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Multimedia Communications (371) Speech and Image Communications (348). John Mason Engineering Swansea University. Features in speech. X 1 . . . . X i. Feature extraction. Acquisition. time. (frame: 20/30 ms & sampling F: 8khz). Features in speech. X 1 . . . . X i . . .
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Multimedia Communications (371)Speech and Image Communications (348)
John Mason
Engineering
Swansea University
EG348_371_09
X1
.
.
.
.
Xi
.
.
.
.
.
Feature extraction
Acquisition
time
(frame: 20/30 ms & sampling F: 8khz)
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X1
.
.
.
.
Xi
.
.
.
.
.
Feature extraction
Acquisition
(frame: 20/30 ms & sampling F: 8khz)
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Air from
the lungs
Vocal fold
Vocal tract
Speech
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Air from
the lungs
Vocal fold
Vocal tract
Speech
H1(z)
H2(z)
synthesised
Speech
noise
Spectral envelop reflects morphological characteristics of the vocal tract
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Features: building of statistical model
T1
T2
T1
T2
T1
T2
T1
T2
T2
T1
T2
T1
T2
T1
T2
T1
T2
T1
T2
T1
T2
T1
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PersontoPerson
PersontoMachine
speech/speaker recognition
MachinetoPerson
speech synthesis
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perhaps separated by long distance
(or in time)
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Acoustic Air Path
l Transmission Path
Acoustic Air Path
Electronic
Link
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Channel Transmission Path
Electronic
Link
Microphone
ADC
Analysis
Coding
Transmitter
Receiver
Decoding
(re)Synthesis
DAC
Loudspeaker
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Human
Acoustic
generation
Transmission
Message
Creation
Language
Coding
hundreds
thousands
Tens of
thousands
tens
Approx. bit rate in bps
Acoustic Space
Human
Hearing
Extraction
Message
Realisation
Language
decoding
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Excellent
Quality
Good
ADPCM
GSM
Fair
CELP
Poor
4
8
16
32
64 kbps
Quality versus Bitrate
4 Quality Measures:
intelligibilityloudness
naturalnesseaseoflistening
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The three main application areas are:
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Dynamic Range  for flexibility
and robustness
Timevarying  to convey
information
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Excitation:
voiced
unvoiced
sn
speech
en
H(z)
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Excitation
Speech
Vocal
Tract
Voiced
Speech
Model
f0
Unvoiced
Time Varying
Parameters
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H(z)
hn
S(z)
E(z)
en
sn
E(z)
S(z)
1/H(z)
sn
en
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S(z)
E(z)
E(z)
S(z)
1/H(z)
H(z)
sn
en
sn
en
Input sn and output sn are identical
(within arithmetic limits)
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Practical Analysis/Synthesis
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S(z)
E(z)
E(z)
S(z)
1/H(z)
H(z)
sn
en
sn
en
Transmission
Sending
Receiving
Practical Analysis/Synthesis
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a
s
s
a
s
a
s
a
s
.
.
.
.
.
.
.
.
n
p
p
n
1
n
1
n
2
3
n
2
3
Linear Predictive Coding  LPC
Principle of linear prediction:
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Transforming to the zdomain gives:
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Error is simply difference between predicted and actual values:
sn
en
+

ˆ
sn
A’(z)
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en
sn
H(z)
Parameters updated at frame rate
sn
en
+
+
A’(z)
NB ‘hat’ of approximation omitted for simplicity
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Synthesis
en
sn
H(z)
Analysis
Analysis
sn
en
S(z)
+
E(z)
1/H(z)
sn

