harmonic plus noise models
TRANSCRIPT
Outline HNM Analysis Synthesis Further Work Thanks References
Harmonic plus Noise models
Yannis Stylianou
University of Crete, Dept of Computer [email protected]
IIT Guwahati, Dec 2016
[email protected] Harmonic plus Noise, HNM 1/59
Outline HNM Analysis Synthesis Further Work Thanks References
1 HNMIntroduction to HNMs
2 AnalysisFrequencyMaximum Voice FrequencyPhase based estimation of pitchAmplitudes and PhasesResidual
3 Synthesis4 Further Work
Error FunctionLeast SquaresTotal Least SquaresNon Linear Least Squares, NLSEnergy modulation function
5 Thanks6 [email protected] Harmonic plus Noise, HNM 2/59
Outline HNM Analysis Synthesis Further Work Thanks References
Background
Mentioning just a few works for speech ...
MBE (Griffin et al.1988 [1])
Sinusoids plus band-pass random signals (Abrantes etal.1991 [2]
Harmonic and Stochastic Model (Laroche et al.1993 [3]
Iterative decomposition of the excitation signal(Yegnayarayana et al.1995 [4])
Harmonic plus Noise Model (Stylianou et al.1995 [5]
Harmonic plus Noise Model 2 (Stylianou, PhD, 1996 [6])
[email protected] Harmonic plus Noise, HNM 3/59
Outline HNM Analysis Synthesis Further Work Thanks References
Background
Mentioning just a few works for speech ...
MBE (Griffin et al.1988 [1])
Sinusoids plus band-pass random signals (Abrantes etal.1991 [2]
Harmonic and Stochastic Model (Laroche et al.1993 [3]
Iterative decomposition of the excitation signal(Yegnayarayana et al.1995 [4])
Harmonic plus Noise Model (Stylianou et al.1995 [5]
Harmonic plus Noise Model 2 (Stylianou, PhD, 1996 [6])
[email protected] Harmonic plus Noise, HNM 3/59
Outline HNM Analysis Synthesis Further Work Thanks References
Background
Mentioning just a few works for speech ...
MBE (Griffin et al.1988 [1])
Sinusoids plus band-pass random signals (Abrantes etal.1991 [2]
Harmonic and Stochastic Model (Laroche et al.1993 [3]
Iterative decomposition of the excitation signal(Yegnayarayana et al.1995 [4])
Harmonic plus Noise Model (Stylianou et al.1995 [5]
Harmonic plus Noise Model 2 (Stylianou, PhD, 1996 [6])
[email protected] Harmonic plus Noise, HNM 3/59
Outline HNM Analysis Synthesis Further Work Thanks References
Background
Mentioning just a few works for speech ...
MBE (Griffin et al.1988 [1])
Sinusoids plus band-pass random signals (Abrantes etal.1991 [2]
Harmonic and Stochastic Model (Laroche et al.1993 [3]
Iterative decomposition of the excitation signal(Yegnayarayana et al.1995 [4])
Harmonic plus Noise Model (Stylianou et al.1995 [5]
Harmonic plus Noise Model 2 (Stylianou, PhD, 1996 [6])
[email protected] Harmonic plus Noise, HNM 3/59
Outline HNM Analysis Synthesis Further Work Thanks References
Background
Mentioning just a few works for speech ...
MBE (Griffin et al.1988 [1])
Sinusoids plus band-pass random signals (Abrantes etal.1991 [2]
Harmonic and Stochastic Model (Laroche et al.1993 [3]
Iterative decomposition of the excitation signal(Yegnayarayana et al.1995 [4])
Harmonic plus Noise Model (Stylianou et al.1995 [5]
Harmonic plus Noise Model 2 (Stylianou, PhD, 1996 [6])
[email protected] Harmonic plus Noise, HNM 3/59
Outline HNM Analysis Synthesis Further Work Thanks References
Background
Mentioning just a few works for speech ...
MBE (Griffin et al.1988 [1])
Sinusoids plus band-pass random signals (Abrantes etal.1991 [2]
Harmonic and Stochastic Model (Laroche et al.1993 [3]
Iterative decomposition of the excitation signal(Yegnayarayana et al.1995 [4])
Harmonic plus Noise Model (Stylianou et al.1995 [5]
Harmonic plus Noise Model 2 (Stylianou, PhD, 1996 [6])
[email protected] Harmonic plus Noise, HNM 3/59
Outline HNM Analysis Synthesis Further Work Thanks References
Why to decompose?
Decomposing speech into (quasi)periodic and non-periodic parthas many applications in:
Speech modification
Speech coding
Pathologic voice detection (i.e., HNR ...)
Psychoacoustic research
[email protected] Harmonic plus Noise, HNM 4/59
Outline HNM Analysis Synthesis Further Work Thanks References
Why to decompose?
Decomposing speech into (quasi)periodic and non-periodic parthas many applications in:
Speech modification
Speech coding
Pathologic voice detection (i.e., HNR ...)
Psychoacoustic research
[email protected] Harmonic plus Noise, HNM 4/59
Outline HNM Analysis Synthesis Further Work Thanks References
Why to decompose?
Decomposing speech into (quasi)periodic and non-periodic parthas many applications in:
Speech modification
Speech coding
Pathologic voice detection (i.e., HNR ...)
