(Click Here updated 7.19.2017 Final Document of this page -"A Planetary Model of Sound with Beat Frequencies in Psychoacoustics" )
We have searched this topic and only found theory of primary order 'beating' frequencies which = difference between two component frequencies played.
Apologies in advance if this subject is already known to you, for it does seem obvious, a surprisingly simple expression but it doesn't seem to be written about much.
This subject deals with the secondary order low frequency oscillations heard with two frequencies sounded together in a integer ratio but slightly off, such as (3+δ)/2, (5+δ)/3 etc...
Considering the visual model of 2 orbiting particles, the conjunction frequency, when the particles align radially = subtraction between the two frequencies.
Here is a basic proof:
T1 = timeperiod of first planet in a coupled orbital system
T2 = timeperiod of second planet in the same system
N = number of orbits it takes for an alignment (conjunction to occur)
T2 = timeperiod of second planet in the same system
N = number of orbits it takes for an alignment (conjunction to occur)
Since sound is matter in vibration, and mutually, matter in vibration is sound, then the orbital motion of planets are really a very low frequency form of sound.
By definition, timeperiod = 1 / frequency so in the diagram in a simple representation of planetary orbits, the lower series of lines associated with T2 (Timeperiod of planet 2) has one more orbit to happen before both planets align.
So (N+1)*(T2) = N*(T1)
To find N, : (N+1)/N = (T1)/(T2)
So 1 + (1/N) = (T1)/(T2)
(1/N) = (T1-T2)/T2
ie: N = T2/(T1-T2)
So we know an expression for N.
We know that in the diagram N*T1 is the timeperiod of conjunction, when the planets align, so letting this = Tc:
Tc = N*T1 = (T1*T2)/(T1-T2)
So since frequency = 1/timeperiod by definition, we can rearrange equation to give:
1/Tc = (T1-T2)/(T1*T2)
Which is 1/Tc = (1/T2)-(1/T1)
So frequencies Fc = F2 - F1
So we now know conjunction frequency, Fc = difference between two planet frequencies F1 & F2
By definition, timeperiod = 1 / frequency so in the diagram in a simple representation of planetary orbits, the lower series of lines associated with T2 (Timeperiod of planet 2) has one more orbit to happen before both planets align.
So (N+1)*(T2) = N*(T1)
To find N, : (N+1)/N = (T1)/(T2)
So 1 + (1/N) = (T1)/(T2)
(1/N) = (T1-T2)/T2
ie: N = T2/(T1-T2)
So we know an expression for N.
We know that in the diagram N*T1 is the timeperiod of conjunction, when the planets align, so letting this = Tc:
Tc = N*T1 = (T1*T2)/(T1-T2)
So since frequency = 1/timeperiod by definition, we can rearrange equation to give:
1/Tc = (T1-T2)/(T1*T2)
Which is 1/Tc = (1/T2)-(1/T1)
So frequencies Fc = F2 - F1
So we now know conjunction frequency, Fc = difference between two planet frequencies F1 & F2
(An Expression For Secondary Order Beat Frequencies Analogy of Planetary Cycles With Psychoacoustics.
So as an example Considering two component frequencies being close to a numerical
integer ratio like 8/5
Letting bottom frequency integer , a = 5 and top frequency integer ,b = 8, and then increment b = 8 to
(8+δ) where δ is a small change
The primary difference frequency (or conjunction alignment frequency) = (8+δ)-5 =
(3+δ)
As shown in the picture before,
the primary angle swept out = 5/(3+δ) which is a/(b-a+δ)
So how many conjunctions will it take to have a conjunction point rotated round a
number of full turns to be back close to the starting point?
Well to get the conjunction angle=5/(3+δ) back to near the start angle, can multiply by 3
to give close to 5 complete full rotations.
To find the angle of displacement θ as shown in the picture:
θ = 5-[3.(5/(3+δ))]
or
θ = a-[(b-a).(a/(b-a+δ))]
Simplifying equation gives :
θ = a.δ/(b+δ-a)
The secondary order time period, Tβ = (primary order time period)/ θ
SO if primary order time period = Tα = (1/fo).[a/((b+δ)-a)]
{because timeperiod =1/frequency}
{f0 = fundamental base frequency in relation to frequency component , fa}
Then expression for Tβ = (1/fo).[a/(b+δ-a)].[(b+δ-a)/(a.δ)]
Terms cancel giving :
Tβ = 1/(δ.fo)
By definition, frequency = 1/timeperiod
So for secondary order beat frequency,Fβ:
-----------
Fβ = δ.fo
-----------
This can be observed as an additional periodicity in the waveform.
