# The Amplitude of the Correlation Function Peaks After Matched Filtering

I have 2 signals: $$s_1(t)$$ and $$s_2(t)$$. The autocorrelation functions are given by $$R_{s_1}(\tau)$$ and $$R_{s_2}(\tau)$$ with $$|R_{s_2}(\tau)|<|R_{s_1}(\tau)|$$.

I would like to know if the previous inequality remains true after transmission in an identical channel for the 2 signals, i.e: $$r_1(t)=s_1(t)\ast h(t)+n(t)$$ and $$r_2(t)=s_2(t)\ast h(t)+n(t)$$. Do we have $$|R_{r_2}(\tau)|<|R_{r_1}(\tau)|$$. And if it's true how to prove it?

• Do we have some info about $h$?
– Royi
Commented Apr 17, 2022 at 19:59
– Royi
Commented May 30, 2023 at 7:29
• Where is the matched filtering that is mentioned in the title of the question? The text of your question refers to transmission of the signals through identical channels, but no filtering of any kind is being done at the receiver. And $h(t)$ cannot be the matched filter for both signals unless the only difference between the two signals is that $s_2 = \alpha s_1$ where $|\alpha|<1$. Commented Dec 5, 2023 at 15:38

Without any restrictions on $$h$$ you may build some very non real world cases.
Especially if we're talking about the correlation function without the removal of the DC component.

For instance, think of the case of $${s}_{2} \left( t \right)$$ is a gaussian function with its peak equals to 1.5. Now, $${s}_{1} \left( t \right)$$ to be a gaussian with a peak of 1 but with a DC of 2.

Make $$h$$ to be an High Pass Filter and you get that their auto correlation won't obey the proposition in your question.