How to calculate the time delay of a signal by a first-order filter

Question

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!

Answer 1

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.