We have experimented with a few ratios like 3/2 , 5/3, 5/2, 8/5 and this equation seems
consistent with results.
And if we want to introduce a small displacement, ζ, from b/a, then we can equate
(b+δ)/a = (b/a)+ζ
Which can be simplified: ζ.a = δ
So Secondary order time period, Tβ = 1/( fo.a.ζ)
and secondary order frequency, Fβ = fo.a.ζ
If we call an Interval, k = b/a + ζ
secondary order frequency, Fβ = fo. (a.k - b) if k > b/a
If we call an Interval, k = b/a - ζ
secondary order frequency, Fβ = fo. (b - a.k) if k < b/a
Experimental Evidence for Secondary Order Beat Frequencies:
So as an example Considering two component frequencies being close to a numerical
integer ratio like 8/5
Letting bottom frequency integer , a = 5 and top frequency integer ,b = 8, and then increment b = 8 to
(8+δ) where δ is a small change
The primary difference frequency (or conjunction alignment frequency) = (8+δ)-5 =
(3+δ)
As shown in the picture before,
the primary angle swept out = 5/(3+δ) which is a/(b-a+δ)
So how many conjunctions will it take to have a conjunction point rotated round a
number of full turns to be back close to the starting point?
Well to get the conjunction angle=5/(3+δ) back to near the start angle, can multiply by 3
to give close to 5 complete full rotations.
To find the angle of displacement θ as shown in the picture:
θ = 5-[3.(5/(3+δ))]
or
θ = a-[(b-a).(a/(b-a+δ))]
Simplifying equation gives :
θ = a.δ/(b+δ-a)
The secondary order time period, Tβ = (primary order time period)/ θ
SO if primary order time period = Tα = (1/fo).[a/((b+δ)-a)]
{because timeperiod =1/frequency}
{f0 = fundamental base frequency in relation to frequency component , fa}
Then expression for Tβ = (1/fo).[a/(b+δ-a)].[(b+δ-a)/(a.δ)]
Terms cancel giving :
Tβ = 1/(δ.fo)
By definition, frequency = 1/timeperiod
So for secondary order beat frequency,Fβ:
-----------
Fβ = δ.fo
-----------
This can be observed as an additional periodicity in the waveform.
We have experimented with a few ratios like 3/2 , 5/3, 5/2, 8/5 and this equation seems
consistent with results.
And if we want to introduce a small displacement, ζ, from b/a, then we can equate
(b+δ)/a = (b/a)+ζ
Which can be simplified: ζ.a = δ
So Secondary order time period, Tβ = 1/( fo.a.ζ)
and secondary order frequency, Fβ = fo.a.ζ
If we call an Interval, k = b/a + ζ
secondary order frequency, Fβ = fo. (a.k - b) if k > b/a
If we call an Interval, k = b/a - ζ
secondary order frequency, Fβ = fo. (b - a.k) if k < b/a
Experimental Evidence for Secondary Order Beat Frequencies:
As shown in this picture of the Secondary Order Beat frequencies that are
heard for two sine waves played together, frequency f & f*k in a range for k from 1 to 2
For example an interval near central point 3/2 has beat lines rising on either side
away and up as heard experimentally when generating the constant beating by making
a .wav file of y = sin(2*pi*f*t)+ (1/k)*(sin(2*pi*f*k*t)
Say k = 1.4967 and f = 216 Hz
When putting the waveform into a DAW workstation and synchronizing an expected
BPM from our previous formula, the beating and the BPM predicted match.
secondary order frequency, Fβ = fo. (b - a.k) if k < b/a
b/a = 3/2 so b = 3, a = 2 , Fβ = 216. (3 - 2*1.4967) = 1.4256 Hz
1.4256 Hz * 60 = 85.536 BPM
when amplifying the waveform for clipping distortion to emphasise the beats, a slight doubling
is heard at 171.072 BPM Other intervals were tried in this way and It appears
so far that if (b-a) is odd numbered, then an overlaying doubling happens probably to do with analogy of planetary orbits and opposition doubling.
The Expression Function:
What is psychoacoustically perceived could be velocity which is analogous to angular frequency, Energy or something else.
