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“The Fourier transform cannot measure two phases at the same frequency.” Why not?

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“The Fourier transform cannot measure two phases at the same frequency.” Why not?

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3

1

$begingroup$

I have read that the Fourier transform cannot distinguish components with the same frequency but different phase. For example, in Mathoverflow, or xrayphysics, where I got the title of my question from: "The Fourier transform cannot measure two phases at the same frequency."

Why is this true mathematically?

fourier-transform fourier

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asked 8 hours ago

ButternutButternut

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add a comment | 

3

1

$begingroup$

I have read that the Fourier transform cannot distinguish components with the same frequency but different phase. For example, in Mathoverflow, or xrayphysics, where I got the title of my question from: "The Fourier transform cannot measure two phases at the same frequency."

Why is this true mathematically?

fourier-transform fourier

share|improve this question

asked 8 hours ago

ButternutButternut

1822 bronze badges

New contributor

Butternut is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

I have read that the Fourier transform cannot distinguish components with the same frequency but different phase. For example, in Mathoverflow, or xrayphysics, where I got the title of my question from: "The Fourier transform cannot measure two phases at the same frequency."

Why is this true mathematically?

fourier-transform fourier

share|improve this question

asked 8 hours ago

ButternutButternut

1822 bronze badges

New contributor

Butternut is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this question

asked 8 hours ago

ButternutButternut

1822 bronze badges

New contributor

Butternut is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

asked 8 hours ago

ButternutButternut

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New contributor

Butternut is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

ButternutButternut

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Butternut is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 8 hours ago

ButternutButternut

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5

$begingroup$

It's because the simultaneous presence of two sinusoidal signals with the same frequency and different phases is actualy equivalent to a single sinusoidal at the same frequency, but, with a new phase and amplitude as follows:

Let the two sinusodial components be summed like this :

$$ x(t) = a cos(omega_0 t + phi) + b cos(omega_0 t + theta) $$

Then from trigionometric manipulations it can be shown that :

$$ x(t) = A cos(omega_0 t + Phi) $$

where
$$A = sqrt{ a^2 + b^2 + 2 a b cos(theta-phi) } $$ and
$$ Phi = tan^{-1}(frac{ a sin(phi) + bsin(theta) } { a cos(phi) + bcos(theta) } ) $$

hence you actually have a single sinusoidal (with a new phase and amplitude), and therefore nothing to distinguish indeed...

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Fat32Fat32

17.2k33 gold badges1212 silver badges3333 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My brain must be on shut-down because I follow the trig stuff but there's still confusion swirling around.. The OP didn't day they were being added so what justifies the initial step where you add them ? In other words, if we just think of them as two signals where one starts "later" than the other but they're not added, can we distinguish those ? Is it that you have to add them because you can't have two data points at one frequency ? Thanks.
  $endgroup$
  – mark leeds
  2 hours ago
  

  
  
  
  

add a comment | 

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5

$begingroup$

It's because the simultaneous presence of two sinusoidal signals with the same frequency and different phases is actualy equivalent to a single sinusoidal at the same frequency, but, with a new phase and amplitude as follows:

Let the two sinusodial components be summed like this :

$$ x(t) = a cos(omega_0 t + phi) + b cos(omega_0 t + theta) $$

Then from trigionometric manipulations it can be shown that :

$$ x(t) = A cos(omega_0 t + Phi) $$

where
$$A = sqrt{ a^2 + b^2 + 2 a b cos(theta-phi) } $$ and
$$ Phi = tan^{-1}(frac{ a sin(phi) + bsin(theta) } { a cos(phi) + bcos(theta) } ) $$

hence you actually have a single sinusoidal (with a new phase and amplitude), and therefore nothing to distinguish indeed...

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Fat32Fat32

17.2k33 gold badges1212 silver badges3333 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My brain must be on shut-down because I follow the trig stuff but there's still confusion swirling around.. The OP didn't day they were being added so what justifies the initial step where you add them ? In other words, if we just think of them as two signals where one starts "later" than the other but they're not added, can we distinguish those ? Is it that you have to add them because you can't have two data points at one frequency ? Thanks.
  $endgroup$
  – mark leeds
  2 hours ago
  

  
  
  
  

add a comment | 

5

$begingroup$

It's because the simultaneous presence of two sinusoidal signals with the same frequency and different phases is actualy equivalent to a single sinusoidal at the same frequency, but, with a new phase and amplitude as follows:

Let the two sinusodial components be summed like this :

$$ x(t) = a cos(omega_0 t + phi) + b cos(omega_0 t + theta) $$

Then from trigionometric manipulations it can be shown that :

$$ x(t) = A cos(omega_0 t + Phi) $$

where
$$A = sqrt{ a^2 + b^2 + 2 a b cos(theta-phi) } $$ and
$$ Phi = tan^{-1}(frac{ a sin(phi) + bsin(theta) } { a cos(phi) + bcos(theta) } ) $$

hence you actually have a single sinusoidal (with a new phase and amplitude), and therefore nothing to distinguish indeed...

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Fat32Fat32

17.2k33 gold badges1212 silver badges3333 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  My brain must be on shut-down because I follow the trig stuff but there's still confusion swirling around.. The OP didn't day they were being added so what justifies the initial step where you add them ? In other words, if we just think of them as two signals where one starts "later" than the other but they're not added, can we distinguish those ? Is it that you have to add them because you can't have two data points at one frequency ? Thanks.
  $endgroup$
  – mark leeds
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

It's because the simultaneous presence of two sinusoidal signals with the same frequency and different phases is actualy equivalent to a single sinusoidal at the same frequency, but, with a new phase and amplitude as follows:

Let the two sinusodial components be summed like this :

$$ x(t) = a cos(omega_0 t + phi) + b cos(omega_0 t + theta) $$

Then from trigionometric manipulations it can be shown that :

$$ x(t) = A cos(omega_0 t + Phi) $$

where
$$A = sqrt{ a^2 + b^2 + 2 a b cos(theta-phi) } $$ and
$$ Phi = tan^{-1}(frac{ a sin(phi) + bsin(theta) } { a cos(phi) + bcos(theta) } ) $$

hence you actually have a single sinusoidal (with a new phase and amplitude), and therefore nothing to distinguish indeed...

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Fat32Fat32

17.2k33 gold badges1212 silver badges3333 bronze badges

$endgroup$

share|improve this answer

edited 7 hours ago

answered 7 hours ago

Fat32Fat32

17.2k33 gold badges1212 silver badges3333 bronze badges

answered 7 hours ago

Fat32Fat32

17.2k33 gold badges1212 silver badges3333 bronze badges

answered 7 hours ago

Fat32Fat32

17.2k33 gold badges1212 silver badges3333 bronze badges