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FROM PTOLEMY TO MP3 DIGITAL MUSIC:

2000 Years of Applications of Harmonic Analysis

Yuri S. Ledyaev

Department of Mathematics

Western Michigan University

7 [email protected]

℡ (269) 387-4557

MATH 5710, Advanced Calculus II, March 23, 2020 – p. 1/13

Ptolemaic Geocentric System

Ancient Greek Astronomer Claudius Ptolemaeus (c.90-169) (inEnglish, Ptolem y)

Famous astronomical treatiseAlma gest ("The Great Treatise")It was preserved (as many AncientGreek texts in Arabic manuscriprs)Translated into Latinin 12th century by Gerard of Cremona

Geocentric model of solarsystem




Moon

Mercury

Venus

Sun

Mars

Jupiter

Saturn

Fixed StarsIdeal motions are circular (Aristotle, Plato)




Moon

Mercury

Venus

Sun

Mars

Jupiter

Saturn

Fixed StarsGeocentric system: planets moving around spherical EarthDeferent-and-epicycle model



Geocentric system: planets moving around spherical EarthDeferent-and-epicycle model




planets moving around spherical Earth



Heliocentric solar system of Copernicus, Galileo and Kepler:All planets (including Earth) moving around Sun




QUESTION: Why geocentric system gave satisfactoryapproximation of observations?




QUESTION: Why geocentric system gave satisfactoryapproximation of observations?

ANSWER: For the same reason that we can listen digitalmusic now



Ancient Greek Astronomer Claudius Ptolemaeus (c.90-169)

Portable media players designed by Apple iPod (c.2001 - )



Ancient Greek Astronomer Claudius Ptolemaeus (c.90-169)

Portable media players designed by Apple iPod (c.2001 - )


Harmonic Analysis during last 2000 years

Harmonic analysis : approximation of functions by using finiteseries of simple periodic functions




sin(t) and cos(t)

Consider function f : [−T, T ] → R

f(t) v a0+a1 cos(πt/T )+b1 sin(πt/T )+a2 cos(2πt/T )+b2 sin(2πt/T )) . . .




sin(t) and cos(t)


f(t) v c0+c1 cos(πt/T )+s1 sin(πt/T )+c2 cos(2πt/T )+s2 sin(2πt/T )) . . .

f(t) v∞∑

k=0

ck cos(πk

Tt) + sk sin(

πk

Tt)




f(t) v c0+c1 cos(πt/T )+s1 sin(πt/T )+c2 cos(2πt/T )+s2 sin(2πt/T )) . . .

f(t) v∞∑

k=0

ck cos(πk

Tt) + sk sin(

πk

Tt)

where c0 =1

2T

∫ T

−Tf(t)dt and for k ≥ 1

ck =1

T

∫ T

−Tf(t) cos(

πk

Tt)dt, sk =

1

T

∫ T

−Tf(t) sin(

πk

Tt)dt



Fourier series (Fourier 1807 , Analytical Theory of Heat )

f(t) v∞∑

k=0

ck cos(πk

Tt) + sk sin(

πk

Tt)

where c0 =1

2T

∫ T

−Tf(t)dt and for k ≥ 1

ck =1

T

∫ T

−Tf(t) cos(

πk

Tt)dt, sk =

1

T

∫ T

−Tf(t) sin(

πk

Tt)dt



Fourier series (Fourier 1807 , Analyt-ical Theory of Heat )For large class of functions f

f(t)! =∞∑

k=0

ck cos(πk

Tt) + sk sin(

2πk

Tt)

Good approximating properties


Examples: Approximation by Fourier Trig. Polynomials

f(t) = t(1 − t). t ∈ [0, 1]



0.1666666667 − 0.1013211836 cos (6.283185308 t) −4.156265120 × 10−11 sin (6.283185308 t)





0.1666666667 − 0.1013211836 cos (6.283185308 t) −4.156265120 × 10−11 sin (6.283185308 t) − 0.0253303 cos (12.5663 t) −0.01125790930 cos (18.84955592 t) MATH 5710, Advanced Calculus II, March 23, 2020 – p. 4/13
























f(t) = 1 − |t − 1|, t ∈ [0, 2]



0.5000000000 − 0.4052847344 cos (3.141592654 t) −0.0000000001677970458 sin (3.141592654 t)



0.5000000000 − 0.4052847344 cos (3.141592654 t) −0.0000000001677970458 sin (3.141592654 t) +

3.469446952 × 10−17 cos (6.283185308 t) +

3.199817670 × 10−12 sin (6.283185308 t)MATH 5710, Advanced Calculus II, March 23, 2020 – p. 5/13


























0

0.2

0.4

0.6

0.8

1

0.5 1 1.5 2 2.5 3

x

f(t) = t2, t ∈ [0, 1]