Imagine a pair of orbiting planets exerting a sinusoidal pressure on the Galactic plane field
In dealing with velocity which has reciprocal of time units matching with angular frequency, ω,
We can construct a displacement wave function ψ(y)
and from ω = v/r velocity over radius, velocity is the differential of displacement with time
Letting r = 1 for convenience.
For when frequency is changing and not constant, a displacement expression
y = sin(2*pi*∫ f.dt) ∫f.dt is the integral of frequency with respect to time.
So the Angular frequency expression function is: ψ(ω) = sin(2*pi*∫ω.dt)
ψ(ω) = sin(2*pi*ψ(y))
It's just a case of finding the possible displacement wave function.
In the meantime we can theorize from our planetary model various types of displacement wave function
formats that yields interesting results.
ψ(y) = e^(i*2*pi*f*t)+(1/k)*e^(i*2*pi*f*k*t)
e is the exponential constant and i is the imaginary unit, the square root of negative 1
e^(i*2*pi*f*t) = cos(2*pi*f*t)+ i*sin(2*pi*f*t)
angular frequency expression function:
ψ(ω) = sin(2*pi*ψ(y)) = sin(2*pi*μ *( cos(2*pi*f*t)+(1/k)*cos(2*pi*f*k*t) ) )
μ is a constant to be varied.
This is frequency modulation but under Fast fourier transform frequency analysis for μ =0.2
, a plethora of constant psychoacoustic frequencies exist. But for this angular frequency expression
function, the Secondary order beat frequencies are missing when (b-a) is an even number from predicted
formula from before:
If we call an Interval, k = b/a + ζ
secondary order frequency, Fβ = fo. (a.k - b) if k > b/a
If we call an Interval, k = b/a - ζ
secondary order frequency, Fβ = fo. (b - a.k) if k < b/a
An alternative displacement wave function could be including the conjunction frequency when two planets
align which is the difference frequency.
ψ(y) = e^(i*2*pi*f*t)+(1/k)*e^(i*2*pi*f*k*t) + (1/k)*i*e^(i*2*pi*f*(k-1)*t)
The orbital conjunction term (1/k)*i*e^(i*2*pi*f*(k-1)*t) has i as a factor since the waveform result is more in
normalized symmetry . And also has (1/k) as a factor for assuming the difference
between radius = 1 for e^(i*2*pi*f*t) term and radius = (1/k) for e^(i*2*pi*f*k*t) term
in circular orbits yields the total distance separating the orbiting particles =1 - (1 - 1/k) = 1/k
angular frequency expression function:
ψ(ω) = sin(2*pi*ψ(y))
= sin(2*pi*μ *(cos(2*pi*f*t)+(1/x)*cos(2*pi*f*x*t)-(1/x)*sin(2*pi*f*t*(x-1))))
μ is a constant to be varied.
In this case under frequency analysis of this alternative angular frequency expression function,
the Zeta frequencies are all visible but with some doubling and tripling anomalies for high value of μ
but for μ < 0.3 the results look good.
heard for two sine waves played together, frequency f & f*k in a range for k from 1 to 2
For example an interval near central point 3/2 has beat lines rising on either side
away and up as heard experimentally when generating the constant beating by making
a .wav file of y = sin(2*pi*f*t)+ (1/k)*(sin(2*pi*f*k*t)
Say k = 1.4967 and f = 216 Hz
When putting the waveform into a DAW workstation and synchronizing an expected
BPM from our previous formula, the beating and the BPM predicted match.
secondary order frequency, Fβ = fo. (b - a.k) if k < b/a
b/a = 3/2 so b = 3, a = 2 , Fβ = 216. (3 - 2*1.4967) = 1.4256 Hz
1.4256 Hz * 60 = 85.536 BPM
when amplifying the waveform for clipping distortion to emphasise the beats, a slight doubling
is heard at 171.072 BPM Other intervals were tried in this way and It appears
so far that if (b-a) is odd numbered, then an overlaying doubling happens probably to do with analogy of planetary orbits and opposition doubling.
The Expression Function:
What is psychoacoustically perceived could be velocity which is analogous to angular frequency, Energy or something else.
Imagine a pair of orbiting planets exerting a sinusoidal pressure on the Galactic plane field
In dealing with velocity which has reciprocal of time units matching with angular frequency, ω,
We can construct a displacement wave function ψ(y)
and from ω = v/r velocity over radius, velocity is the differential of displacement with time
Letting r = 1 for convenience.
