<p>Here is the <strong>10 KHz signal</strong> (the <strong>maximum frequency</strong> of the signal is 10 KHz):</p>

<p><a href="https://i.sstatic.net/4t7v4.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/4t7v4.png" alt="enter image description here"></a></p>

<p>Now, if we sample the signal at 5 KHz and 10 KHz, the signal will look like 
as shown below (the brown points):</p>

<p><a href="https://i.sstatic.net/yC3S0.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/yC3S0.png" alt="enter image description here"></a></p>

<p>It is clear that we can't get any <strong>useful information</strong> from the sampled signal. 
Now, to get some information about the signal, Nyquist said that we must sample the signal at least <strong>2*max signal frequency</strong>. Now, here are the signals sampled at the <strong>Nyquist frequency</strong> and more than it:</p>

<p><a href="https://i.sstatic.net/FgRci.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/FgRci.png" alt="enter image description here"></a></p>

<p>It can be seen that by sampling at the Nyquist rate, we can get the frequency information about the signal. However, to faithfully reconstruct the signal, we have to increase the sampling rate even more. </p>

<p>For more details please visit <a href="https://www.gaussianwaves.com/tag/sampling-theorem/" rel="nofollow noreferrer">https://www.gaussianwaves.com/tag/sampling-theorem/</a></p>

CC BY-SA 4.0
2018-08-27T05:07:17.007Z
2018-09-02T07:34:12.303Z
2018-09-02T07:34:12.303Z
Pankaj Singh
<p><strong>Short practical answer</strong></p>

<blockquote>
  <p>for a signal to be properly sampled, it has to be done [at] the Nyquist sampling rate</p>
</blockquote>

<p>Up front: sampling at the Nyquist frequency is the <em>bare minimum rate</em> to reproduce the <em>frequency</em> of the input signal. Dependent on your demands, I would advise to go at least 2 times that, if not more, to reproduce amplitude and shape of the signal. In other words, higher sampling rates yield better reconstructed signals.</p>

<p><strong>Background</strong><br>
The problem is that in this digital age, analogue signals are sampled digitally, <em>i.e.</em> at fixed sampling rates. The <a href="https://users.cs.cf.ac.uk/Dave.Marshall/Multimedia/node149.html" rel="nofollow noreferrer"><strong>Nyquist criterion</strong></a> states that the sampling rate should be <em>at least</em> twice the target frequency of the signal. </p>

<p>Suppose the target signal is a simple sinusoid with a frequency <em>f</em> (Fig. 1). And let us start with the worst-case scenario, namely a digital sampling rate (SR) equaling <em>f</em>. In this scenario, we end up with a straight line in the digitally recorded signal (upper panel). </p>

<p>In case we slightly increase the SR to <sup>4</sup>/<sub>3</sub> the result is not much better (lower panel).</p>

<p>In fact, it is not until we we go to doubling the SR to 2 times <em>f</em>, that we obtain a saw-tooth with a frequency equaling the input signal. The shape is, however, unlike the input signal, but <strong>at least we have the target input frequency right</strong>. However, <em>this does <strong>not</strong> mean we can faithfully reproduce the signal</em>. For one thing, the <strong>amplitude</strong> of the input signal will depend on the phase shift between SR and the signal. From here on, you can imagine that going to 4 <em>f</em> will substantially improve the reconstructed signal in terms of shape and amplitude (see, <em>e.g.</em> this web page of <a href="https://users.cs.cf.ac.uk/Dave.Marshall/Multimedia/node149.html" rel="nofollow noreferrer">Cardif Universiy</a>).  </p>

<p><a href="https://i.sstatic.net/6At4y.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6At4y.jpg" alt="Nyquist"></a><br>
<sup>Fig. 1. Discrete sampling of an analogue target signal. source: <a href="http://www.ni.com/white-paper/2709/en/" rel="nofollow noreferrer">National Instruments</a></sup></p>

CC BY-SA 4.0
2019-07-29T10:29:49.517Z
as a digital signal processing person: uuuh, these graphs are just wrong! The linear interpolation is simply not what you do when you've got a sampled signal. Never. It's never bandwidth-limited, so you know that this interpolation *must* be wrong. The statement is still true – a sampling rate too low can't unambiguously allow you to recreate your signal – but you can't use linear interpolation as an argument.
Marcus Müller
CC BY-SA 4.0
2019-07-29T11:02:13.687Z
@MarcusMüller - thanks for the comment. What do you mean with interpolation? It's not mentioned anywhere. In the image the points are simply connected, not interpolated in any way. It's just an illustrative graph to visualize undersampling graphically.
AliceD
CC BY-SA 4.0
2019-07-29T21:48:49.243Z
"Connecting with a line" means "interpolating the path that the signal takes between the original points", i.e. "connected" = "interpolated".
Marcus Müller
CC BY-SA 3.0
2017-10-07T19:11:51.930Z
2017-10-07T19:34:26.067Z
2017-10-07T19:34:26.067Z
AliceD
<p>I do understand that for a signal to be properly sampled, it has to be done on the Nyquist sampling rate. What I do not understand is, what happens at sampling rates lower rate than the Nyquist rate, and at rates higher than the Nyquist rate, for that matter.</p>

CC BY-SA 3.0
2017-10-07T19:27:48.850Z
thanks @AliceD, that means the higher the sampling rate the better a signal is been reconstructed.
wokoro douye samuel
CC BY-SA 3.0
2017-10-07T19:33:51.160Z
Yes exactly! I have made that more explicit, thanks. By the way - if you think the answer is useful a '*thanks*' is certainly appreciated, but the convention is to let others know it is helpful by upvoting the answer (use the up-arrow). If the answer answers your question properly, you can also accept it by hitting the check mark.
AliceD
CC BY-SA 3.0
2017-10-07T20:26:35.973Z
i would love to upvote, but i don't have enough reputation to do that @AliceD.
wokoro douye samuel
CC BY-SA 3.0
2017-10-07T21:13:08.660Z
That's interesting... Anyway - glad to help :-)
AliceD
CC BY-SA 3.0
2017-10-09T01:36:46.180Z
'Aliasing' is the keyword for what happens... https://en.wikipedia.org/wiki/Aliasing
Memming
<measurement><methodology><signal-processing><sampling>
What happens at sampling rates lower or higher that the Nyquist rate?
wokoro douye samuel
# What happens at sampling rates lower or higher that the Nyquist rate?

