I am asking this question sorta as a surrogate for a friend at comp.dsp who posted a similar one.

Even though I did it for a quarter century, laying out math (using "ASCII-math") is crappy, which is why I think the traffic at comp.dsp is in decline (and being displaced by traffic here).

So here's the question, but I am gonna frame it differently than Bob Adams did, making it more about the sampling and reconstruction theorem.

Suppose we have an analog signal that is a collection of simple sinusoids:

$$ x(t) = \sum\limits_{m=1}^{M} A_m \cos(2 \pi f_m t + \phi_m) $$

Without loss of generality, we can order the terms w.r.t. frequency, $0 < f_m < f_{m+1}$, so that $$f_M = \max\{ f_m \} \ .$$

We can uniformly sample $x(t)$ if the sample rate, $f_\text{s} \triangleq \tfrac{1}{T} > 2 \, f_M$, is sufficiently high

$$\begin{align} x_\text{s}(t) &= x(t) \cdot T \, \mathbf{III}_T(t) \\ &= x(t) \cdot T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ &= T \sum\limits_{n=-\infty}^{\infty} x(t) \, \delta(t - nT) \\ &= T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \delta(t - nT) \\ &= T \sum\limits_{n=-\infty}^{\infty} x[n] \, \delta(t - nT) \\ \end{align}$$

It is also true that the sampling function is periodic and has a Fourier series.

$$\begin{align} T \, \mathbf{III}_T(t) &\triangleq T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ &= \sum\limits_{k=-\infty}^{\infty} e^{j 2 \pi k f_\text{s} t} \\ \end{align}$$

Turns out that all of the Fourier series coefficients are 1. This means that the uniform sampled function is

$$\begin{align} x_\text{s}(t) &= x(t) \cdot T \, \mathbf{III}_T(t) \\ &= x(t) \cdot T \sum\limits_{n=-\infty}^{\infty} \delta(t - nT) \\ &= x(t) \sum\limits_{k=-\infty}^{\infty} e^{j 2 \pi k f_\text{s} t} \\ &= \sum\limits_{k=-\infty}^{\infty} x(t) \, e^{j 2 \pi k f_\text{s} t} \\ \end{align}$$

Accordingly, taking the continuous Fourier Transform, the spectrum of the sampled signal is

$$\begin{align} X_\text{s}(f) & \triangleq \mathscr{F} \Big\{ x_\text{s}(t) \Big\} \\ &= \mathscr{F} \left\{ \sum\limits_{k=-\infty}^{\infty} x(t) \, e^{j 2 \pi k f_\text{s} t} \right\} \\ &= \sum\limits_{k=-\infty}^{\infty} \mathscr{F} \Big\{ x(t) \, e^{j 2 \pi k f_\text{s} t} \Big\} \\ &= \sum\limits_{k=-\infty}^{\infty} X(f - k f_\text{s}) \\ \end{align}$$

And we know, as long as $f_M < \tfrac12 f_\text{s}$, that there is no overlap in the adjacent shifted spectra of $X(f)$ and the original $X(f)$ can be recovered from the $k=0$ term of the summation.

$$\begin{align} X(f) &= \Pi\left( \tfrac{f}{f_\text{s}} \right) \, \sum\limits_{k=-\infty}^{\infty} X(f - k f_\text{s}) \\ &= \Pi\left( \tfrac{f}{f_\text{s}} \right) \, X_\text{s}(f) \\ \end{align}$$

where $\Pi(u)$ (sometimes "$\operatorname{rect}(u)$") is the rectangular function

$$\Pi(u) \triangleq \begin{cases} 1 \qquad & \text{ if } |u| < \tfrac12 \\ \tfrac12 \qquad & \text{ if } |u| = \tfrac12 \\ 0 \qquad & \text{ if } |u| > \tfrac12 \\ \end{cases}$$

And we know that the inverse Fourier transform is

$$ \mathscr{F}^{-1} \left\{ \Pi\left( \tfrac{f}{f_\text{s}} \right) \right\} = f_\text{s} \, \operatorname{sinc}(f_\text{s} t) $$

where the sinc function is

$$\operatorname{sinc}(u) \triangleq \begin{cases} \frac{\sin(\pi u)}{\pi u} \qquad & \text{ if } |u| \ne 0 \\ 1 \qquad & \text{ if } |u| = 0 \\ \end{cases}$$

