“The Fourier transform cannot measure two phases at the same frequency.” Why not?Fourier Transform IdentitiesHow do I explain a complex exponential intuitively?3D wiggle plot for an analytic signal: Heyser corkscrew/spiralExample of Fourier Transform not existing for real-life signals?Fourier Transform Form: two sin components & a phase shift & a magnitude for only one termIntuitive explanation of the Fourier Transform for some of the functionsintuitive interpretation Fourier Transform of two different Rectangular pulsesWhy Fourier transform is not sufficient and we have to use Laplace transform?Does the inverse Fourier transform only produce stationary time signals?Why is not Fourier Transform Good for Non-linear ProcessesPeriodicity of the discrete-time Fourier TransformWhy do the two methods give different answers for the Fourier transform of $Y = cos(omega_0 t + phi)$?How do I obtain the fourier series coefficients for a signal obtained by multiplication of two signals of different frequency?

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


Fourier Transform IdentitiesHow do I explain a complex exponential intuitively?3D wiggle plot for an analytic signal: Heyser corkscrew/spiralExample of Fourier Transform not existing for real-life signals?Fourier Transform Form: two sin components & a phase shift & a magnitude for only one termIntuitive explanation of the Fourier Transform for some of the functionsintuitive interpretation Fourier Transform of two different Rectangular pulsesWhy Fourier transform is not sufficient and we have to use Laplace transform?Does the inverse Fourier transform only produce stationary time signals?Why is not Fourier Transform Good for Non-linear ProcessesPeriodicity of the discrete-time Fourier TransformWhy do the two methods give different answers for the Fourier transform of $Y = cos(omega_0 t + phi)$?How do I obtain the fourier series coefficients for a signal obtained by multiplication of two signals of different frequency?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








15












$begingroup$


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?










share|improve this question









$endgroup$









  • 5




    $begingroup$
    Can you distinguish the components of, say $sin(x) + sin(x+c)$? I bet you can't.
    $endgroup$
    – Ilmari Karonen
    Aug 5 at 9:14











  • $begingroup$
    The FT finds components that could be added together to reconstruct a given signal. But that doesn't mean that those components somehow actually were present in the original. There are infinite different ways that a given signal could have been "constructed," but the signal will have only one unique FT.
    $endgroup$
    – Solomon Slow
    Aug 6 at 20:09

















15












$begingroup$


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?










share|improve this question









$endgroup$









  • 5




    $begingroup$
    Can you distinguish the components of, say $sin(x) + sin(x+c)$? I bet you can't.
    $endgroup$
    – Ilmari Karonen
    Aug 5 at 9:14











  • $begingroup$
    The FT finds components that could be added together to reconstruct a given signal. But that doesn't mean that those components somehow actually were present in the original. There are infinite different ways that a given signal could have been "constructed," but the signal will have only one unique FT.
    $endgroup$
    – Solomon Slow
    Aug 6 at 20:09













15












15








15


6



$begingroup$


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?










share|improve this question









$endgroup$




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













share|improve this question











share|improve this question




share|improve this question










asked Aug 4 at 21:42









ButternutButternut

811 silver badge4 bronze badges




811 silver badge4 bronze badges










  • 5




    $begingroup$
    Can you distinguish the components of, say $sin(x) + sin(x+c)$? I bet you can't.
    $endgroup$
    – Ilmari Karonen
    Aug 5 at 9:14











  • $begingroup$
    The FT finds components that could be added together to reconstruct a given signal. But that doesn't mean that those components somehow actually were present in the original. There are infinite different ways that a given signal could have been "constructed," but the signal will have only one unique FT.
    $endgroup$
    – Solomon Slow
    Aug 6 at 20:09












  • 5




    $begingroup$
    Can you distinguish the components of, say $sin(x) + sin(x+c)$? I bet you can't.
    $endgroup$
    – Ilmari Karonen
    Aug 5 at 9:14











  • $begingroup$
    The FT finds components that could be added together to reconstruct a given signal. But that doesn't mean that those components somehow actually were present in the original. There are infinite different ways that a given signal could have been "constructed," but the signal will have only one unique FT.
    $endgroup$
    – Solomon Slow
    Aug 6 at 20:09







5




5




$begingroup$
Can you distinguish the components of, say $sin(x) + sin(x+c)$? I bet you can't.
$endgroup$
– Ilmari Karonen
Aug 5 at 9:14





$begingroup$
Can you distinguish the components of, say $sin(x) + sin(x+c)$? I bet you can't.
$endgroup$
– Ilmari Karonen
Aug 5 at 9:14













$begingroup$
The FT finds components that could be added together to reconstruct a given signal. But that doesn't mean that those components somehow actually were present in the original. There are infinite different ways that a given signal could have been "constructed," but the signal will have only one unique FT.
$endgroup$
– Solomon Slow
Aug 6 at 20:09




$begingroup$
The FT finds components that could be added together to reconstruct a given signal. But that doesn't mean that those components somehow actually were present in the original. There are infinite different ways that a given signal could have been "constructed," but the signal will have only one unique FT.
$endgroup$
– Solomon Slow
Aug 6 at 20:09










4 Answers
4






active

oldest

votes


















30












$begingroup$

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











$endgroup$










  • 1




    $begingroup$
    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
    Aug 5 at 3:57






  • 2




    $begingroup$
    @markleeds, The OP didnt say that he was referring to the windowed Fourier transform, and the links given clearly indicate the regular non-windowed version. In the regular version of Fourier analysis, signals are assumed composed as a weighted sum of sinusoidals with different phase. The analysis consists of getting these weights and phases. The collection of them is the spectrum. If you concatenate 2 sinusoids, this global Fourier analysis can not distinguish their phase either. However, the windowed Fourier transform is designed for such a job... not that it does it remarkable well.
    $endgroup$
    – Stefan Karlsson
    Aug 5 at 7:08







  • 1




    $begingroup$
    As my comment suggested, it could be informative to add a mention of the windowed Fourier transform. If @Fat32 has the time, he could mention the discontinuity involved with concatenating 2 sinusoids of different frequency, and why we get a range of seemingly random frequencies added to the global fourier transform if we try to analyze that.
    $endgroup$
    – Stefan Karlsson
    Aug 5 at 7:14






  • 2




    $begingroup$
    Hi @markleeds, as StefanKarlsson already indicated, the question was about the case of superposition (simultaneous additive presence) of those two sinusoidals of the same frequency. Note very carefully that phase is a relative term and not absolute; i.e., it's measured with respect to a chosen common (time) origin, which is $t=0$ above. The concatenation (like in Phase Shift Keying) allows windowed discrimination but you should still refer to a common time origin to tell the phase differences anyway. That's why PSK receivers require strict pulse time syncronisation ;-)
    $endgroup$
    – Fat32
    Aug 5 at 12:27






  • 1




    $begingroup$
    @smsc feels like repeating myself but if the output of those two cables are added and then analysed via FT, then you will see a single sine wave with a composite phase & ampl... But if you do not add them and analyse separately, then you will be able to tell their relative phases ... And this is not related to DFT.
    $endgroup$
    – Fat32
    Aug 6 at 15:14



















1












$begingroup$

If you read further, down to "The simplified version of the Fourier transform we discussed above can't account for phase shifts - how does the Fourier transform actually do it?" you'll note a slightly better explanstion, they use sines and cosines.




