Formants = (.5) average(f n+ F n ) The spacing between neighboring formant frequencies is approximately average(f n+ F n ) ß c l (.) where c = 500cm/s is the speed of sound, and l is the length of the vocal tract. The length of a male vocal tract is close to 7.7cm, so there is approximately one formant per 000Hz: p ß Fs 000Hz +(to) (.7) The length of a female vocal tract is close to.75cm, so there is approximately one formant per 00Hz: Fs p ß +(to) (.8) 00Hz

7 Lecture Notes in Speech..., DRAFT COPY. 9.. Choosing the LPC Gain The LPC excitation is dened by the formula: The LPC error is dened by the formula: If we dene then s(n) = e(n) s(n) G = a k s(n k)+gu(n) (.9) N X. Frequency-Domain Interpretation of LPC The LPC error is so the error spectrum is Using Parseval's theorem, we get that k=0 e(n) =s(n) E(z) =S(z)( E n = X m Substituting in the form of E(z), we get E n = ß Z ß ß a k s(n k) =Gu(n) (.0) u (n) (.) P e (n) P u (n) = E min (.) a k s(n k) (.) a k z k )=S(z)A(z) (.) Z e (m) = ß je n (e j! )j d! (.5) ß ß jx n (e j! )j ja(e j! )j d! = G ß Z ß ß jx n (e j! )j jh(e j! d! (.) )j Since X is in the numerator inside the integral, any algorithm which minimizes E n will automatically try to produce an H(e j! )which does a good job of modeling X at frequencies where X is large. In other words, LPC models spectral peaks better than spectral valleys.. Lattice Filtering Lattice ltering analyzes or synthesizes speech usingith-order forward prediction error e (i) (m) and backward prediction error b (i) (m): e (i) (m) =s(m) b (i) (m) =s(m i) ix ix a k s(m k) =e (i ) (m) k i b (i ) (m ) (.7) a k s(m + k i) =b (i ) (m ) k i e (i ) (m) (.8) In LPC synthesis, s(n) andb (p) (m) are calculated recursively from e (p) (m), as shown in gure..

8 70 M. Hasegawa-Johnson. DRAFT COPY. (p) E (z) (p) B (z) -kp kp -k k /z /z /z -k k - G S(z) Figure.: LPC synthesis using a lattice lter structure... How to Calculate Reflection Coefcients. Reflection coefcients can be estimated directly from the speech signal by minimizing an error metric of the form ~E (i) = some function of e (i) (m); b (i) (m) (.9) Direct calculation of reflection coefcients is computationally expensive, but sometimes leads to better estimates of the transfer function (e.g. Burg's method).. Any LPC synthesis lter can be implemented as a lattice lter. The relationship between a i and k i is given by the Levinson-Durbin recursion... Why Use Lattice Filter Instead of Direct-Form LPC? Tests for stability of =A(z): ffl Roots of A(z): jr i j <. ffl Reflection coefcients: jk i j <. ffl Direct form coefcients a k : No simple test... Equivalence of Lattice and Concatenated-Tube Models Reflection coefcients in the lattice lter are equivalent to reflection coefcients in a lossless tube model of the vocal tract. There are many ways to make the two structures correspond. One convenient formulation numbers the tube areas A i backward from the lips: k i = A i A i+ A i + A i+ ; A i =Areaoth tube section (.50) A 0 = (Area of the space beyond the lips lossless termination) (.5) A p+ = A p (Area of the glottis lossy termination) (.5) The length l of each tube section is determined by the sampling period T and the speed of sound c: T = l c (.5).5 Stability of the LPC Filter.5. Stability of the Unquantized Filter H(z) py H(z) = P p i= ff iz i = i= r i z (.5)