Then, remembering that $f_\text{s}=\tfrac1T $, we know that the output of the brickwall reconstruction filter is

$$\begin{align} X(f) &= \Pi\left( \tfrac{f}{f_\text{s}} \right) \, X_\text{s}(f) \\ & \iff \\ x(t) &= f_\text{s} \, \operatorname{sinc}(f_\text{s} t) \ \circledast \ x_\text{s}(t) \\ &= f_\text{s} \, \operatorname{sinc}(f_\text{s} t) \ \circledast \ T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \delta(t - nT) \\ &= f_\text{s} \, T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \big( \operatorname{sinc}(f_\text{s} t) \ \circledast \ \delta(t - nT) \big) \\ &= \sum\limits_{n=-\infty}^{\infty} x(nT) \, \operatorname{sinc}\big( f_\text{s} (t - nT) \big) \\ &= \sum\limits_{n=-\infty}^{\infty} x(nT) \, \operatorname{sinc}\big( f_\text{s} t - n \big) \\ \end{align}$$

That's how we reconstruct out original $x(t)$ out of the samples $x(nT)$. So much for the sampling and reconstruction theorem. remember, that so the spectra of adjacent shifted copies of $X(f)$, which are $X(f-k f_\text{s})$, do not overlap, it is necessary that $f_M < \tfrac12 f_\text{s}$.

What if $x(t)$ is oversampled?? Even grossly oversampled? That is

$$ f_M \ll \tfrac12 f_\text{s} $$

While it continues to be true that

$$ X(f) = \Pi\left( \tfrac{f}{f_\text{s}} \right) \, X_\text{s}(f) $$

it is also true that

$$ X(f) = \Pi\left( \tfrac{f}{2 f_M + \Delta f} \right) \, X_\text{s}(f) $$

where $\Delta f$ is any tiny frequency guard bandwidth greater than zero

$$ 0 < \Delta f $$

This means that

$$ x(t) = \sum\limits_{n=-\infty}^{\infty} x(nT) \, \tfrac{2 f_M + \Delta f}{f_\text{s}}\operatorname{sinc}\big( (2 f_M + \Delta f) (t - nT) \big) $$

In fact it means moreover

$$ x(t) = \sum\limits_{n=-\infty}^{\infty} x(nT) \, \tfrac{f_W}{f_\text{s}} \operatorname{sinc}\big( f_\text{W} (t - nT) \big) $$

for any brickwall rect width $f_\text{W}$ such that

$$ 2 f_M < f_\text{W} < 2 f_\text{s} - 2 f_M $$

So, quoting and paraphrasing Bob (because i swapped frequency and time domain)

... so obviously the [Fourier Transform] result will be identical [because of the multiplication with the rectangular function $$ \Pi\left( \tfrac{f}{f_\text{W}} \right) \, X_\text{s}(f) \ = \ \Pi\left( \tfrac{f}{f_\text{s}} \right) \, X_\text{s}(f)$$ ], but if you force yourself to use the time-domain convolution method, the width of the sinc signal will vary continually as you change the [rectangular function] width, and yet somehow you must get an identical convolution result for that entire range of sinc widths [, $ f_\text{W} $,] (since the [frequency-]domain signal doesn't change). Can anyone explain this without resorting to the [frequency] domain ?

I mean, this is an ugly way to put it, but if

$$ f_M \ll \tfrac12 f_\text{s} $$


$$ \sum\limits_{m=1}^{M} A_m \cos(2 \pi f_m t + \phi_m) = \sum\limits_{n=-\infty}^{\infty} \sum\limits_{m=1}^{M} A_m \cos\big(2 \pi f_m nT + \phi_m \big) \, \tfrac{f_W}{f_\text{s}} \operatorname{sinc}\big( f_\text{W} (t - nT) \big) $$

for any $A_m, \phi_m, f_m \le f_M$ and

$$ 2 f_M < f_\text{W} < 2 f_\text{s} - 2 f_M $$

doesn't matter what $f_\text{W}$ is, if it's constrained to that range of values.

