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Notation

Type Font Examples
Variables (scalars) italics $a, b, x, y$
Functions upright $\mathrm{f}, \mathrm{g}(x), \mathrm{max}(x)$
Vectors bold, elements row-wise $\mathbf{a}, \mathbf{b}= \begin{pmatrix}x\\y\end{pmatrix} = (x, y)^\top,$ $\mathbf{B}=(x, y, z)^\top$
Matrices Typewriter $\mathtt{A}, \mathtt{B}= \begin{bmatrix}a & b\\c & d\end{bmatrix}$
Sets calligraphic $\mathcal{A}, B=\{a, b\}, b \in \mathcal{B}$
Number systems, Coordinate spaces double-struck $\mathbb{N}, \mathbb{Z}, \mathbb{R}^2, \mathbb{R}^3$

What's happening?

And here?

Sampling Theorem

Sampling Theorem

  • The sampling theorem states how an analog signal must be sampled
  • If the theorem is fulfilled, the original signal can be reconstructed with almost any accuracy
whitakker_kotelnikov_shannon
E. T. Whittaker
W. A. Kotelnikow
C. E. Shannon
Image source: W. A. Kotelnikow von www.kremlin.ru (CC-BY), Wikipedia

Sampling of an Analog Signal

sin_analog
Analog time signal
Samples at discrete points in time
$\mathrm{f}(t)$
$t$
$\mathrm{f}_a[n]$
$n$

Sampling Theorem

  • Sampling theorem: An analog signal must be sampled at a frequency $F_a$ that is larger than twice the maximum signal frequency $F_{\mathrm{max}}$
    $F_a > 2 \,F_{\mathrm{max}}$

Sampling Theorem

cos_sample_speed
$F_a = 5 \,F_{\mathrm{max}}\Rightarrow$ Sampling theorem fulfilled
$F_a = \frac{5}{6} \,F_{\mathrm{max}}\Rightarrow$ Sampling theorem not fulfilled
$F_a = \frac{5}{4} \,F_{\mathrm{max}}\Rightarrow$ Sampling theorem not fulfilled
$F_a > 2 \,F_{\mathrm{max}}$

Example: Audio signal with increasing frequency

chrip
  • Original: Frequency of approx. 100 Hz to 8000 Hz
  • Reconstruction after sampling with 8000 Hz

Mathematical Description of the Ideal Sampling Process

  • Sampling an analog signal with frequency $F_a = \frac{1}{T_a}$
    $\mathrm{x}_a(t) = \mathrm{x}(t) \, \mathrm{p}_s(t)\quad$ mit $\quad\mathrm{p}_s(t) = \sum\limits_{k=-\infty}^{+\infty} T_a\, \delta(t - k \,T_a)$
func_sample_analog
$t$
$t$
$\mathrm{x}_a(t)$
$\mathrm{x}(t)$
$\mathrm{p}_s(t)$

Analysis in the Frequency Domain

  • The Fourier transform of a Dirac sequence is also a Dirac sequence
    $\mathrm{p}_s(t) = \sum\limits_{k=-\infty}^{+\infty} T_a\, \delta(t - k \,T_a) \quad \stackrel{\operatorname{FT}}{\longrightarrow} \quad \mathrm{P}_s(f) = \sum\limits_{k=-\infty}^{+\infty} \delta(f - k \,F_a) $
  • Thus, it follows for the Fourier transform of $\mathrm{x}_a(t)$
    $\begin{array}{lll} \mathrm{x}_a(t) = \mathrm{x}(t) \, \mathrm{p}_s(t) & \stackrel{\operatorname{FT}}{\longrightarrow} & \mathrm{X}_a(f) = \mathrm{X}(f) \ast \mathrm{P}_s(f)\\ \mathrm{x}_a(t) = \mathrm{x}(t) \, \sum\limits_{k=-\infty}^{+\infty} T_a\, \delta(t - k \,T_a) & \stackrel{\operatorname{FT}}{\longrightarrow} & \mathrm{X}_a(f) = \mathrm{X}(f) \ast \sum\limits_{k=-\infty}^{+\infty} \delta(f - k \,F_a)\\ && \mathrm{X}_a(f) = \sum\limits_{k=-\infty}^{+\infty} \mathrm{X}(f - k \,F_a) \end{array}$

