Converting a lowpass filter to a highpass filter. FIR filter-type 1

Question

I was told that in the FIR filter type I, the impulse response
of a highpass filter $h_H[n] $can be obtained as follows,
$$h_H[n] = delta[n − m] − h_L[n]$$
where $h_L[n]$ is the impulse response of a lowpass filter, $delta$ is a Dirac pulse (Dirac delta function), which delayed by an appropriate value of $m$.
I've computed the impulse response according to the formula in MATLAB and observed the frequency response. I noticed that when $m=frac{M+1}{2}$, where $M$ is the number of filter taps, the formula is indeed correct. However, I don't understand Why and How it possible.

Matt L. · Answer

If you have a real-valued amplitude function $A_{LP}(omega)$ of a low pass filter with unity gain in the passband, then it is easy to see that
$$A_{HP}(omega)=1-A_{LP}(omega)tag{1}$$
is the (real-valued) amplitude function of a high pass filter.
A type I length $N$ linear phase low pass filter's frequency response is given by
$$H_{LP}(omega)=A_{LP}(omega)e^{-jomega(N-1)/2}tag{2}$$
Consequently, according to $(1)$, the desired high pass amplitude function is obtained by computing
$$begin{align}H_{HP}(omega)&=big(1-A_{LP}(omega)big) e^{-jomega(N-1)/2}\&=e^{-jomega(N-1)/2}-H_{LP}(omega)tag{3}end{align}$$
The inverse discrete-time Fourier transform of $(3)$ is
$$h_{HP}[n]=deltaleft[n-frac{N-1}{2}right]-h_{LP}[n]tag{4}$$
Note that for type I linear phase filters the filter length $N$ is odd, so $(N-1)/2$ is an integer shift. Eq. $(4)$ shows that apart from sign inversion we only modify the center bin of the low pass filter in order to obtain a type I linear phase high pass filter.

Converting a lowpass filter to a highpass filter. FIR filter-type 1

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