week 4 february 2, 2012 time-frequency - penn engineeringese250/week4/day4-s'12.pdf · 2012. 2....

ESE 250 – S'12 Kod & DeHon 1

ESE250:

Digital Audio Basics

Week 4

February 2, 2012

Time-Frequency

2

Course Map

Numbers correspond to course weeks

2,5 6

11

13

12

Today

ESE 250 – S'12 Kod & DeHon


Where Are We Heading After Today? • Week 2

Received signal is o discrete-time-stamped

o 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” -„ed

o then linearly transformed

o into frequency domain

Q = DFT[ q ]

[Painter & Spanias. Proc.IEEE, 88(4):451–512, 2000]

q

Sample Code Store/

Transmit Decode Produce r(t) p(t)

Generic Digital Signal Processor

q c

c

Q Psychoacoustic Audio Coder


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


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

H0 0.6 , 0.6 , 0.6

H1 0.7 , 0 , 0.7

H2 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


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-

synthesis

perspective


A Sampled (Real) Signal

Sample Data: Sampled Signal: (D e b u g ) O u t[3 7 7 ]=t v

4

5

1

41 5 2 5 5

2

5

1

41 5 10 2 5

0 1

2

5

1

41 5 10 2 5

4

5

1

41 5 2 5 5


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 values

o 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) ¢


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 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] }


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

0 p(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)


Fourier Analysis

Time-Values (D e b u g ) O u t[3 7 7 ]=t v

4

5

1

41 5 2 5 5

2

5

1

41 5 10 2 5

0 1

2

5

1

41 5 10 2 5

4

5

1

41 5 2 5 5

(D e b u g ) O u t[3 8 4 ]=

f A

Cos 0 t 0

Sin t 0

Cos t5

2

Sin 2 t5

2

Cos 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

S

a

m

p

l

e

d

Q

u

a

n

ti

z

e

d

DFT (D e b u g ) O u t[3 9 5 ]=

f A

0 0

1s 0

1c 1.6

2s 1.6

2c 0

(“closed form”)

(computation)


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 samples

o and using successively higher harmonics


Another Sampled (Real) Signal

t

v

Sample Data: Sampled Signal:

O u t[2 9 5 ]=

t v6

7

3

281 4 Cos

7Sin

7

4

7

1

142 Sin

7

2

7

1

282 Cos

14

0 0

2

7

1

281 2 Cos

14

4

7

1

141 2 Sin

7

6

7

3

284 Cos

7Sin

7


• 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 using any “good” set of harmonics

Approximating the Sampled Signal

{ Cos[0t], Sin[1t], Cos[1t] , Sin[2t], Cos[2t] , Sin[3t], Cos[3t] }


Approximate

Reconstruction

(D e b u g ) O u t[3 6 0 ]=

Cos 0 tC o s

1 42 1 3 C o s

7S in

7

7 7

Sin t2 C o s

1 4C o s

3

1 43 S in

7

7 1 4

Cos t1 4 7 5 4 1 1 1 4 1 3 1 4 5 1 5 1 4 1 9 1 4 4 1 1 1 1 4

1 4 1 4

Sin 2 tC o s

1 43 C o s

3

1 42 S in

7

7 1 4

Cos 2 t1 9 1 4 1 2 1 1 7 3 1 2 7 3 1 3 7 2 1 4 7 1 5 7

1 4 1 4

Sin 3 t3 C o s

1 42 C o s

3

1 4S in

7

7 1 4

Cos 3 t1 9 1 4 1 2 1 1 7 3 1 2 7 3 1 3 7 2 1 4 7 1 5 7

1 4 1 4

(Successively Thinner

Green Dashed Curves

Denote Successively

Fewer Harmonic

Components)

Sum up the (black) harmonics using

the (green) coefficients:


More Harmonics are Better

(D e b u g ) O u t[3 6 9 ]=

Cos 0 t2 C o s

2 23 C s c

2 2S in

1 14 C o s

3

2 2S in

3

2 25 S in

2

1 12 C o s

2

1 1S in

2

1 1

1 1 1 1

Sin t2 0 C s c

2 26 4 C s c

3

2 2C s c

5

2 2S in

1 14 S in

2

1 1

4 4 2 2

Cos t1

1 1

2

1 13 Sin

2 2Sin

1 14 Cos

3

2 2Sin

3

2 2

25 Cos

1 1Sin

2

1 12 Cos

2

1 1

2Sin

2

1 12 Cos

2 2Sin

5

2 24

Cos5

2 2Sin

5

2 2

2

Sin 2 t2 4 C s c

2 21 6 4 C s c

3

2 2C s c

5

2 2S in

1 14 0 S in

2

1 1

8 8 2 2

Cos 2 t3 2 C s c

2 28 2 5 C s c

3

2 2C s c

5

2 2S in

1 12 4 S in

2

1 1

8 8 2 2

Sin 3 t1 6 C s c

2 24 1 0 3 C s c

3

2 2C s c

5

2 2S in

1 13 2 S in

2

1 1

8 8 2 2

Cos 3 t1

1 2 22 6 1

1 1 11

2 1 13 1

3 1 14 1

4 1 14 1

6 1 13 1

7 1 11

8 1 16 1

9 1 12 1

1 0 1 1

2 2 2 2

Sin 4 t3 2 C s c

2 28 2 5 C s c

3

2 2C s c

5

2 2S in

1 12 4 S in

2

1 1

8 8 2 2

Cos 4 t8 C s c

5

2 2S in

1 1C s c

2 2C s c

3

2 2S in

1 12 3 C s c

2

1 1C s c

5

2 2S in

1 11 6 S in

2

1 18 0 S in

3

2 2S in

2

1 1

8 8 2 2

Sin 5 t4 C s c

2 21 0 3 2 C s c

3

2 2C s c

5

2 2S in

1 18 S in

2

1 1

4 4 2 2

Cos 5 t1 1 2 2 1 8 1 1 1 1 6 1 2 1 1 7 1 3 1 1 2 1 4 1 1 2 1 6 1 1 7 1 7 1 1 6 1 8 1 1 8 1 9 1 1 1 1 0 1 1

2 2 2 2

(D e b u g ) O u t[3 6 3 ]=

t v10

11

5

441 2 Sin

2

11

8

11

1

114 Cos

5

22Sin

5

22

6

11

1

443 6 Sin

11

4

11

1

224 Cos

3

22Sin

3

22

2

11

1

444 Cos

2

11Sin

2

11

0 0

2

11

1

441 4 Cos

2

11Sin

2

11

4

11

1

221 4 Cos

3

22Sin

3

22

6

11

3

442 Sin

11

8

11

1

111 2 Cos

22

10

11

5

442 Sin

2

11

7 Samples; 7 Harmonics 11 Samples 15 Samples; 15 Harmonics ; 11 Harmonics


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 A

0 0.4

1s 0.6

1c 0.8

2s 0.3

2c 0.2

3s 0.2

3c 0.4

4s 0.2

4c 0.1

5s 0.2

5c 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 A

0 0.5

1s 0.7

1c 0.9

2s 0.4

2c 0.2

3s 0.2

3c 0.5

4s 0.2

4c 0.2

5s 0.2

5c 0.1

6s 0.2

6c 0

7s 0.1

7c 0

(D e b u g ) O u t[4 0 4 ]=

t v2.7 0.2

1.8 0

0.9 0.3

0 0

0.9 0.1

1.8 0.4

2.7 0.9

(D e b u g ) O u t[4 0 5 ]=

f A

0 0.4

1s 0.5

1c 0.7

2s 0.3

2c 0.2

3s 0.2

3c 0.2

DFT DFT

the “spectrum” is often plotted as a function of frequency


Yet Another Sampled

(Real) Signal

t

v

Measured Data:

Sampled Signal:

