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lets assume following signal $$x=a_1 \sin(2\pi f_1 t) + a_2 \sin(2\pi f_2 t).$$ This is passed through a simple single-pole filter with cut-off frequency $f_c$. How is $x$ delayed by the filter (peak of the cross-correlation)?

In the case $a_2=0$, the time delay between the original and the filtered frequency is $$T_1=\arctan(f_1/f_c)/(2\pi f_1).$$

In the case $a_1=0$, the time delay between the original and the filtered frequency is $$T_2=\arctan(f_2/f_c)/(2\pi f_2).$$

But how can we calculate the time delay $T_{12}$ in the case $a_1\neq0$ and $a_2\neq0$?

It is not the group delay: It must depend in some way on the values of $a_1$ and $a_2$, because if $a_1>>a_2$, then $T_{12}\approx T_1$.

Any light on this would be greatly appreciated!

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The answer is that the output signal for the case $a_1\neq 0$ and $a_2\neq 0$ is not simply a delayed version of the input signal anymore, as it would be if the input were just a single sinusoid. With $T_1$ and $T_2$ as given in your question, the response to the combined input signal is

$$y(t)=\frac{a_1}{\sqrt{1+(f_1/f_c)^2}}\sin(2\pi f_1(t-T_1))+\frac{a_2}{\sqrt{1+(f_2/f_c)^2}}\sin(2\pi f_2(t-T_2))$$

which cannot be written as $x(t-T_x)$ with some unknown delay $T_x$.

But as you suggested, you could define some 'global delay' by the first maximum of the cross-correlation between the input signal and the output signal. This cross-correlation is

$$R_{xy}(\tau)=\frac{a_1^2}{2\sqrt{1+(f_1/f_c)^2}}\cos(2\pi f_1(\tau-T_1))+\\+ \frac{a_2^2}{2\sqrt{1+(f_2/f_c)^2}}\cos(2\pi f_2(\tau-T_2))\tag{1}$$

For $a_2=0$, there is obviously a maximum of $R_{xy}(\tau)$ at $\tau=T_1$. The same is true for $a_1=0$, which gives $\tau=T_2$ as the location of the maximum. However, for $a_1\neq 0$ and $a_2\neq 0$ I don't see how to compute the first maximum of (1) analytically. A numerical solution is of course possible.

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  • $\begingroup$ Thanks Matt. But when we calculate the cross correlation, the peak will be somewhere between $T_1$ and $T_2$. Is there no analytic solution to find that peak? Btw, shouldn't the new amplitudes be $1/\sqrt(1+f_1^2/f_c^2)$ ? $\endgroup$ – Ankaios Argo Aug 16 '14 at 22:27
  • $\begingroup$ @AnkaiosArgo: Thanks for pointing out the magnitudes, I was just focusing on the delays. Fixed it. I'll need to think about what you say about the cross-correlation. $\endgroup$ – Matt L. Aug 16 '14 at 22:33
  • $\begingroup$ Sorry, I didn't want to be picky :-) Would be great, if you find a solution... In order to find the peak in the cross-correlation, maybe we can calculate the cross-spectra? $\endgroup$ – Ankaios Argo Aug 16 '14 at 22:43
  • $\begingroup$ @AnkaiosArgo: I added some info about cross-correlation to my answer. Unfortunately, I can't see how to find an analytical solution. $\endgroup$ – Matt L. Aug 17 '14 at 0:10
  • $\begingroup$ Thanks, this already helps... would you mind to elaborate, how you determined the $R_{xy}(\tau)$? The auto-correlation of a sum of sinus is a sum of cosines, see e.g. chem.purdue.edu/courses/chm621/lecture/electronics/… , but in the case of the cross-correlation and the time-shift $T_i$ it does not seems very trivial to me. $\endgroup$ – Ankaios Argo Aug 17 '14 at 15:15

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