Question

HₗₚF[n] is the impulse response of a low-pass filter for discrete time signals with frequency response HₗₚF

(eʲω). Show that a time-discrete filter with the impulse response h[n]=(−1)ⁿ hₗₚF
​[n] is a high-pass filter with the freq response H(eʲω)=HₗₚF(eʲ⁽ω−π⁾)

Answer 1

The modulation of a low-pass filter's impulse response by (-1)^n transforms it into a high-pass filter due to a phase shift of π in the frequency domain, thus reversing the frequency spectrum.

Step-by-step explanation:

Signal processing in the realm of discrete-time systems, where the impulse response of a filter defines its properties. Given a low-pass filter with an impulse response h₁ₚF[n], the transformation into a high-pass filter is achieved by modulating the impulse response with (-1)ⁿ resulting in the new impulse response h[n]. This modulation introduces a shift in the frequency response by π in the frequency domain, changing the characteristics from low-pass to high-pass.

To show this formally, consider the frequency response H₁ₚF (eᵗω) of the original low-pass filter. When the impulse response is multiplied by (-1)ⁿ, it yields a new frequency response H(eᵗω) = H₁ₚF(eᵗ(ω-π)), which can be interpreted as a high-pass filter. This is because the modulation in time domain corresponds to a phase shift in the frequency domain, which basically reverses the spectrum, hence transforming a low-pass filter into a high-pass filter.