(D e b u g ) O u t[4 2 0 ]=

t v14

15

1

2

4

5

1

2

2

3

1

2

8

15

1

2

2

5

1

2

4

15

1

2

2

15

1

2

01

2

2

15

1

2

4

15

1

2

2

5

1

2

8

15

1

2

2

3

1

2

4

5

1

2

14

15

1

2


• 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




t 2.1

t 0

t 2.1

H0 0.6 , 0.6 , 0.6

H1 0.7 , 0 , 0.7

H2 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 exactly

o with one or two pulse-train harmonics

but “sound-like” signals o would typically require a very large number

o 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


Intuitive Concept Inventory

(D e b u g ) O u t[3 6 9 ]=

Cos 0 t2 C o s

2 23 C s c

2 2S in

1 14 C o s

3

2 2S in

3

2 25 S in

2

1 12 C o s

2

1 1S in

2

1 1

1 1 1 1

Sin t2 0 C s c

2 26 4 C s c

3

2 2C s c

5

2 2S in

1 14 S in

2

1 1

4 4 2 2

Cos t1

1 1

2

1 13 Sin

2 2Sin

1 14 Cos

3

2 2Sin

3

2 2

25 Cos

1 1Sin

2

1 12 Cos

2

1 1

2Sin

2

1 12 Cos

2 2Sin

5

2 24

Cos5

2 2Sin

5

2 2

2

Sin 2 t2 4 C s c

2 21 6 4 C s c

3

2 2C s c

5

2 2S in

1 14 0 S in

2

1 1

8 8 2 2

Cos 2 t3 2 C s c

2 28 2 5 C s c

3

2 2C s c

5

2 2S in

1 12 4 S in

2

1 1

8 8 2 2

Sin 3 t1 6 C s c

2 24 1 0 3 C s c

3

2 2C s c

5

2 2S in

1 13 2 S in

2

1 1

8 8 2 2

Cos 3 t1

1 2 22 6 1

1 1 11

2 1 13 1

3 1 14 1

4 1 14 1

6 1 13 1

7 1 11

8 1 16 1

9 1 12 1

1 0 1 1

2 2 2 2

Sin 4 t3 2 C s c

2 28 2 5 C s c

3

2 2C s c

5

2 2S in

1 12 4 S in

2

1 1

8 8 2 2

Cos 4 t8 C s c

5

2 2S in

1 1C s c

2 2C s c

3

2 2S in

1 12 3 C s c

2

1 1C s c

5

2 2S in

1 11 6 S in

2

1 18 0 S in

3

2 2S in

2

1 1

8 8 2 2

Sin 5 t4 C s c

2 21 0 3 2 C s c

3

2 2C s c

5

2 2S in

1 18 S in

2

1 1

4 4 2 2

Cos 5 t1 1 2 2 1 8 1 1 1 1 6 1 2 1 1 7 1 3 1 1 2 1 4 1 1 2 1 6 1 1 7 1 7 1 1 6 1 8 1 1 8 1 9 1 1 1 1 0 1 1

2 2 2 2

(D e b u g ) O u t[3 6 3 ]=

t v10

11

5

441 2 Sin

2

11

8

11

1

114 Cos

5

22Sin

5

22

6

11

1

443 6 Sin

11

4

11

1

224 Cos

3

22Sin

3

22

2

11

1

444 Cos

2

11Sin

2

11

0 0

2

11

1

441 4 Cos

2

11Sin

2

11

4

11

1

221 4 Cos

3

22Sin

3

22

6

11

3

442 Sin

11

8

11

1

111 2 Cos

22

10

11

5

442 Sin

2

11

11 Samples;

Q = FT(q)

11 Harmonics

Time Domain Frequency Domain

r (received signal)

q Q


(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 Inventory 11 Samples;

Q = DFT(q)

11 Harmonics


Flo

ating P

oin

t Flo

atin

g P

oin

t

r (received signal)

Sampling & Quantization

q Q

this

week’s

idea

Perceptual coding

(D e b u g ) O u t[3 9 7 ]=

f A

0 0.4

1s 0.6

1c 0.8

2s 0.3

2c 0.2

3s 0.2

3c 0.4

4s 0.2

4c 0.1

5s 0.2

5c 0


Interlude: Audio Communications

Close Encounters

../../RepositoryMaterial/2010/week4/interlude.close_encouters.avi


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


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 · E

o 0 < di < b

Non-uniform quantization o bp different “mantissas”

o 2E different exponents

o ~ Log2[2E] + Log2[bp] bits

Associated Flop Arithmetic op 2 { +, -, *, /} [ { Sqrt, Mod, Flint}

) Flop(x,y) = Float[ op(x,y) ]

