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ESE 250 – S'12 Kod & DeHon
1
ESE250:Digital Audio Basics
Week 4 February 2, 2012
Time-Frequency
ESE 250 – S'12 Kod & DeHon
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Course Map
Numbers correspond to course weeks
2,5 6
11
13
12
Today
ESE 250 – S'12 Kod & DeHon
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Teaser: Musical Representation
• With this compact notation Could communicate a sound to pianist Much more compact than 44KHz time-sample
amplitudes (fewer bits to represent) Represent frequencies
ESE 250 – S'12 Kod & DeHon
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Week 4: Time-Frequency
• There are other ways to represent Frequency representation particularly efficient
http://en.wikipedia.org/wiki/File:Lead_Sheet.png
t 2 .1
t 0
t 2 .1
H 0 0 .6 , 0 .6 , 0 .6
H 1 0 .7 , 0 , 0 .7 H 2 0 .4 , 0 .8 , 0 .4
In this lecture we will learn that the frequency domain entails representing time-sampled signals using a conveniently rotated coordinate system
ESE 250 – S'12 Kod & DeHon
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Prelude: Harmonic Analysis• Fourier Transform ( FT )
Fourier (& other 19th Century Mathematicians) discovered that (real) signals can always (if they are smooth enough) be expressed as the sum of harmonics
• Defn: “Harmonics” (Fourier Series) collections of periodic signals (e.g., cos, sin) whose frequencies are related by integer
multiples arranged in order of increasing frequency summed in a linear combination whose coefficients provide an alternative representation
the job of this lecture is to replace this signals-analysis perspective with a symbols-synthesisperspective
ESE 250 – S'12 Kod & DeHon
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A Sampled (Real) Signal
Sample Data: Sampled Signal:(D e b u g ) O u t [ 3 7 7 ] =
t v
4
5
1
4 1 5 2 5 5
2
5
1
4 1 5 1 0 2 5
0 1
2
5
1
4 1 5 1 0 2 5
4
5
1
4 1 5 2 5 5
ESE 250 – S'12 Kod & DeHon
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Reconstructing the Sampled Signal• Exact Reconstruction
May be possible Under the right
assumptions Given the right model
• This example A “harmonic” signal Sampled in time Can be reconstructed
o exactly o from the time-sampled
valueso given knowledge of the
harmonics:
Cos[1t]/p (5/2)
p(5/2) ¢ Sin[2t]/ p(5/2)
+=
{ Cos[0t], Sin[1t], Cos[1t], Sin[2t], Cos[2t], Sin[3t], Cos[3t] }
p(5/2) ¢
ESE 250 – S'12 Kod & DeHon
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Reconstructing the Sampled Signal• Exact Reconstruction
May be possible Under the right
assumptions Given the right model
• This example A “harmonic” signal Sampled in time Can be reconstructed
o exactly o from the time-sampled
valueso given knowledge of the
harmonics:
p (5/2) ¢ Cos[1t]/p (5/2)
p(5/2) ¢ Sin[2t]/p (5/2)
+=
{ Cos[0t], Sin[1t], Cos[1t], Sin[2t], Cos[2t], Sin[3t], Cos[3t] }
ESE 250 – S'12 Kod & DeHon
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Sequence of Analysis• Given
Fundamental frequency: f = 1/2 Sampling Rate: ns = 5 Measured Data:
• Compute “basis” functions coefficients
• Reconstruct exact function from linear combination of
o “basis elements”(known)o coefficients(computed)
{r(-4/5), r(-2/5), r(02/5) , r(2/5), r(4/5) }
h0(t) = Cos[0t] / p 5
h1s(t) = Sin[1t] / p
(5/2)
h1c(t) = Cos[1t]/ p
(5/2)
h2s(t) = Sin[2t] / p
(5/2)
h2c(t) = Cos[2t]/ p
(5/2)
0 0p(5/2)
p(5/2)
0
r(t) = Cos[t] + Sin[2t] = 0 ¢ h0(t)
+ 0 ¢ h1s(t)
+ p(5/2) ¢ h1c(t)
+ p
(5/2) ¢ h2s(t)
+ 0 ¢ h2c(t)
ESE 250 – S'12 Kod & DeHon
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Fourier AnalysisTime-Values
(D e b u g ) O u t [ 3 7 7 ] =
t v
4
514
1 5 2 5 5
2
514
1 5 1 0 2 5
0 1
2
514
1 5 1 0 2 5
4
514
1 5 2 5 5
(D e b u g ) O u t [ 3 8 4 ] =
f AC os 0 t 0
S in t 0
C os t 5
2
S in 2 t 5
2
C os 2 t 0
Frequency-Amplitudes
FT
(D e b u g ) O u t [ 3 9 0 ]=
t v 3 0 .1
1 0 .3
0 1 .