(Asked by: username_0 on 2017-10-07)

I do understand that for a signal to be properly sampled, it has to be done on the Nyquist sampling rate. What I do not understand is, what happens at sampling rates lower rate than the Nyquist rate, and at rates higher than the Nyquist rate, for that matter.

**Tags:** measurement, methodology, signal-processing, sampling

> **Comments:**
> **username_0**: thanks @username_1, that means the higher the sampling rate the better a signal is been reconstructed.  
> **username_1**: Yes exactly! I have made that more explicit, thanks. By the way - if you think the answer is useful a '*thanks*' is certainly appreciated, but the convention is to let others know it is helpful by upvoting the answer (use the up-arrow). If the answer answers your question properly, you can also accept it by hitting the check mark.  
> **username_0**: i would love to upvote, but i don't have enough reputation to do that @username_1.  
> **username_1**: That's interesting... Anyway - glad to help :-)  
> **username_2**: 'Aliasing' is the keyword for what happens... https://en.wikipedia.org/wiki/Aliasing  

---

## 2 answers

### Answer by username_1 on 2017-10-07. Score: 3 | Accepted answer

**Short practical answer**

> for a signal to be properly sampled, it has to be done [at] the Nyquist sampling rate

Up front: sampling at the Nyquist frequency is the *bare minimum rate* to reproduce the *frequency* of the input signal. Dependent on your demands, I would advise to go at least 2 times that, if not more, to reproduce amplitude and shape of the signal. In other words, higher sampling rates yield better reconstructed signals.

**Background**  
The problem is that in this digital age, analogue signals are sampled digitally, *i.e.* at fixed sampling rates. The [**Nyquist criterion**](https://users.cs.cf.ac.uk/Dave.Marshall/Multimedia/node149.html) states that the sampling rate should be *at least* twice the target frequency of the signal.

Suppose the target signal is a simple sinusoid with a frequency *f* (Fig. 1). And let us start with the worst-case scenario, namely a digital sampling rate (SR) equaling *f*. In this scenario, we end up with a straight line in the digitally recorded signal (upper panel).

In case we slightly increase the SR to 4/3 the result is not much better (lower panel).

In fact, it is not until we we go to doubling the SR to 2 times *f*, that we obtain a saw-tooth with a frequency equaling the input signal. The shape is, however, unlike the input signal, but **at least we have the target input frequency right**. However, *this does **not** mean we can faithfully reproduce the signal*. For one thing, the **amplitude** of the input signal will depend on the phase shift between SR and the signal. From here on, you can imagine that going to 4 *f* will substantially improve the reconstructed signal in terms of shape and amplitude (see, *e.g.* this web page of [Cardif Universiy](https://users.cs.cf.ac.uk/Dave.Marshall/Multimedia/node149.html)).

[![Nyquist](https://i.sstatic.net/6At4y.jpg)](https://i.sstatic.net/6At4y.jpg)  
Fig. 1. Discrete sampling of an analogue target signal. source: [National Instruments](http://www.ni.com/white-paper/2709/en/)

> **Comments:**
> **username_4**: as a digital signal processing person: uuuh, these graphs are just wrong! The linear interpolation is simply not what you do when you've got a sampled signal. Never. It's never bandwidth-limited, so you know that this interpolation *must* be wrong. The statement is still true – a sampling rate too low can't unambiguously allow you to recreate your signal – but you can't use linear interpolation as an argument.  
> **username_1**: @MarcusMüller - thanks for the comment. What do you mean with interpolation? It's not mentioned anywhere. In the image the points are simply connected, not interpolated in any way. It's just an illustrative graph to visualize undersampling graphically.  
> **username_4**: "Connecting with a line" means "interpolating the path that the signal takes between the original points", i.e. "connected" = "interpolated".  

---

### Answer by username_3 on 2018-08-27. Score: 2

Here is the **10 KHz signal** (the **maximum frequency** of the signal is 10 KHz):

[![enter image description here](https://i.sstatic.net/4t7v4.png)](https://i.sstatic.net/4t7v4.png)

Now, if we sample the signal at 5 KHz and 10 KHz, the signal will look like
as shown below (the brown points):

[![enter image description here](https://i.sstatic.net/yC3S0.png)](https://i.sstatic.net/yC3S0.png)

It is clear that we can't get any **useful information** from the sampled signal.
Now, to get some information about the signal, Nyquist said that we must sample the signal at least **2\*max signal frequency**. Now, here are the signals sampled at the **Nyquist frequency** and more than it:

[![enter image description here](https://i.sstatic.net/FgRci.png)](https://i.sstatic.net/FgRci.png)

It can be seen that by sampling at the Nyquist rate, we can get the frequency information about the signal. However, to faithfully reconstruct the signal, we have to increase the sampling rate even more.

For more details please visit <https://www.gaussianwaves.com/tag/sampling-theorem/>


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