3-point highpass filter

Question

I came across an interesting question from Strang & Nguyen$^\color{magenta}{\dagger}$ where it asks to design a $3$-point (or 4-point) high-pass filter with better performance than the $2$-point moving difference filter, which have the impulse response $$ h(n) = 0.5\delta(n) - 0.5\delta(n-1)$$

Please see the question below.

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Invent a highpass filter $K$ with three or four taps (coefficients) that is better than the moving difference $H_1$: the goal is $$ | K (\omega) | < | H_1 (\omega) | \quad \text{ for } 0 < | \omega | < \frac{\pi}{2}$$ and $$ | H_1 (\omega) | < | K (\omega) | < 1 \quad \text{ for } \frac{\pi}{2} < | \omega| < \pi.$$

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I found a 4-point solution here, however I could not find a 3-point solution. I am also interested in the methodology of finding the solution.

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_{$\color{magenta}{\dagger}$ Gilbert Strang, Truong Nguyen, Wavelets and Filter Banks, Wellesley-Cambridge Press, Wellesley, 1997.}

Answer 1

Cannot be done as written, at least not with a real impulse response and the requirements as stated.

Let's call the 3 coefficients for the impulse response $[k_0,k_1,k_2]$ We assume that "3 points" implies that $k_0 \neq 0$ and $k_2 \neq 0$, otherwise it would be 2-point or 1-point solution

From the requirements it follows directly that

$$K(0) = 0 \tag{1}$$ $$K(\pi) = 1 \tag{2}$$ $$|K(\pi/2)| = \sqrt{0.5} \tag3 $$

We can derive the following equations from the z-Transform $H(\omega) = k_0 + e^{-j\omega}k_1 + e^{-2j\omega}k_2$

$$K(0) = k_0+k_1+k_2 = 0 \tag{4}$$ $$K(\pi) = k_0-k_1+k_2 = 1 \tag{5}$$ $$|K(\pi/2)| = |k_0-j\cdot k_1-k_2| = \sqrt{0.5} \tag{6}$$

From equations (4) and (5) we get $$k_0+k_2 = -k_1 \tag{7}$$ $$k_0+k_2 = 1 + k_1 \tag{8} $$

We can solve for $k_1$ as

$$k_1 = -0.5 \tag{9}$$

And from equation (1) we get

$$k_0 -.5 + k_2 = 0 \rightarrow k_2 = 0.5-k_0 \tag{10}$$

We can pop this into equation (3) $$|2\cdot k_0 - .5 -j0.5| = \sqrt{0.5}\tag{11}$$

Assuming a real impulse response, the absolute value of the imaginary part is 0.5. In order for the magnitude of the complex expression to be $\sqrt{0.5}$, the absolute value of the real part also has to be 0.5. We get

$$ 2\cdot k_0 - .5 = .5 \tag{12} $$

or $$ 2\cdot k_0 - .5 = -.5 \tag{13} $$

There are two solutions: from eq (12) we would get $k_{0,12} = 0.5$ and $k_{2,12} = 0 $. From eq (13) we get $k_{0,13} = 0$ and $k_{2,13} = 0.5$

Both solutions turn out to be "2-point" solutions: the original high pass and the original highpass delayed and phase flipped.

Answer 2

I think Hilmar's answer is the "correct" one to select (although to clarify the requirements cannot be met with any 2 coefficients real or complex). This post is to demonstrate why the attempted solution with complex coefficients does not work when all relevant frequencies are included.

Even with complex coefficients the requirement can not be met completely: with complex coefficients the frequency range must extend from $-\pi/2$ to $+\pi/2$ since the positive and negative frequencies will no longer be related as they are with real coefficients and the requirements are specified for both positive and negative frequencies given the $|\omega|$ in the formulation.

Using the OP's answer as an example, with $z_1=20e^{-j1.25\pi}$ and $z_0=1$ we would get the following plot when we review the full spectrum, compared to the original differencer:

$$K(z) = A(1- 20e^{-j1.25\pi}z)(1-z^{-1}) = A(1 - 20e^{-j1.25\pi}z^{-1} + 20e^{-j1.25\pi}z^{-2})$$

Where the scaling $A$ is determined by setting $|K(z)|=1$ when $z = -1$.

$$H(z) = 0.5 - 0.5z^{-1}$$

Answer 3

In this answer I give an alternative proof that a filter with $3$ taps cannot outperform the solution with two taps for the given design specifications.

