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!!!
Here is a demonstration that confirms what @Hilmar said:
We recall the initial relationship: \begin{equation} s_{hp}[n]=(−1)^ns_{lp}[n] \end{equation}
Let us calculate its discrete Fourier transform:
\begin{equation} DFT\{s_{hp}[n]\}=S_{hp}[k]=\sum_{n=0}^{N-1}\{(−1)^ns_{lp}[n]\}e^{-j2\pi \frac{nk}{N}} \end{equation}
But $(-1)^n$ is also equal to $e^{j\pi n}$, so: \begin{equation} S_{hp}[k]=\sum_{n=0}^{N-1}\{(e^{j\pi n})s_{lp}[n]\}e^{-j2\pi \frac{nk}{N}}\\ =\sum_{n=0}^{N-1}s_{lp}[n]e^{j\pi n}e^{-j2\pi \frac{nk}{N}}\\ =\sum_{n=0}^{N-1}s_{lp}[n]e^{j2\pi \frac{n}{N}(k-\frac{N}{2})} \end{equation}
And so with a change of variable, we get:
\begin{equation} \sum_{n=0}^{N-1}s_{lp}[n]e^{-j2\pi n\frac{k'}{N}}=S_{hp}[k']=S_{hp}[k-N/2] \end{equation}
As a conclusion, we have the properties stated by @himar:
What happens here is that SHIFT your signal frequency spectrum by half the sample rate 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.
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.
