German: Sie betrachten das zeitkontinuierliche Signal $s(t)$ mit einem digitalen Oszilloskop mit Abtastperiode $T_\mathrm{s}$ und messen das zeitdiskrete Signal $\tilde{s}[n] = s(n\:T_s)$

enter image description here

Nehmen Sie an, dass das Abtasttheorem eingehalten wird.

  1. Können Sie $s(t)$ aus $\tilde{s}[n]$ rekonstruieren? Begründen Sie!
  2. Skizzieren sie $s(t)$ für einen Bereich von $- 3\:T_\mathrm{s}\leq t \leq 3 \:T_\mathrm{s}$

English: You consider the time-continuous signal $ s (t) $ with a digital oscilloscope with sampling period $ T_\mathrm{s} $ and measure the time-discrete signal $\tilde{s}[n] = s(n\:T_s)$

Assume that the sampling theorem is met.

  1. Can you reconstruct $ s (t) $ from $\tilde{s}[n]$? Explain!
  2. Sketch $ s (t) $ for a range from $- 3\:T_\mathrm{s}\leq t \leq 3 \:T_\mathrm{s}$

Sadly, that's all that's given. This is an exam problem from university and I don't have any clue how to get a useful solution.

Maybe some one is able to reconstruct it for me and tell me how it works?

  • 1
    $\begingroup$ Well, that's plenty info. Take this sentence: "Nehmen Sie an, ... eingehalten wird.". Remember what that theorem says about the answer to question 1.! $\endgroup$ Commented Sep 7, 2020 at 18:01
  • 1
    $\begingroup$ By the way, questions asking for homework solutions are off-topic, unless you demonstrate your own attempt and try to ask a precise question of your own that helps you proceed. $\endgroup$ Commented Sep 7, 2020 at 18:14

2 Answers 2


German: Nach einiger zeit der recherche bin ich auf folgende lösung gekommen, danke an alle die hier Hinweise gegeben haben. Hält diese lösung jemand für Korrekt?

English: After some time of research, I came up with the following solution, thanks to everyone who gave me tips. Does anyone think this solution is correct?


German: Ja das Signal $s(t)$ kann aus $\tilde{s}(t)$ rekonstruiert werden, da das Abtasttheorem eingehalten wurde. Es gilt also für die Abtastfrequenz: $f_\mathrm{s} > 2 \cdot f_\mathrm{Signal}$. Da das Signal mit der Frequenz 2 gegeben ist, muss das Signal mindestens 4 mal abgetastet werden. Um das Signal zurück zu gewinnen, wird die Sinc-Interpolation verwendet.

English: Yes, the signal $ s (t) $ can be reconstructed from $ \tilde{s} (t) $, since the sampling theorem was adhered to. The following applies for the sampling frequency: $ f_\mathrm{s} > 2 \cdot f_\mathrm{signal} $. Since the signal is given with frequency 2, the signal must be sampled at least 4 times. Sinc interpolation is used to recover the signal.

\begin{align*} \textbf{ Sinc-Interpolation:} ~~~~ s(t)& =\sum_{n=-\infty}^{\infty}\: s[n\:T_\mathrm{s}] \: \mathrm{sinc}\left( \frac{t-n\:T_\mathrm{s}}{T_s}\right) \\ s(t)& = \sum_{n=-\infty}^{\infty}\: \delta[n\:T_\mathrm{s}-1] \: \frac{\sin\left(\left( \frac{t-n\:T_\mathrm{s}}{T_s} \right) \: \pi\right) }{\left( \frac{t-n\:T_\mathrm{s}}{T_s} \right)\pi} \end{align*}

enter image description here

  • $\begingroup$ There you go! All it took was looking at what you already knew. $\endgroup$
    – Max
    Commented Sep 14, 2020 at 8:31

Consider things you know about the sampling theorem:

  • If it is met, the original signal can always be reconstructed perfectly!
  • The signal has to be bandlimited for the theorem to be met!

Further, think about the spectral and time domains and their relation to each other:

  • If a signal is bandlimited in spectral domain, it has to be infinite in time domain.

Then, thinking about the given signal:

  • The given sampled signal has just zeros before and after the one none-zero value.
  • It is not periodic.
  • One can conclude, that in the original signal, there have to be none zero values between all (or at least some) sampling points, otherwise it would not be infinite.

So we have to ask ourselves the question:

  • Which signal is bandlimited, has zero crossings at regular intervals, a non zero value in the middle and is not periodic?

If you had any lectures on signal theory, the answer should be pretty obvious.

  • $\begingroup$ Ok i am new to this Forum can you have a look on my solution above. Thanks $\endgroup$
    – Dansi
    Commented Sep 10, 2020 at 11:12

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