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On a Generalization of the GCD for Intervals in R +. or how can a camera see at least 1 tone for unkown T exp. Stan Baggen June 4, 2014. Contents . Introduction Cameras, exposure times and problem definition Introduction to Solution using GCD for Integer Frequencies - PowerPoint PPT Presentation
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On a Generalization of the GCD for Intervals in R+
Stan Baggen June 4, 2014
orhow can a camera see at least 1 tone for unkown Texp
Philips Research Stan Baggen
Contents
• Introduction
• Cameras, exposure times and problem definition
• Introduction to Solution using GCD for Integer Frequencies
• Extension of GCD to intervals over R+
• Application to the Original Problem
• Discussion
• Yet another generalization
2
Philips Research Stan Baggen 3
Introduction
• Transmit digital information from a luminaire to a smartphone or tablet using Visible Light Communication (VLC)
– Bits are encoded in small intensity variations of the emitted light
– Detect bits using the camera of a smartphone
• We consider an FSK-based system
– Symbols correspond to frequencies (tones)
– Emitted light variations are sinusoidal
• Problem: camera may be “blind” for certain frequencies
Philips Research Stan Baggen 4
Camera Image divided into lines and pixels
original image sequence
lines covering source
lines per frame
hidden lines
active lines
• Each line consists of a row of pixels
Philips Research Stan Baggen
• A camera can set its exposure time Texp • typically, Texp ranges from 1/30 to 1/2500 [s]
• Each pixel “sees” the average light during Texp seconds before read-out– smearing of intensity variations of received light
• If an integer number of periods of a sinusoid fit into Texp, the camera cannot detect such a sinusoid
Exposure Time
time
Texp
ISI filter (moving average)
0 1 2 3 4 5 6 7 8 9 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
f
|sin
c(f)|
Transfer Function of Exposure Time
|sinc(f)||sinc(0.7f)|
f1 f2
Philips Research Stan Baggen
• Due to the exposure time Texp of a camera, certain frequencies cannot be detected by it (multiples of fexp = 1/Texp)
• Can we have sets of 2 frequencies each, such that not both can be blocked for any fexp ≥ 30 Hz
• Each set then forms an fexp-independentdetection set for a light source that emitsboth frequencies
0 1 2 3 4 5 6 7 8 9 100
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
f
|sin
c(f)|
Transfer Function of Exposure Time
|sinc(f)||sinc(0.7f)|
Exposure Time
6
f1 f2
time
Texp
ISI filter
Philips Research Stan Baggen
Discrete Solution
• If the involved frequencies can only take on integer values, we can find solutions using the GCD (Greatest Common Divisor) from number theory
• We would like to have 2 frequencies f1 and f2, such that not both can be integer multiples of any fexp ≥ 30
• Suppose that both f1 and f2 are integer multiples of fexp
• If GCD(f1,f2) < 30 no solution possible for fexp ≥ 30
pair (f1,f2) is a good choice7
),GCD(||
|21exp
2exp
1exp fffff
ff
Philips Research Stan Baggen
Discrete Solution: Example
• f1 = 290; f2 = 319
• Largest integer that divides both f1 and f2 equals GCD(f1,f2) = 29
• No integer fexp ≥ 30 exists for which multiples are simultaneously equal to f1 and f2
8
Philips Research Stan Baggen
Problem with Discrete Solution
• GCD(300,301) = 1; GCD(300,300) = 300
• Physically: due to the nature of the Texp-filter and detection algorithms, if a pair of frequencies (f1,f2) is bad for detection, then a real interval (f1±ε,f2 ±ε) is also bad
• We need a method that allows us to eliminate bad intervals over R+
90 1 2 3 4 5 6 7 8 9 10
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
f
|sin
c(f)|
Transfer Function of Exposure Time
|sinc(f)||sinc(0.7f)|
f1 f2
Philips Research Stan Baggen
GCD for intervals in R+
• Consider 2 half-open intervals I1 and I2 in R+
• Definition:
• Note that the concept I1,I2: GCD(I1,I2) < 30 solves our original problem:
• There can be no real fexp≥30 such that integer multiples are simultaneously close to F1 and F2 10
21,21 |max:),GCD( ImaInaRaII Nmn
0( ] ( ]
I1 I2
0( ] ( ]
1F 2F
30
Philips Research Stan Baggen
GCD for intervals in R+
• How to find GCD(I1,I2)?
