Given the transfer function
$$H(z) = \frac{1}{4}(1+z^{-1}+z^{-2}+z^{-3})$$
I understand I have to switch $z$ to $e^{j\Omega}$, $0\le\Omega\le\pi$, in order to find out how it interacts with frequencies.
$$H(\Omega)=\frac{1}{4}(1+e^{-j\Omega}-e^{-j2\Omega}-e^{-j3\Omega})$$
And now the amplitude function, $|H(\Omega)|$, would tell us how much of each frequency there would be in our output signal.
However, is it possible to see this is a low-pass filter without the help of a computer or tedious calculations?
For $z=1$ (the zero frequency) we have $|H(1)|=1$ which is clearly the largest possible amplitude since $|H(z)|\le 1$. For other $z$ on the unit circle we can calculate the geometric sum $$ H(z)=\frac14\cdot\frac{1-z^{-4}}{1-z^{-1}}. $$ Denoting $z^{-1}=\cos t+i\sin t$ we get $$ |H(z)|^2=\frac{1}{16}\cdot\frac{1-\cos 4t}{1-\cos t}. $$ As $t$ grows to $\pi$ the numerator oscillates and stays bounded while the denominator grows, so the latter is damping more and more at higher frequencies.
Intuition: in the time domain the filter is acting very simple - it averages the last four input samples. The average of low frequency components remains almost unchanged while the high frequency ones almost cancel each other.
P.S. The filter is called the moving average filter. It is one of the simplest low pass filters that can be realized as a FIR filter.