Psychoacoustic research
[email protected] Harmonic plus Noise, HNM 4/59
Outline HNM Analysis Synthesis Further Work Thanks References
Why to decompose?
Decomposing speech into (quasi)periodic and non-periodic parthas many applications in:
Speech modification
Speech coding
Pathologic voice detection (i.e., HNR ...)
Psychoacoustic research
[email protected] Harmonic plus Noise, HNM 4/59
Outline HNM Analysis Synthesis Further Work Thanks References
Motivation for HNM
0 200 400 600−2
−1
0
1
2x 10
4
(a) Time in samples
Am
plitu
deOriginal speech signal
0 2000 4000 6000 8000−50
0
50
100
(b) Frequency (Hz)
db
Original magnitude spectrum
0 200 400 600−2
−1
0
1
2x 10
4
(c) Time in samples
Am
plitu
de
Harmonic part (0−5000Hz)
0 200 400 600−2000
−1000
0
1000
2000
(d) Time in samples
Am
plitu
de
Noise part (5000−8000Hz)
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Outline HNM Analysis Synthesis Further Work Thanks References
Brief overview of HNM
HNM is a pitch-synchronous harmonic plus noiserepresentation of the speech signal.
Speech spectrum is divided into a low and a high banddelimited by the so-called maximum voiced frequency.
The low band of the spectrum (below the maximum voicedfrequency) is represented solely by harmonically related sinewaves.
The upper band is modeled as a noise component modulatedby a time-domain amplitude envelope.
HNM allows high-quality copy synthesis and prosodicmodifications.
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HNM in equations
Harmonic part:
h(t) =
L(t)∑k=−L(t)
Ak(t)e j kω0(t) t
Noise part:n(t) = e(t) [v(τ, t) ? b(t)]
Speech:s(t) = h(t) + n(t)
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Models for periodic part
HNM1: Sum of exponential functions without slope
h1[n] =
L(nia)∑k=−L(nia)
ak(nia)e j2πkf0(nia)(n−nia)
HNM2: Sum of exponential function with complex slope
h2[n] = <
L(nia)∑k=1
Ak(n) expj2πkf0(nia)(n−nia)
where
Ak(n) = ak(nia) + (n − nia)bk(nia)
with ak(nia), bk(nia) to be complex numbers (amplitude andslope respectively). < denotes taking the real part.
[email protected] Harmonic plus Noise, HNM 8/59
Outline HNM Analysis Synthesis Further Work Thanks References
Models for periodic part
HNM1: Sum of exponential functions without slope
h1[n] =
L(nia)∑k=−L(nia)
ak(nia)e j2πkf0(nia)(n−nia)
HNM2: Sum of exponential function with complex slope
h2[n] = <
L(nia)∑k=1
Ak(n) expj2πkf0(nia)(n−nia)
where
Ak(n) = ak(nia) + (n − nia)bk(nia)
with ak(nia), bk(nia) to be complex numbers (amplitude andslope respectively). < denotes taking the real part.
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Outline HNM Analysis Synthesis Further Work Thanks References
Models for periodic part continuing
HNM3: Sum of sinusoids with time-varying real amplitudes
h3[n] =
L(nia)∑k=0
ak(n)cos(ϕk(n))
where
ak(n) = ck0 + ck1 (n − nia)1 + · · ·+ ckp (n − nia)p(n)
ϕk(n) = εk + 2π kζ (n − nia)
where p(n) is the order of the amplitude polynomial, which is, ingeneral, a time-varying parameter.
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Residual (Noise) part
The non-periodic part is just the residual signal obtained bysubtracting the periodic-part (harmonic part) from the originalspeech signal in the time-domain
r [n] = s[n]− h[n]
where h[n] is either h1[n], h2[n], or h3[n].
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Initial fundamental frequency
Get an initial estimation of fundamental frequency f0 [7]
Determine the voicing of the frame:
E =
∫ 4.3f0
0.7f0
(|S(f )| − |S(f )|)2
∫ 4.3f0
0.7f0
|S(f )|2
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Initial fundamental frequency
Get an initial estimation of fundamental frequency f0 [7]
Determine the voicing of the frame:
E =
∫ 4.3f0
0.7f0
(|S(f )| − |S(f )|)2
∫ 4.3f0
0.7f0
|S(f )|2
[email protected] Harmonic plus Noise, HNM 11/59
Outline HNM Analysis Synthesis Further Work Thanks References
Maximum Voiced Frequency
Let’s assume that for a voiced frame, a frequency fd has alreadybeen declared as voiced or unvoiced, then
In [fd + f0/2, fd + 3f0/2] find the fc frequency with thehighest magnitude A.For fc compute the cumulative amplitude Ac
Determine all other peaks fi in [fc − f0/2, fc + f0/2], and thecorresponding amplitudes, A(fi ) and cumulative amplitudesAc(fi )Compute the average cumulative amplitude for all fi : Ac(fi )Pass fc through the voicing testSet fd equal to fc and repeat the steps until fs/2.Determine voiced and unvoiced spectral areasMaximum voiced frequency is the maximum frequency of thefirst voiced spectral area.
