Sampling Theorem in Reverse

The sampling theorem states that if a continuous-time function, $x(t)$, is ideally sampled, by multiplying by a string of appropriately scaled dirac impulse functions spaced in time at discrete intervals, $T$, the spectrum, $X(f)$, of that continuous-time function is copied, repeated, and overlapped at regular intervals in the frequency domain with frequency interval $1/T$. Since the Fourier transform is an invertible (one-to-one) operator, one could restate the sampling theorem and say that if one somehow copies, repeats, and overlaps at regular intervals (of $1/T$) the spectrum of some continous time function, it would have the effect of sampling that time function at regular discrete intervals spaced by $T$.

$$ \mathscr{F}\left\{ x(t) \cdot \left( T \sum\limits_{n=-\infty}^{\infty} \delta(t-nT) \right) \right\} = \sum\limits_{k=-\infty}^{\infty} X\Big(f - k\frac1T \Big) $$

where

$$ X(f) = \int\limits_{-\infty}^{+\infty} x(t) e^{-j 2 \pi f t} \ \mathrm{d}t $$

$$ x(t) = \int\limits_{-\infty}^{+\infty} X(f) e^{+j 2 \pi f t} \ \mathrm{d}f $$

is the Fourier Transform and inverse. $\delta(t)$ is the Dirac unit impulse function.

The duality theorem of the Fourier Transform says that the roles of time $t$ and frequency $f$ can be exchanged, sometimes requiring a sign change in either $t$ or $f$. What this means is:

$$\begin{align} \mathscr{F} \Big\{ y(t) \Big\} &= \mathscr{F}\left\{ \sum\limits_{k=-\infty}^{\infty} \hat{x}(t-kP) \right\} \\ \\ &= \hat{X}(f) \cdot \left(\frac1P \sum\limits_{n=-\infty}^{\infty} \delta\Big(f -n\frac1P\Big) \right) \\ \end{align}$$

where

$$ y(t) \triangleq \sum\limits_{k=-\infty}^{\infty} \hat{x}(t-kP) $$

and

$$ \hat{X}(f) \triangleq \mathscr{F} \Big\{ \hat{x}(t) \Big\} $$

This says that copying, repeating, and overlapping a time function (call it a "grain" or a "grain") $\hat{x}(t)$ with repeat period of $P$ (or repeat rate or $1/P$) essentially samples the grain's spectrum at regular discrete-frequency intervals of $1/P$. This converts the continuous spectrum, $\hat{X}(f)$ of the nonrepeating grain function $\hat{x}(t)$ into a line spectrum, which is the Fourier Transform $Y(f)$ of the periodic function $y(t)$. The amplitudes of the discrete lines are the coefficients of the Fourier series of the repeated wave function $y(t)$.

The original grain spectrum $\hat{X}(f)$ can now be considered to be the "spectral envelope" of the line spectrum $Y(f)$. Note that if the repeat rate $1/P$ of the grains is changed, this changes the spacing of the line spectrum $Y(f)$, but not the spectral envelope $\hat{X}(f)$, which does not depend of P.

Sampling Theorem in Reverse

The sampling theorem states that if a continuous-time function, $x(t)$, is ideally sampled, by multiplying by a string of appropriately scaled dirac impulse functions spaced in time at discrete intervals, $T$, the spectrum, $X(f)$, of that continuous-time function is copied, repeated, and overlapped at regular intervals in the frequency domain with frequency interval $1/T$. Since the Fourier transform is an invertible (one-to-one) operator, one could restate the sampling theorem and say that if one somehow copies, repeats, and overlaps at regular intervals (of $1/T$) the spectrum of some continous time function, it would have the effect of sampling that time function at regular discrete intervals spaced by $T$.

$$ \mathscr{F}\left\{ x(t) \cdot \left( T \sum\limits_{k=-\infty}^{\infty} \delta(t-kT) \right) \right\} = \sum\limits_{n=-\infty}^{\infty} X\Big(f - n\frac1T \Big) $$

where

$$ X(f) = \int\limits_{-\infty}^{+\infty} x(t) e^{-j 2 \pi f t} \ \mathrm{d}t $$

$$ x(t) = \int\limits_{-\infty}^{+\infty} X(f) e^{+j 2 \pi f t} \ \mathrm{d}f $$

is the Fourier Transform and inverse. $\delta(t)$ is the Dirac unit impulse function.

The duality theorem of the Fourier Transform says that the roles of time $t$ and frequency $f$ can be exchanged, sometimes requiring a sign change in either $t$ or $f$. What this means is:

$$\begin{align} \mathscr{F} \Big\{ y(t) \Big\} &= \mathscr{F}\left\{ \sum\limits_{k=-\infty}^{\infty} \hat{x}(t-kP) \right\} \\ \\ &= \hat{X}(f) \cdot \left(\frac1P \sum\limits_{n=-\infty}^{\infty} \delta\Big(f -n\frac1P\Big) \right) \\ \end{align}$$

where

$$ y(t) \triangleq \sum\limits_{k=-\infty}^{\infty} \hat{x}(t-kP) $$

and

$$ \hat{X}(f) \triangleq \mathscr{F} \Big\{ \hat{x}(t) \Big\} $$

This says that copying, repeating, and overlapping a time function (call it a "wavelet" or a "grain") $\hat{x}(t)$ with repeat period of $P$ (or repeat rate or $1/P$) essentially samples the grain's spectrum at regular discrete-frequency intervals of $1/P$. This converts the continuous spectrum, $\hat{X}(f)$ of the nonrepeating grain function $\hat{x}(t)$ into a line spectrum, which is the Fourier Transform $Y(f)$ of the periodic function $y(t)$. The amplitudes of the discrete lines are the coefficients of the Fourier series of the repeated wave function $y(t)$.

The original grain spectrum $\hat{X}(f)$ can now be considered to be the "spectral envelope" of the line spectrum $Y(f)$. Note that if the repeat rate $1/P$ of the grains is changed, this changes the spacing of the line spectrum $Y(f)$, but not the spectral envelope $\hat{X}(f)$, which does not depend of P.