1 0 .9
3 1 .8
Sampled
Quantized
DFT (D e b u g ) O u t [ 3 9 5 ] =
f A0 01s 01c 1.62s 1.62c 0
(“closed form”)
(computation)
ESE 250 – S'12 Kod & DeHon
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Reconstruction vs Approximation• Previous Example
received function was “in the span” of the harmonics reconstruction achieves exact match at all times
• More General Case received function is “close” to the “span” reconstruction achieves exact match only at the sampled times get successively better approximation at all times
o by taking successively more sampleso and using successively higher harmonics
ESE 250 – S'12 Kod & DeHon
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Another Sampled (Real) Signal
t
v
Sample Data: Sampled Signal:
O u t [ 2 9 5 ] =
t v 6
7328
1 4 Cos 7
Sin 7
4
7114
2 Sin 7
2
7128
2 Cos 14
0 02
7128
1 2 Cos 14
4
7 114
1 2 Sin 7
6
7 328
4 Cos 7
Sin 7
ESE 250 – S'12 Kod & DeHon
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• Approximate Reconstruction is always achievable and more relevant to our problem
• Example A roughly “harmonic” signal Sampled in time Can be approximated
o “arbitrarily” closely o from the time-sampled valueso using any “good” set of harmonics
Approximating the Sampled Signal
{ Cos[0t], Sin[1t], Cos[1t] , Sin[2t], Cos[2t] , Sin[3t], Cos[3t] }
ESE 250 – S'12 Kod & DeHon
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ApproximateReconstruction
(D e b u g ) O u t [ 3 6 0 ] =
C os 0 t C os 14
2 1 3 C os 7 Sin
7
7 7
S in t 2 C os
14 C os 3
14 3 Sin
7
7 14
C os t 147 5 4 1114 1314 5 1514 1914 4 11114
14 14
S in 2 t C os
14 3 C os 3
14 2 Sin
7
7 14
C os 2 t 1914 1 2 117 3 127 3 137 2 147 157 14 14
S in 3 t 3 C os
14 2 C os 3
14 Sin
7
7 14
C os 3 t 1914 1 2 117 3 127 3 137 2 147 157 14 14
(Successively Thinner Green Dashed Curves Denote Successively Fewer Harmonic Components)
Sum up the (black) harmonics using the (green) coefficients:
ESE 250 – S'12 Kod & DeHon
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More Harmonics are Better
(D e b u g ) O u t [ 3 6 9 ] =
C os 0 t 2 C os
22 3 C sc
22 Sin
11 4 C os 3
22 Sin 3
22 5 Sin 2
11 2 C os 2
11 Sin 2
11
11 11
S in t 20 C sc
22 6 4 C sc 3
22 C sc 5
22 Sin
11 4 Sin 2
11
44 22
C os t 1
11
2
11 3 S in
22S in
11 4 C os 3
22S in 3
222
5 C os
11S in 2
11 2 C os 2
112
S in 2
11 2 C os
22S in 5
22 4
C os 5
22S in 5
222
S in 2 t 24 C sc 22
16 4 C sc 3 22
C sc 5 22
Sin 11
40 Sin 2 11
88 22
C os 2 t 32 C sc 22
8 2 5 C sc 3 22
C sc 5 22
Sin 11
24 Sin 2 11
88 22
S in 3 t 16 C sc 22
4 10 3 C sc 3 22
C sc 5 22
Sin 11
32 Sin 2 11
88 22