en
A’(z)
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Recall:
where ai are the pprediction coefficients.The principle
behind LPC is to find a set of pcoefficients, a1, a2, a3, ...
ap, which in some sense minimizes the error signal en,
over a frame of speech, N. This leads to a set p
coefficients for each frame.
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for i = 1, 2, .… p
From which:
where:
In matrix form:
or
Minimisation of En is achieved by setting the ppartial derivatives to zero:
The matrix [R] is Toepliz symmetric, offering numerically efficient inversion techniques  Durbin’s recursion algorithm being one of the most popular.
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Harmonic Structures and Periodicities
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Harmonic Structures and Periodicities
voiced
or
unvoiced
sn
speech
en
H(z)
Vocal tract
Short Term
Tp
p
Short term prediction
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Harmonic Structures and Periodicities
voiced
unvoiced
epn
sn
speech
Hlt(z)
Hst(z)
en
Pitch
Vocal tract
Tp
P
Long term prediction
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k
Gain
en
epn
sn
Hlt(z)
Hst(z)
Harmonic Structures and Periodicities
Two Structures: shortterm (formants) & longterm  pitch (excitation)
eg 20ms frame
160 samples @ 8Khz
ai eg p=3
ai eg p=10
NB Representations of these parameters are transmitted
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en
epn
sn
Hlt(z)
Hst(z)
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S(z)
E(z)
E(z)
S(z)
1/H(z)
H(z)
sn
en
sn
en
Input sn and output sn are identical
(within arithmetic limits)
EG348_371_09
S(z)
E(z)
E(z)
E(z)
S(z)
S(z)
1 – A’(z)
1/H(z)
H(z)
sn
sn
en
sn
en
en
S(z)
E(z)
1/(1–A’(z))
en
sn
en
sn
sn
en
1/(1–A’(z))
1 – A’(z)
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sn
sn1
a1
ai
sni
snp
sn
en
sn
en
1/(1–A’(z))
1 – A’(z)
Original Speech
Residual
sn
en
+

sn
Z1
Z1
Note – minus sign:
in Matlab combined with ai What determines p?
Z1
ap
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sn
en
sn
en
1/(1–A’(z))
1 – A’(z)
sn
sn1
a1
a1
ai
ai
sni
snp
Residual
ReSynth.
Original Speech
en
en
sn
+
+

sn
sn
Z1
Z1
Note
No minus
sn1
Z1
Z1
sni
Z1
Z1
snp
ap
ap
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S(z)
E(z)
E(z)
S(z)
1/H(z)
H(z)
sn
en
sn
en
Input sn and output sn are “similar”
Transmitted
Data Frame
What does the Transmitted Data Frame Contain?
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Integrated encoder & decoder at the encoder

sn
Basic
decoder
Adaptive
encoder
+
Weighted error
LPAS Encoder
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In Comms, compuation is expensive and parameter vector approximations to D are used
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GSMEuropean CellularRPELTP13kb/s
FS1016Secure VoiceCELP4.8
IS54NA CellularVSELP7.95
IS96“QCELP18
JDCFRJapanese CellularVSELP6.7
JDCHR“PSICELP3.67
G.728(terrestrial)LDCELP16
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Excellent
Quality
Good
ADPCM
GSM
Fair
CELP
Poor
4
8
16
32
64 kbps
Quality versus Bitrate
4 Quality Measures:
intelligibilityloudness
naturalnesseaseoflistening
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Shortterm coefficients
(formants)
Longterm coefficients
(pitch)
CB
Index
Gain
en
sn
Hlt(z)
Hst(z)
Excitation is
represented
by address
ie CB Index
en
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Shortterm coefficients
(formants)
Longterm coefficients
(pitch)
CB
Index
Gain
sn
en
en
sn
sn
Hlt(z)
Hst(z)
Excitation is
represented
by address
ie CB Index
en