For when frequency is changing and not constant, a displacement expression
y = sin(2*pi*∫ f.dt) ∫f.dt is the integral of frequency with respect to time.
So the Angular frequency expression function is: ψ(ω) = sin(2*pi*∫ω.dt)
ψ(ω) = sin(2*pi*ψ(y))
It's just a case of finding the possible displacement wave function.
In the meantime we can theorize from our planetary model various types of displacement wave function
formats that yields interesting results.
ψ(y) = e^(i*2*pi*f*t)+(1/k)*e^(i*2*pi*f*k*t)
e is the exponential constant and i is the imaginary unit, the square root of negative 1
e^(i*2*pi*f*t) = cos(2*pi*f*t)+ i*sin(2*pi*f*t)
angular frequency expression function:
ψ(ω) = sin(2*pi*ψ(y)) = sin(2*pi*μ *( cos(2*pi*f*t)+(1/k)*cos(2*pi*f*k*t) ) )
μ is a constant to be varied.
This is frequency modulation but under Fast fourier transform frequency analysis for μ =0.2
, a plethora of constant psychoacoustic frequencies exist. But for this angular frequency expression
function, the Secondary order beat frequencies are missing when (b-a) is an even number from predicted
formula from before:
If we call an Interval, k = b/a + ζ
secondary order frequency, Fβ = fo. (a.k - b) if k > b/a
If we call an Interval, k = b/a - ζ
secondary order frequency, Fβ = fo. (b - a.k) if k < b/a
An alternative displacement wave function could be including the conjunction frequency when two planets
align which is the difference frequency.
ψ(y) = e^(i*2*pi*f*t)+(1/k)*e^(i*2*pi*f*k*t) + (1/k)*i*e^(i*2*pi*f*(k-1)*t)
The orbital conjunction term (1/k)*i*e^(i*2*pi*f*(k-1)*t) has i as a factor since the waveform result is more in
normalized symmetry . And also has (1/k) as a factor for assuming the difference
between radius = 1 for e^(i*2*pi*f*t) term and radius = (1/k) for e^(i*2*pi*f*k*t) term
in circular orbits yields the total distance separating the orbiting particles =1 - (1 - 1/k) = 1/k
angular frequency expression function:
ψ(ω) = sin(2*pi*ψ(y))
= sin(2*pi*μ *(cos(2*pi*f*t)+(1/x)*cos(2*pi*f*x*t)-(1/x)*sin(2*pi*f*t*(x-1))))
μ is a constant to be varied.
In this case under frequency analysis of this alternative angular frequency expression function,
the Zeta frequencies are all visible but with some doubling and tripling anomalies for high value of μ
but for μ < 0.3 the results look good.
Experiment to Tangibly detect Phantom Psychoacoustic Frequencies
when Interacting via Physical Matter.
Abstract:
It has been quite common consensus in the past to believe that the phantom frequencies can not be detected with frequency analysis such as the FFT ( fast fourier transform) Spectrometer. But in general that is only true if
the component frequencies are generated electronically and directly analysed electronically, bi-passing the actual interaction with matter in the Physical world that makes these Psychoacoustic frequencies become tangible.
Method:
We shall compose an experiment to actually sound the vibration of two electronically created component
sinusoidal frequencies y = A*(sin(2*pi*fo*t)+(1/k)*sin(2*pi*fo*k*t) of frequencies fo & fo*k
via a small Loud-speaker (headphones for purity without added harmonics) to vibrate the medium of matter physically to then be received
by a microphone to test by electronic computerized frequency analysis if the phantom frequencies are tangible.
Results:
Specific frequency spikes appear at predicted locations in the Audio spectrum but with underlying noise at low amplitude and at the low frequency End due to using very Basic tools.
The frequency spikes will be given the notation [b/a] as well as it's frequency in Hz
[b/a] is integers b and a in the formula mentioned, that are used to predict where:
secondary order frequency, Fβ = fo. (a.k - b)
In this new notation let secondary order frequency , F[b/a] = fo. (a.k - b) if k > (b/a)
F[b/a] = fo. (b - a.k) if k < (b/a)
Experimental Results:
Base frequency , fo = 1440 Hz , k = 1.074
when Interacting via Physical Matter.