Analysis in the Frequency Domain

time2spec
Input signal
Spectrum of the input signal
f
$\mathrm{x}(t)$
$\mathrm{X}(f)$
$t$
$-F_{\mathrm{max}}$
$F_{\mathrm{max}}$

Spectrum of the sampled input signal: $\mathrm{X}_a(f) = \sum\limits_{k=-\infty}^{+\infty} \mathrm{X}(f - k \,F_a)$

spec_copies
$\mathrm{X}_a(f)$
$-2\,F_a$
$-F_a$
0
$F_a$
$2\,F_a$

Reconstruction of the Analog Signal by Low-pass Filtering

sample_reconstruct
$\mathrm{X}(f)$
$-F_{\mathrm{max}}$
$F_{\mathrm{max}}$
$f$
$\mathrm{X}_a(f)$
$-2\,F_a$
$-F_a$
$0$
$F_a$
$2\,F_a$
$f$
$\mathrm{H}(f)$
Low-pass filter
$\mathrm{X}_a(f)$
$f$
$\mathrm{X}_r(f)$
$f$

If the sampling theorem is fulfilled: perfect reconstruction of the analog signal by low-pass filtering

sample_reconstruct_part
$\mathrm{H}(f)$
$\mathrm{X}_a(f) = \sum\limits_{k=-\infty}^{+\infty} \mathrm{X}(f - k \,F_a)$
$-2\,F_a$
$-F_a$
$0$
$F_a$
$2\,F_a$
$f$
$-F_{\mathrm{max}}$
$F_{\mathrm{max}}$
Low-pass filter

Ideal low-pass filter: $\mathrm{H}(f) = \begin{cases} 1 & \,\,:\,\, |f| < F_{\mathrm{max}} \\ 0.5 & \,\,:\,\, |f| = F_{\mathrm{max}} \\ 0 & \,\,:\,\, |f| > F_{\mathrm{max}} \\ \end{cases}$

$\begin{array}{lll} \mathrm{X}_r(f) &=& \mathrm{H}(f) \, \mathrm{X}_a(f)\\ &=& \mathrm{H}(f) \, \sum\limits_{k=-\infty}^{+\infty} \mathrm{X}(f - k \,F_a)\\ &=& \sum\limits_{k=-\infty}^{+\infty} \mathrm{H}(f) \, \mathrm{X}(f - k \,F_a)\\ &=& \mathrm{X}(f) \quad \mathrm{wenn} \quad F_a > 2 \,F_{\mathrm{max}} \end{array}$

In case of violation of the sampling theorem: Aliasing

alias
$\mathrm{X}_a(f)$
$-F_{\mathrm{max}}$
$F_{\mathrm{max}}$
$f$
$\mathrm{X}_a(f)$
$-2\,F_a$
$-F_a$
$0$
$F_a$
$2\,F_a$
$\mathrm{X}_a(f)$
$f$
$\mathrm{X}_a(f)$
$-2\,F_a$
$-F_a$
$0$
$F_a$
$2\,F_a$
$\mathrm{H}(f)$
Tiefpass
$\mathrm{X}_a(f)$
$\mathrm{X}_r(f)$
$f$
$F_a > 2 \,F_{\mathrm{max}}$
$F_a \le 2 \,F_{\mathrm{max}}$
$F_a \le 2 \,F_{\mathrm{max}}$

Aliasing effect: If the sampling theorem is violated, additional frequencies can appear that may not have been present in the original spectrum

Low-pass filtering before sampling the input signal helps against aliasing

sampling_image_example
Low-pass filter
Sampling
Sampling
Reconstruction
Reconstruction

Summary: Sampling Theorem

  • Sampling theorem: An analog signal must be sampled at a frequency $F_a$ that is larger than twice the maximum signal frequency $F_{\mathrm{max}}$
    $F_a > 2 \,F_{\mathrm{max}}$
  • If the theorem is fulfilled, the original signal can be reconstructed with almost any accuracy
  • For some applications, e.g. digital images, a certain degree of aliasing is acceptable and preferred to very strong low-pass filtering

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