C os 3 t 1122 2 6 1111 1211 3 1311 4 1411 4 1611 3 1711 1811 6 1911 2 11011 22 22
S in 4 t 32 C sc 22
8 2 5 C sc 3 22
C sc 5 22
Sin 11
24 Sin 2 11
88 22
C os 4 t 8 C sc 5
22 Sin
11 C sc
22 C sc 3
22 Sin
11 2 3 C sc 2
11 C sc 5
22 Sin
11 16 Sin 2
11 80 Sin 3
22 Sin 2
11
88 22
S in 5 t 4 C sc 22
10 3 2 C sc 3 22
C sc 5 22
Sin 11
8 Sin 2 11
44 22
C os 5 t 1122 1 8 1111 6 1211 7 1311 2 1411 2 1611 7 1711 6 1811 8 1911 11011
22 22
(D e b u g ) O u t [ 3 6 3 ] =
t v 10
11 544
1 2 Sin 2 11
8
11111
4 Cos 5 22
Sin 5 22
6
11144
3 6 Sin 11
4
11122
4 Cos 3 22
Sin 3 22
2
11144
4 Cos 2 11
Sin 2 11
0 02
11144
1 4 Cos 2 11
Sin 2 11
4
11122
1 4 Cos 3 22
Sin 3 22
6
11 344
2 Sin 11
8
11 111
1 2 Cos 22
10
11 544
2 Sin 2 11
7 Samples; 7 Harmonics 11 Samples 15 Samples; 15 Harmonics; 11 Harmonics
ESE 250 – S'12 Kod & DeHon
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Usually Computed, Not “Solved”7 Samples; 7 Harmonics 11 Samples 15 Samples; 15 Harmonics; 11 Harmonics
(D e b u g ) O u t [ 3 9 8 ] =
t v 2 .9 0
2 .3 0 .3
1 .7 0 .1
1 .1 0 .4
0 .6 0 .2
0 0
0 .6 0 .1
1 .1 0 .1
1 .7 0 .3
2 .3 0 .9
2 .9 0 .7
DFT
(D e b u g ) O u t [ 3 9 7 ] =
f A0 0.41s 0.61c 0.82s 0.32c 0.23s 0.23c 0.44s 0.24c 0.15s 0.25c 0
(D e b u g ) O u t [ 4 0 0 ]=
t v 2 .9 0 .1
2 .5 0 .3
2 .1 0 .2
1 .7 0 .1
1 .3 0 .3
0 .8 0 .3
0 .4 0 .1
0 0
0 .4 0
0 .8 0 .1
1 .3 0
1 .7 0 .3
2 .1 0 .7
2 .5 0 .9
2 .9 0 .7
(D e b u g ) O u t [ 3 9 9 ] =
f A0 0.51s 0.71c 0.92s 0.42c 0.23s 0.23c 0.54s 0.24c 0.25s 0.25c 0.16s 0.26c 07s 0.17c 0
(D e b u g ) O u t [ 4 0 4 ] =
t v2.7 0.21.8 00.9 0.30 00.9 0.11.8 0.42.7 0.9
(D e b u g ) O u t [4 0 5 ]=
f A0 0.41s 0.51c 0.72s 0.32c 0.23s 0.23c 0.2
DFTDFT
the “spectrum” is often plotted as a function of frequency
ESE 250 – S'12 Kod & DeHon
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Yet Another Sampled (Real) Signal
t
v
Measured Data:
Sampled Signal:
(D e b u g ) O u t [ 4 2 0 ] =
t v 14
15 12
4 5
12
2 3
12
8 15
12
2 5
12
4 15
12
2 15
12
0 12
2
1512
4
1512
2
5 12
8
15 12
2
312
4
512
14
15 12
ESE 250 – S'12 Kod & DeHon
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• Approximate Reconstruction although always achievable may require a lot of samples to get good performance from “poorly chosen”
harmonics
• Different “bases” match different “data” better or worse
(sometimes time is better than frequency)
Some Signals Dislike Some Harmonics
15 Samples & Harmonics
21 Samples & Harmonics
31 Samples & Harmonics
ESE 250 – S'12 Kod & DeHon
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t 2 .1
t 0
t 2 .1
H 0 0 .6 , 0 .6 , 0 .6
H 1 0 .7 , 0 , 0 .7 H 2 0 .4 , 0 .8 , 0 .4
Choice of Basis• What is a “harmonic”?