0.3333333333 + 0.1013211839 cos (6.283185308 t) −0.3183098860 sin (6.283185308 t)



0.3333333333 + 0.1013211839 cos (6.283185308 t) −0.3183098860 sin (6.283185308 t) + 0.02533029678 cos (12.56637062 t)−0.1591549429 sin (12.56637062 t)























f(t) =

{1, t ∈ [0, 1]

0, t ∈ [1, 2)MATH 5710, Advanced Calculus II, March 23, 2020 – p. 7/13





























Fourier Series: Approximation and Ptolemaic Geocentric System


f(t)! =∞∑

k=0

ck cos(πk

Tt) + sk sin(

πk

Tt)

Good approximating properties – Ptole-maic Geocentric System gave satisfac-tory approximation of observation data



Good approximating properties – Ptolemaic Geocentric Systemgave satisfactory approximation of observation data





Good approximating properties – Ptolemaic Geocentric Systemgave satisfactory approximation of observation dataDescription of rotation with period T1

r1(t) = Ω1(t)r1(0)

where Ω1(t) is the rotation matrix

Ω1(t) =

[cos(2π

T1t) sin(2π

T1t)

− cos(2πT1

t) cos(2πT1

t)

]



Why motion of planet is approximated by the function r(t)

r(t) =

[cos(2π

T1t) sin(2π

T1t)

− cos(2πT1

t) cos(2πT1

t)

]

r1(0) +

[cos(2π

T2t) sin(2π

T2t)

− cos(2πT2

t) cos(2πT2

t)

]

r2(0)



Fourier series, Cosine and Sine Series, Fourier Transform


f(t)! =∞∑

k=0

ck cos(πk

Tt) + sk sin(

πk

Tt)

Good approximating properties – Ptole-maic Geocentric System gave satisfac-tory approximation of observation data



Why in

f(t)=∞∑

k=0

ck cos(πk

Tt) + sk sin(

πk

Tt)

we have

ck =1

T

∫ T

−Tf(t) cos(

πk

Tt)dt, sk =

1

T

∫ T

−Tf(t) sin(

πk

Tt)dt

Consider R3 with orthogonal basis: vec-tors −→e1 ,−→e2 ,−→e3 ( −→ei ∙ −→ej = 0). Vector

−→f :

−→f = c1

−→e1 + c2−→e2 + c3

−→e3

−→e1 ∙−→f = c1

−→e1 ∙ −→e1 c1 =−→e1 ∙

−→f

−→e1 ∙ −→e1MATH 5710, Advanced Calculus II, March 23, 2020 – p. 9/13


Cosine series for f : [0, T ] → R

f(t)=∞∑

k=0

ck cos(πk

Tt)

Sine series for f : [0, T ] → R

f(t)=∞∑

k=1

sk sin(πk

Tt)



Cosine series for f : [0, T ] → R

f(t)=∞∑

k=0

ck cos(πk

Tt)

Sine series for f : [0, T ] → R

f(t)=∞∑

k=1

sk sin(πk

Tt)

Integral representation for f : (−∞, +∞) → R

f(t) =

∫ +∞

0C(ω) cos ωt dω +

∫ +∞

0S(ω) sin ωt dω




f(t) =

∫ +∞


∫ +∞

0S(ω) sin ωt dω

where Fourier transform

C(ω) =1

π

∫ +∞

−∞f(t) cos ωt dt, S(ω) =

1

π

∫ +∞

−∞f(t) sin ωt dt




f(t) =

∫ +∞


∫ +∞

0S(ω) sin ωt dω


C(ω) =1

π

∫ +∞


1

π

∫ +∞


Hundreds applications : Electric and Control engineering, Signalprocessing, Optics and Spectroscopy, X-ray Crystallography(Protein structure, DNA,. . . ), Computerized Tomography,Radioastronomy, . . .




f(t) =

∫ +∞


∫ +∞

0S(ω) sin ωt dω


C(ω) =1

π

∫ +∞


1

π

∫ +∞


Hundreds applications : Electric and Control engineering, Signalprocessing, Optics and Spectroscopy, X-ray Crystallography(Protein structure, DNA,. . . ),At WMU : Math 5710 - Analysis (Fourier series), Math 5740Advanced Differential Equations (basics of Harmonic Analysis,applications to Partial Differential Equations)


Harmonic Analysis and Signal Processing

Assume that we have an audio signal given by a continuousfunction f(t).How to reproduce such function? Fourier series?

f(t)=∞∑

k=0

ck cos(2πk

Tt) + sk sin(

2πk

Tt)



Assume that we have an audio signal given by a continuousfunction f(t).How to reproduce such function? Fourier series?

f(t)=∞∑

k=0

ck cos(2πk

Tt) + sk sin(

2πk

Tt)

