# How to separate a low amplitude signal of same frequency but different phase

I have a signal consisting of two sine waves of same frequency but have different phase (and also a lot of HF noise). Their amplitudes differ by about $120\textrm{ dB}$. How to I go about separating them? SNR is about $130\textrm{ db}$ but its frequency is far higher than my signal, i.e. $f(\textrm{noise}) \gg f(\textrm{signal})$.

Any special technique? Also I am really just a beginner, so extra points for those who give a detailed explanation.

• Are these signals acquired from some kind of real hardware, or are they generated in a simulation? In many cases I would be surprised for your equipment to have >120 dB of dynamic range, meaning your weaker signal of interest might be dwarfed by the system's noise floor. – Jason R Feb 20 '14 at 13:38
• Yeah generated in simulation. No hardware – Andy Stow Away Feb 20 '14 at 13:47
• do you have one of the signals or the phase information? either the real phase of each signal or the phase difference with your dB-difference should suffice to calculate the two signals. – user7981 Feb 20 '14 at 21:38
• You can try Blind Source Separation Technique . – user22624 Jun 24 '16 at 9:18

$E_1=E_{10}sin (\omega t)$

$E_2=E_{20}sin (\omega t + \delta)$

$E_{\theta} = E_1 + E_2 = E_{\theta0}sin (\omega t + \phi )$

This can be described as the figure below:

Now given the $E_{\theta 0}$, you can rotate $E_1$ and $E_2$ arbitrarily as long as $E_1$ , $E_2$, and $E_{\theta 0}$ form the triangular. As a result, your given sine wave can be decomposed into an infinite numbers of pairs of sine waves of the same frequency, which is not necessarily the two original ones.

• Shouldnt E10 be the magnitude of the phasor and not the projection on the vertical axis? same for E20. – Andy Stow Away Feb 21 '14 at 14:27
• correct. Thanks for pointing out. I removed that label – lennon310 Feb 21 '14 at 14:30

Two exact sine waves of the same frequency but different phases sum to another exact sine wave. That resultant sine wave can be decomposed into an infinite numbers of pairs (or any other greater number) of sine waves of the same frequency, not just the two original ones. Thus more information is required to reduce the possible solution space below infinite.

There are many ways to add two sine waves of the same frequency and different phase and get back a sine wave. These cant be separated without more information.

The noise seems to be irrelevant to this problem since you say that it is a frequency band far removed from the sinusoids of interest. Thus, with a relatively simple bandpass filter, you should be able to recover a signal $$s(t) = C\cos(2\pi f_c t + \theta) = A\cos(2\pi f_ct + \theta_1) + B\cos(2\pi f_ct + \theta_2), \tag{1}$$ where $C$ and $\theta$ are presumably easily measured. Assume that all the amplitude parameters in $(1)$ have positive value (negative values having been absorbed into the phase terms). Now, all we know about $A$ and $B$ is that $A \gg B$, in fact, $A \approx 10^6B$ or $A \approx 10^{12}B$ depending on whether that $120$ dB that you specify is a voltage ratio or a power ratio. Let's just say that $A \approx 10^nB$ for some largish $n$. Now, apply @lennon310's phasor diagram with the additional knowledge that $A \approx C$ and $\theta_1 \approx \theta$.

• The maximum value of $A$ is $C(1+10^{-n})$ and this occurs when $\theta_1=\theta$, $\theta_2 = \theta+\pi$, that is, $(1)$ reduces to $$s(t) = C\cos(2\pi f_c t + \theta) = C(1+10^{-n})\cos(2\pi f_ct + \theta) - C\cdot10^{-n}\cos(2\pi f_c t+\theta).$$

• The minimum value of $A$ is $C(1-10^{-n})$ and this occurs when $\theta_1= \theta_2 = \theta$, that is, $(1)$ reduces to $$s(t) = C\cos(2\pi f_c t + \theta) = C(1-10^{-n})\cos(2\pi f_ct + \theta) + C\cdot10^{-n}\cos(2\pi f_c t+\theta).$$

More pictorially, lennon310's $E_{\theta 0}$ and $E_1$ are very nearly the same phasor, with the tip of the arrowhead of $E_1$ lying inside a small circle centered at the tip of the arrowhead of $E_{\theta 0}$. The special cases above occur when all three phasors are collinear with $E_2$ reducing (or increasing) the amplitude of $E_1$ by a very small amount. But, as hotpaw2 and Aaron have already pointed out to you, we cannot say much more that this about the relative amplitudes and phases of the two sinusoids that sum to $C\cos(2\pi f_ct+\theta)$.

It is impossible separate two sinusoidal, as all previous answer pointed out. But if you know precise parameters of first signal you can estimate parameters of second signal. If it is possible you can try. Your signal:

$y(t)=A\exp(i(\omega t+\phi_0))+ b\exp(i(\omega t + \phi_1)) + n(t)$, and you know A, $\omega$ and $\phi_0$

Calculate:

\begin{align} C(t) &= y(t)\exp(-i(\omega t+\phi_0)) \\ &= A\exp[i(\omega t+\phi_0)-i(\omega t+\phi_0)] +b\exp[i(\omega t +\phi_1)-i(\omega t+\phi_0)] + n(t)\exp(-i(\omega t+\phi_0)) \\ &= A + b\exp(i(\phi_1- \phi_0)) + n(t)\exp(-i(\omega t+\phi_0)) \end{align}

For eliminating noise try to average over $k*2 \pi/\omega$

$$<C(t)> = A+b\exp(i(\phi_1-\phi_0))$$

As you know A and $\phi_0$ you can calculate $b$ and $\phi_1$. But you must know A and $\phi_0$ with very high precision.