# Shannon interpolation formula for downsampled data with an "almost ideal" low pass filter

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

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

with

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

Am I correct?

• 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? Mar 28 '17 at 11:00
• 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.
– msm
Mar 28 '17 at 11:38

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