But

ck =2

T

∫ T

0f(t) cos(

2πk

Tt)dt, sk =

2

T

∫ T

0f(t) sin(

2πk

Tt)dt

and f(t) should be T−periodic and it cannot be transmitted in realtime



Practical approach based on the Sampling Theorem :

any practical audio- (or video-) signal has bounded bandwidth




any practical audio- (or video-) signal has bounded bandwidth(namely, human ear can hear frequencies bounded fromabove by 20 KHz)





We can assume that there exists a frequency ωm such that

C(ω) = 0, S(ω) = 0 ∀ ω ≥ ωm






C(ω) = 0, S(ω) = 0 ∀ ω ≥ ωm






C(ω) = 0, S(ω) = 0 ∀ ω ≥ ωm

Thus, the signal f(t)

f(t) =

∫ ωm


∫ ωm

0S(ω) sin ωt dω



Sampling TheoremTHEOREM: Let f(t) have bounded bandwidth ωm andT > 0 satisfy

ω0 :=2π

T> 2 ωm

Then

f(t) =∞∑

k=−∞

f(kT )sin π(t/T − k)

π(t/T − k)



THEOREM: Let f(t) have bounded bandwidth ωm andT > 0 satisfy

ω0 :=2π

T> 2 ωm

Then

f(t) =+∞∑

k=−∞


π(t/T − k)

CONCLUSION: To reconstruct the entire signal f(t) it is enoughto know values of this signal at sampling moments kT , T shouldbe "small"

f(0), f(T ), f(2T ), f(3T ), . . . , f (kT ), . . .




Sampling Theorem

f(t) =+∞∑

k=−∞


π(t/T − k)

1

T> 2νm

Sampling rate is greater than the double of bandwidth


Sampling Theorem

f(t) =+∞∑

k=−∞


π(t/T − k)

1

T> 2νm

Sampling rate is greater than the double of bandwidthEXAMPLE: Audio CD (Red Book standard): channel bandwidth22.05 KHz, sampling rate 44.1 KHZHardware: Analog-to-Digital Converter (ADC),to reconstruct signal use Digital-To-Analog Converter (DAC)


Sampling Theorem

Aliasing


Sampling Theorem

Aliasing


Sampling Theorem

NO Aliasing if1

T> 2νm


Sampling Theorem

Anti-aliasing filters (usually Low-pass filter added to ADC)


Sampling Theorem

Shannon 1949 Communication inthe presence of noiseVery elegant and short proof using Diracdelta -function


Sampling Theorem

Kotelnikov 1933 On the transmis-sion capacity of wireless and ca-bles in electrical communicationsShort proof using traditional toolsOther names associated with this result:Whittaker 1915 Interpolation TheoremNyquist 1928Someya circa 1949Weston circa 1949


Proof of the Sampling Theorem ( after Kotelnikov)

Assume that ω0

2 := πT > ωm

STEP1: show that for any signal f(t) with bandwidth bounded byωm

f(t) =∞∑

k=−∞

Dksin π(t/T − k)

t − kT

for some coefficients Dk

f(t) =∞∑

k=−∞


π(t/T − k)



Assume that ω0

2 := πT > ωm


f(t) =∞∑

k=−∞

Dksin π(t/T − k)

t − kT


We start with

f(t) =

∫ ω02


∫ ω02

0S(ω) sin ωt dω

f(t) =∞∑

k=−∞


π(t/T − k)



Assume that ω0

2 := πT > ωm


f(t) =∞∑

k=−∞

Dksin π(t/T − k)

t − kT


We start with

f(t) =

∫ ω02


∫ ω02

0S(ω) sin ωt dω

REMINDER:

C(ω) = 0, S(ω) = 0 for all ω > ωm (Signal with a bounded spectrum)

f(t) =∞∑

k=−∞


π(t/T − k)



f(t) =∞∑

k=−∞

Dksin π(t/T − k)

t − kT


We start with

f(t) =

∫ ω02


∫ ω02

0S(ω) sin ωt dω

TRICK: Write Cosine series forC(ω) and Sine series for S(ω)(remember ω0

2 > ωm)

C(ω) =∞∑

k=0

Ak cos(2πk

ω0ω), S(ω) =

∞∑

k=0

Bk sin(2πk

ω0ω)

f(t) =∞∑

k=−∞


π(t/T − k)



f(t) =∞∑

k=−∞

Dksin π(t/T − k)

π(t/T − k)

We start with

f(t) =

∫ ω02


∫ ω02

0S(ω) sin ωt dω

TRICK: Write Cosine series forC(ω) and Sine series for S(ω)(remember ω0

2 > ωm)

C(ω) =∞∑

k=0

Ak cos(2πk

ω0ω), S(ω) =

∞∑

k=0

Bk sin(2πk

ω0ω)