[email protected] Harmonic plus Noise, HNM 12/59
Outline HNM Analysis Synthesis Further Work Thanks References
Maximum Voiced Frequency
Let’s assume that for a voiced frame, a frequency fd has alreadybeen declared as voiced or unvoiced, then
In [fd + f0/2, fd + 3f0/2] find the fc frequency with thehighest magnitude A.For fc compute the cumulative amplitude Ac
Determine all other peaks fi in [fc − f0/2, fc + f0/2], and thecorresponding amplitudes, A(fi ) and cumulative amplitudesAc(fi )Compute the average cumulative amplitude for all fi : Ac(fi )Pass fc through the voicing testSet fd equal to fc and repeat the steps until fs/2.Determine voiced and unvoiced spectral areasMaximum voiced frequency is the maximum frequency of thefirst voiced spectral area.
[email protected] Harmonic plus Noise, HNM 12/59
Outline HNM Analysis Synthesis Further Work Thanks References
Maximum Voiced Frequency
Let’s assume that for a voiced frame, a frequency fd has alreadybeen declared as voiced or unvoiced, then
In [fd + f0/2, fd + 3f0/2] find the fc frequency with thehighest magnitude A.For fc compute the cumulative amplitude Ac
Determine all other peaks fi in [fc − f0/2, fc + f0/2], and thecorresponding amplitudes, A(fi ) and cumulative amplitudesAc(fi )Compute the average cumulative amplitude for all fi : Ac(fi )Pass fc through the voicing testSet fd equal to fc and repeat the steps until fs/2.Determine voiced and unvoiced spectral areasMaximum voiced frequency is the maximum frequency of thefirst voiced spectral area.
[email protected] Harmonic plus Noise, HNM 12/59
Outline HNM Analysis Synthesis Further Work Thanks References
Maximum Voiced Frequency
Let’s assume that for a voiced frame, a frequency fd has alreadybeen declared as voiced or unvoiced, then
In [fd + f0/2, fd + 3f0/2] find the fc frequency with thehighest magnitude A.For fc compute the cumulative amplitude Ac
Determine all other peaks fi in [fc − f0/2, fc + f0/2], and thecorresponding amplitudes, A(fi ) and cumulative amplitudesAc(fi )Compute the average cumulative amplitude for all fi : Ac(fi )Pass fc through the voicing testSet fd equal to fc and repeat the steps until fs/2.Determine voiced and unvoiced spectral areasMaximum voiced frequency is the maximum frequency of thefirst voiced spectral area.
[email protected] Harmonic plus Noise, HNM 12/59
Outline HNM Analysis Synthesis Further Work Thanks References
Maximum Voiced Frequency
Let’s assume that for a voiced frame, a frequency fd has alreadybeen declared as voiced or unvoiced, then
In [fd + f0/2, fd + 3f0/2] find the fc frequency with thehighest magnitude A.For fc compute the cumulative amplitude Ac
Determine all other peaks fi in [fc − f0/2, fc + f0/2], and thecorresponding amplitudes, A(fi ) and cumulative amplitudesAc(fi )Compute the average cumulative amplitude for all fi : Ac(fi )Pass fc through the voicing testSet fd equal to fc and repeat the steps until fs/2.Determine voiced and unvoiced spectral areasMaximum voiced frequency is the maximum frequency of thefirst voiced spectral area.
[email protected] Harmonic plus Noise, HNM 12/59
Outline HNM Analysis Synthesis Further Work Thanks References
Maximum Voiced Frequency
Let’s assume that for a voiced frame, a frequency fd has alreadybeen declared as voiced or unvoiced, then
In [fd + f0/2, fd + 3f0/2] find the fc frequency with thehighest magnitude A.For fc compute the cumulative amplitude Ac
Determine all other peaks fi in [fc − f0/2, fc + f0/2], and thecorresponding amplitudes, A(fi ) and cumulative amplitudesAc(fi )Compute the average cumulative amplitude for all fi : Ac(fi )Pass fc through the voicing testSet fd equal to fc and repeat the steps until fs/2.Determine voiced and unvoiced spectral areasMaximum voiced frequency is the maximum frequency of thefirst voiced spectral area.
[email protected] Harmonic plus Noise, HNM 12/59
Outline HNM Analysis Synthesis Further Work Thanks References
Maximum Voiced Frequency
Let’s assume that for a voiced frame, a frequency fd has alreadybeen declared as voiced or unvoiced, then
In [fd + f0/2, fd + 3f0/2] find the fc frequency with thehighest magnitude A.For fc compute the cumulative amplitude Ac
Determine all other peaks fi in [fc − f0/2, fc + f0/2], and thecorresponding amplitudes, A(fi ) and cumulative amplitudesAc(fi )Compute the average cumulative amplitude for all fi : Ac(fi )Pass fc through the voicing testSet fd equal to fc and repeat the steps until fs/2.Determine voiced and unvoiced spectral areasMaximum voiced frequency is the maximum frequency of thefirst voiced spectral area.
[email protected] Harmonic plus Noise, HNM 12/59
Outline HNM Analysis Synthesis Further Work Thanks References
Maximum Voiced Frequency
Let’s assume that for a voiced frame, a frequency fd has alreadybeen declared as voiced or unvoiced, then
In [fd + f0/2, fd + 3f0/2] find the fc frequency with thehighest magnitude A.For fc compute the cumulative amplitude Ac
Determine all other peaks fi in [fc − f0/2, fc + f0/2], and thecorresponding amplitudes, A(fi ) and cumulative amplitudesAc(fi )Compute the average cumulative amplitude for all fi : Ac(fi )Pass fc through the voicing testSet fd equal to fc and repeat the steps until fs/2.Determine voiced and unvoiced spectral areasMaximum voiced frequency is the maximum frequency of thefirst voiced spectral area.