sn
Basic
decoder
Adaptive
encoder
+
Weighted error
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LSF = ws . /2
zplane jy
x
ws
x
LSF’s
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Consider one pair of complex roots, A1(z) :
A1(z) = 1 + a1 z 1 + a2 z 2
P1(z) = 1 + a1 z 1 + a2 z 2 + z 3(1 + a1 z1 + a2 z2 )
= (z2 + (a1+ a2 1)z + 1 )( z + 1 ) z –3
Q1(z) = 1 + a1 z 1 + a2 z 2  z 3(1 + a1 z1 + a2 z2 )
= (z2 + (a1  a2 + 1)z + 1 )( z  1 ) z 3
The roots at 0 and 1 are discarded
It follows that the LSF’s, 1 & 2 , are given by:
cos (1) =  (a1 + a2 1)/2
andcos (2) =  (a1  a2+ 1)/2
Show:
a1 = (cos (1) + cos (2) ) and
a2 = (cos (2)  cos (1) +1 )
EG348_371_09
A1(z) = 1 + a1 z 1 + a2 z  2
= (z2 + a1 z+ a2 )z  2
= (z2 + 2 cos() wn z+ wn2 ) z  2
where wn is radius and is angle from . So: radius = a2 & = 
Note: in P & Q all w n2 terms (of the multiple 2nd orders) are unity
EG 1: a2 = 1 then cos (1) =  (a1 + a2 1)/2 = (a1)/2
roots already on circle and do not move (unstable system – not practical)
EG 2: a1 = 0 then cos (1) =  (a1+ a21)/2 =  (a2  1)/2
cos (2) =  (a1 a2+ 1)/2 =  (a2 + 1)/2
so LSF’s are symmetric about /4
EG348_371_09
LSF Review & Example (1)
LSF’s/LSP’s are defined as:
P(z) = A(z) + z(n+1) A(z1 )
and
Q(z) = A(z)  z(n+1) A(z1 )
thus
A(z) = {P(z) + Q(z)} / 2
EG348_371_09
LSF Review & Example (2)
For a second order A(z)= 1 + a1 z1 + a2 z2
P (z) = 1 + a1 z1 + a2 z2 + (1 + a1 z1 + a2 z2)z3
= (z2 + (a1 + a2  1)z + 1)(z + 1)z–3
Q (z) = 1 + a1 z1 + a2 z2  (a1 z1 + a2 z2)z3
= (z2 + (a1  a2 + 1)z + 1)(z  1 )z–3
cf: (s2 + ( 2cos()wn )s + wn2)
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Q(z)
P(z)
Q(z)
P(z)
2
1
LSF Review & Example (3)
For a second order A(z)= 1 + a1 z1 + a2 z2 :
P (z) = (z2 + (a1 + a2  1)z + 1)(z + 1)z–3
Q (z)= (z2 + (a1  a2 + 1)z + 1)(z  1 )z–3
cf: (s2 + ( 2cos()wn )s + wn2)
Thus:(a1 + a2  1) = 2cos(1)
=  2cos(1)
&
(a1  a2 + 1) =  2cos(2 )
So, given:
i) LPC coeffs., a1 and a2 , then LSFs 1 & 2can be found
ii) LSFs, 1 & 2 , then the LPC coeffs. a1 and a2be found
2
1
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LSF Review & Example (4)
For a second order and with P(z) corresponding to the first root, Q(z) to the second root,
and so
P (z) = 1 + a1 z1 + a2 z2 + (1 + a1 z1 + a2 z2)z3
= (z2 + (a1 + a2  1)z + 1)(z + 1)z–3
for the second pair of qi, 1.37 and 1.77
= (z2  2cos(1.37) z + 1 )(z + 1) z–3
= (z3 +(1  2cos(1.37) z2+ (1  2cos(1.37))z + 1)z–3
Likewise
Q (z) = 1 + a1 z1 + a2 z2  (a1 z1 + a2 z2)z3
= (z2 + (a1  a2 + 1)z + 1)(z  1 )z–3
= (z2  2cos(1.77) z + 1 )(z  1) z–3
= (z3 +(1  2cos(1.77) z2+ (1 + 2cos(1.77))z  1)z–3
Then
A(z) = {P(z) + Q(z)} / 2)
= (z3 + (cos(1.37) + cos(1.77))z2 + (1  cos(1.37) + cos(1.77))z)z–3
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LSF Examples
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LSF Examples
A(z)= 1 + a1 z1 + a2 z2
P (z) = 1 + a1 z1 + a2 z2 + (1 + a1 z1 + a2 z2)z3
= (z2 + (a1 + a2  1)z + 1)(z + 1)z–3
= (z2 + (1.8 + 0.9  1)z + 1)(z + 1)z–3
= (z2  1.9 z + 1) (z + 1)z–3
cf: (z2 + ( 2cos()wn )z + wn2)
thus cos() =  1.9/2 or = 2.824 and 1 = π 
= 0.318
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Example Bit Allocation
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Codebooks & VQ
