I am confused about this Q&A : Interpolation by factor of 2
If my input signal $$ x[n]= x0,x1,x2,x3 $$ then according to the threads explaination my $v[n]$ will be
$$ v[n]=x0,0,x1,0,x2,0,x3,0 $$
$x[n] \rightarrow$ zeropad $\rightarrow v[n]$
$v[n]\rightarrow$ LPF $\rightarrow y[m]$
My $y[m]$ is the output of LPF which removes aliases. So basically LPF transfer function is given as
$$ y[m] = v[n]*h[n] $$
So here $h[m]$ is my filter coefficients and $v[n]$ is my zero padded input signal... so, if I multiply zero with my filter coefficients, then my output will also be zero...
If I send 10 input samples, after zero padding I will get 20 samples, out of which 10 will be zero....
So what is the use of getting zero as my output? Correct me if am wrong (I know I am wrong, but please give me some idea how zero padding & LPF is helping in this regard).
For FIR filter: $y(n)=h(0)x(n)+h(1)x(n−1)+....+h(N−1)x(n−N+1)$, where $x(n)$ is the input signal, $y(n)$ is the output signal, $h(n)$ is the impulse response, and $N$ is the filter length.
For simplicity, I am taking order $N=4$,and some random numbers
(time) n --> [ 0 1 2 3] x[n]--> [ 2 4 6 8] h[n]--> [.1 .2 .3 .4]
what is $y[2.5]$???
n-->[0 0.5 1 1.5 2 2.5 3 3.5] v[n]-->[2 0 4 0 6 0 8 0 ] (after zero padding)
Lets calculate for $y[2.5]=h(0)x(2.5)+h(1)x(1.5)+h(2)x(0.5)+h(3)x(-0.5)$
so here $x(2.5)=x(1.5)=x(0.5)=0$
Please answer this to clear my confusion???