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One technique to make a high pass filter is to subtract a lowpass filter from a wire (meaning subtract the output of the filter from the input as I have demonstrated in this post).
What lowpass can you subtract from a wire that will result in an all-pass? Specifically pass all frequencies with unity gain and modify the phase only, such that the group delay is 1 sample?
This is a “DSP Puzzle”, please preface your answer with spoiler notation by typing ythe following two characters first ">!"
Not only a lowpass filter but also a highpass filter is possible: $$\delta[n]-h[n]=\pm\delta[n-1]$$ which implies $$h[n]=\delta[n]\pm\delta[n-1]\tag{1}$$ where the positive sign in $(1)$ results in a simple first order FIR lowpass filter, and the negative sign corresponds to a highpass filter. Note that both filters, when subtracted from a wire, result in an allpass filter with a group delay of one sample. Of course, the phase responses are different.
Let APF $A(z)$ be $$ A(z) = \frac{z^{-N} + a_1z^{1-N} + ... + a_{N-1}z^{-1} + a_N}{1 + a_1z^{-1} + ... + a_{N-1}z^{1-N} + a_Nz^{-N}} $$ Then define your complementary filter as $$\begin{align} H(z) &= 1 - A(z) \\ \\ &= 1 - \frac{z^{-N} + a_1z^{1-N} + ... + a_{N-1}z^{-1} + a_N}{1 + a_1z^{-1} + ... + a_{N-1}z^{1-N} + a_Nz^{-N}} \\\\&= \frac{1 + a_1z^{-1} + ... + a_{N-1}z^{1-N} + a_Nz^{-N}}{1 + a_1z^{-1} + ... + a_{N-1}z^{1-N} + a_Nz^{-N}} - \frac{z^{-N} + a_1z^{1-N} + ... + a_{N-1}z^{-1} + a_N}{1 + a_1z^{-1} + ... + a_{N-1}z^{1-N} + a_Nz^{-N}} \\ \\&= \frac{1-a_N + (a_1-a_{N-1})z^{-1} + ... + (a_{N-1}-a_1)z^{1-N} + (a_N-1)z^{-N}}{1 + a_1z^{-1} + ... + a_{N-1}z^{1-N} + a_Nz^{-N}} \\\end{align}$$
Let's consider $H[k]$ is the frequency response(DFT) of the low pass filter that is of interest - satisfies the conditions mentioned in the problem statement.
Let $x[n]$ represent the original signal and $x_{f}[n]$ be the output signal obtained by passing the signal $x[n]$ through the filter $H[k]$.Let $X[k]$ represent the N-point DFT of $x[n]$, then $$x[n] = \frac{1}{N}\sum_{k=0}^{N-1}X[k]. e^\frac{j2\pi nk}{N}$$ The filtered signal can be expressed as$$x_f[n]= \frac{1}{N}\sum_{k=0}^{N-1}H[k]X[k]. e^\frac{j2\pi nk}{N} $$ From the problem statement the following relation can be established. Let $y[n]$ represent the signal obtained by subtracting the filtered signal $x_f[n]$ from wire. Then,$$y[n] = x[n] - x_{f}[n] = \frac{1}{N}\sum_{k=0}^{N-1}(1-H[k]).X[k]. e^\frac{j2\pi nk}{N}$$ The resulting signal has a constant group delay equal to $1$ and the gain of all frequencies is unity. Hence $$|1-H[k]| = 1$$ and $$\angle{(1-H[k])} = -\frac{2\pi}{N}k $$
Two trivial solutions for $H[k]$ satisfying the above conditions are as given below:$$H[k]= 1 + e^\frac{-j2\pi k}{N} \; and \; H[k] = 1 - e^\frac{-j2\pi k}{N}$$Applying inverse Fourier transforms to the trivial solutions gives the filter response as below$$h[n] = \delta[n] + \delta[n - 1] \; and\; h[n] = \delta[n] - \delta[n - 1] $$The result is same as what Matt already has posted but the derivation is from frequency domain approach.
