Küpfmüller's uncertainty principle

Template:Use dmy dates Küpfmüller's uncertainty principle by Karl Küpfmüller in the year 1924 states that the relation of the rise time of a bandlimited signal to its bandwidth is a constant.

$\mathrm{\Delta }f\mathrm{\Delta }t\ge k$

with $k$ either $1$ or $\frac{1}{2}$

Proof

Template:Unreferenced section Template:Improve A bandlimited signal $u(t)$ with fourier transform $\hat{u}(f)$ is given by the multiplication of any signal $\underset{\_}{\hat{u}}(f)$ with a rectangular function of width $\mathrm{\Delta }f$ in frequency domain:

$\hat{g}(f)=\mathrm{rect}\left(\frac{f}{\mathrm{\Delta }f}\right)={\chi }_{[-\mathrm{\Delta }f/2,\mathrm{\Delta }f/2]}(f):=\{\begin{array}{cc}1& |f|\le \mathrm{\Delta }f/2\\ 0& \text{else}\end{array}.$

This multiplication with a rectangular function acts as a Bandlimiting filter and results in $\hat{u}(f)=\hat{g}(f)\underset{\_}{\hat{u}}(f)=:\underset{\_}{\hat{u}}(f){|}_{\mathrm{\Delta }f}.$

Applying the convolution theorem, we also know

$\hat{g}(f)\cdot \hat{u}(f)=ℱ((g*u)(t))$

Since the fourier transform of a rectangular function is a sinc function $\mathrm{si}$ and vice versa, it follows directly by definition that

$g(t)={ℱ}^{-1}(\hat{g})(t)=\frac{1}{\sqrt{2\pi }}\underset{-\frac{\mathrm{\Delta }f}{2}}{\overset{\frac{\mathrm{\Delta }f}{2}}{\int }}1\cdot e^{j2\pi ft}df=\frac{1}{\sqrt{2\pi }}\cdot \mathrm{\Delta }f\cdot \mathrm{si}\left(\frac{2\pi t\cdot \mathrm{\Delta }f}{2}\right)$

Now the first root $g(\mathrm{\Delta }t)=0$ is at $\mathrm{\Delta }t=\pm \frac{1}{\mathrm{\Delta }f}$. This is the rise time $\mathrm{\Delta }t$ of the pulse $g(t)$. Since the rise time influences how fast g(t) can go from 0 to its maximum, it affects how fast the bandwidth limited signal transitions from 0 to its maximal value.

We have the important finding, that the rise time is inversely related to the frequency bandwidth:

$\mathrm{\Delta }t=\frac{1}{\mathrm{\Delta }f},$

the lower the rise time, the wider the frequency bandwidth needs to be.

Equality is given as long as $\mathrm{\Delta }t$ is finite.

Regarding that a real signal has both positive and negative frequencies of the same frequency band, $\mathrm{\Delta }f$ becomes $2\cdot \mathrm{\Delta }f$, which leads to $k=\frac{1}{2}$ instead of $k=1$

See also

• Heisenberg's uncertainty principle
• Nyquist theorem

Further reading

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