Archetypal Computation: Inner product o 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]


• 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


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 6.28

t 0.69

t 4.9

t 6.28

t 0.69

t 4.9


• 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


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 Orthonormality

o 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 ¢ b

2

q

b1 b2

where

hx,yi = x1y1 + x2y2 hq,b1i = 0.8 ¢ 1 + (-0.9) ¢ 0 = 0.8

hq,b2i = 0.8 ¢ 0 + (-0.9) ¢ 1 = - 0.9

(computational definition):


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,b

1i = 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


Change of Coordinates

[Google Maps]

Vs.

Independence Hall

500 Chestnut St.

http://maps.google.com/maps?f=q&source=s_q&hl=en&geocode=&q=19104&sll=37.0625,-95.677068&sspn=31.013085,45.703125&ie=UTF8&hq=&hnear=Philadelphia,+Pennsylvania+19104&ll=39.948964,-75.150647&spn=0.01464,0.022316&z=15&pw=2


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


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

H1 H2

Length(H1)2 = h H1, H1 i = 1/

p (2 ¢ 2) + 1/

p (2 ¢ 2)

= ½ + ½

= 1

Length(H2)2 = h H2, H2 i = 1/

p (2 ¢ 2) + 1/

p (2 ¢ 2)

= ½ + ½

= 1

hH1, H2i = h11 h2

1 + h1

2 h2

2 = - 1/

p 2 ¢ 2 + 1/

p 2 ¢ 2

= 0

t = - 6.28

t = 2.5

b2

b1


Linear Algebra: Change of Basis • Goal

Re-express q = (q1, q2) o specified by coordinate

representation

o in terms of the old basis, BT

As Q= [Q1, Q2] o Specified by coordinate

representation

o 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,H1i

o scale H2 by Q2 = hq,H2i

o form the resultant vector

q = (q1, q2) = q1 ¢ b

1 + q2 ¢ b2

= hq, b1i¢ b1 + hq, b2i ¢ b2

) Q1 = hq , H1i =h (0.8, - 0.9), (1/

p2, 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/

p2, 1/p

2)i

= - (0.8/1.1 + 0.9/1.1) ¼ - 1.6

t = - 6.28

t = 2.5

b2

b1

Q

H1 H2

- Q2

-Q1


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 .00 .50 .00 .51 .0

t 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 .0

t 0 , v 0

1 .0

0 .5

0 .0

0 .5

1 .0

t 2.1 , v 0.7

Generalize to ns = 3 Samples h0(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 .50 .0 0 .5 1 .0

t 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


Generalize to ns = 3 Samples h0(t) = Cos[0t]/p3

h1(t) = 2 Sin[t]/p3

h2(t) = 2 Cos[t]/p3 t 2.1

t 0

t 2.1

H0 0.6 , 0.6 , 0.6

H1 0.7 , 0 , 0.7

H2 0.4 , 0.8 , 0.4


(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


Flo

ating P

oin

t Flo

atin

g P

oin

t

r (received signal)

Sampling & Quantization

q Q

this

week’s

idea

Perceptual coding

(D e b u g ) O u t[3 9 7 ]=

f A

0 0.4

1s 0.6

1c 0.8

2s 0.3

2c 0.2

3s 0.2

3c 0.4

4s 0.2

4c 0.1

5s 0.2

5c 0

Generalize to Arbitrary Samples


… 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 Transformations

o G. Strang. The discrete cosine transform. SIAM Review, 41(1):135–

147, 1999


ESE250:

Digital Audio Basics

End Week 4 Lecture

Time-Frequency

week 4 february 2, 2012 time-frequency - penn engineeringese250/week4/day4-s'12.pdf · 2012. 2....

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