Subsections

Sampling

The analogue signal $X_a$ is sampled at intervals $T$ to get the sampled signal $X(n)$.
  $\displaystyle x(n) = x_{a}(nT), \qquad -\infty < n < +\infty
$ (2)

Normalised frequency

Note that the sampling frequency
  $\displaystyle F_{s} = \frac{1}{T_s}
$ (3)
is lost and needs to be stored additionally so that the analogue signal can be reconstructed:
  $\displaystyle x_{a}(t) = x_{a}(nT) = x_{a}(\frac{n}{F_{s}})
$ (4)

Figure 1: Maximum frequency of a sampled signal is half the normalised frequency.
\includegraphics[width=0.75\textwidth]{max_sampl}

What is the frequency range of our digital signal? What is the maximum frequency in digital terms? Fig 1 shows that the max frequency is $0.5$ because we need two samples to represent a “wave”. The frequency range $0\ldots 0.5$ is called the normalised frequency range. What is the relation between normalised frequency and sampling rate?

Let's have a look at the analog signal $X_{a}(t)$ with the frequency $F$:

  $\displaystyle x_{a}(t) = A\cos (2\pi F t)
$ (5)
and we sample it only at:
  $\displaystyle t = n/F_s
$ (6)
Then the digital signal is:
  $\displaystyle x_{a}(n) = A \cos (2 \pi F/F_s n)
$ (7)

Now, we can define the normalised frequency as:

  $\displaystyle \mbox{Normalised frequency:} \qquad f = \frac {F}{F_s}
$ (8)
which has its max value at $0.5$ which represents one period of a sine wave within two samples1

Nyquist frequency

Recall that the max value for the normalised frequency $f$ is $0.5$ and that:
  $\displaystyle x_{a}(n) = A \cos (2 \pi n f)
$ (9)
with $n$ as an integer because we are sampling.

What happens above $f>0.5$? Imagine $f = 1$

  $\displaystyle x_{a}(n) = \cos (2 \pi n) = 1 \qquad f=1
$ (10)
which gives us DC or zero frequency. The same is true for $f = 2, 3, 4, \ldots$. We see that above $f=0.5$ we never get higher frequencies. Instead they will always stay between $0\ldots 0.5$ for the simple reason that it is not possible to represent higher frequencies. This will be discussed later in greater detail.

The ratio $F/F_s$ must be lower than $0.5$ to avoid ambiguity or in other words the maximum frequency in a signal must be lower than $\frac{1}{2} F_s$. This is the Nyquist frequency.

If there are higher frequencies in the signal then these frequencies are “folded down” into the frequency range of $0\ldots \frac{1}{2} F_s$ and creating an alias of its original frequency in the so called “baseband” ($f=0\ldots 0.5$). As long as the alias is not overlapping with other signal components in the baseband this can be used to downmix a signal. This leads to the general definition of the sampling theorem which states that the bandwidth $B$ of the input signal must be half of the sampling rate $F_s$:

  $\displaystyle B < \frac{1}{2}F_s
$ (11)
The frequency $\frac{1}{2} F_s$ is called the Nyquist frequency.

Figure 2: Anti alias filtering. A) with a lowpass filter. B) with a sigma delta converter with very high sampling rate.
\includegraphics[width=0.75\textwidth]{anti_alias}

What do we do if the signal contains frequencies above $\frac{1}{2} F_s$? There are two ways to tackle this problem: The classical way is to use a lowpass filter (see Fig. 2A) which filters out all frequencies above the Nyquist frequency. However this might be difficult in applications with high resolution A/D converters. Alternatively one can use a much higher sampling rate to avoid aliasing. This is the idea of the sigma delta converter which operates at sampling rates hundred times higher than the Nyquist frequency.

Sampling theorem

Is it possible to reconstruct an analogue signal from a digital signal which contains only frequencies below the Nyquist frequency?

  $\displaystyle F_s > 2 F_{\mbox{max}}
$ (12)
where $F_{\mbox{max}}$ is max frequency in the signal which we represent by sine waves:
  $\displaystyle x_{a}(t) = \sum_{i=1}^{n} A_i \cos (2 \pi F_{i}t + \Theta_{i})
$ (13)

The analogue signal $x_{a}(t)$ can be completely reconstructed if:

  $\displaystyle g(t) = \frac{\sin 2\pi Bt}{2 \pi Bt}
$ (14)

with

  $\displaystyle B = F_{\mbox{max}}
$ (15)

  $\displaystyle x_{a}(t) = \sum_{h= -a}^{a}x_{a}(\frac{n}{F_s})g(t - \frac{h}{F_s})
$ (16)

The problem is that $g(t)$ runs from negative time to positive time and as we see later is a-causal so that this cannot be implemented for real but approximations of $g(t)$ are possible and are analogue lowpass filters which smooth out the step like outout of an digital to analogue converter.

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