• Define divisor sets D1,D2 in R:
• Theorem 1:
• Proof: □ 11
21,21 |max:),GCD( ImaInaRaII Nmn
0( ] ( ]
I1 I2
2121 max),GCD( DDII
22
11
||
ImdRdDIndRdD
Nm
Nn
21
2121
|||ImdIndRd
ImdRdIndRdDD
NmNn
NmNn
Philips Research Stan Baggen
Example
12
0 50 100 150 200 250 300-1
0
1
2
3
4
5
frequency
divisor sets, their intersection and the GCD, f1 = 240, f2 = 256, interval = 16
f1 = 240f2 = 256GCD = 28.4444
]240,16240(1 I
]256,16256(2 I
21 DD
),GCD( 21 II
11 | IndRdD Nn
22 | ImdRdD Nm
Philips Research Stan Baggen 13
10 15 20 25 30
0
0.5
1
1.5
2
2.5
frequency
divisor sets, their intersection and the GCD, f1 = 240, f2 = 256, interval = 16
f1 = 240f2 = 256GCD = 28.4444
Enlargement of Example
1D
21 DD ),GCD( 21 II
2D
Philips Research Stan Baggen
Overlap of Intervals in Divisor Sets• Consider divisor set
• Let where
• Theorem 2: for w>0, D consists of a finite number n0 of disjunct intervals, where
• Proof: overlap of consecutive intervals happens if
• Corollary: 14
0( ]
I
],(,| fwfIIndRdD Nn
2n3nfwf
NnnIIndRdDn ,/|
1D2/12 DD
n
nDD
wfn0
wfn
wfn
nwf
nf
0.1
00
,0,0,,0nfwD
nf
3/13 DD
Philips Research Stan Baggen
Another Theorem
• Suppose that we have 2 intervals I1=(f1-w1,f1] and I2=(f2-w2,f2]
• Theorem 3: For w1,w2> 0, GCD(f1,f2 ; w1,w2) equals an integer sub-multiple of either f1, f2 or both
• Proof:
equals a right limit point of for some i and j.
Each is the intersection of 2 half-open intervals (...], where the right limit point of each half-open interval is an integer sub-multiple of either f1 or f2 or both. □
• Note: f1 and f2 are real numbers 15
ji
ji
j
j
i
i DDDDDD,
212121
21max DD ji DD 21
ji DD 21
Philips Research Stan Baggen
Some Interesting Examples
• Numbers in N+
– For w sufficiently small, we find the classical solutions for f1, f2 in N+
– GCD(15,21; w≤1) = 3
– GCD(15,21; w=1.1) = 7 • w too large for finding the classical solution
• Numbers in Q+
– GCD(0.9,1.2; w=0.1) = 0.3
• Numbers in R+ (computed with finite precision)– GCD(7π,8π; w=0.1) = 3.1416– GCD(6π,8π; w=0.1) = 6.2832
16
Philips Research Stan Baggen
Application to the Original Problem• Suppose that we find that for a certain (f1, f2; w1,w2) :
GCD (f1, f2; w1,w2) < 30
• Then there exists no real number fexp≥30 such that integer multiples of fexp fall simultaneously in (f1-w1, f1] and (f2-w2, f2]
• By picking F1= f1-w1/2 and F2= f2-w2/2, we can insure that if one multiple of fexp≥30 falls within a range of wi/2 of Fi for some i, then the other interval is free from any multiple of fexp
17
2f22 wf
0( ] ( ]
30f
11 wf 1f1F 2F
Philips Research Stan Baggen 18
Numerical Examples (1)
100 200 300 400 500 600 7000
200
400
600
800
1000
1200
frequency
acce
ptab
le fr
eque
ncy
pairs
Acceptable integer frequency pairs for w = 15, GCD < 30, 100 < f < 700
acceptable_frequencies_2012_10_20_1
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100 120 140 160 180 200 220
100
150
200
250
300
frequency
acce
ptab
le fr
eque
ncy
pairs
Acceptable integer frequency pairs for w = 15, GCD < 30, 100 < f < 700
acceptable_frequencies_2012_10_20_1
typical solutions: (f1,f2) = (f1, f1+15)
Numerical Examples (1) detail
Philips Research Stan Baggen 20
100 200 300 400 500 600 7000
500
1000
1500
2000
2500
frequency
acce
ptab
le fr
eque
ncy
pairs
Acceptable integer frequency pairs for w = 14, GCD < 30, 100 < f < 700
Numerical Examples (2)
acceptable_frequencies_2012_10_18_2
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90 100 110 120 130 140 150 160 170 180 190
0
50
100
150
200
250
300
350
frequency
acce
ptab
le fr
eque
ncy
pairs
Acceptable integer frequency pairs for w = 14, GCD < 30, 100 < f < 700
Numerical Examples (2) detail
acceptable_frequencies_2012_10_18_2
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100 200 300 400 500 600 7000
500
1000
1500
2000
2500
3000
3500
4000
4500
5000
frequency
acce
ptab
le fr
eque
ncy
pairs
Acceptable integer frequency pairs for w = 12, GCD < 30, 100 < f < 700
Numerical Examples (3)
acceptable_frequencies_2012_10_18_3
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100 200 300 400 500 600 7000
1000
2000
3000
4000
5000
6000
7000
8000
9000
frequency
acce
ptab
le fr
eque
ncy
pairs
Acceptable integer frequency pairs for w = 10, GCD < 30, 100 < f < 700
Numerical Examples (4)
acceptable_frequencies_2012_10_18_4
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100 120 140 160 180 200 220 240 260 280 3000
100
200
300
400
500
600
700
800
900
frequency
acce
ptab
le fr
eque
ncy
pairs
Acceptable integer frequency pairs for w = 10, GCD < 30, 100 < f < 700
Numerical Examples (4) detail
acceptable_frequencies_2012_10_18_4
typical solutions: (f1,f2) = (f1, f1+15), (f1, 2f1-20), ), (f1, 2f1+15)
Philips Research Stan Baggen
Discussion (1)
• It is convenient to use half open intervals (…] and have the right limit point as a characterizing number, since then
– We can reproduce the familiar results from number theory– The maximum in the definition of GCD exists– We do not obtain subsets in having measure 0
• The concept of GCD can be generalized to an arbitrary number of K intervals over R+
• Theorem 2 shows that the complexity of the computation of a GCD is reasonable
• Can we have an efficient algorithm like Euclid’s algorithm for computing the GCD of real intervals?