[email protected] Harmonic plus Noise, HNM 12/59
Outline HNM Analysis Synthesis Further Work Thanks References
Voicing Test
Voicing Test:
IfAc
Ac(fi )> 2
or|A−max {A(fi )}| > 13db
,then
if fc is really close to the closest harmonic lf0, then
declare fc as voiced frequency. Otherwise, declare fc asunvoiced frequency.
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Maximum voiced frequency example
0 1000 2000 3000 4000 5000 6000 7000 80000
50
100
(a) Frequency (Hz)
Mag
nitu
de (
db)
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 10000−2
0
2
4x 10
4
(b) Number of samples
0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100000
5000
10000
(c) Number of samples
Voi
cing
Fre
q. in
(H
z)
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Fundamental frequency refinement
Using the initial f0 value and the L detected voiced frequencies fi ,then the refined fundamental frequency, f0 is defined as the valuethat minimizes the error:
E (f0) =L∑
i=1
|fi − i · f0|2
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Refinement frequency example
0 500 1000 1500 2000 2500 3000 3500 400020
40
60
80
100
(a) Frequency in Hz
Mag
inut
e (d
b)
0 500 1000 1500 2000 2500 3000 3500 400020
40
60
80
100
(b) Frequency in Hz
Mag
inut
e (d
b)
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Another way to find the initial pitch [8]
Ifx(n) ≈ A e j(ω n+θ+u(n))
withφ(n) = ω n + θ + u(n)
, then by differentiating phase
∆(n) = φ(n + 1)− φ(n) = ω + u(n + 1)− u(n)
we may get an estimation of ω.
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Testing estimation
−10 −5 0 5 10 15 20 25 300
20
40
60
80
100
120
SNR (dB)
Mea
n
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Applying it to speech ...
HilbertBand Pass filts(n) Transform
ωEst.
w(n)
Arg{x*(n) x(n+1)}
x(n)
where
w(n) =32N
N2 − 1
{1−
[n − (N/2− 1)
N/2
]2}
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Results with noisy speech
Tests with noisy speech [8]
0 0.5 1 1.5 2 2.50
100
200
0 0.5 1 1.5 2 2.50
100
200
Time(sec)
0 0.5 1 1.5 2 2.50
100
200
Pitc
h F
requ
ency
(H
z)
Clean Speech
SNR 10 dB
SNR 0 dB
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Amplitudes and phases estimation
Having f0 estimated for voiced frames, amplitudes and phases areestimated by minimizing the criterion:
ε =
nia+N∑n=nia−N
w2[n](s[n]− h[n])2
where nia = ni−1a + P(ni−1
a ), and P(ni−1a ) denotes the pitch period
at ni−1a .
for HNM1 and HNM2, this criterion has a quadratic form andis solved by inverting an over-determined system of linearequations.
For HNM3, however,a non-linear system of equations has tobe solved.
[email protected] Harmonic plus Noise, HNM 21/59
Outline HNM Analysis Synthesis Further Work Thanks References
Amplitudes and phases estimation
Having f0 estimated for voiced frames, amplitudes and phases areestimated by minimizing the criterion:
ε =
nia+N∑n=nia−N
w2[n](s[n]− h[n])2
where nia = ni−1a + P(ni−1
a ), and P(ni−1a ) denotes the pitch period
at ni−1a .
for HNM1 and HNM2, this criterion has a quadratic form andis solved by inverting an over-determined system of linearequations.
For HNM3, however,a non-linear system of equations has tobe solved.
[email protected] Harmonic plus Noise, HNM 21/59
Outline HNM Analysis Synthesis Further Work Thanks References
Harmonic Amplitudes and Phases: HNM1
We recall that:
h1[n] =
L(nia)∑k=−L(nia)
ak(nia)e j2πkf0(nia)(n−nia)
or in matrices [9]h1 = P1x1
where:
P1 =
[b−L
... b−L+1... b−L+2
... · · ·... bL
]bk =
[e j2πkf0(t ia−N) e j2πkf0(t ia−N+1) · · · e j2πkf0(t ia+N)
]Tx1 =
[a∗−L a∗−L+1 a∗−L+2 · · · aL
][email protected] Harmonic plus Noise, HNM 22/59
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Harmonic Amplitudes and Phases: HNM1
We recall that:
h1[n] =
L(nia)∑k=−L(nia)
ak(nia)e j2πkf0(nia)(n−nia)
or in matrices [9]h1 = P1x1
where:
P1 =
[b−L
... b−L+1... b−L+2
... · · ·... bL
]bk =
[e j2πkf0(t ia−N) e j2πkf0(t ia−N+1) · · · e j2πkf0(t ia+N)
]Tx1 =
[a∗−L a∗−L+1 a∗−L+2 · · · aL
][email protected] Harmonic plus Noise, HNM 22/59
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Harmonic Amplitudes and Phases: HNM2
We recall (with n′
= (n − nia)):
h2[n] = <
L(nia)∑k=1
(ak(nia) + n
′bk(nia)
)expj2πkf0(nia)n
′
or in matrices [9]
h2 = P2x2
where
P2 = [B1|B2]
(B1)nk = E (n−N)(k)
(B2)nk = (n − N) (B1)nkx2 = [<{a0} a1...aL<{b0} b1...bL]
and E = exp(j2πf0(nia))[email protected] Harmonic plus Noise, HNM 23/59
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Harmonic Amplitudes and Phases: HNM2
We recall (with n′
= (n − nia)):
h2[n] = <
L(nia)∑k=1
(ak(nia) + n
′bk(nia)
)expj2πkf0(nia)n
′
or in matrices [9]
h2 = P2x2
where
P2 = [B1|B2]
(B1)nk = E (n−N)(k)
(B2)nk = (n − N) (B1)nkx2 = [<{a0} a1...aL<{b0} b1...bL]
and E = exp(j2πf0(nia))[email protected] Harmonic plus Noise, HNM 23/59
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Solutions for HNM1 and HNM2
HNM1:
x1 =(P1
hWTWP1
)−1P1
hWTWs
Matrix to invert is Toeplitz.