A solution

$$H_3(e^{j\omega})=h_0+h_1e^{-j\omega}+h_2e^{-2j\omega}\tag{1}$$

with three (real-valued) taps must satisfy the constraints at DC and at Nyquist:

$$H_3(1)=0,\qquad H_3(-1)=1\tag{2}$$

leading to the equations

\begin{align*} h_0+h_1+h_2 &= 0\\ h_0-h_1+h_2 &= 1 \tag{3} \end{align*}

These equations leave one degree of freedom for the design of the filter with three taps:

$$H_3(e^{j\omega})=\frac12-h_2-\frac12 e^{-j\omega}+h_2e^{-2j\omega}\tag{4}$$

With the frequency response of the two-tap filter

$$H_2(e^{j\omega})=\frac12\left(1-e^{-j\omega}\right)=je^{-j\omega/2}\sin\omega/2\tag{5}$$

the frequency response of the three-tap filter $(4)$ can be written as

\begin{align*} H_3(e^{j\omega}) &= H_2(e^{j\omega})-h_2\left(1-e^{-2j\omega}\right)\\ &= H_2(e^{j\omega}) - 2je^{-j\omega}h_2\sin\omega \tag{6} \end{align*}

Its squared magnitude is

\begin{align*} \big|H_3(e^{j\omega})\big|^2 &= \big|H_2(e^{j\omega})\big|^2 - 2\textrm{Re}\left\{H_2(e^{j\omega})2(-j)e^{j\omega}h_2\sin\omega\right\} + 4h_2^2\sin^2\omega \\ & = \big|H_2(e^{j\omega})\big|^2 - 4h_2\textrm{Re}\left\{e^{j\omega/2}\sin\omega\sin\omega/2\right\} + 4h_2^2\sin^2\omega \\ &= \big|H_2(e^{j\omega})\big|^2 - 4h_2\cos\omega/2\sin\omega/2\sin\omega+ 4h_2^2\sin^2\omega \\ &= \big|H_2(e^{j\omega})\big|^2 - 2h_2\sin^2\omega+ 4h_2^2\sin^2\omega \\ &= \big|H_2(e^{j\omega})\big|^2 + 2h_2(2h_2-1)\sin^2\omega\tag{7} \end{align*}

We can see from $(7)$ that the difference term on the right-hand side is either positive or negative for all $\omega$, depending on the value of $h_2$, but it cannot change sign. However, for $H_3(e^{j\omega})$ to be superior to $H_2(e^{j\omega})$, its magnitude would need to be smaller than $|H_2(e^{j\omega})|$ for $0<\omega <\pi/2$, and it would need to be greater than $|H_2(e^{j\omega})|$ for $\pi/2<\omega<\pi$. But for this to be the case, the difference term in $(7)$ would need to change sign. This shows that it is impossible for a three-tap filter to satisfy the specifications better than a two-tap filter. It takes at least four taps to improve on the two-tap solution (see this question).

Answer 4

As @Dan stated, to get a 3-point filter the transfer function has the from

$$K(z) = A(z-z_0)(z-z_1)$$

In order to satisfy the requirements, we set the first zero $z_0 = 1$ and makes sure that $K(1) = 0$, i.e. a good highpass filter. As for the second zero, I tried with real values but couldn't get a performance meeting the requirements. So I turned to a complex zero (there is no room for a conjugate) hence we end up with a complex impulse response. After trying many candidates, I got the zero $z_1 = 20 e^{-j 1.25\pi}$. I got other good candidates, which I included in the code below.

enter code here
 clc
 clearvars
 %% 2 points HPF
 hn = [0.5, -0.5];
 [Hmag, omega] = freqResponse(hn);

 %% 3 points HPF
 z0 = -1;
 z1 = 20*exp(-1j*pi*1.9);
 % z1 = 0.05*exp(-1j*pi*1.25);
 % z1 = 20*exp(1j*pi*0.75);

 kn3 = conv([1 z0], [1 z1]);
 Kz3 = freqResponse(kn3);

 close all
 plot(omega/pi, Hmag)
 grid on, hold on
 plot(omega/pi, Kz3)

 xlabel('\omega/ \pi')
 ylabel('| K(\omega) |')
 legend('2 points', '3 points', 'location','northwest')

 % Helper function
 function [Hmag, omega] = freqResponse(hn)
 % Compute the frequency response magnitude over [-pi, pi].

 N = 200;
 [Hz, w] = freqz(hn,1,N,'whole');
 Hmag = abs(Hz);
 omega = w - pi;
 Hmag = [Hmag(omega>=0)' Hmag(omega<0)'];
 Hmag= Hmag/max(Hmag);

EDIT: As per the discussion by the community members, the above suggested filter does not meet the requirements in the negative frequency part. Still this is the closest perform ance I could get using a 3-point filter, which has complex values and as such the frequency response is not symmetric. I updated the code and the figure as well.