9 Lecture Notes in Speech..., DRAFT COPY. 7 ffl H(z) is stable iff jr i j <. This is equivalent tosaying that the reflection coefcients are jk i j <. ffl Autocorrelation and lattice methods guarantee a stable lter, because they choose values of jk i j <. ffl Covariance method does not guarantee a stable lter..5. Stability of the Quantized Filter ^H(z) ^H(z) = ffl Filter is stable iff all j^r i j <,j^k i j <. P p i= ^ff iz i = py i= ^r i z (.55) ffl If the synthesis lter ^H(z) is unstable, energy of the decoded speech will rapidly MAX OUT. ffl Since lter state is carried forward from frame to frame, this erroneously high energy can last for a long time after the offending frame. ffl This means that a single error can wipe out a whole second or more of speech, and maybe the D/A and the speakers as well..5. Quantizing Direct-Form Coefcients Leads to Unstable Filters A small quantization error in one of the direct-form coefcients ^ff i can easily make ^H(z) unstable. example, this lter is stable: H(z) = (.5) 0:z +0:z +0:8z +0:9z If ff is changed from 0.9 to 0.5, the lter is unstable: ^H(z) = (.57) 0:z +0:z +0:8z +0:5z If ff is different, however, the same change in ff leaves a stable lter: ^H(z) =.5. THE SOLUTION +0:z +0:z +0:8z +0:5z (.58) Quantize either k i or r i, and design the quantizer levels so that j^k i j or j^r i j is always less than.0.. Log Area Ratios Goal: Quantize k i such that j^k i j <... The Problem Changes in k i have amuch larger effect on the synthesized speech spectrum if jk i jßthanifjk i j <<, as shown in gure..... The Solution: Companded Quantization k i! Expand! g i! Linear PCM! ^g i! Compress! ^k i (.59) g i =log ki (.0) +k i For

10 7 M. Hasegawa-Johnson. DRAFT COPY. Spectral Sensitivity of k (solid) and of k (dashed) in Second Order LPC 0.9 Dependence of Resonant Frequency on k, dω/dk Reflection Coefcients k (for k=0) and k (for k =) Figure.: Spectral sensitivity to changes in the reflection coefcients. 0 Log Area Ratio Companding 8 Log Area Ration g(i) Reflection Coefcient k(i) Figure.: LAR companding.

11 Lecture Notes in Speech..., DRAFT COPY. 7 Glottal Excitation Area=A0 Loss Signal A A A Radiated Speech Vocal Tract Resonator: Unyielding Walls, Discrete Area Function SPEECH S(z) EXCITATION E(z) LOSS B(z) k k k -k -k -k Figure.: Acoustic resonator and lattice model with a matched impedance termination at the glottis... Interpretation: Log Area Ratios Remember that the reflection coefcients k i can be used to dene a stylized vocal tract model, with reflection coefcients of r L = at the lips, r g = 0 at the glottis, and r i = k p i (i = 0;;p ), as shown in gure... If the cross-sectional areas are A i, A i+, then - and r i = k p i = A i+ A i A i+ + A i (.) A i+ A i So the expanded" reflection coefcient is really a log area ratio:".7 Line Spectral Frequencies Goal: Quantize r i such that jr i j <. = k p i +k p i (.) g i =log ki =log Ap i+ +k i A p i.7. The Impedance Interpretation of the LSFs (.) There is no regular relationship between the real and imaginary parts of r i (the frequencies and bandwidths of the roots), so they are difcult to quantize. Solution Instead, we nd two adjunct polynomials Q(z) andp (z) whose roots e jqn and e jpn are on the unit circle if and only if the roots of A(z) are inside the unit circle (jr i j < ). If the real numbers p n and q n have nice properties, then they might be easier to quantize than the complex numbers r i. Concise Denition The reflection coefcients k i can be used to dene a stylized vocal tract model, with reflection coefcients of r L = at the lips, r g = 0 at the glottis, and r i = k p i. The LSFs are the zero and pole frequencies of the impedance of this vocal tract model as seen from the glottis, which is:

12 7 M. Hasegawa-Johnson. DRAFT COPY. Z T (z) = Q(z) P (z) = ( ej0 z p= ) Y ( e jqn z )( e jqn z ) ( e jß z ) ( e jpn z )( e jpn z ) n= (.) Since the termination at the lips is lossless, and every tube section is lossless, the impedance Z T (z) must also be lossless. This means that the poles and zeros e jpn and e jqn are on the unit circle. Nice Characteristics of LSFs. The pole frequencies p n (Z T (e jpn )=) approximately equal the formant frequencies.. The zeros and poles alternate: 0 <p <q <p <q < (.5) ) Range(q n ;p n ) is limited. (.) ) q n ;p n easy to quantize. (.7). q n ;p n are correlated with each other,sointra-frame prediction and block quantization are possible.. q n ;p n change slowly from frame to frame, so inter-frame prediction is also possible..7. Derivation of the LSF Polynomials Interpretation Lattice lter is like a vocal tract resonator with no reflections at the glottis, as shown in gure... In the gure, the glottal excitation signal" is the forward LPC prediction error E (p) (z): E (p) (z) = ψ i= ff i z i! X(z) (.8) and the glottal loss signal" is the backward LPC prediction error z (p+) B (p) (z), which can be obtained by time-reversing the LPC analysis lter (as proven in R&S 8..): z (p+) B (p) (z) =z (p+) ψ i= ff i z i! X(z) (.9) Notice that, since r L =, the only loss in this vocal tract model is at the glottis. The impedance of the rest of the vocal tract, as seen from the glottis, is pure imaginary: That means the zeros and poles of Z T (e j! ) are on the unit circle: <fz T (e j! )g =0 (.70)...where q n and p n are real numbers. Y T (e jpn )= Z T (e jqn )=0 (.7) =0 (.7) Z T (e jpn )

13 Lecture Notes in Speech..., DRAFT COPY. 75 Closed Glottis A0 A A A Radiated Speech Closed-Glottis Vocal Tract Model (Unyielding Walls) D(z)=E(z)+B(z) SPEECH S(z) + k k k -k -k -k - Figure.5: Acoustic resonator and lattice lter model with a zero-admittance termination at the glottis. Finding the Poles where Y g (e j! ) is the admittance of a rigid wall, Y T (e jpn )=0) Y g (e jpn )+Y T (e jpn )=0 (.7) Y g (e j! )=0 8! (.7) This is the same as saying that p n are the resonant frequencies of the system shown in gure.7.. Find the resonances of the augmented lattice lter: = ψ i= D(z) =E (p) (z)+z (p+) B (p) (z) (.75) ff i z i! X(z)+z (p+) ψ i= ff i z i! X(z) (.7) X(z) = D(z) A(z)+z (p+) A(z ) So the resonances e jpn are the roots of the polynomial (.77) P (z) =A(z)+z (p+) A(z ) (.78) Finding the Zeros where Z g (e j! )istheimpedanceofopenspace, Z T (e jqn )=0) Z g (e jqn )+Z T (e jqn )=0 (.79) Z g (e j! )=0 8! (.80) This is the same as saying that q n are the resonant frequencies of the system shown in gure.7.. Find the resonances of the augmented lattice lter: = ψ i= D(z) =E (p) (z) z (p+) B (p) (z) (.8) ff i z i! X(z) z (p+) ψ i= ff i z i! X(z) (.8)

14 7 M. Hasegawa-Johnson. DRAFT COPY. Open Glottis A0 A A A Radiated Speech Open-Glottis Vocal Tract Model (Unyielding Walls) D(z)=E(z)-B(z) SPEECH S(z) - k k k -k -k -k - Figure.: Acoustic resonator and lattice lter model with a zero-impedance termination at the glottis. X(z) = D(z) A(z) z (p+) A(z ) So the resonances e jqn are the roots of the polynomial (.8) Q(z) =A(z) z (p+) A(z ) (.8).7. The Augmented-Tube Interpretation of Line Spectral Frequencies Setting A p+ = 0 in the concatenated tube model yields: Setting A p+ = yields: S(z) U(z) = S(z) E (p) (z)+z (p+) B (p) (z) = P (z) ; P (z) A(z)+z (p+) A(z ) (.85) S(z) U(z) = S(z) E (p) (z) z (p+) B (p) (z) = Q(z) ; Q(z) A(z) z (p+) A(z ) (.8) Because of symmetry, the roots of both P (z) andq(z) are on the unit circle. If p is even: P (z) = ( + z ) Q(z) =( z ) p= Y n= p= Y n= ( e jpn z )( e jpn z ); p n real (.87) ( e jqn z )( e jqn z ); q n real (.88) The frequencies p n and q n,for» n» p=, are called the line spectral frequencies (LSFs). The LSFs have the following useful characteristics: ffl LSFs are real, so they are easier to quantize than the LPC roots r i,which are complex. ffl If and only if =A(z) is stable, the LSFs satisfy: 0 <p <q <p <q <<ß ffl The LSFs tend to track the LPC root frequencies arg(r i ), but... ffl The LSFs vary more slowly and smoothly than the LPC roots r i. ffl Efcient algorithms exist for calculating the LSFs.

