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I'm wondering if I have calculated a nulling filter coefficients correctly:

$\ 1, -2\cos(0.44\pi), 1$

I have to make a nulling filter that filters out the frequencies $\theta = 0.44\pi$. I've been told that the formula for nulling filters is $y[n] = x[n] - 2\cos(\theta)*x[n-1]+x[n-2]$.

If I'm correct the two 1s come from the fact that $\ x[n]$ and $\ x[n-2]$ both have a coefficient of 1 and no other variable and the $\ -2\cos(0.44\pi)$ comes from the $\ - 2\cos(\theta)*x[n-1]$ and the fact that $\theta$ = 0.44$\pi$.

If the coefficients are correct, is all I have to do to find the output signal just multiply the input signal by each coefficient and add them together (convolution sum)?

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    $\begingroup$ an FIR with those three coefficients will null out a sinusoid with angular frequency of 0.44 $\pi$ radians per sample. this puts a pair of zeros on the unit circle at that angle. you could make a much sharper notch filter with those three coefficients in the numerator (we still need zeros there) and a pair of poles that are very close to the zeros. $\endgroup$ – robert bristow-johnson Dec 26 '16 at 22:43
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You are looking at an FIR filter whose transfer function is

$$ \begin{align} H(z)&=(1-e^{j\omega_0}z^{-1})(1-e^{-j\omega_0}z^{-1})\\[10pt] &=1-(e^{-j\omega_0}+e^{j\omega_0})z^{-1}+z^{-2}\\[10pt] &=1-2\cos(\omega_0)z^{-1}+z^{-2} \end{align}$$

So as you can also see in the comment, there are two zeros at $e^{\pm j\omega_0}$ that null the frequencies $\pm\omega_0$.

Obviously, here we have $\omega_0=0.44\pi$. The filter's response is

enter image description here

The impulse response is $$h[n]=\delta[n]-2\cos(0.44\pi)\delta[n-1]+\delta[n-2]$$

Assume an arbitrary input $x[n]$. The output $y[n]$ is given by:

$$y[n]=x[n]*h[n]$$ Therefore $$y[n]=x[n]-2\cos(0.44\pi)x[n-1]+x[n-2]$$

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