# 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.

For the frequency-domain approach, the key observation is that the DTFT of the expanded sequence $$y[n]$$ of the middle figure, includes frequency-scaled and shifted copies (a.k.a images) of the original DTFT of the input $$x[n]$$. The images are centered at $$\omega = 2\pi k/M$$, where $$M$$ is the oversampling factor and $$k=0,1,...,M-1$$.
The LPF, whose passband includes only the base copy of the expanded spectrum, removes all the remaining surplus images, and retains only the baseband. This is accomplished in the time-domain by convolving filter's impulse reponse and signal $$y[n]$$, to produce $$z[n]$$. The cutoff frequency of the ideal LPF is given by $$\omega_c = \pi/M$$. In your case $$w_c = \pi/2$$ radians per sample, and the filter also must have a passpand gain of $$M=2$$ if it's to be used as an interpolation filter.