"Mathematics of Phase Shifts (optional).



In order to see how a phase shift can be broken down into non-shifted sines and cosines, we need a trigonometric identity: sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b).



A*sin(2 * π * f * t + φ) = A*cos(φ) * sin(2 * π * f * t) + A*sin(φ) * cos(2 * π * f * t)



As you can see, the phase shift moves some of the amplitude (energy) of the sine signal into a cosine signal, but the frequency doesn't change. If you use the complex number representation of the Fourier transform, the phase shift simply represents a rotation of the value in the complex plane, with the magnitude unchanged. The fact that phase shifts only move amplitude from sine to cosine means that adding two signals with the same frequency and different phase gives a signal with an overall (average) phase shift at that frequency - and no memory of the components.".




In practice it's more complicated, see "Partial Fourier Techniques", "Phase-conjugate Symmetry", and "FOV and k-space". In the "Intro to Phase-encoding - I" they explain:




"... when two sine waves (A and B) with the same frequency but different phases are added together, the result is another sine wave with the same frequency but a different phase. When the sine waves are close together in phase they constructively interfere, and when out of phase they destructively interfere.



... Looking only at their sum, you simply see a sine wave of a certain frequency and phase. It is impossible from this single observation to sort out the individual contributions made by waves A and B.



However, by making two observations with A and B shifted by different phases, it is possible to determine their individual contributions by looking only at their sums. This is illustrated below in an MR image, where A and B are two pixels in the same vertical column resonating at the same encoded frequency (ω). Specifically, at Step 0 (baseline, when no phase-encoding gradient is has been applied) the total signal from A&B together can be written: So(t) = A sin ωt + B sin ωt = (A+B) sin ωt.



Apply a phase encoding gradient to dephase spins along the vertical axis







...







From this single measurement in Step 1, we still do not know the individual amplitudes A and B, only their difference (A−B). Using information from both Step 0 and Step 1 together, we are able to extract the unique signal contributions by simple algebra:



½ [So + S1] = ½ [(A+B) + (A−B)] = A    and    ½ [So − S1] = ½ [(A+B) − (A−B)] = B



".




Otherwise it would look like this (image A):



Effect of SDPS on PFI image



PFI showing artifacts from various algorithms: (A) basic algorithm, (B) BAX algorithm, (C) zero-fill algorithm, (D) basic algorithm using data that had prior constant, linear SDPS correction, illustrating artifacts from higher order SDPS.






share|improve this answer









$endgroup$






















    1












    $begingroup$

    It might be slightly clearer to write $ccos(omega t+phi)$ as $Re(ce^(omega t+phi)i)$. Then, since $Re$ distributes over addition, $c_1cos(omega t+phi_1)+c_2cos(omega t+phi_2)=Re(c_1e^(omega t+phi_1)i+c_2e^(omega t+phi_2)i)$. We can factor out $ae^omega t i$, and we get $Re(e^omega t i(c_1e^phi_1i+c_2e^phi_2i))$. This shows that when we're dealing with two signals of the same frequency, we can factor out the time-dependent part, leaving each signal as being characterized by a constant term, and of course when taking the Fourier transform, constant terms can be factored out. We can further note that $c e^phi i$ can be interpreted as a vector in the complex plane where the magnitude is $c$ and the angle is given by $phi$. And we can perform the addition in this vector space: the vector representing the sum is the sum of the vectors representing the summands.



    So while both signals affect the magnitude of the output, an additional signal will not affect where in phase space the output is.






    share|improve this answer









    $endgroup$






















      1












      $begingroup$

      I would like to take the path of a geometrical version of the question, using sums of circles.



      Sines and cosines are "just" the real and imaginary parts of cisoids, or complex exponentials (some references can be found at How do I explain a complex exponential intuitively?, 3D wiggle plot for an analytic signal: Heyser corkscrew/spiral, Fourier Transform Identities).



      If you take $s_omega,phi(t) = e^2pi i (omega t+phi)$, then $mathrmRe(s_omega,0(t))=cos(2pi omega t)$, or $mathrmIm(s_omega,pi/2(t))=cos(2pi omega t)$, and you can do a lot of combinations. The advantage of a cisoid is that it better uses the 2D space, as it can be depicted as a circle (a wheel) on which a point moves at different speeds driven by $omega$. A sum of "frequencies with different amplitudes" can be represented at "sums of spinning wheels" (borrowed from harmonic circles, or the Fourier Series Animation) with different radii and speeds, as depicted here:



      Harmonic circles



      Going back to a sum of two harmonics at the same frequency, the problem reads as: can we
      separate or measure the combination:
      $$a_1 s_omega,phi_1(t) +a_2 s_omega,phi_2(t),?$$



      Constants $a_1$ and $a_2$ could be complex, so let us simplify the problem a bit before. Since Fourier has shift invariance properties, we can factorize either $e^2 pi iphi_1$ or $e^2 pi iphi_2$, and keep only one phase difference. We can also factorize an amplitude (the biggest for instance), and reduce the question to the behavior of the simplified problem:



      $$ s_omega,0(t) +a s_omega,phi(t),, $$



      with $|a|<1$. This simplification can be written as:



      $$e^2pi i (omega t)+ae^2pi i (omega t+phi)labelatag1$$



      and thus as:



      $$(1+ae^2pi iphi)e^2pi i (omega t),,labelbtag2$$



      which is another harmonic component with same frequency, but a different phase and amplitude.
      The complex number $(1+ae^2pi iphi)$ could be rewritten as $alpha e^2pi ivarphi$, with trigonometric rules as detailed by @Fat32 (which I could detail later if needed). Now, let us geometrize the intuition. The unit circle is the motion of a point (say the tip of the valve) on a running bicycle wheel. The $a$-radius circle is like a small spinning wheel attached to the valve (like the blue and red circles only from the picture above). An now, we look at the motion of a dot on the perimeter of the small wheel.



