I understand why there need an low pass filter before down-sampling, the sample frequency need to be at least twice of the max frequency signal, else there will be alias issue.

But base on the Wiki page, we need an lowpass filter after up-sampling.

Upsampling requires a lowpass filter after increasing the data rate, and downsampling requires a lowpass filter before decimation.

I don't understand that, can anyone explain it?

  • 1
    $\begingroup$ The filter does the interpolation to compute the new sample values (after up-sampling) from the old sample values (before up-sampling). If you don't interpolate, you're just left with zero values between the original samples. $\endgroup$
    – Matt L.
    Mar 19 '14 at 9:43

Let's say you've sampled an analog signal $x(t)$ with spectrum $X(\omega)$ at rate $1/T$ that is high enough to satisfy the sampling theorem. The spectrum (i.e. the discrete time Fourier transform (DTFT)) of the sampled signal $x_{1,k}$ will be a periodic repetition of $X(\omega)$. The repetition period is $1/T$.

Now you sample the same signal with a higher rate $L/T,\, L\in\mathbb N$ yielding $x_{L,k}$. Again the spectrum will be a periodic repetition of $X(\omega)$ but this time the repetition period is $L/T$, so the spectral images have greater frequency distance than before.

The task of upsampling consists in calculating $x_{L,k}$ from $x_k$. First, $L-1$ zeros are inserted after every sample of $x_k$. Actually this just changes the basic frequency support of the DTFT to $-L/(2T)\ldots L/(2T)$ containing $L$ copies of the original (analog) spectrum. Therefore the unwanted copies are filtered out with a lowpass filter so that only the original spectrum in range $-1/(2T)\ldots 1/(2T)$ remains. This is identical to $x_{L,K}$

The above steps are quite well explained in the figure of the Wiki article you quoted (in the same order).


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