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I know that multiple a lowpass filter signal by $(-1)^n$ can transform it to a highpass filter. $$ s_{hp}[n]=(-1)^n s_{lp}[n] $$

But exactly what happened to the unit sample response and difference equation? Assume that the original lowpass filter signal has a difference equation. Can someone explain to me?? Thanks!!!

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I know that multiple a lowpass filter signal by (−1)n can transform it to a highpass filter.

Depends a bit what you mean by that. If you assume that this will turn 1 kHz lowpass signal into a 1kHz high pass signal, your assumption is plain wrong. What happens here is that SHIFT your signal frequency spectrum by half the sample rate, $f_s$. Due to the periodicity of the spectrum, this looks like the spectrum has been mirrored at $fs_/4$ (at least for real valued signals), i.e. DC become Nyquist, Nyquist becomes DC.

So this operation flips the entire signal spectrum, regardless of whether it's been high-pass, low-pass, pink, white, etc.

Loosely speaking, if you have signal that's been low passed and you flip it, the spectrum may look a bit like that of a high passed signal, but it's completely different operation than filtering the original signal with a complimentary high pass.

But exactly what happened to the unit sample response and difference equation? Assume that the original lowpass filter signal has a difference equation.

This is NOT an LTI operation, so difference equations and unit sample response do not apply here. Specifically, it's time variant.

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