      What does your question ask: if the angular rotation of the small an the big wheel are the same, you cannot tell whether the motion of the dot results from the combination of the motion of two wheels of radii $1$ and $a$ (with some initial angle) or from a single bigger wheel (of radius $alpha$), with some other starting angle. This is what is mean by $refa$ and $refb$.



      In other words, neither a Fourier transform, nor a human eye, can distinguish components with the same frequency but different phase.



      [[I'll add animations if I find the time]]






      share|improve this answer











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        4 Answers
        4






        active

        oldest

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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        30












        $begingroup$

        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











        $endgroup$










        • 1




          $begingroup$
          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
          Aug 5 at 3:57






        • 2




          $begingroup$
          @markleeds, The OP didnt say that he was referring to the windowed Fourier transform, and the links given clearly indicate the regular non-windowed version. In the regular version of Fourier analysis, signals are assumed composed as a weighted sum of sinusoidals with different phase. The analysis consists of getting these weights and phases. The collection of them is the spectrum. If you concatenate 2 sinusoids, this global Fourier analysis can not distinguish their phase either. However, the windowed Fourier transform is designed for such a job... not that it does it remarkable well.
          $endgroup$
          – Stefan Karlsson
          Aug 5 at 7:08







        • 1




          $begingroup$
          As my comment suggested, it could be informative to add a mention of the windowed Fourier transform. If @Fat32 has the time, he could mention the discontinuity involved with concatenating 2 sinusoids of different frequency, and why we get a range of seemingly random frequencies added to the global fourier transform if we try to analyze that.
          $endgroup$
          – Stefan Karlsson
          Aug 5 at 7:14






        • 2




          $begingroup$
          Hi @markleeds, as StefanKarlsson already indicated, the question was about the case of superposition (simultaneous additive presence) of those two sinusoidals of the same frequency. Note very carefully that phase is a relative term and not absolute; i.e., it's measured with respect to a chosen common (time) origin, which is $t=0$ above. The concatenation (like in Phase Shift Keying) allows windowed discrimination but you should still refer to a common time origin to tell the phase differences anyway. That's why PSK receivers require strict pulse time syncronisation ;-)
          $endgroup$
          – Fat32
          Aug 5 at 12:27






        • 1




          $begingroup$
          @smsc feels like repeating myself but if the output of those two cables are added and then analysed via FT, then you will see a single sine wave with a composite phase & ampl... But if you do not add them and analyse separately, then you will be able to tell their relative phases ... And this is not related to DFT.
          $endgroup$
          – Fat32
          Aug 6 at 15:14
















        30












        $begingroup$

        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











        $endgroup$










        • 1




          $begingroup$
          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
          Aug 5 at 3:57






        • 2




          $begingroup$
          @markleeds, The OP didnt say that he was referring to the windowed Fourier transform, and the links given clearly indicate the regular non-windowed version. In the regular version of Fourier analysis, signals are assumed composed as a weighted sum of sinusoidals with different phase. The analysis consists of getting these weights and phases. The collection of them is the spectrum. If you concatenate 2 sinusoids, this global Fourier analysis can not distinguish their phase either. However, the windowed Fourier transform is designed for such a job... not that it does it remarkable well.
          $endgroup$
          – Stefan Karlsson
          Aug 5 at 7:08







        • 1




          $begingroup$
          As my comment suggested, it could be informative to add a mention of the windowed Fourier transform. If @Fat32 has the time, he could mention the discontinuity involved with concatenating 2 sinusoids of different frequency, and why we get a range of seemingly random frequencies added to the global fourier transform if we try to analyze that.
          $endgroup$
          – Stefan Karlsson
          Aug 5 at 7:14






        • 2




          $begingroup$
          Hi @markleeds, as StefanKarlsson already indicated, the question was about the case of superposition (simultaneous additive presence) of those two sinusoidals of the same frequency. Note very carefully that phase is a relative term and not absolute; i.e., it's measured with respect to a chosen common (time) origin, which is $t=0$ above. The concatenation (like in Phase Shift Keying) allows windowed discrimination but you should still refer to a common time origin to tell the phase differences anyway. That's why PSK receivers require strict pulse time syncronisation ;-)
          $endgroup$
          – Fat32
          Aug 5 at 12:27






        • 1




          $begingroup$
          @smsc feels like repeating myself but if the output of those two cables are added and then analysed via FT, then you will see a single sine wave with a composite phase & ampl... But if you do not add them and analyse separately, then you will be able to tell their relative phases ... And this is not related to DFT.
          $endgroup$
          – Fat32
          Aug 6 at 15:14














        30












        30








        30





        $begingroup$

        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











        $endgroup$



        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














        share|improve this answer



        share|improve this answer








        edited Aug 4 at 22:55

























        answered Aug 4 at 22:50









        Fat32Fat32

        17.7k3 gold badges14 silver badges34 bronze badges




        17.7k3 gold badges14 silver badges34 bronze badges










        • 1




          $begingroup$
          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
          Aug 5 at 3:57






        • 2




          $begingroup$
          @markleeds, The OP didnt say that he was referring to the windowed Fourier transform, and the links given clearly indicate the regular non-windowed version. In the regular version of Fourier analysis, signals are assumed composed as a weighted sum of sinusoidals with different phase. The analysis consists of getting these weights and phases. The collection of them is the spectrum. If you concatenate 2 sinusoids, this global Fourier analysis can not distinguish their phase either. However, the windowed Fourier transform is designed for such a job... not that it does it remarkable well.
          $endgroup$
          – Stefan Karlsson
          Aug 5 at 7:08







        • 1




          $begingroup$
          As my comment suggested, it could be informative to add a mention of the windowed Fourier transform. If @Fat32 has the time, he could mention the discontinuity involved with concatenating 2 sinusoids of different frequency, and why we get a range of seemingly random frequencies added to the global fourier transform if we try to analyze that.
          $endgroup$
          – Stefan Karlsson
          Aug 5 at 7:14






        • 2




          $begingroup$
          Hi @markleeds, as StefanKarlsson already indicated, the question was about the case of superposition (simultaneous additive presence) of those two sinusoidals of the same frequency. Note very carefully that phase is a relative term and not absolute; i.e., it's measured with respect to a chosen common (time) origin, which is $t=0$ above. The concatenation (like in Phase Shift Keying) allows windowed discrimination but you should still refer to a common time origin to tell the phase differences anyway. That's why PSK receivers require strict pulse time syncronisation ;-)
          $endgroup$
          – Fat32
          Aug 5 at 12:27






        • 1




          $begingroup$
          @smsc feels like repeating myself but if the output of those two cables are added and then analysed via FT, then you will see a single sine wave with a composite phase & ampl... But if you do not add them and analyse separately, then you will be able to tell their relative phases ... And this is not related to DFT.
          $endgroup$
          – Fat32
          Aug 6 at 15:14