HNM2:
x2 =(P2
hWTWP2
)−1P2
hWTWs
Matrix to invert is block Toeplitz.
where s denotes the vector of the original speech samples
s = [s[−N] s[−N + 1] · · · s[N]]T
and W denotes a diagonal matrix of weights
w = [w [−N]w [−N + 1] · · · w [N]]T
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Solutions for HNM1 and HNM2
HNM1:
x1 =(P1
hWTWP1
)−1P1
hWTWs
Matrix to invert is Toeplitz.
HNM2:
x2 =(P2
hWTWP2
)−1P2
hWTWs
Matrix to invert is block Toeplitz.
where s denotes the vector of the original speech samples
s = [s[−N] s[−N + 1] · · · s[N]]T
and W denotes a diagonal matrix of weights
w = [w [−N]w [−N + 1] · · · w [N]]T
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Solutions for HNM1 and HNM2
HNM1:
x1 =(P1
hWTWP1
)−1P1
hWTWs
Matrix to invert is Toeplitz.
HNM2:
x2 =(P2
hWTWP2
)−1P2
hWTWs
Matrix to invert is block Toeplitz.
where s denotes the vector of the original speech samples
s = [s[−N] s[−N + 1] · · · s[N]]T
and W denotes a diagonal matrix of weights
w = [w [−N]w [−N + 1] · · · w [N]]T
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Solutions for HNM1 and HNM2
HNM1:
x1 =(P1
hWTWP1
)−1P1
hWTWs
Matrix to invert is Toeplitz.
HNM2:
x2 =(P2
hWTWP2
)−1P2
hWTWs
Matrix to invert is block Toeplitz.
where s denotes the vector of the original speech samples
s = [s[−N] s[−N + 1] · · · s[N]]T
and W denotes a diagonal matrix of weights
w = [w [−N]w [−N + 1] · · · w [N]]T
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Harmonic Amplitudes and Phases: HNM3
An iterative approach is adopted:
Step 1: Determine phase parameters using HNM1
Step 2: Determine amplitude parameters using least squaressimilar to that for HNM1 and HNM2
back to Step 1
Usually 1-2 iterations are enough.
[email protected] Harmonic plus Noise, HNM 25/59
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Avoiding ill-conditioning
For HNM1 there is no problem if window length is twice thelocal pitch period
Same thing for HNM2
For HNM3 stands the same in case the maximum voicedfrequency is less than 3/4 of the sampling frequency and orderof amplitude polynomial is 2
[email protected] Harmonic plus Noise, HNM 26/59
Outline HNM Analysis Synthesis Further Work Thanks References
Avoiding ill-conditioning
For HNM1 there is no problem if window length is twice thelocal pitch period
Same thing for HNM2
For HNM3 stands the same in case the maximum voicedfrequency is less than 3/4 of the sampling frequency and orderof amplitude polynomial is 2
[email protected] Harmonic plus Noise, HNM 26/59
Outline HNM Analysis Synthesis Further Work Thanks References
Avoiding ill-conditioning
For HNM1 there is no problem if window length is twice thelocal pitch period
Same thing for HNM2
For HNM3 stands the same in case the maximum voicedfrequency is less than 3/4 of the sampling frequency and orderof amplitude polynomial is 2
[email protected] Harmonic plus Noise, HNM 26/59
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Residual Signal
The residual signal r [n] is estimated by
r [n] = s[n]− h[n]
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Time domain characteristics of r [n]
0 100 200−1
−0.5
0
0.5
1
1.5x 10
4
(a) Number of samples
Original speech signal
0 100 200−3000
−2000
−1000
0
1000
2000
3000
(a) Number of samples
Error signal using HNM1
0 100 200−3000
−2000
−1000
0
1000
2000
3000
(c) Number of samples
Error signal using the HNM2
0 100 200−3000
−2000
−1000
0
1000
2000
3000
(d) Number of samples
Error signal using HNM3
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Spectral domain characteristics of r [n]
0 1000 2000 3000 4000 5000 6000 7000 8000−20
0
20
40
60
80
Mag
nitu
de (
db)
Residual from HNM2
0 1000 2000 3000 4000 5000 6000 7000 8000−20
0
20
40
60
80
Frequency in Hz
Mag
nitu
de (
db)
Residual from HNM3
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... and after adding noise
0 1000 2000 3000 4000 5000 6000 7000 8000−20
0
20
40
60
80
(a) Frequency in Hz
dB(a) Residual from HNM2
0 1000 2000 3000 4000 5000 6000 7000 8000−20
0
20
40
60
80
(a) Frequency in Hz
dB
(a) Residual from HNM3
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Modeling error
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−2
−1
0
1
2
3x 10
4
Time in sec
(a) Original signal
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1−60
−50
−40
−30
−20
−10
0
Time in sec
Err
or in
dB
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Variance of the residual signal
The variance of the residual signal is given as:
E (rrh) = I−WP(PhWhWP)−1PhWh
0 10 20 30 40 50 60 70 80 900
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Time in samples
Var
ianc
e
Variance using the three harmonic models
