Let $x[n]$ be a discrete time signal with DFT given by $X(f)=\sum_n x[n]e^{-2\pi inf}$ supported on $[-1/2M,1/2M]$ with $f\in[-1/2,1/2]$.

I can then down-sample to get $y[n]:=x[nM]$. Then, let

$$\widetilde{x}[n]=\begin{cases}My[n/M],& M|n,\\0,&\text{otherwise}.\end{cases}$$

Then its DFT is given by

$$ \begin{aligned}\widetilde{X}(f)&=\sum_{n\in\mathbb{Z}}\widetilde{x}[n]e^{-2\pi inf}\\&=\begin{cases}M\sum_{n\in\mathbb{Z}}y[n/M]e^{-2\pi inf},& M|n,\\0,&\text{otherwise}\end{cases}\\&=\begin{cases}M\sum_{n}x[n]e^{-2\pi inf},&M|n,\\0,&\text{otherwise}.\end{cases}\end{aligned} $$

Now, let $\hat{x}[n]=(\tilde{x}\ast h)[n]$ be the discrete Hilbert transform of $\tilde{x}$, with $h$ an "almost" ideal low-pass filter with cut-off frequency $f_c=1/M$.

My question is, how do I then apply the Shannon interpolation formula to reconstruct $x(t)$?

Intuitively, I would guess that it would be something along the lines of

$$x(t)=\left(\sum_{n\in\mathbb{Z}}x[n]\cdot\delta(t-n\Delta t)\right)\ast H(f),$$


$$H(f)=\begin{cases}\frac{DTFT\{\hat{x}[\cdot]\}(f)}{M\cdot X(f)},&\text{if }M|n, \\ 0,&\text{otherwise}. \end{cases}$$

Am I correct?

  • 1
    $\begingroup$ Your formulation is not correct. In your second equation, n is the summation index, you cannot put a case depending on it, outside of the sum. Furthermore, what do you mean by "almost" ideal? How ideal is it? Also, do you really mean discrete Hilbert transform or just a low-passed filtered signal? Is the cutoff frequency two-sided or one-sided (i.e. does the filter have same bandwidth of the signal?) What makes the big problem here, why can't you just do sinc-interpolation of $\tilde{x}[n]$ using a Dirichlet kernel since you are in the discrete periodic setting? $\endgroup$ Commented Mar 28, 2017 at 11:00
  • $\begingroup$ You said nothing about $x(t)$ and whether $x[n]$ is its sampled version. If yes, is the sampling frequency above the Nyquist rate? if yes, why to bother about $\hat{x}[n]$? Your $$x(t)=\left(\sum_{n\in\mathbb{Z}}x[n]\cdot\delta(t-n\Delta t)\right)\ast H(f),$$ implies that $x[n]$ is assumed known. You can reconstruct $x(t)$ directly from $x[n]$. It is unclear. $\endgroup$
    – msm
    Commented Mar 28, 2017 at 11:38

1 Answer 1


I don't get your downsample step when you downsampled by factor $M$.

Let me go from scratch with the spectrum visualization below, with time domain, continuous frequency domain and discrete frequency domain from left to right.

DownSampling process in time, continuous freq and discrete freq

When we reduce the sampling frequency by a factor $k$, the signal spectrum is copied to new replicas at $f_s/k$. The discrete spectrum is a snapshot of continuos spectrum at $[-f_s, f_s]$ and $[-f_{s,down} = f_s/k, f_{s,down} = f_s/k]$, so it is expanded by the factor $k$.

The DFT works on the discrete frequency domain. The downsample $k$ must be chosen so that the expanded spectrum (in discrete frequency, or the new replicas in continuous version) does not overlap with their copies centered at $-2\pi$ and $2\pi$, or $-1/2$ and $1/2$ if you take discrete instantenous frequency by dividing $2\pi$; otherwise aliasing happens and the reconstruction of $x(t)$ is impossible. If there is not alias, a low pass filter to take the spectrum part centered at 0 is suffice. This filter does not need to be "ideal", just to be sure that the filter takes only the center part.

In your calculation, if I understand well, you are talking about the discrete frequency domain and the original spectrum takes $1/M$ the normalized band (you said $f$ is in $[-1/2, 1/2]$ and your signal has support in $[-1/2M, 1/2M]$). In this case your downsample factor $k$ must be less than $M/2$ and the cutoff frequency of your ideal filter must be at $1/4$ if $k=M/2$ (and yes, in this case $k=M/2$, we need "ideal filter" assumption).


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