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?

  • 6
    $\begingroup$ Can you distinguish the components of, say $\sin(x) + \sin(x+c)$? I bet you can't. $\endgroup$ Commented Aug 5, 2019 at 9:14
  • 1
    $\begingroup$ 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. $\endgroup$ Commented Aug 6, 2019 at 20:09

5 Answers 5


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, by trigonometric 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 $$ \Phi = \tan^{-1}\left(\frac{ a \sin(\phi) + b\sin(\theta) } { a \cos(\phi) + b\cos(\theta) } \right) $$

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

  • 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
    Commented Aug 5, 2019 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. $\endgroup$ Commented Aug 5, 2019 at 7:08
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Aug 5, 2019 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
    Commented Aug 5, 2019 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
    Commented Aug 6, 2019 at 15:14

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.


It might be slightly clearer to write $c\cos(\omega t+\phi)$ as $Re(ce^{(\omega t+\phi)i})$. Then, since $Re$ distributes over addition, $c_1\cos(\omega t+\phi_1)+c_2\cos(\omega t+\phi_2)=Re(c_1e^{(\omega t+\phi_1)i}+c_2e^{(\omega t+\phi_2)i})$. 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.


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 $s_{\omega,\phi}(t) = e^{2\pi i (\omega t+\phi)}$, then $\mathrm{Re}(s_{\omega,0}(t))=\cos(2\pi \omega t)$, or $\mathrm{Im}(s_{\omega,\pi/2}(t))=\cos(2\pi \omega t)$, 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 $e^{2 \pi i\phi_1}$ or $e^{2 \pi i\phi_2}$, 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:

$$e^{2\pi i (\omega t)}+ae^{2\pi i (\omega t+\phi)}\label{a}\tag{1}$$

and thus as:

$$(1+ae^{2\pi i\phi})e^{2\pi i (\omega t)}\,,\label{b}\tag{2}$$

which is another harmonic component with same frequency, but a different phase and amplitude. The complex number $(1+ae^{2\pi i\phi})$ could be rewritten as $\alpha e^{2\pi i\varphi}$, 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 $\ref{a}$ and $\ref{b}$.

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]]


If you change the phase at a constant frequency, FFT calculates a different result. Unfortunately, it is common practice to erase the phase by forming and displaying the magnitude (see line 7 in the following program). The phase can be calculated with line 8.

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.