we could have used periodic “pulse trains”o previous signal would be reconstructed exactlyo with one or two pulse-train harmonics
but “sound-like” signals o would typically require a very large numbero of “pulse-train” harmonics
• Fourier Theory (and generalizations) permits very broad choice of harmonics such choices amount to the selection of a model
• Today’s Lecture interprets the choice of harmonics
o as a selection of coordinate reference frame o in the space of received (sampled,quantized) data
lends (geometric) insight to high-dimensional phenomena introduces arsenal of linear algebraic computation encourages “learning” data-driven models
ESE 250 – S'12 Kod & DeHon
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Intuitive Concept Inventory
(D e b u g ) O u t [ 3 6 9 ] =
C os 0 t 2 C os
22 3 C sc
22 Sin
11 4 C os 3
22 Sin 3
22 5 Sin 2
11 2 C os 2
11 Sin 2
11
11 11
S in t 20 C sc
22 6 4 C sc 3
22 C sc 5
22 Sin
11 4 Sin 2
11
44 22
C os t 1
11
2
11 3 S in
22S in
11 4 C os 3
22S in 3
222
5 C os
11S in 2
11 2 C os 2
112
S in 2
11 2 C os
22S in 5
22 4
C os 5
22S in 5
222
S in 2 t 24 C sc 22
16 4 C sc 3 22
C sc 5 22
Sin 11
40 Sin 2 11
88 22
C os 2 t 32 C sc 22
8 2 5 C sc 3 22
C sc 5 22
Sin 11
24 Sin 2 11
88 22
S in 3 t 16 C sc 22
4 10 3 C sc 3 22
C sc 5 22
Sin 11
32 Sin 2 11
88 22
C os 3 t 1122 2 6 1111 1211 3 1311 4 1411 4 1611 3 1711 1811 6 1911 2 11011 22 22
S in 4 t 32 C sc 22
8 2 5 C sc 3 22
C sc 5 22
Sin 11
24 Sin 2 11
88 22
C os 4 t 8 C sc 5
22 Sin
11 C sc
22 C sc 3
22 Sin
11 2 3 C sc 2
11 C sc 5
22 Sin
11 16 Sin 2
11 80 Sin 3
22 Sin 2
11
88 22
S in 5 t 4 C sc 22
10 3 2 C sc 3 22
C sc 5 22
Sin 11
8 Sin 2 11
44 22
C os 5 t 1122 1 8 1111 6 1211 7 1311 2 1411 2 1611 7 1711 6 1811 8 1911 11011
22 22
(D e b u g ) O u t [3 6 3 ] =
t v 10
11 544
1 2 Sin 2 11
8
11111
4 Cos 5 22
Sin 5 22
6
11144
3 6 Sin 11
4
11122
4 Cos 3 22
Sin 3 22
2
11144
4 Cos 2 11
Sin 2 11
0 02
11144
1 4 Cos 2 11
Sin 2 11
4
11122
1 4 Cos 3 22
Sin 3 22
6
11 344
2 Sin 11
8
11 111
1 2 Cos 22
10
11 544
2 Sin 2 11
11 Samples;
Q = FT(q)
11 Harmonics
Time Domain Frequency Domain
r (received signal)
(sam
pling) q Q
ESE 250 – S'12 Kod & DeHon
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(D e b u g ) O u t [ 3 9 6 ] =
t v 3 0
2 0 .3
2 0 .1
1 0 .4
1 0 .2
0 0
1 0 .1
1 0 .1
2 0 .3
2 0 .9
3 0 .7
Intuitive Concept Inventory11 Samples;
Q = DFT(q)
11 Harmonics
Time Domain Frequency Domain
Floating P
oint Flo
atin
g P
oint
r (received signal)
Sampling &Quantization
q Qthis week’s idea
Perceptual coding
(D e b u g ) O u t [ 3 9 7 ] =
f A0 0.41s 0.61c 0.82s 0.32c 0.23s 0.23c 0.44s 0.24c 0.15s 0.25c 0
ESE 250 – S'12 Kod & DeHon
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Where Are We Heading After Today?• Week 2
Received signal iso discrete-time-stampedo quantized
q = PCM[ r ]
= quantL [SampleTs[r] ]
• Week 3 Quantized Signal is
Coded c =code[ q ]
• Week 4 Sampled signal
o not coded directly o but rather, “Float” -‘edo then linearly
transformed o into frequency domain
Q = DFT[ q ]
[Painter & Spanias. Proc.IEEE, 88(4):451–512, 2000]
q
Sample CodeStore/
Transmit Decode Producer(t) p(t)
Generic Digital Signal Processor
q c
c
Q
Psychoacoustic Audio Coder
ESE 250 – S'12 Kod & DeHon
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Interlude: Audio Communications