        • 1




          $begingroup$
          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
          Aug 5 at 3:57






        • 2




          $begingroup$
          @markleeds, The OP didnt say that he was referring to the windowed Fourier transform, and the links given clearly indicate the regular non-windowed version. In the regular version of Fourier analysis, signals are assumed composed as a weighted sum of sinusoidals with different phase. The analysis consists of getting these weights and phases. The collection of them is the spectrum. If you concatenate 2 sinusoids, this global Fourier analysis can not distinguish their phase either. However, the windowed Fourier transform is designed for such a job... not that it does it remarkable well.
          $endgroup$
          – Stefan Karlsson
          Aug 5 at 7:08







        • 1




          $begingroup$
          As my comment suggested, it could be informative to add a mention of the windowed Fourier transform. If @Fat32 has the time, he could mention the discontinuity involved with concatenating 2 sinusoids of different frequency, and why we get a range of seemingly random frequencies added to the global fourier transform if we try to analyze that.
          $endgroup$
          – Stefan Karlsson
          Aug 5 at 7:14






        • 2




          $begingroup$
          Hi @markleeds, as StefanKarlsson already indicated, the question was about the case of superposition (simultaneous additive presence) of those two sinusoidals of the same frequency. Note very carefully that phase is a relative term and not absolute; i.e., it's measured with respect to a chosen common (time) origin, which is $t=0$ above. The concatenation (like in Phase Shift Keying) allows windowed discrimination but you should still refer to a common time origin to tell the phase differences anyway. That's why PSK receivers require strict pulse time syncronisation ;-)
          $endgroup$
          – Fat32
          Aug 5 at 12:27






        • 1




          $begingroup$
          @smsc feels like repeating myself but if the output of those two cables are added and then analysed via FT, then you will see a single sine wave with a composite phase & ampl... But if you do not add them and analyse separately, then you will be able to tell their relative phases ... And this is not related to DFT.
          $endgroup$
          – Fat32
          Aug 6 at 15:14








        1




        1




        $begingroup$
        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
        Aug 5 at 3:57




        $begingroup$
        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
        Aug 5 at 3:57




        2




        2




        $begingroup$
        @markleeds, The OP didnt say that he was referring to the windowed Fourier transform, and the links given clearly indicate the regular non-windowed version. In the regular version of Fourier analysis, signals are assumed composed as a weighted sum of sinusoidals with different phase. The analysis consists of getting these weights and phases. The collection of them is the spectrum. If you concatenate 2 sinusoids, this global Fourier analysis can not distinguish their phase either. However, the windowed Fourier transform is designed for such a job... not that it does it remarkable well.
        $endgroup$
        – Stefan Karlsson
        Aug 5 at 7:08





        $begingroup$
        @markleeds, The OP didnt say that he was referring to the windowed Fourier transform, and the links given clearly indicate the regular non-windowed version. In the regular version of Fourier analysis, signals are assumed composed as a weighted sum of sinusoidals with different phase. The analysis consists of getting these weights and phases. The collection of them is the spectrum. If you concatenate 2 sinusoids, this global Fourier analysis can not distinguish their phase either. However, the windowed Fourier transform is designed for such a job... not that it does it remarkable well.
        $endgroup$
        – Stefan Karlsson
        Aug 5 at 7:08





        1




        1




        $begingroup$
        As my comment suggested, it could be informative to add a mention of the windowed Fourier transform. If @Fat32 has the time, he could mention the discontinuity involved with concatenating 2 sinusoids of different frequency, and why we get a range of seemingly random frequencies added to the global fourier transform if we try to analyze that.
        $endgroup$
        – Stefan Karlsson
        Aug 5 at 7:14




        $begingroup$
        As my comment suggested, it could be informative to add a mention of the windowed Fourier transform. If @Fat32 has the time, he could mention the discontinuity involved with concatenating 2 sinusoids of different frequency, and why we get a range of seemingly random frequencies added to the global fourier transform if we try to analyze that.
        $endgroup$
        – Stefan Karlsson
        Aug 5 at 7:14




        2




        2




        $begingroup$
        Hi @markleeds, as StefanKarlsson already indicated, the question was about the case of superposition (simultaneous additive presence) of those two sinusoidals of the same frequency. Note very carefully that phase is a relative term and not absolute; i.e., it's measured with respect to a chosen common (time) origin, which is $t=0$ above. The concatenation (like in Phase Shift Keying) allows windowed discrimination but you should still refer to a common time origin to tell the phase differences anyway. That's why PSK receivers require strict pulse time syncronisation ;-)
        $endgroup$
        – Fat32
        Aug 5 at 12:27




        $begingroup$
        Hi @markleeds, as StefanKarlsson already indicated, the question was about the case of superposition (simultaneous additive presence) of those two sinusoidals of the same frequency. Note very carefully that phase is a relative term and not absolute; i.e., it's measured with respect to a chosen common (time) origin, which is $t=0$ above. The concatenation (like in Phase Shift Keying) allows windowed discrimination but you should still refer to a common time origin to tell the phase differences anyway. That's why PSK receivers require strict pulse time syncronisation ;-)
        $endgroup$
        – Fat32
        Aug 5 at 12:27




        1




        1




        $begingroup$
        @smsc feels like repeating myself but if the output of those two cables are added and then analysed via FT, then you will see a single sine wave with a composite phase & ampl... But if you do not add them and analyse separately, then you will be able to tell their relative phases ... And this is not related to DFT.
        $endgroup$
        – Fat32
        Aug 6 at 15:14





        $begingroup$
        @smsc feels like repeating myself but if the output of those two cables are added and then analysed via FT, then you will see a single sine wave with a composite phase & ampl... But if you do not add them and analyse separately, then you will be able to tell their relative phases ... And this is not related to DFT.
        $endgroup$
        – Fat32
        Aug 6 at 15:14














        1












        $begingroup$

        If you read further, down to "The simplified version of the Fourier transform we discussed above can't account for phase shifts - how does the Fourier transform actually do it?" you'll note a slightly better explanstion, they use sines and cosines.




        "Mathematics of Phase Shifts (optional).



        In order to see how a phase shift can be broken down into non-shifted sines and cosines, we need a trigonometric identity: sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b).