Abstract:
It has been quite common consensus in the past to believe that the phantom frequencies can not be detected with frequency analysis such as the FFT ( fast fourier transform) Spectrometer. But in general that is only true if
the component frequencies are generated electronically and directly analysed electronically, bi-passing the actual interaction with matter in the Physical world that makes these Psychoacoustic frequencies become tangible.
Method:
We shall compose an experiment to actually sound the vibration of two electronically created component
sinusoidal frequencies y = A*(sin(2*pi*fo*t)+(1/k)*sin(2*pi*fo*k*t) of frequencies fo & fo*k
via a small Loud-speaker (headphones for purity without added harmonics) to vibrate the medium of matter physically to then be received
by a microphone to test by electronic computerized frequency analysis if the phantom frequencies are tangible.
Results:
Specific frequency spikes appear at predicted locations in the Audio spectrum but with underlying noise at low amplitude and at the low frequency End due to using very Basic tools.
The frequency spikes will be given the notation [b/a] as well as it's frequency in Hz
[b/a] is integers b and a in the formula mentioned, that are used to predict where:
secondary order frequency, Fβ = fo. (a.k - b)
In this new notation let secondary order frequency , F[b/a] = fo. (a.k - b) if k > (b/a)
F[b/a] = fo. (b - a.k) if k < (b/a)
Experimental Results:
Base frequency , fo = 1440 Hz , k = 1.074
Using [1/1] for [b/a] which is b = 1 & a = 1 , F[1/1] = fo. (1.k - 1) = 106.56 Hz
|
eg: Using [2/1] for [b/a] which is b = 2 & a = 1 , F[2/1] = fo. (2 - 1.k ) = 1333.4Hz
eg: Using [-1/1] for [b/a] which is b = -1 & a = 1 , F[-1/1] = fo. (1.k -(-1)) = 2986.56Hz
eg: [-1/2] for [b/a] which is b = -1 & a = 2 , F[-1/2] = fo. (2*k +1 ) = 4533.12Hz
Base frequency , fo = 1440 Hz , k = 1.034
Base frequency , fo = 1440 Hz , k = 1.034
eg: [1/1] for [b/a] which is b = 1 & a = 1 , F[1/1] = fo. (1*k - 1 ) = 48.96 Hz
eg: [4/3] for [b/a] which is b = 4 & a = 3 , F[4/3] = fo. (4-3*k ) = 1293.12 Hz
Base frequency , fo = 1440 Hz , k = 2.053
eg: [2/1] for [b/a] which is b = 2 & a = 1 , F[2/1] = fo. (1*k -2 ) = 76.32 Hz
eg: [5/2] for [b/a] which is b = 5 & a = 2 , F[5/2] = fo. (5 - 2*k ) = 1287.36 Hz
eg: [4/3] for [b/a] which is b = 4 & a = 3 , F[4/3] = fo. (3*k -4 ) = 3108.96 Hz
eg: [1/2] for [b/a] which is b = 1 & a = 2 , F[1/2] = fo. (2k -1 ) = 4472.64 Hz
Base frequency , fo = 8640 Hz , k = 1.5188
Base frequency , fo = 8640 Hz , k = 1.5188
[3/2] for [b/a] which is b = 3 & a = 2 , F[3/2] = fo. (2k -3 ) = 324.86 Hz
Base frequency fo = 4320 Hz, k = 1.529
Base frequency fo = 4320 Hz, k = 1.529
Base frequency , fo = 1440 Hz , k = 1.618034 (Golden Ratio)
Conclusion:
This Experiment was meditated upon because of a recognition that if we are using some type of instrument to detect a property, whether that be mathematical, a Physical or metaphysical construct,then that previously used instrument to make observations is to be recognized as Old to be substituted by a New observational instrument.
Modern Science and technology has recently passed through a revolutionary phase with the advent of 32-Bit Computing which enables the magnification of quiet sound to very high detail.16-bit would have had a wall of Noise. Future plans are to repeat this experiment in a Sound proofed professional Music Studio to obtain clearer data results and the theory is steadily developing to be released at a Later date with updates.
Taking the Realist approach or the experimentalist approach or both is a trinity of positional points which is perhaps the way in Scientific Observation, even being void of senses of sight and hearing to merely feel vibration of sound through the self's bodily environment.
Or just Pure mind to feel and wonder what the world is about.