Close Encounters
ESE 250 – S'12 Kod & DeHon
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Technical Concept Inventory• Floating Point Quantization
a symbolic representation admitting a mimic of continuous arithmetic
• Vectors sampled signals are points in a (high dimensional) vector space
• Linear Algebra the “Swiss Army Knife” of high dimensions provides a logical, geometric, and computational toolset for manipulating vectors
• Change of Basis DFT is a high dimensional rotation in the vector space of time-sampled signals
ESE 250 – S'12 Kod & DeHon
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Technical Concept Inventory• Floating Point Quantization
a symbolic representation admitting a mimic of continuous arithmetic
• Vectors sampled signals are points in a (high dimensional) vector space
• Linear Algebra the “Swiss Army Knife” of high dimensions provides a logical, geometric, and computational toolset for manipulating vectors
• Change of Basis DFT is a high dimensional rotation in the vector space of time-sampled signals
ESE 250 – S'12 Kod & DeHon
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6 4 2 2 4 6
2
1
1
2
r(t)
q1
q2 q3
q4 q5
Float-Quantized Symbols Act “Real”• q = PCM[ r(t) ] = Float(b,p,E) [SampleTs
[r(t)] ] eliminates continuous time dependence discretizes continuous values
o cannot represent an uncountable collection of functions o with a countable (of course, in fact, finite!) set of “symbols”
• Floating Point Representation and Computer Arithmetic Choose: Base (b), Precision (p), Magnitude (E)
o q = be ¢ [d0 + d1 ¢ b-1 + … + dp-1 ¢ b-(p-1)]o - E · e · Eo 0 < di < b
Non-uniform quantizationo bp different “mantissas”o 2E different exponentso ~ Log2[2E] + Log2[bp] bits
Associated Flop Arithmeticop 2 { +, -, *, /} [ { Sqrt, Mod, Flint} ) Flop(x,y) = Float[ op(x,y) ]
Archetypal Computation: Inner producto x = (x1, .., xn), y = (y1, … , yn)o hx,yi = x1¢y1 + x2¢y2 + … + xn¢ yn
Crucially important operation for signal processing applications !
[Widrow, et al., IEEE TIM’96]
ESE 250 – S'12 Kod & DeHon
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Technical Concept Inventory• Floating Point Quantization
a symbolic representation admitting a mimic of continuous arithmetic
• Vectors sampled signals are points in a (high dimensional) vector space
• Linear Algebra the “Swiss Army Knife” of high dimensions provides a logical, geometric, and computational toolset for manipulating vectors
• Change of Basis DFT is a high dimensional rotation in the vector space of time-sampled signals
ESE 250 – S'12 Kod & DeHon
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• Sampled received signal
• Is a discrete sequence of time-stamped floats q = (q1, q2, … qns
)
= Float( r(T0+Ts), r(T0 + 2Ts), …. , r(T0 + nsTs)) of “real” (i.e. Float’ed) values
at each of the ns time-stamps
• Think of each of the time-stamps as an “axis” of “real” (float) values
6 4 2 2 4 6
2
1
1
2
Time Functions are Vectors
r(t)
q1
q2
q3
q4
q5
ESE 250 – S'12 Kod & DeHon
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Time Functions are Vectors• Think of each of the time-
stamps as an “axis” of “real” (float) values
• E.g., for three time stamps, ns = 3, we can record the values arrange each axis located
