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I'm learning about filters and obviously the first kind of filters I came across were low pass and high pass filters:
$$\frac{1}{2}(1+z^{-1})$$ is a low pass filter and $$\frac{1}{2}(1-z^{-1})$$ is a high pass filter
which got me thinking are high order low pass and high pass fir filters possible?
Yes, you can design FIR filter of any finite order.
You can re-write the examples by multiplying Z in the numerator and denominator to get:
$$\frac{1}{2}\frac{(z+1)}{z}$$
Where it is clearer that you have only one zero and one pole. Therefore, this filter is of first order. In stable and causal FIR filters the order is given by the number of poles, and these poles are always located at the origin. You may add as many poles as you wish and get any order. And you may also locate as many zeros as you can (for causal filter the number of zeros should be less-equal to the number of poles) in any arbitrary position.
The FIR filter design theory explains where exactly these zeroes should be placed to obtain certain desirable features such as lineal phase, real impulse response, or to actually have a low, high, band-pass or band-reject filter.