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Modeling the residual signal
Full bandwidth representation using a low-order (10th) ARfilter
Time-domain characteristics of the residual signal are modeledusing deterministic functions
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For all HNMs
nis ←→ nia
For the periodic part: Overlap-and-Add
For the stochastic (noise) part):
Instead of AR coefficients we use reflection coefficientsSample-by-sample filtering of Gaussian noise using normalizedlattice filteringModulation in time with a deterministic function (i.e.,triangular)
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For all HNMs
nis ←→ nia
For the periodic part: Overlap-and-Add
For the stochastic (noise) part):
Instead of AR coefficients we use reflection coefficientsSample-by-sample filtering of Gaussian noise using normalizedlattice filteringModulation in time with a deterministic function (i.e.,triangular)
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For all HNMs
nis ←→ nia
For the periodic part: Overlap-and-Add
For the stochastic (noise) part):
Instead of AR coefficients we use reflection coefficientsSample-by-sample filtering of Gaussian noise using normalizedlattice filteringModulation in time with a deterministic function (i.e.,triangular)
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For all HNMs
nis ←→ nia
For the periodic part: Overlap-and-Add
For the stochastic (noise) part):
Instead of AR coefficients we use reflection coefficientsSample-by-sample filtering of Gaussian noise using normalizedlattice filteringModulation in time with a deterministic function (i.e.,triangular)
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For all HNMs
nis ←→ nia
For the periodic part: Overlap-and-Add
For the stochastic (noise) part):
Instead of AR coefficients we use reflection coefficientsSample-by-sample filtering of Gaussian noise using normalizedlattice filteringModulation in time with a deterministic function (i.e.,triangular)
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For all HNMs
nis ←→ nia
For the periodic part: Overlap-and-Add
For the stochastic (noise) part):
Instead of AR coefficients we use reflection coefficientsSample-by-sample filtering of Gaussian noise using normalizedlattice filteringModulation in time with a deterministic function (i.e.,triangular)
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For HNM1 specifically
for Periodic part (as an alternative to OLA)
Direct frequency matching
Linear amplitude interpolation
Linear phase interpolation using average pitch value
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Reformulate the error function
Cost function:
ε(a−L, ..., aL, f0) =1
2
N∑n=−N
(e[n])2 =1
2ehe
wheree[n] = w [n](s[n]− h[n])
ore =
[e[−N] e[−N + 1] ... e[N]
]T
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Reformulate the error function
ε(a) =1
2(s− Ea)hW2(s− Ea)
wherea =
[a−L ... a0 ... aL
]Tand
E =
e j2π(−L)f0(−N)/fs ... e j2πLf0(−N)/fs
e j2π(−L)f0(−N+1)/fs ... e j2πLf0(−N+1)/fs
......
...
e j2π(−L)f0N/fs ... e j2πLf0N/fs
T
(2L+1×2N+1)
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Least Squares
Setting:
∂ε(a)
∂a= 0⇒ EhW2Ea− EhW2s = 0
Solution:aLS = (EhW2E)−1EhW2s
Properties:
Asymptotically efficient even when the noise is colored.Rather fast, O(L(N + L)).Assumes no errors in E matrix.
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Least Squares
Setting:
∂ε(a)
∂a= 0⇒ EhW2Ea− EhW2s = 0
Solution:aLS = (EhW2E)−1EhW2s
Properties:
Asymptotically efficient even when the noise is colored.Rather fast, O(L(N + L)).Assumes no errors in E matrix.
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Least Squares
Setting:
∂ε(a)
∂a= 0⇒ EhW2Ea− EhW2s = 0
Solution:aLS = (EhW2E)−1EhW2s
Properties:
Asymptotically efficient even when the noise is colored.Rather fast, O(L(N + L)).Assumes no errors in E matrix.
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Least Squares
Setting:
∂ε(a)
∂a= 0⇒ EhW2Ea− EhW2s = 0
Solution:aLS = (EhW2E)−1EhW2s
Properties:
Asymptotically efficient even when the noise is colored.Rather fast, O(L(N + L)).Assumes no errors in E matrix.
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Least Squares
Setting:
∂ε(a)
∂a= 0⇒ EhW2Ea− EhW2s = 0
Solution:aLS = (EhW2E)−1EhW2s
Properties:
Asymptotically efficient even when the noise is colored.Rather fast, O(L(N + L)).Assumes no errors in E matrix.
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Least Squares
Setting:
∂ε(a)
∂a= 0⇒ EhW2Ea− EhW2s = 0
Solution:aLS = (EhW2E)−1EhW2s
Properties:
Asymptotically efficient even when the noise is colored.Rather fast, O(L(N + L)).Assumes no errors in E matrix.