perpendicular to the other two in space mark their values and interpret them as a vector
t 6 .2 8
t 0 .6 9
t 4 .9
t 6 .2 8
t 0 .6 9
t 4 .9
ESE 250 – S'12 Kod & DeHon
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• Think of each of the time-stamps as an “axis” of “real” (float) values E.g., for two time stamps, ns = 2,
o we can draw both axes o on “graph paper”
… for a greater number of time stamps …
o we can “imagine” arranging each axis
o in a mutually perpendicular direction
o in space of appropriately high dimension
t = - 6.28
t = 2.5
q1
q2
q
b1
b2
Time Functions are Vectors
ESE 250 – S'12 Kod & DeHon
31
Technical Concept Inventory• Floating Point Quantization
a symbolic representation admitting a mimic of continuous arithmetic
• Vectors sampled signals are points in a (high dimensional) vector space
• Linear Algebra the “Swiss Army Knife” of high dimensions provides a logical, geometric, and computational toolset for manipulating vectors
• Change of Basis DFT is a high dimensional rotation in the vector space of time-sampled signals
ESE 250 – S'12 Kod & DeHon
32
Linear Algebra: “Swiss Army Knife”• We cannot “see” in high
dimensions• Linear Algebra enables us
in high dimensions to reason precisely think geometrically compute
• Essential Ideas Basis expansion Change of basis Ingredients
o Orthonormalityo Inner Product h ¢ , ¢ i
t = - 6.28
t = 2.5
q1
r(t)q1
q2
BT = { b1 , b2 } = { (1,0), (1,0)}
q2
q = (q1, q2) = (0.8, - 0.9) = 0.8 ¢ (1,0) – 0.9 ¢ (1,0) = 0.8 ¢ b1 + (– 0.9) ¢ b2
= hq,b1i¢ b1 + hq,b2i ¢ b2
= q1 ¢ b1 + q2 ¢ b2
q
b1b2
wherehx,yi = x1y1 + x2y2
hq,b1i = 0.8 ¢ 1 + (-0.9) ¢ 0 = 0.8hq,b2i = 0.8 ¢ 0 + (-0.9) ¢ 1 = - 0.9
(computational definition):
ESE 250 – S'12 Kod & DeHon
33
Linear Algebra: “Swiss Army Knife”• Orthonormal Basis
set of unit length vectors
each “perpendicular” to all the others
total number given by dimension of the space
• Inner Product (scaled) cosine of
relative angle scales unit length
t = - 6.28
t = 2.5
q
b1
b2
q1 = hq,b1i = Length(q ) ¢ Cos [Å(q,b1)]
Å(q,b1)
Å(q,b2)
q2 = hq,b2i = Length(q ) ¢ Cos [Å(q,b2)]
Generally: hr, si = Length(r) ¢ Length(s) ¢ Cos [Å(r,s)] ) hr, ri = Length(r)2
geometric re-interpretation of computational definition: hx,yi = x1y1 + x2y2
ESE 250 – S'12 Kod & DeHon
34
Technical Concept Inventory• Floating Point Quantization
a symbolic representation admitting a mimic of continuous arithmetic
• Vectors sampled signals are points in a (high dimensional) vector space
• Linear Algebra the “Swiss Army Knife” of high dimensions provides a logical, geometric, and computational toolset for manipulating vectors
• Change of Basis DFT is a high dimensional rotation in the vector space of time-sampled signals
ESE 250 – S'12 Kod & DeHon
35
Change of Coordinates
[Google Maps]
Vs.
Independence Hall500 Chestnut St.
ESE 250 – S'12 Kod & DeHon
36
Why Change Basis ?• Efficiency
data sets often lie along lower dimensional
subspaces Of high dimensional data
space• Decoupling
receiver model may “prefer”
a specific basis
ESE 250 – S'12 Kod & DeHon
37
Linear Algebra: Change of Basis• Goal
Re-express q In terms of BH
• Notation use new symbol, Q denoting different computational
representation even though vector is geometrically
unchanged• Check: “good” basis?
both unit length? mutually perpendicular vectors?