15 Lecture Notes in Speech..., DRAFT COPY LPC Distance Measures Suppose we want to calculate the distance between two all-pole spectra, S (!) =jx (!)j ß We can: G A (!) ; S (!) =jx (!)j ß G A (!) (.89) ffl Calculate the spectral L norm by integrating log jg =A (!)j log jg =A (!)j over!. ffl Find the LPC cepstrum, and calculate a weighted cepstral distance. ffl Calculate an LPC likelihood distortion..8. Itakura-Saito Distortion Suppose that S (!) is a random spectrum, produced by ltering a unit-energy noise process U(!) through an unknown all-pole lter A (!): S (!) = E[S (!)] = G A (!) G A (!) ju(!)j (.90) (.9) Suppose we don't know A, but wehave a spectrum A whichmightormightnotbeagoodapproximation to A. One question worth asking is, what is the probability that the signal x (n) was generated using lter G =A? This probability is related to a distance called the Itakura-Saito measure of the distortion between spectra G =A and G =A, named after the engineers who rst derived it: G d IS ja j ; G ja j Z = ß ß ß jx (!)j ja (!)j G d! log G (.9) G ο log (p x ([x (0) x (L )] j G ;A (!))) (.9) The rst term in the I-S distortion is the residual energy of the random signal x (n), ltered through the inverse lter A (z)=g : x (n)! A (z)=g! e (n) (.9) Z ß Z je (!)j d! = ß ß ß ß ß X = G = G n ψ jx (!)j ja (!)j d! (.95) (x (n) R (0) G a k; x (n k)) (.9) a k; R (k)+ i= a i; a k; R (ji kj)! (.97) = a0 R p; a G (.98) where a is the LPC coefcient vector representing the polynomial A (z), and R p; is the autocorrelation matrix built out of samples of signal x (n). Using this notation, the I-S distortion can be written in the easy-to-compute form:

16 78 M. Hasegawa-Johnson. DRAFT COPY. G d IS ja j ; G = a0 R p; a log G ja j G G (.00) (.00) Characteristics: ffl Asymmetric! G d IS ja j ; G G = d IS ja j ja j ; G ja j (.0) ffl Minimum value of zero is obtained when A = A, G = G. G d IS ja j ; G = E min log G =0 (.0) ja j G G.8. Likelihood-Ratio Distortion, Itakura Distortion Many times, we don't care about differences in the spectral energy of S and S. The likelihood-ratio distortion measure is obtained by normalizing both S and S to unit energy, and then calculating an Itakura-Saito distortion: d LR ja j ; Z = d IS ja j ja j ; = ß ja j ß ß ja (!)j S (!) d! = a0 R p; a (.0) G G (.0) A similar distortion measure called the Itakura distortion is often used instead of the likelihood-ratio distortion: d I ja j ; Z ß log ja j ß ß ja (!)j S (!) d! G a 0 = log R p; a G (.05)