  • $\begingroup$ i have tried to prune the question down and have asked the math SE about it. $\endgroup$ Jul 18, 2017 at 7:34
  • $\begingroup$ By the way, I don't really like HTML emails myself, but if you're explaining signal processing hardware for a living, you might not be picky about that; have you tried the LaTeX! It plugin for Thunderbird? Allows you to place formulas in-mail and with a click replace them with rendered PNGs. $\endgroup$ Jul 19, 2017 at 6:56
  • $\begingroup$ of course we can swap the order of summation: $$ \sum\limits_{m=1}^{M} A_m \cos(2 \pi f_m t + \phi_m) = \sum\limits_{m=1}^{M} A_m \sum\limits_{n=-\infty}^{\infty} \cos\big(2 \pi f_m nT + \phi_m \big) \, f_W T \operatorname{sinc}\big( f_\text{W} (t - nT) \big) $$ and, if we can prove it for just one term $$ \cos(2 \pi f_M t + \phi_M) = \sum\limits_{n=-\infty}^{\infty} \cos\big(2 \pi f_M nT + \phi_M \big) \, f_W T \operatorname{sinc}\big( f_\text{W} (t - nT) \big) $$ with $$ 2 f_M < f_\text{W} < \tfrac{2}{T} - 2 f_M $$ we've proven it in general. $\endgroup$ Jul 19, 2017 at 8:42
  • $\begingroup$ May interest you $\endgroup$ May 15, 2022 at 20:34

1 Answer 1


There's a key step in your argument where you did this:

$$\begin{align} x(t) &= f_\text{s} \, \operatorname{sinc}(f_\text{s} t) \ \circledast \ x_\text{s}(t) \\ &= f_\text{s} \, \operatorname{sinc}(f_\text{s} t) \ \circledast \ T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \delta(t - nT) \\ &= f_\text{s} \, T \sum\limits_{n=-\infty}^{\infty} x(nT) \, \big( \operatorname{sinc}(f_\text{s} t) \ \circledast \ \delta(t - nT) \big) \\ &= \sum\limits_{n=-\infty}^{\infty} x(nT) \, \operatorname{sinc}\big( f_\text{s} (t - nT) \big) \end{align}$$

At the 2nd-last line, you have $f_s T$ which has been eliminated in the the last line (since $f_s T=1$). However, later on you've decided to change the sample-rate ($f_s\rightarrow f_W$), and at this point, you should have revised the 2nd-last line of this block with $f_sT\rightarrow f_WT$, and this cannot be eliminated.

As a result, the last line of this block could instead be:

$$x(t) = \frac{f_W}{f_s} \sum\limits_{n=-\infty}^{\infty} x(nT) \, \operatorname{sinc}\big( f_\text{s} (t - nT) \big)$$

and your sinc-interpolation equation:

$$x(t)=\sum_{n=−\infty}^\infty x(nT) \operatorname{sinc}(f_W(t−nT))$$

should really be:

$$x(t)= \frac{f_W}{f_s} \sum_{n=−\infty}^\infty x(nT) \operatorname{sinc}(f_W(t−nT))$$

What this tells us is that, as the width of the sinc function increases (as $f_W$ decreases), then we must also decrease the amplitude of the sinc function. Intuitively, this means that the average (DC) level of the sinc function must remain equal to 1.

  • $\begingroup$ i did not change the sample rate. i only changed the bandwidth of the reconstruction filter. the sample rate is still the same $f_\text{s}=\tfrac1T$ but with $f_M \ll \tfrac12 f_\text{s}$, i don't need the reconstruction filter to have a passband that extents from $-\tfrac12 f_\text{s}$ to $+\tfrac12 f_\text{s}$. it can, instead, extend from $-f_M-\Delta f$ to $+f_M+\Delta f$ for any tiny $\Delta f > 0$. $\endgroup$ Jul 19, 2017 at 8:20
  • $\begingroup$ but you have a point there, about decreasing the amplitude of the sinc function. i might have to fix my question. $\endgroup$ Jul 19, 2017 at 8:24
  • 1
    $\begingroup$ i corrected the question. thank you Dave. i missed it and you caught it. $\endgroup$ Jul 19, 2017 at 8:31

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