Define Dk := Ak+Bk

2 for k ≥ 0, Dk := A−k−B−k

2 for k < 0

f(t) =∞∑

k=−∞


π(t/T − k)



C(ω) =∞∑

k=0

Ak cos(2πk

ω0ω), S(ω) =

∞∑

k=0

Bk sin(2πk

ω0ω)

Define Dk := Ak+Bk

2 for k ≥ 0, Dk := A−k−B−k

2 for k < 0. Then

C(ω) =∞∑

k=−∞

Dk cos(2πk

ω0ω), S(ω) =

∞∑

k=−∞

Dk sin(2πk

ω0ω)

f(t) =∞∑

k=−∞


π(t/T − k)



C(ω) =∞∑

k=0

Ak cos(2πk

ω0ω), S(ω) =

∞∑

k=0

Bk sin(2πk

ω0ω)

Define Dk := Ak+Bk

2 for k ≥ 0, Dk := A−k−B−k

2 for k < 0. Then

C(ω) =∞∑

k=−∞

Dk cos(2πk

ω0ω), S(ω) =

∞∑

k=−∞

Dk sin(2πk

ω0ω)

Plug C(ω) and S(ω) in

f(t) =

∫ ω02


∫ ω02

0S(ω) sin ωt dω

f(t) =∞∑

k=−∞


π(t/T − k)



C(ω) =∞∑

k=−∞

Dk cos(2πk

ω0ω), S(ω) =

∞∑

k=−∞

Dk sin(2πk

ω0ω)


f(t) =

∫ ω02


∫ ω02

0S(ω) sin ωt dω

to obtain

f(t) =

∫ ω02

0

∞∑

k=−∞

Dk cos(2πk

ω0ω) cos ωt dω+

∫ ω02

0

∞∑

k=−∞

Dk sin(2πk

ω0ω) sin ωt dω

f(t) =∞∑

k=−∞


π(t/T − k)




f(t) =

∫ ω02


∫ ω02

0S(ω) sin ωt dω

to obtain

f(t) =

∫ ω02

0

∞∑

k=−∞

Dk cos(2πk

ω0ω) cos ωt dω+

∫ ω02

0

∞∑

k=−∞

Dk sin(2πk

ω0ω) sin ωt dω

We can rewrite it

f(t) =∞∑

k=−∞

Dk

∫ ω02

0[cos(

2πk

ω0ω) cos ωt + sin(

2πk

ω0ω) sin ωt] dω

f(t) =∞∑

k=−∞


π(t/T − k)



We can rewrite it

f(t) =∞∑

k=−∞

Dk

∫ ω02

0[cos(

2πk

ω0ω) cos ωt + sin(

2πk

ω0ω) sin ωt] dω

Now we use trig identity

f(t) =∞∑

k=−∞

Dk

∫ ω02

0cos(ωt −

2πk

ω0ω) dω

f(t) =∞∑

k=−∞


π(t/T − k)




f(t) =∞∑

k=−∞

Dk

∫ ω02

0cos(ωt −

2πk

ω0ω) dω

Evaluate integrals to obtain

f(t) =∞∑

k=−∞

Dk

sin ω0

2 (t − 2πkω0

)

t − 2πkω0

f(t) =∞∑

k=−∞


π(t/T − k)




f(t) =∞∑

k=−∞

Dk

∫ ω02

0cos(ωt −

2πk

ω0ω) dω

Evaluate integrals to obtain

f(t) =∞∑

k=−∞

Dk

sin ω0

2 (t − 2πkω0

)

t − 2πkω0

Recall that ω0 = 2πT then

f(t) =∞∑

k=−∞

Dksin π(t/T − k)

t − kT

f(t) =∞∑

k=−∞


π(t/T − k)



But this is a result fromSTEP 1: show that for any signal f(t) with bandwidth bounded byωm

f(t) =∞∑

k=−∞

Dksin π(t/T − k)

t − kT


f(t) =∞∑

k=−∞


π(t/T − k)



STEP 1: We demonstrated for any signal f(t)

f(t) =∞∑

k=−∞

Dksin π(t/T − k)

t − kTfor some coefficients Dk

STEP 2: how to find Dk?Take limit t → nT

f(nT ) = limt→nT

f(t) =+∞∑

k=−∞

Dk limt→nT

sin π(t/T − k)

t − kT

Then for k 6= n limt→nT sin π(t/T − k) = 0.For k = n use l’Hopital rule to obtain

f(nT ) = Dnπ

TDk = f(kT )

T

π

f(t) =∞∑

k=−∞


π(t/T − k)



f(t) =∞∑

k=−∞


π(t/T − k)


The Last Slide

Example: application of the Sampling Theorem


The Last Slide


The Last Slide


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