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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.

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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

$$H_{LP}(\omega)=A_{LP}(\omega)e^{-j\omega(N-1)/2}\tag{2}$$

Consequently, according to $(1)$, the desired high pass amplitude function is obtained by computing

$$\begin{align}H_{HP}(\omega)&=\big(1-A_{LP}(\omega)\big) e^{-j\omega(N-1)/2}\\&=e^{-j\omega(N-1)/2}-H_{LP}(\omega)\tag{3}\end{align}$$

The inverse discrete-time Fourier transform of $(3)$ is

$$h_{HP}[n]=\delta\left[n-\frac{N-1}{2}\right]-h_{LP}[n]\tag{4}$$

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.

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