25
21 DD
K
kkK DII
11 max),...,GCD(
Philips Research Stan Baggen
Discussion (2)
• It can be shown that GCD(f1, f2;w) is non-decreasing as w increases
• For rational numbers a/b and p/q, where a,b,p,q are in N+, we find for sufficiently small w:
where LCM(.) is the Least Common Multiple.
How small must w be as a function of a,b,p and q to find this solution?
• Conjecture: for incommensurable numbers a and b
• Effects of finite precision computations
26
,),LCM(
),LCM(,),LCM(;,
qbqpqb
baqbGCD
wqp
baGCD
0;,lim0
wbaGCDw
Philips Research Stan Baggen
Yet Another Generalization
• GCD(f1,f2;w) on intervals still makes hard decisions on frequencies being in or out of intervals
• Can we make some sensible reasoning that leads to “smooth” decisions concerning acceptable frequency pairs
• We have to use a more friendly measure on the intervals
• We start by re-phrasing the previous approach in a different manner
27
Philips Research Stan Baggen
• GCD(f1,f2;w) on intervals as discussed previously, effectively uses indicator functions as a measure of membership:
• Divisor Measure DM1,DM2 in R+
28
0( ] ( ]
I1 I2
1
1)()(max:),( 2121
fDMfDMIIGCDRf
2,1,),(max:)(
iRfnfMfDM
iINni
1IM
2IM
RffMfM IIII ,1)(;1)(2211
f
Philips Research Stan Baggen 29
0 2 4 6 8 10 12 140
1
2
3
4
5
6
7
f
cost
s
Cost Functions
D1D2D1 D2Example
f1 = 9; f2 = 12
w = 0.5
GCD(f1,f1;w) = 3
0 0.5 1 1.5 2 2.5 3 3.50
1
2
3
4
5
6
f
cost
s
Cost Functions
D1D2D1 D2
9
12
3
Philips Research Stan Baggen
Using a Different Measure
• Suppose that we change the definition of the measure of membership for the fundamental interval
• Divisor Measure:
• Common Divisor Measure:
30
),;(),;(:),,,;( 22112121 ffDMffDMfffCDM
2,1,),,;(max:),;(
iRffnfMffDM iiNnii
2,1,,
2exp:),;( 2
2
iRf
ffffM
i
iii
0 2 4 6 8 10 12 14
0
0.5
1
1.5
2
2.5
3
f
cost
s
Cost Functions of Fundamental Intervals I1 and I2
)25.0,12;( fM
)25.0,9;( fM
example
Philips Research Stan Baggen 31
0 2 4 6 8 10 12 140
1
2
3
4
5
6
7
f
cost
s
Cost Functions
D1D2D1 D2
0 0.5 1 1.5 2 2.5 3 3.5
0
0.2
0.4
0.6
0.8
1
1.2
1.4
f
cost
s
Cost Functions
D1D2D1 D2
)25.0,25.0,12,9;( fCDM
)25.0,9;( fDM
)25.0,12;( fDM
Example
Multiples of frequencies in the neighborhood of 3(and 3/n) also end up both near 9 and 12
For frequencies f>3.2, no multiples end up both near 9 and 12 according to the measure
Multiples of 1.1, 1.3 and 1.7 come somewhat close to both 9 and 12 (c.f. other measure)
)25.0,25.0,12,9;( fCDM
Philips Research Stan Baggen 32
0 2 4 6 8 10 12 140
1
2
3
4
5
6
7
f
cost
s
Cost Functions
D1D2D1 D2
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5
0
0.2
0.4
0.6
0.8
1
1.2
f
cost
s
Cost Functions
D1D2D1 D2
)5.0,5.0,12,9;( fCDM
)5.0,9;( fDM
)5.0,12;( fDM
Example
If we increase σ, it becomes more difficult to“avoid” the intervals around 9 and 12 for integer multiples of f
For σ=0.5, some multiples of 4.16 also come close to both 9 and 12 according to the measure
Philips Research Stan Baggen 33
0 2 4 6 8 10 12 14
0
0.5
1
1.5
2
2.5
3
fco
sts
Cost Functions of Fundamental Intervals I1 and I2
f1 = 9; f2 = 12σ = 0.5fexp = 4.16
Example
samples taken at integer multiples of 4.16
CDM(4.16;.) equals product of largest “red” sample (n=3) and largest “blue” sample (n=2)
)5.0,12;( fM
)5.0,9;( fM
Philips Research Stan Baggen 34