        A*sin(2 * π * f * t + φ) = A*cos(φ) * sin(2 * π * f * t) + A*sin(φ) * cos(2 * π * f * t)



        As you can see, the phase shift moves some of the amplitude (energy) of the sine signal into a cosine signal, but the frequency doesn't change. If you use the complex number representation of the Fourier transform, the phase shift simply represents a rotation of the value in the complex plane, with the magnitude unchanged. The fact that phase shifts only move amplitude from sine to cosine means that adding two signals with the same frequency and different phase gives a signal with an overall (average) phase shift at that frequency - and no memory of the components.".




        In practice it's more complicated, see "Partial Fourier Techniques", "Phase-conjugate Symmetry", and "FOV and k-space". In the "Intro to Phase-encoding - I" they explain:




        "... when two sine waves (A and B) with the same frequency but different phases are added together, the result is another sine wave with the same frequency but a different phase. When the sine waves are close together in phase they constructively interfere, and when out of phase they destructively interfere.



        ... Looking only at their sum, you simply see a sine wave of a certain frequency and phase. It is impossible from this single observation to sort out the individual contributions made by waves A and B.



        However, by making two observations with A and B shifted by different phases, it is possible to determine their individual contributions by looking only at their sums. This is illustrated below in an MR image, where A and B are two pixels in the same vertical column resonating at the same encoded frequency (ω). Specifically, at Step 0 (baseline, when no phase-encoding gradient is has been applied) the total signal from A&B together can be written: So(t) = A sin ωt + B sin ωt = (A+B) sin ωt.



        Apply a phase encoding gradient to dephase spins along the vertical axis







        ...







        From this single measurement in Step 1, we still do not know the individual amplitudes A and B, only their difference (A−B). Using information from both Step 0 and Step 1 together, we are able to extract the unique signal contributions by simple algebra:



        ½ [So + S1] = ½ [(A+B) + (A−B)] = A    and    ½ [So − S1] = ½ [(A+B) − (A−B)] = B



        ".




        Otherwise it would look like this (image A):



        Effect of SDPS on PFI image



        PFI showing artifacts from various algorithms: (A) basic algorithm, (B) BAX algorithm, (C) zero-fill algorithm, (D) basic algorithm using data that had prior constant, linear SDPS correction, illustrating artifacts from higher order SDPS.






        share|improve this answer









        $endgroup$



















          1












          $begingroup$

          If you read further, down to "The simplified version of the Fourier transform we discussed above can't account for phase shifts - how does the Fourier transform actually do it?" you'll note a slightly better explanstion, they use sines and cosines.




          "Mathematics of Phase Shifts (optional).



          In order to see how a phase shift can be broken down into non-shifted sines and cosines, we need a trigonometric identity: sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b).



          A*sin(2 * π * f * t + φ) = A*cos(φ) * sin(2 * π * f * t) + A*sin(φ) * cos(2 * π * f * t)



          As you can see, the phase shift moves some of the amplitude (energy) of the sine signal into a cosine signal, but the frequency doesn't change. If you use the complex number representation of the Fourier transform, the phase shift simply represents a rotation of the value in the complex plane, with the magnitude unchanged. The fact that phase shifts only move amplitude from sine to cosine means that adding two signals with the same frequency and different phase gives a signal with an overall (average) phase shift at that frequency - and no memory of the components.".




          In practice it's more complicated, see "Partial Fourier Techniques", "Phase-conjugate Symmetry", and "FOV and k-space". In the "Intro to Phase-encoding - I" they explain:




          "... when two sine waves (A and B) with the same frequency but different phases are added together, the result is another sine wave with the same frequency but a different phase. When the sine waves are close together in phase they constructively interfere, and when out of phase they destructively interfere.



          ... Looking only at their sum, you simply see a sine wave of a certain frequency and phase. It is impossible from this single observation to sort out the individual contributions made by waves A and B.



          However, by making two observations with A and B shifted by different phases, it is possible to determine their individual contributions by looking only at their sums. This is illustrated below in an MR image, where A and B are two pixels in the same vertical column resonating at the same encoded frequency (ω). Specifically, at Step 0 (baseline, when no phase-encoding gradient is has been applied) the total signal from A&B together can be written: So(t) = A sin ωt + B sin ωt = (A+B) sin ωt.



          Apply a phase encoding gradient to dephase spins along the vertical axis







          ...







          From this single measurement in Step 1, we still do not know the individual amplitudes A and B, only their difference (A−B). Using information from both Step 0 and Step 1 together, we are able to extract the unique signal contributions by simple algebra:



          ½ [So + S1] = ½ [(A+B) + (A−B)] = A    and    ½ [So − S1] = ½ [(A+B) − (A−B)] = B



          ".




          Otherwise it would look like this (image A):



          Effect of SDPS on PFI image



          PFI showing artifacts from various algorithms: (A) basic algorithm, (B) BAX algorithm, (C) zero-fill algorithm, (D) basic algorithm using data that had prior constant, linear SDPS correction, illustrating artifacts from higher order SDPS.






          share|improve this answer









          $endgroup$

















            1












            1








            1





            $begingroup$

            If you read further, down to "The simplified version of the Fourier transform we discussed above can't account for phase shifts - how does the Fourier transform actually do it?" you'll note a slightly better explanstion, they use sines and cosines.




            "Mathematics of Phase Shifts (optional).



            In order to see how a phase shift can be broken down into non-shifted sines and cosines, we need a trigonometric identity: sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b).



            A*sin(2 * π * f * t + φ) = A*cos(φ) * sin(2 * π * f * t) + A*sin(φ) * cos(2 * π * f * t)



            As you can see, the phase shift moves some of the amplitude (energy) of the sine signal into a cosine signal, but the frequency doesn't change. If you use the complex number representation of the Fourier transform, the phase shift simply represents a rotation of the value in the complex plane, with the magnitude unchanged. The fact that phase shifts only move amplitude from sine to cosine means that adding two signals with the same frequency and different phase gives a signal with an overall (average) phase shift at that frequency - and no memory of the components.".




            In practice it's more complicated, see "Partial Fourier Techniques", "Phase-conjugate Symmetry", and "FOV and k-space". In the "Intro to Phase-encoding - I" they explain:




            "... when two sine waves (A and B) with the same frequency but different phases are added together, the result is another sine wave with the same frequency but a different phase. When the sine waves are close together in phase they constructively interfere, and when out of phase they destructively interfere.



            ... Looking only at their sum, you simply see a sine wave of a certain frequency and phase. It is impossible from this single observation to sort out the individual contributions made by waves A and B.



            However, by making two observations with A and B shifted by different phases, it is possible to determine their individual contributions by looking only at their sums. This is illustrated below in an MR image, where A and B are two pixels in the same vertical column resonating at the same encoded frequency (ω). Specifically, at Step 0 (baseline, when no phase-encoding gradient is has been applied) the total signal from A&B together can be written: So(t) = A sin ωt + B sin ωt = (A+B) sin ωt.



