# What is the role of a LPF in oversampling?

Lets assume that a signal $x(n)$ is up-sampled by adding 1 zero between two adjacent samples to form a signal $y(n)$. How does a digital LPF give an oversampled version $z(n)$, with more data points? (As show in the image below) I know that the question might seem stupid but the books that I am using don't really explain that part.

Effects of an ideal low pass filter (LPF) on the output of an upsampling interpolator can be described in both of the domains of time and frequency. I believe, a frequency domain explanation provides a conceptually more solid understanding of the role played by the LPF. Whereas the time domain operation explicitly defines the generation of the output interpolated samples $z[n]$.
For the frequency domain approach, the key observation is that, the DTFT $Y(e^{j \omega})$ of the expanded sequence $y[n]$ of the middle figure, includes what is called as the frequency-scaled images $X(e^{j\omega M})$ of the original DTFT spectrum $X(e^{j\omega})$ of the input sequence $x[n]$ at locations $\omega = 2\pi k/M$, where $M$ is the oversampling factor and $k=0,1,...,M-1$.
The LPF whose passband includes only one copy of the input spectrum at the baseband, removes all the remaning surplus images and retains only the baseband spectrum, which happens to create the signal $z[n]$ in time domain by the mechanics of convolution operation of the filter's impulse reponse and signal $y[n]$. The cutoff radian frequency $\omega_c$ of the ideal LPF depends on the oversampling ratio $M$ as $\omega_c = \pi/M$ and in your case it will be $w_c = \pi/2$ radians per sample, (with a passpand gain of $M=2$).