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Total Least Squares, TLS
Using SVD on matrix Es = W[E|s] we have:
Es = USVh ,
Solution:
aTLS = − 1
v2L+1,2L+1
[v0,2L+1 v1,2L+1 ... v2L,2L+1
]TErrors in matrix E are taken into accountSVD is computationally demandingand ... result is not satisfactory.
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Total Least Squares, TLS
Using SVD on matrix Es = W[E|s] we have:
Es = USVh ,
Solution:
aTLS = − 1
v2L+1,2L+1
[v0,2L+1 v1,2L+1 ... v2L,2L+1
]TErrors in matrix E are taken into accountSVD is computationally demandingand ... result is not satisfactory.
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Total Least Squares, TLS
Using SVD on matrix Es = W[E|s] we have:
Es = USVh ,
Solution:
aTLS = − 1
v2L+1,2L+1
[v0,2L+1 v1,2L+1 ... v2L,2L+1
]TErrors in matrix E are taken into accountSVD is computationally demandingand ... result is not satisfactory.
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Total Least Squares, TLS
Using SVD on matrix Es = W[E|s] we have:
Es = USVh ,
Solution:
aTLS = − 1
v2L+1,2L+1
[v0,2L+1 v1,2L+1 ... v2L,2L+1
]TErrors in matrix E are taken into accountSVD is computationally demandingand ... result is not satisfactory.
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Total Least Squares, TLS
Using SVD on matrix Es = W[E|s] we have:
Es = USVh ,
Solution:
aTLS = − 1
v2L+1,2L+1
[v0,2L+1 v1,2L+1 ... v2L,2L+1
]TErrors in matrix E are taken into accountSVD is computationally demandingand ... result is not satisfactory.
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Non Linear Least Squares, NLS
Estimate both complex amplitudes and local fundamentalfrequency
Two iterative algorithms
Steepest DescentNewton-Gauss
Iteration step:
x(new) = x(old) − ηf ′(x(old)) ,
where η is the rate of correction.
The difference is in the iteration rate
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NLS- Steepest Descent
Iteration step: [a(new)
f(new)
0
]=
[a(old)
f(old)
0
]+ ηJe
where J is the Jacobian of the error vector e
Jacobian:
J =
∂e[−N]∂a−L
∂e[−N+1]∂a−L
... ∂e[N]∂a−L
......
......
∂e[−N]∂aL
∂e[−N+1]∂aL
... ∂e[N]∂aL
∂e[−N]∂f0
∂e[−N+1]∂f0
... ∂e[N]∂f0
(2L+2×2N+1)
The rate, η, is manually adjusted.
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NLS- Steepest Descent
Iteration step: [a(new)
f(new)
0
]=
[a(old)
f(old)
0
]+ ηJe
where J is the Jacobian of the error vector e
Jacobian:
J =
∂e[−N]∂a−L
∂e[−N+1]∂a−L
... ∂e[N]∂a−L
......
......
∂e[−N]∂aL
∂e[−N+1]∂aL
... ∂e[N]∂aL
∂e[−N]∂f0
∂e[−N+1]∂f0
... ∂e[N]∂f0
(2L+2×2N+1)
The rate, η, is manually adjusted.
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NLS- Steepest Descent
Iteration step: [a(new)
f(new)
0
]=
[a(old)
f(old)
0
]+ ηJe
where J is the Jacobian of the error vector e
Jacobian:
J =
∂e[−N]∂a−L
∂e[−N+1]∂a−L
... ∂e[N]∂a−L
......
......
∂e[−N]∂aL
∂e[−N+1]∂aL
... ∂e[N]∂aL
∂e[−N]∂f0
∂e[−N+1]∂f0
... ∂e[N]∂f0
(2L+2×2N+1)
The rate, η, is manually adjusted.
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NLS-Newton-Gauss
Iteration step:[a(new)
f(new)
0
]=
[a(old)
f(old)
0
]+ (JhJ)−1Jhe
The correction rate, η, is adjusted by the inverse of the JhJ
Expression, (JhJ)−1Jhe, is similar to LS solution.
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NLS-Newton-Gauss
Iteration step:[a(new)
f(new)
0
]=
[a(old)
f(old)
0
]+ (JhJ)−1Jhe
The correction rate, η, is adjusted by the inverse of the JhJ
Expression, (JhJ)−1Jhe, is similar to LS solution.
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NLS-Newton-Gauss
Iteration step:[a(new)
f(new)
0
]=
[a(old)
f(old)
0
]+ (JhJ)−1Jhe
The correction rate, η, is adjusted by the inverse of the JhJ
Expression, (JhJ)−1Jhe, is similar to LS solution.
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NLS - Initialization
First voiced frame: Through Least Squares
Subsequent frames: Using the parameters of the previousframe
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NLS - Initialization
First voiced frame: Through Least Squares
Subsequent frames: Using the parameters of the previousframe
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Comparisons
Average segmental SNR
LS TLS SD NG
26.70 26.70 26.95 28.40
Average number of iterations
SD NG
17.3 3.8
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Explanation on TLS
Whenf0 + ∆f0 = f0
then(E)N+n,L+l = e j2πl f0n
= e j2πlf0ne j2πl∆f0n
TLS expects an additive error, not a multiplicative
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Again on the energy modulation
0 200 400 600 800 1000 1200 1400 1600 1800 2000−0.4
−0.2
0
0.2
0.4
Time in samples
0 200 400 600 800 1000 1200 1400 1600 1800 2000−0.1
−0.05
0
0.05
0.1
Time in samples
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So far, mainly
So far we mainly use the Triangular Envelope:
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Signal envelope
There are many ways to obtain the “envelope” of a signal, as:
Hilbert Transform (analytic signal)
Low-pass local energy (energy envelope):
e[n] =1
2N + 1
N∑k=−N
|r [n − k]|
where r [n] denotes the residual signal.