            Apply a phase encoding gradient to dephase spins along the vertical axis







            ...







            From this single measurement in Step 1, we still do not know the individual amplitudes A and B, only their difference (A−B). Using information from both Step 0 and Step 1 together, we are able to extract the unique signal contributions by simple algebra:



            ½ [So + S1] = ½ [(A+B) + (A−B)] = A    and    ½ [So − S1] = ½ [(A+B) − (A−B)] = B



            ".




            Otherwise it would look like this (image A):



            Effect of SDPS on PFI image



            PFI showing artifacts from various algorithms: (A) basic algorithm, (B) BAX algorithm, (C) zero-fill algorithm, (D) basic algorithm using data that had prior constant, linear SDPS correction, illustrating artifacts from higher order SDPS.






            share|improve this answer









            $endgroup$



            If you read further, down to "The simplified version of the Fourier transform we discussed above can't account for phase shifts - how does the Fourier transform actually do it?" you'll note a slightly better explanstion, they use sines and cosines.




            "Mathematics of Phase Shifts (optional).



            In order to see how a phase shift can be broken down into non-shifted sines and cosines, we need a trigonometric identity: sin(a+b) = sin(a)*cos(b) + cos(a)*sin(b).



            A*sin(2 * π * f * t + φ) = A*cos(φ) * sin(2 * π * f * t) + A*sin(φ) * cos(2 * π * f * t)



            As you can see, the phase shift moves some of the amplitude (energy) of the sine signal into a cosine signal, but the frequency doesn't change. If you use the complex number representation of the Fourier transform, the phase shift simply represents a rotation of the value in the complex plane, with the magnitude unchanged. The fact that phase shifts only move amplitude from sine to cosine means that adding two signals with the same frequency and different phase gives a signal with an overall (average) phase shift at that frequency - and no memory of the components.".




            In practice it's more complicated, see "Partial Fourier Techniques", "Phase-conjugate Symmetry", and "FOV and k-space". In the "Intro to Phase-encoding - I" they explain:




            "... when two sine waves (A and B) with the same frequency but different phases are added together, the result is another sine wave with the same frequency but a different phase. When the sine waves are close together in phase they constructively interfere, and when out of phase they destructively interfere.



            ... Looking only at their sum, you simply see a sine wave of a certain frequency and phase. It is impossible from this single observation to sort out the individual contributions made by waves A and B.



            However, by making two observations with A and B shifted by different phases, it is possible to determine their individual contributions by looking only at their sums. This is illustrated below in an MR image, where A and B are two pixels in the same vertical column resonating at the same encoded frequency (ω). Specifically, at Step 0 (baseline, when no phase-encoding gradient is has been applied) the total signal from A&B together can be written: So(t) = A sin ωt + B sin ωt = (A+B) sin ωt.



            Apply a phase encoding gradient to dephase spins along the vertical axis







            ...







            From this single measurement in Step 1, we still do not know the individual amplitudes A and B, only their difference (A−B). Using information from both Step 0 and Step 1 together, we are able to extract the unique signal contributions by simple algebra:



            ½ [So + S1] = ½ [(A+B) + (A−B)] = A    and    ½ [So − S1] = ½ [(A+B) − (A−B)] = B



            ".




            Otherwise it would look like this (image A):



            Effect of SDPS on PFI image



            PFI showing artifacts from various algorithms: (A) basic algorithm, (B) BAX algorithm, (C) zero-fill algorithm, (D) basic algorithm using data that had prior constant, linear SDPS correction, illustrating artifacts from higher order SDPS.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Aug 5 at 18:06









            RobRob

            1111 gold badge1 silver badge4 bronze badges




            1111 gold badge1 silver badge4 bronze badges
























                1












                $begingroup$

                It might be slightly clearer to write $ccos(omega t+phi)$ as $Re(ce^(omega t+phi)i)$. Then, since $Re$ distributes over addition, $c_1cos(omega t+phi_1)+c_2cos(omega t+phi_2)=Re(c_1e^(omega t+phi_1)i+c_2e^(omega t+phi_2)i)$. We can factor out $ae^omega t i$, and we get $Re(e^omega t i(c_1e^phi_1i+c_2e^phi_2i))$. This shows that when we're dealing with two signals of the same frequency, we can factor out the time-dependent part, leaving each signal as being characterized by a constant term, and of course when taking the Fourier transform, constant terms can be factored out. We can further note that $c e^phi i$ can be interpreted as a vector in the complex plane where the magnitude is $c$ and the angle is given by $phi$. And we can perform the addition in this vector space: the vector representing the sum is the sum of the vectors representing the summands.



                So while both signals affect the magnitude of the output, an additional signal will not affect where in phase space the output is.






                share|improve this answer









                $endgroup$



















                  1












                  $begingroup$

                  It might be slightly clearer to write $ccos(omega t+phi)$ as $Re(ce^(omega t+phi)i)$. Then, since $Re$ distributes over addition, $c_1cos(omega t+phi_1)+c_2cos(omega t+phi_2)=Re(c_1e^(omega t+phi_1)i+c_2e^(omega t+phi_2)i)$. We can factor out $ae^omega t i$, and we get $Re(e^omega t i(c_1e^phi_1i+c_2e^phi_2i))$. This shows that when we're dealing with two signals of the same frequency, we can factor out the time-dependent part, leaving each signal as being characterized by a constant term, and of course when taking the Fourier transform, constant terms can be factored out. We can further note that $c e^phi i$ can be interpreted as a vector in the complex plane where the magnitude is $c$ and the angle is given by $phi$. And we can perform the addition in this vector space: the vector representing the sum is the sum of the vectors representing the summands.



                  So while both signals affect the magnitude of the output, an additional signal will not affect where in phase space the output is.






                  share|improve this answer









                  $endgroup$

















                    1












                    1








                    1





                    $begingroup$

                    It might be slightly clearer to write $ccos(omega t+phi)$ as $Re(ce^(omega t+phi)i)$. Then, since $Re$ distributes over addition, $c_1cos(omega t+phi_1)+c_2cos(omega t+phi_2)=Re(c_1e^(omega t+phi_1)i+c_2e^(omega t+phi_2)i)$. We can factor out $ae^omega t i$, and we get $Re(e^omega t i(c_1e^phi_1i+c_2e^phi_2i))$. This shows that when we're dealing with two signals of the same frequency, we can factor out the time-dependent part, leaving each signal as being characterized by a constant term, and of course when taking the Fourier transform, constant terms can be factored out. We can further note that $c e^phi i$ can be interpreted as a vector in the complex plane where the magnitude is $c$ and the angle is given by $phi$. And we can perform the addition in this vector space: the vector representing the sum is the sum of the vectors representing the summands.