N = 2L
Identical book
i (0 … N1)
p
p
Data reduction: (p x B) to L
time
time
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N = 2 k
i
M
index, i
A(z)
en
sn
H(z)
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sn
H(z)
Codebook of timedomain samples
start point
en
y ms
y ms
y ms
en are time domain samples (integers)
R samples per second (eg 8000 Hz)
Frame rate governs vector size
P = 2 j
Bit rate = j/y bits/ms
P
NB en also includes gain
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x ms
N = 2 k
time
i
M
index, i
A[z] at time t
Vector with M elements, every x ms
Codebook with N = 2 kvectors
Bit rate = k/x bits per ms (not a function of M)
In practice A[z] is converted to LSF’s.
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1) Initialise:
form a single centroid of all training data, N=1
2) Repeat
Split centroids: N > 2N
Repeat
Cluster data to nearest centroid
until convergence
until N large enough
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VQ Performance on Unseen Data
Ramachandran & Mamone (eds) ‘Modern Methods of Speech Processing’ Kluer Academic, 1995
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VQ Performance on Unseen Data
Ramachandran & Mamone (eds) ‘Modern Methods of Speech Processing’ Kluer Academic, 1995
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1
0.5
0
Waveform
0.5
1
0
3.2
6.4
9.6
12.8
16
19.2
22.4
25.6
Time (ms)
LPC & FFT Spectra
LPC Roots
0.6651 ± 0.6695i
0.0560 ± 0.9709i
0.7228 ± 0.6225i
0.8714 ± 0.3694i
0.5758
0.4200
LSFs
40
20
0
Magnitude (dB)
20
40
0
1
2
3
4
5
Frequency (KHz) ( 0toFs/2)
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40
20
Magnitude (dB)
0
20
40
0
1
2
3
4
5
LPC Spectra & LSF’s
LPC Roots
0.6651 ± 0.6695i
0.0560 ± 0.9709i
0.7228 ± 0.6225i
0.8714 ± 0.3694i
0.5758
0.4200
LSFs
Frequency (KHz) ( 0toFs/2)
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LPC & FFT Spectra  2nd Order
A(z):
1.5537 0.8276
Roots:
0.7769 ± 0.4733i
1
0.5
0
0.5
H(0) = K
(1 (1.5537 0.8276))
H(ws/2) = K
(1 (1.5537 0.8276))
H(0)K/0.274
= = 21.8dB
H(ws /2) K/ 3.38
1
0
3.2
6.4
9.6
12.8
16
19.2
22.4
25.6
Time (ms)
40
20
0
20
40
0
1
2
3
4
5
Frequency (KHz) ( 0toFs/2)
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Other Related Topics
Spectral Lifting: H(z) = (1az1)
Codebook Training
Spectral Differences between 2 frames
Cepstra
Modeling Speech Space  HMM’s
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1
 1
1
 1
30ms
(a)
(b)
Figure Q1
PreEmphasis Example
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PreEmphasis Example
zplane jy
G(ws/2) = 1 + a
G(0) = 1  a
a
For G(ws/2 ) > G(0) then
a must be > 0
1+a = 2
ws/2
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Zplane to Magnitude Spectrum
1
0.5
0
Imaginary Part
0.5
1
1
0.5
0
0.5
1
Real Part
50
40
30
1+a = 2
20
10
Magnitude (dB)
0
10
ws/2
20
30
0
1
2
3
4
5
Frequency (KHz) ( 0toFs/2)
EG348_371_09
Air from
the lungs
Vocal fold
Vocal tract
Speech
H1(z)
H2(z)
synthesised
Speech
noise
Spectral envelop reflects morphological characteristics of the vocal tract
EG348_371_09
+

Z1
Z1
a1
a1
Z1
ai
ai
ap
ap
Speech
Residual
e`n
sn
en
1 – A’(z)
1 – A’(z)
sn
+

Z1
sn
sn1
Z1
STP
sni
Z1
LTP
ai
Z1
snp
EG348_371_09