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Signal envelope
There are many ways to obtain the “envelope” of a signal, as:
Hilbert Transform (analytic signal)
Low-pass local energy (energy envelope):
e[n] =1
2N + 1
N∑k=−N
|r [n − k]|
where r [n] denotes the residual signal.
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Hilbert envelope
We may also use the Hilbert envelope, computed as:
eH [n] =L∑
k=L−M+1
ake2πk(f0/fs)n
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Example of energy envelope
Example of Energy Envelope, with N = 7
0 200 400 600 800 1000 1200 1400 1600 1800 2000−0.1
−0.05
0
0.05
0.1
Time in samples
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Energy envelope
The energy envelope can be efficiently parameterized with a fewFourier coefficients:
e[n] =Le∑
k=−Le
Akej2πk(f0/fs)n
where Le is set to be 3 to 4
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Looking at time domain properties
200 400 600 800 1000 1200 1400 1600 1800 2000−0.1
0
0.1
Noise Part
200 400 600 800 1000 1200 1400 1600 1800 2000−0.1
0
0.1
Triangular Envelope
200 400 600 800 1000 1200 1400 1600 1800 2000−0.1
0
0.1
Hilbert Envelope
200 400 600 800 1000 1200 1400 1600 1800 2000−0.1
0
0.1
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Results from Listening test I
Triangular No pref. Hilbert
Male 8 (8.3%) 43 (44.8%) 45 (46.9%)
Female 40 (41.7%) 47 (48.9%) 9 (9.4%)
Hilbert No pref. Energy
Male 22 (22.9%) 47 (49.0%) 27 (28.1%)
Female 22 (22.9%) 54 (56.3%) 20 (20.8%)
Energy No pref. Triangular
Male 43 (44.8%) 50 (52.0%) 3 (3.2%)
Female 16 (16.7%) 67 (69.8%) 13 (13.5%)
Table: Results from the listening test for the English sentences.
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Results from Listening test II
Triangular No pref. Hilbert
Male 10 (10.4%) 47 (49.0%) 39 (40.6%)
Female 8 (8.3%) 71 (74.0%) 17 (17.7%)
Hilbert No pref. Energy
Male 11 (11.5%) 58 (60.4%) 27 (28.1%)
Female 13 (13.5%) 58 (60.4%) 25 (26.1%)
Energy No pref. Triangular
Male 42 (43.7%) 48 (50.0%) 6 (6.3%)
Female 16 (16.7%) 68 (70.8%) 12 (12.5%)
Table: Results from the listening test for the French sentences.
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Sound Examples using HNM1
Examples with HNM1 and Maximum Voiced frequency fixed at4000 Hz
Original Triangular Hilbert Energy
Male
Male
Female
Female
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Sound Examples using HNM2
Examples with HNM2 and Maximum Voiced frequency fixed at4000 Hz
Original Triangular Hilbert Energy
Male
Male
Female
Female
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Acknowledgments
My ex-PhD: Yannis Pantazis for his contributions to thefurther work on HNMs
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THANK YOU
for your attention
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D. Griffin and J. Lim, “Multiband-excitation vocoder,” IEEE Trans. Acoust., Speech, Signal Processing,
vol. ASSP-36, pp. 236–243, Fev 1988.
A. Abrantes, J. Marques, and I. Transcoso, “Hybrid sinusoidal modeling of speech without voicing decision,”
Eurospeech-91, pp. 231–234, 1991.
J. Laroche, Y. Stylianou, and E. Moulines, “HNS: Speech modification based on a harmonic + noise
model.,” Proc. IEEE ICASSP-93, Minneapolis, pp. 550–553, Apr 1993.
B.Yegnanarayana, C. d’Alessandro, and V. Darsinos, “An iterative algorithm for decomposition of speech
signals into periodic and aperiodic components,” vol. 6, no. 1, 1998.
Y. Stylianou, J. Laroche, and E. Moulines, “High-Quality Speech Modification based on a Harmonic +
Noise Model.,” Proc. EUROSPEECH, 1995.
Y. Stylianou, Harmonic plus Noise Models for Speech, combined with Statistical Methods, for Speech and
Speaker Modification.PhD thesis, Ecole Nationale Superieure des Telecommunications, Jan 1996.
W. Hess, Pitch determination of Speech Signals: Algorithmes and Devices.
Berlin: Springer, 1983.
I. Zeljkovic and Y. Stylianou, “Single complex sinusoid and arhe model based pitch extractors,” Proc.
EUROSPEECH, 1999.
Y. Stylianou, “Modeling speech based on harmonic plus noise models.,” in Nonlinear Speech Modeling and
Aplications (G. Chellot, A. Esposito, M. Faundez, and M. M, eds.), pp. 244–260, Springer-Verlag, 2005.
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