Given a discrete signal $ r(nT_c)$ specified at a rate $T_c$, assume we want to resample with a 3/2 rate change to match the sampling rate of some other signal.

We would like to upsample it by 3, then filter it by filter of length $K$ and then downsample by 2.

Also assume we have the following relationship exists $$ \frac{3}{2}T_s = T_c$$

Step 1: Upsample then we have

$r'(n\frac{Ts}{2}) = \left\{ \begin{array}{ll} r(n\frac{Ts}{3}) & \mbox{if $n = 0,3,6$};\\ 0 & \mbox{if OW}.\end{array} \right.$

Step 2: Filter

$\hat{r}(n\frac{Ts}{2})= \sum_{k=0}^{K-1}r(\frac{(n-k)T_s}{2})h_{filter}(k)\,\,\,\,\,\,\,n=0,1....$

Step 3:

$r_o(nT_s)= \hat{r}(n\frac{Ts}{2}-\underbrace{\frac{K-1}{2}\frac{T_s}{2}}{???}) \,\,\,\,\,n=0,1,$

I don't understand two of the above steps

1) why do we need use a filter to resample and

2) why isnt it the part underbrace $$\hat{r}(n\frac{Ts}{2}) \,\,\,\,\,n=0,1,....$$

Thank you very much.


1 Answer 1

  1. Your goal is to create a signal with 3 times more samples. The first step is to insert 2 zeros between each sample.
    Technically, you now have a signal with 3 time more samples, as you require. However most of the samples are zeros. What you need to do next is to interpolate the 2 samples between each 2 nonzero samples. And this is exactly what the low-pass filter does (or, at least, this one way to look at it).

  2. Digital filters introduce delay to the signal by half the filter's length. I believe this is the reason for the time shifting

  • $\begingroup$ why do digital filters introduce delay to the signal by half filters length? Thank you $\endgroup$
    – Tyrone
    May 29, 2015 at 15:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.