                    So while both signals affect the magnitude of the output, an additional signal will not affect where in phase space the output is.






                    share|improve this answer









                    $endgroup$



                    It might be slightly clearer to write $ccos(omega t+phi)$ as $Re(ce^(omega t+phi)i)$. Then, since $Re$ distributes over addition, $c_1cos(omega t+phi_1)+c_2cos(omega t+phi_2)=Re(c_1e^(omega t+phi_1)i+c_2e^(omega t+phi_2)i)$. We can factor out $ae^omega t i$, and we get $Re(e^omega t i(c_1e^phi_1i+c_2e^phi_2i))$. This shows that when we're dealing with two signals of the same frequency, we can factor out the time-dependent part, leaving each signal as being characterized by a constant term, and of course when taking the Fourier transform, constant terms can be factored out. We can further note that $c e^phi i$ can be interpreted as a vector in the complex plane where the magnitude is $c$ and the angle is given by $phi$. And we can perform the addition in this vector space: the vector representing the sum is the sum of the vectors representing the summands.



                    So while both signals affect the magnitude of the output, an additional signal will not affect where in phase space the output is.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Aug 5 at 20:26









                    AcccumulationAcccumulation

                    1111 bronze badge




                    1111 bronze badge
























                        1












                        $begingroup$

                        I would like to take the path of a geometrical version of the question, using sums of circles.



                        Sines and cosines are "just" the real and imaginary parts of cisoids, or complex exponentials (some references can be found at How do I explain a complex exponential intuitively?, 3D wiggle plot for an analytic signal: Heyser corkscrew/spiral, Fourier Transform Identities).



                        If you take $s_omega,phi(t) = e^2pi i (omega t+phi)$, then $mathrmRe(s_omega,0(t))=cos(2pi omega t)$, or $mathrmIm(s_omega,pi/2(t))=cos(2pi omega t)$, and you can do a lot of combinations. The advantage of a cisoid is that it better uses the 2D space, as it can be depicted as a circle (a wheel) on which a point moves at different speeds driven by $omega$. A sum of "frequencies with different amplitudes" can be represented at "sums of spinning wheels" (borrowed from harmonic circles, or the Fourier Series Animation) with different radii and speeds, as depicted here:



                        Harmonic circles



                        Going back to a sum of two harmonics at the same frequency, the problem reads as: can we
                        separate or measure the combination:
                        $$a_1 s_omega,phi_1(t) +a_2 s_omega,phi_2(t),?$$



                        Constants $a_1$ and $a_2$ could be complex, so let us simplify the problem a bit before. Since Fourier has shift invariance properties, we can factorize either $e^2 pi iphi_1$ or $e^2 pi iphi_2$, and keep only one phase difference. We can also factorize an amplitude (the biggest for instance), and reduce the question to the behavior of the simplified problem:



                        $$ s_omega,0(t) +a s_omega,phi(t),, $$



                        with $|a|<1$. This simplification can be written as:



                        $$e^2pi i (omega t)+ae^2pi i (omega t+phi)labelatag1$$



                        and thus as:



                        $$(1+ae^2pi iphi)e^2pi i (omega t),,labelbtag2$$



                        which is another harmonic component with same frequency, but a different phase and amplitude.
                        The complex number $(1+ae^2pi iphi)$ could be rewritten as $alpha e^2pi ivarphi$, with trigonometric rules as detailed by @Fat32 (which I could detail later if needed). Now, let us geometrize the intuition. The unit circle is the motion of a point (say the tip of the valve) on a running bicycle wheel. The $a$-radius circle is like a small spinning wheel attached to the valve (like the blue and red circles only from the picture above). An now, we look at the motion of a dot on the perimeter of the small wheel.



                        What does your question ask: if the angular rotation of the small an the big wheel are the same, you cannot tell whether the motion of the dot results from the combination of the motion of two wheels of radii $1$ and $a$ (with some initial angle) or from a single bigger wheel (of radius $alpha$), with some other starting angle. This is what is mean by $refa$ and $refb$.



                        In other words, neither a Fourier transform, nor a human eye, can distinguish components with the same frequency but different phase.



                        [[I'll add animations if I find the time]]






                        share|improve this answer











                        $endgroup$



















                          1












                          $begingroup$

                          I would like to take the path of a geometrical version of the question, using sums of circles.



                          Sines and cosines are "just" the real and imaginary parts of cisoids, or complex exponentials (some references can be found at How do I explain a complex exponential intuitively?, 3D wiggle plot for an analytic signal: Heyser corkscrew/spiral, Fourier Transform Identities).



                          If you take $s_omega,phi(t) = e^2pi i (omega t+phi)$, then $mathrmRe(s_omega,0(t))=cos(2pi omega t)$, or $mathrmIm(s_omega,pi/2(t))=cos(2pi omega t)$, and you can do a lot of combinations. The advantage of a cisoid is that it better uses the 2D space, as it can be depicted as a circle (a wheel) on which a point moves at different speeds driven by $omega$. A sum of "frequencies with different amplitudes" can be represented at "sums of spinning wheels" (borrowed from harmonic circles, or the Fourier Series Animation) with different radii and speeds, as depicted here:



                          Harmonic circles



                          Going back to a sum of two harmonics at the same frequency, the problem reads as: can we
                          separate or measure the combination:
                          $$a_1 s_omega,phi_1(t) +a_2 s_omega,phi_2(t),?$$



                          Constants $a_1$ and $a_2$ could be complex, so let us simplify the problem a bit before. Since Fourier has shift invariance properties, we can factorize either $e^2 pi iphi_1$ or $e^2 pi iphi_2$, and keep only one phase difference. We can also factorize an amplitude (the biggest for instance), and reduce the question to the behavior of the simplified problem:



                          $$ s_omega,0(t) +a s_omega,phi(t),, $$



                          with $|a|<1$. This simplification can be written as:



                          $$e^2pi i (omega t)+ae^2pi i (omega t+phi)labelatag1$$



                          and thus as:



                          $$(1+ae^2pi iphi)e^2pi i (omega t),,labelbtag2$$



                          which is another harmonic component with same frequency, but a different phase and amplitude.
                          The complex number $(1+ae^2pi iphi)$ could be rewritten as $alpha e^2pi ivarphi$, with trigonometric rules as detailed by @Fat32 (which I could detail later if needed). Now, let us geometrize the intuition. The unit circle is the motion of a point (say the tip of the valve) on a running bicycle wheel. The $a$-radius circle is like a small spinning wheel attached to the valve (like the blue and red circles only from the picture above). An now, we look at the motion of a dot on the perimeter of the small wheel.