17 Lecture Notes in Speech..., DRAFT COPY. 79 LPC Parameter Sets Containing Equivalent Information R n (i) = Direct-Form Coefcients (a k ) a k R n (ji kj) (.0) A(z) = LPC Roots (r i ) r i =roots(a(z)) (.08) A(z) = PARCOR Coefcients (k i ) py i= a k z k (.07) ( r i z ) (.09) k i = a (i) i a (i ) j = a(i) j + k i a (i) k i i j ;» j» i a (i) j = ρ ki a (i ) j j = i k i a (i ) i j» j» i (.0) g i =log ki +k i Log Area Ratios (g i ) (.) k i = egi +e gi (.) Line Spectral Frequencies (p n ;q n ) P (z) =A(z)+z (p+) A(z ) (.) Q(z) =A(z) z (p+) A(z ) (.) p n =arg(roots(p (z))) s.t. 0 <p n <ß q n = arg(roots(q(z))) s.t. 0 <q n <ß P (z) = ( + z ) Q(z) =( z ) p= Y n= p= Y n= ( e jpn z )( e jpn z ) (.5) ( e jqn z )( e jqn z ) (.) LPC Cepstrum (c m ) c 0 = log G c m = a m + c m = m X m X k m k c k a m k ; m c k a m k ; m>p» m» p G = e c0= a m = c m m X k m c k a m k ;» m» p

18 80 M. Hasegawa-Johnson. DRAFT COPY..9 Exercises. Likelihood Ratio Distance Suppose that the autocorrelation coefcients of signal x (n) arer(0)=500andr() = 00. (a) Find the parameters G and A (z) of a rst-order autocorrelation LPC model of x (n). (b) Suppose that the spectrum of another signal, x (n), can be modeled using the following transfer function: G A (z) = 0 (.7) 0:9z What is the likelihood ratio distance d LR ja j ; ja j?. Vocal Tract Area Function Remember that the PARCOR coefcients k i can be viewed as reflection coefcients in a concatenated-tube model of the vocal tract, with tube sections numbered backward from the lips. Calculate the area A i and length l of each tube section; you may assume that the area of the lip opening is A =5cm. Plot the area of the vocal tract A(x) as a function of position x, where x is measured in centimeters. Turn in code and/or equations showing how you calculated l and A(x).. LPC Analysis and Synthesis Write a function which reads in a frame of speech, calculates LPC coefcients, lters the input by A(z) to nd an LPC residual, and then lters the residual by H(z) ==A(z) tosynthesize a speech waveform. Verify that the output is identical to the non-overlapping part of the input. (a) It is important to carry the state of all digital lters forward from frame to frame. The state of digital lters in matlab can be accessed using the Zf and Zi arguments to lter." What happens if the state of the LPC synthesis lter is carried forward from frame to frame, but not the state of the analysis lter? What happens if the state of the analysis lter is carried forward, but not the state of the synthesis lter? What happens if neither lter state is carried forward? (Hint: compare the LPC residual with and without lter state carry-over.) (b) Quantize the LPC coefcients using LAR quantization. Aim for an average of -5 bits per LPC coefcient. Verify that when you use the same quantized lter ^A(z) for both analysis and synthesis, LPC quantization does not introduce any errors into the reconstructed signal. How many bits per second are you using to quantize the LPC coefcients?. Formant Tracking Write a matlab function of the form A = lpcana(x, P, N)which performs LPC analysis of order P on the waveform X using frames of length N samples. The matrix A should contain one row for each frame; each row should contain the LPC lter coefcients for one frame. Write a matlab function [F,BW] = formants(a,fs) which nds the roots of the LPC polynomials stored in A, and calculates up to p= analog formant bandwidths BW i and frequencies F i per frame, such that r i = e ßBW i +jßf i Fs (.8) where r i is one of the roots of A(z), calculated using the matlab roots function. Plot the formant frequencies as a function of time, and compare your plot to a spectrogram of the utterance. Which formants are tracked during voiced segments? What happens when there are less than p= trackable formants? What happens to the LPC-based formant estimates during unvoiced speech segments?