• Further geometric Interpretation if old basis is orthonormal then new basis is also if and only if it is
o A “rotation” o Away from the old
BH = { H1 , H2 } = { (1/p2 , 1/p2), (- 1/p2 , 1/p2)}
Q
H1H2
Length(H1)2 = h H1, H1 i
= 1/p (2 ¢ 2) + 1/
p (2 ¢ 2)
= ½ + ½ = 1Length(H2)2 = h H2, H2 i
= 1/p (2 ¢ 2) + 1/
p (2 ¢ 2)
= ½ + ½ = 1
hH1, H2i = h11 h2
1 + h12 h2
2
= - 1/p 2 ¢ 2 + 1/
p 2 ¢ 2
= 0
t = - 6.28
t = 2.5
b2
b1
ESE 250 – S'12 Kod & DeHon
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Linear Algebra: Change of Basis• Goal
Re-express q = (q1, q2)o specified by coordinate
representationo in terms of the old basis, BT
As Q= [Q1, Q2] o Specified by coordinate
representationo In terms of rotate basis, BH
• Idea: recall geometric meaning of q = (q1, q2)
o scale b1 by q1 = h b1, q i o scale b2 by q2 = h b2, q i o form the resultant vector
• Compute Q= [Q1, Q2] using same geometric idea reveals how to obtain [Q1, Q2]
o scale H1 by Q1 = hq,H1io scale H2 by Q2 = hq,H2io form the resultant vector
q = (q1, q2) = q1 ¢ b1 + q2 ¢ b2
= hq, b1i¢ b1 + hq, b2i ¢ b2
) Q1 = hq , H1i
=h (0.8, - 0.9), (1/p
2, 1/p
2)i = (0.8/1.1 - 0.9/1.1) ¼ - 0.11
Q = [Q1, Q2] = hQ,H1i¢ H1 + hQ,H2i ¢ H2
= hq,H1i¢ H1 + hq,H2i ¢ H2
) Q2 = hq , H2i
=h (0.8, - 0.9), (-1/p
2, 1/p
2)i = - (0.8/1.1 + 0.9/1.1) ¼ - 1.6
- Q2
t = - 6.28
t = 2.5
-Q1
Q
H1b2
b1H2
ESE 250 – S'12 Kod & DeHon
39
1 . 0
0 . 5
0 . 0
0 . 5
1 . 0
t 2 .1 , v 0 .4
1 . 0
0 . 5
0 . 0
0 . 5
1 . 0
t 0 , v 0 .8
1 . 0 0 . 50 . 00 . 51 . 0t 2 .1 , v 0 .4
1 . 0
0 . 5
0 . 0
0 . 5
1 . 0
t 2 .1 , v 0 .7
1 . 0 0 . 5 0 . 0 0 . 5 1 . 0t 0 , v 0
1 . 0
0 . 5
0 . 0
0 . 5
1 . 0
t 2 .1 , v 0 .7
Generalize to ns = 3 Samplesh0(t) = Cos[0t]/p3
h1(t) = 2 Sin[t]/p3
h2(t) = 2 Cos[t]/p3
1 . 0
0 . 5
0 . 0
0 . 5
1 . 0
t 2 .1 , v 0 .6
1 . 0 0 . 5 0 . 0 0 . 5 1 . 0t 0 , v 0 .6
1 . 0
0 . 5
0 . 0
0 . 5
1 . 0
t 2 .1 , v 0 .6 H0 = Float[ h0(-2/3), h0(0/3), h0(2/3)]
H1 = Float[ h1(-2/3), h1(0/3), h1(2/3)]
H2 = Float[ h2(-2/3), h2(-0/3), h2(2/3)]
The 3-sample DFT:• take inner products• of sampled signal• with each harmonic
ESE 250 – S'12 Kod & DeHon
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Generalize to ns = 3 Samplesh0(t) = Cos[0t]/p3
h1(t) = 2 Sin[t]/p3
h2(t) = 2 Cos[t]/p3 t 2 .1
t 0
t 2 .1
H 0 0 .6 , 0 .6 , 0 .6
H 1 0 .7 , 0 , 0 .7 H 2 0 .4 , 0 .8 , 0 .4
ESE 250 – S'12 Kod & DeHon
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(D e b u g ) O u t [ 3 9 6 ] =
t v 3 0
2 0 .3
2 0 .1
1 0 .4
1 0 .2
0 0
1 0 .1
1 0 .1
2 0 .3
2 0 .9
3 0 .7
11 Samples;
Q = DFT(q)
11 Harmonics
Time Domain Frequency Domain
Floating P
oint Flo
atin
g P
oint
r (received signal)
Sampling &Quantization
q Qthis week’s idea
Perceptual coding
(D e b u g ) O u t [ 3 9 7 ] =
f A0 0.41s 0.61c 0.82s 0.32c 0.23s 0.23c 0.44s 0.24c 0.15s 0.25c 0
Generalize to Arbitrary Samples
ESE 250 – S'12 Kod & DeHon
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… for more understanding….• Courses ESE 325 ! (Math 240) ) Math 312 !!!
• Reading Quantization
B. Widrow, I. Kollar, and M. C. Liu. Statistical theory of quantization. IEEE Transactions on Instrumentation and Measurement, 45(2):353–361, 1996.
Floating Point D. Goldberg. What every computer scientist should know about
floating-point arithmetic. ACM Computing Surveys, 23(1), 1991.
Linear Algebra for Frequency Transformationso G. Strang. The discrete cosine transform. SIAM Review, 41(1):135–
147, 1999
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ESE250:Digital Audio Basics
End Week 4 Lecture
Time-Frequency