                          What does your question ask: if the angular rotation of the small an the big wheel are the same, you cannot tell whether the motion of the dot results from the combination of the motion of two wheels of radii $1$ and $a$ (with some initial angle) or from a single bigger wheel (of radius $alpha$), with some other starting angle. This is what is mean by $refa$ and $refb$.



                          In other words, neither a Fourier transform, nor a human eye, can distinguish components with the same frequency but different phase.



                          [[I'll add animations if I find the time]]






                          share|improve this answer











                          $endgroup$

















                            1












                            1








                            1





                            $begingroup$

                            I would like to take the path of a geometrical version of the question, using sums of circles.



                            Sines and cosines are "just" the real and imaginary parts of cisoids, or complex exponentials (some references can be found at How do I explain a complex exponential intuitively?, 3D wiggle plot for an analytic signal: Heyser corkscrew/spiral, Fourier Transform Identities).



                            If you take $s_omega,phi(t) = e^2pi i (omega t+phi)$, then $mathrmRe(s_omega,0(t))=cos(2pi omega t)$, or $mathrmIm(s_omega,pi/2(t))=cos(2pi omega t)$, and you can do a lot of combinations. The advantage of a cisoid is that it better uses the 2D space, as it can be depicted as a circle (a wheel) on which a point moves at different speeds driven by $omega$. A sum of "frequencies with different amplitudes" can be represented at "sums of spinning wheels" (borrowed from harmonic circles, or the Fourier Series Animation) with different radii and speeds, as depicted here:



                            Harmonic circles



                            Going back to a sum of two harmonics at the same frequency, the problem reads as: can we
                            separate or measure the combination:
                            $$a_1 s_omega,phi_1(t) +a_2 s_omega,phi_2(t),?$$



                            Constants $a_1$ and $a_2$ could be complex, so let us simplify the problem a bit before. Since Fourier has shift invariance properties, we can factorize either $e^2 pi iphi_1$ or $e^2 pi iphi_2$, and keep only one phase difference. We can also factorize an amplitude (the biggest for instance), and reduce the question to the behavior of the simplified problem:



                            $$ s_omega,0(t) +a s_omega,phi(t),, $$



                            with $|a|<1$. This simplification can be written as:



                            $$e^2pi i (omega t)+ae^2pi i (omega t+phi)labelatag1$$



                            and thus as:



                            $$(1+ae^2pi iphi)e^2pi i (omega t),,labelbtag2$$



                            which is another harmonic component with same frequency, but a different phase and amplitude.
                            The complex number $(1+ae^2pi iphi)$ could be rewritten as $alpha e^2pi ivarphi$, with trigonometric rules as detailed by @Fat32 (which I could detail later if needed). Now, let us geometrize the intuition. The unit circle is the motion of a point (say the tip of the valve) on a running bicycle wheel. The $a$-radius circle is like a small spinning wheel attached to the valve (like the blue and red circles only from the picture above). An now, we look at the motion of a dot on the perimeter of the small wheel.



                            What does your question ask: if the angular rotation of the small an the big wheel are the same, you cannot tell whether the motion of the dot results from the combination of the motion of two wheels of radii $1$ and $a$ (with some initial angle) or from a single bigger wheel (of radius $alpha$), with some other starting angle. This is what is mean by $refa$ and $refb$.



                            In other words, neither a Fourier transform, nor a human eye, can distinguish components with the same frequency but different phase.



                            [[I'll add animations if I find the time]]






                            share|improve this answer











                            $endgroup$



                            I would like to take the path of a geometrical version of the question, using sums of circles.



                            Sines and cosines are "just" the real and imaginary parts of cisoids, or complex exponentials (some references can be found at How do I explain a complex exponential intuitively?, 3D wiggle plot for an analytic signal: Heyser corkscrew/spiral, Fourier Transform Identities).



                            If you take $s_omega,phi(t) = e^2pi i (omega t+phi)$, then $mathrmRe(s_omega,0(t))=cos(2pi omega t)$, or $mathrmIm(s_omega,pi/2(t))=cos(2pi omega t)$, and you can do a lot of combinations. The advantage of a cisoid is that it better uses the 2D space, as it can be depicted as a circle (a wheel) on which a point moves at different speeds driven by $omega$. A sum of "frequencies with different amplitudes" can be represented at "sums of spinning wheels" (borrowed from harmonic circles, or the Fourier Series Animation) with different radii and speeds, as depicted here:



                            Harmonic circles



                            Going back to a sum of two harmonics at the same frequency, the problem reads as: can we
                            separate or measure the combination:
                            $$a_1 s_omega,phi_1(t) +a_2 s_omega,phi_2(t),?$$



                            Constants $a_1$ and $a_2$ could be complex, so let us simplify the problem a bit before. Since Fourier has shift invariance properties, we can factorize either $e^2 pi iphi_1$ or $e^2 pi iphi_2$, and keep only one phase difference. We can also factorize an amplitude (the biggest for instance), and reduce the question to the behavior of the simplified problem:



                            $$ s_omega,0(t) +a s_omega,phi(t),, $$



                            with $|a|<1$. This simplification can be written as:



                            $$e^2pi i (omega t)+ae^2pi i (omega t+phi)labelatag1$$



                            and thus as:



                            $$(1+ae^2pi iphi)e^2pi i (omega t),,labelbtag2$$



                            which is another harmonic component with same frequency, but a different phase and amplitude.
                            The complex number $(1+ae^2pi iphi)$ could be rewritten as $alpha e^2pi ivarphi$, with trigonometric rules as detailed by @Fat32 (which I could detail later if needed). Now, let us geometrize the intuition. The unit circle is the motion of a point (say the tip of the valve) on a running bicycle wheel. The $a$-radius circle is like a small spinning wheel attached to the valve (like the blue and red circles only from the picture above). An now, we look at the motion of a dot on the perimeter of the small wheel.



                            What does your question ask: if the angular rotation of the small an the big wheel are the same, you cannot tell whether the motion of the dot results from the combination of the motion of two wheels of radii $1$ and $a$ (with some initial angle) or from a single bigger wheel (of radius $alpha$), with some other starting angle. This is what is mean by $refa$ and $refb$.



                            In other words, neither a Fourier transform, nor a human eye, can distinguish components with the same frequency but different phase.



                            [[I'll add animations if I find the time]]







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Aug 9 at 10:37

























                            answered Aug 8 at 22:19









                            Laurent DuvalLaurent Duval

                            18.1k3 gold badges21 silver badges70 bronze badges




                            18.1k3 gold badges21 silver badges70 bronze badges






























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