19 Lecture Notes in Speech..., DRAFT COPY Line Spectral Frequencies Write a matlab function [PF, QF] = tflsf(a) which computes ordered line spectral frequencies corresponding to the LPC coefcients in A. The LSF matrices PF and QF should each contain as many rows as A, and p= columns. (a) Plot the analog P-frequencies, p n F s =ß, and compare to a spectrogram of the utterance. Do the P- frequencies track the formants during voiced segments? What happens if there are 5 P-frequencies, but only trackable formants? What do the P-frequencies do during unvoiced segments? Do you think this behavior is likely to make the P-frequencies more or less quantizable than the LPC-based formant frequencies? Why? Plot the Q-frequencies q n F s =ß. How do the Q-frequencies relate to the P-frequencies? Is there ever a time when a formant frequency is tracked more closely by q n than by p n? How rapidly do the line spectral frequencies change as a function of time? What is the range of each individual line spectral frequency? What is the range of the difference between neighboring line spectral frequencies? Can you think of an efcient way of quantizing the line spectral frequencies? (b) Quantize the LPC coefcients using LSF quantization, then convert the quantized LSFs back into direct-form LPC coefcients, and synthesize speech. How does the speech sound? How can you guarantee that the quantized LPC synthesis lter will be stable?. Pitch Tracking Write a matlab function [N0, B] = pitch(x, N, N0MIN, N0MAX). For each frame, set B and N0 according to the following formulas: B = max r x (m); N 0;min» m» N 0;max (.9) N 0 = arg max r x (m); N 0;min» m» N 0;max (.0) (.) (a) Try different values of N 0;min and N 0;max. For each value you test, plot B and N0 as a function of time, and compare them to the spectrogram. What values give the best pitch tracking? Is B always larger for voiced segments than unvoiced segments? What is the threshold value of B which best divides voiced and unvoiced segments? (b) Try pitch tracking using the autocorrelation of the LPC residual, rather than the signal autocorrelation. Do you nd any improvement? Why or why not? 7. Lattice Filter and Concatenated-Tube Model In this problem, you will explore the relationship between the PARCOR lattice lter structure of R&S Fig. 8. and the lossless tube models of R&S Fig..0. Recall that the lattice lter in Fig. 8. iteratively removes the redundancy from the speech signal S(z) in order to calculate the forward prediction error, E (i) (z), and the backward prediction error, B (i) (z): E (i) (z) =E (i ) (z) k i z B (i ) (z) (.) B (i) (z) =z B (i ) (z) k i E (i ) (z) (.) E (0) (z) =B (0) (z) =S(z) (.) (a) Suppose you are given as inputs the forward error of order (i) and the backward error of order (i ), E (i) (z) andb (i ) (z), and you are asked to synthesize E (i ) (z) andb (i) (z). By re-arranging the equations above, devise a lter structure to accomplish this. Draw the lter structure. (b) Suppose you are given only the forward error of order (), E () (z), and asked to synthesize the speech signal S(z) and the backward error B () (z). Devise a lattice lter structure to accomplish this.

20 8 M. Hasegawa-Johnson. DRAFT COPY. (c) Fig. shows a section of a digital concatenated tube model, similar to the model shown in R&S Fig..0, but with a delay on the backward arc rather than the forward arc. Suppose that the left-hand nodes are the (i)th order LPC prediction errors, A(z) =E (i) (z) and B(z) = B (i) (z). Show that the right-hand nodes are C(z) =fle (i ) (z) andd(z) =flb (i ) (z), for some constant fl. What is the value of fl? A(z) +ki - h - C(z) B(z) ff -ki ff? h -ki ki z - Figure : Section of a digital concatenated tube model. ff D(z) (d) Find values of the reflection coefcients r G, r, r,andr L such that the transfer function of the concatenated tube model in Fig..0c of your text is U l (z) U g (z) = z = G S(z) E () (z) (.5) where S(z) ande () (z) are as dened in part (b) of this problem, and G is a constant. (e) In the concatenated tube model of part (d), what is the acoustic impedance at the lips? at the glottis? (you may express one or both impedances in terms of the cross-sectional tube areas, if necessary). (f) The concatenated tube elements in your model of part (d) are lossless elements, and yet the impulse response of the system decays gradually over time, implying that at least one element in the system must be lossy. Where are the losses occurring? In a real vocal tract, where else would losses occur?