# Nyquist sampling theorem

Started by August 17, 2005
```Hello,
I read in the text book ("Digital signal Processing Principles,
Algorithm and APplications" J. G. Proakis % D. G. Manolakis, page 30)
that the Nyquist frequency (rate) is a double of the highest frequency
of the signal.
In the web, I learn that it is the double of the bandwith. Moreover in
the web they state that there is a mistake in many textbooks where
authors teach that it is the double of the highest frequency. The
theorem in textbook is correct only when the signal is a baseband type.

In my  opinion, if it is a theorem so Nyquist must state clearly the
definition of his Nyquist frequency, and if so there should not be such
a confusing argument.

So which one do you agree? in my textbook or in the web.

Thanks

```
```double the highest frequency of the signal is always correct and
precise

if the signal is baseband type then bandwidth = highest frequency

on the other hand, for example, if the "signal" consists of a
modulating signal and carrier wave then the modulating signal bandwidth
is not equal to highest frequency of the "signal", but the general
statement of the double the highest frequency of the "signal", is of
course, still true

```
```VijaKhara@gmail.com wrote:

> Hello,
> I read in the text book ("Digital signal Processing Principles,
> Algorithm and APplications" J. G. Proakis % D. G. Manolakis, page 30)
> that the Nyquist frequency (rate) is a double of the highest frequency
> of the signal.
> In the web, I learn that it is the double of the bandwith. Moreover in
> the web they state that there is a mistake in many textbooks where
> authors teach that it is the double of the highest frequency. The
> theorem in textbook is correct only when the signal is a baseband type.
>
>
> In my  opinion, if it is a theorem so Nyquist must state clearly the
> definition of his Nyquist frequency, and if so there should not be such
> a confusing argument.
>
> So which one do you agree? in my textbook or in the web.
>
> Thanks
>
There's never been a theorem written that hasn't been proof to rewording
in a textbook, and expansion in further papers to "clarify".

The lower bound for the sampling rate is double the bandwidth, which for
a baseband signal is exactly equal to the highest frequency.  Your
textbook was probably not treating signals with significant shift up in
frequency; in such a case it would be appropriate to leave the full
details for a later course.

For the case of simple sampling where you are taking a single vector of
samples at a constant sampling rate it's pretty easy to use aliasing to
figure out why Nyquist is Nyquist.

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
```
```In order to achieve perfect reconstruction you have to sample at twice the
highest frequency.

In most cases with a non-baseband signal you're not interested in the
frequency content of the carrier wave.  In this case I think you need to
first bring that signal down to baseband and THEN sample at twice the
bandwidth (which is now the highest frequency at baseband).  Someone correct
me if I'm wrong, but I would think that without first bringing the signal to
baseband you could run into problems due to any phase shift between your
carrier signal and your sampling (the worst case being that the sampling
occurs exactly at all the zero crossings!).

<VijaKhara@gmail.com> wrote in message
> Hello,
> I read in the text book ("Digital signal Processing Principles,
> Algorithm and APplications" J. G. Proakis % D. G. Manolakis, page 30)
> that the Nyquist frequency (rate) is a double of the highest frequency
> of the signal.
> In the web, I learn that it is the double of the bandwith. Moreover in
> the web they state that there is a mistake in many textbooks where
> authors teach that it is the double of the highest frequency. The
> theorem in textbook is correct only when the signal is a baseband type.
>
>
> In my  opinion, if it is a theorem so Nyquist must state clearly the
> definition of his Nyquist frequency, and if so there should not be such
> a confusing argument.
>
> So which one do you agree? in my textbook or in the web.
>
> Thanks
>

```
```Brad Griffis wrote:
> In order to achieve perfect reconstruction you have to sample at twice the
> highest frequency.
>
> In most cases with a non-baseband signal you're not interested in the
> frequency content of the carrier wave.  In this case I think you need to
> first bring that signal down to baseband and THEN sample at twice the
> bandwidth (which is now the highest frequency at baseband).  Someone correct
> me if I'm wrong, but I would think that without first bringing the signal to
> baseband you could run into problems due to any phase shift between your
> carrier signal and your sampling (the worst case being that the sampling
> occurs exactly at all the zero crossings!).

Sub-band sampling works, but not every sampling frequency greater than
twice the bandwidth if the modulation is suitable. Rick Lyons' second
edition has the most lucid exposition of that topic I've seen. (That's
not saying much.) He responded here a month or so ago to a thread
similar to this. Google for it.

Jerry
--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```<VijaKhara@gmail.com> wrote in message
> I read [...] that the Nyquist frequency (rate) is a double of the
> highest frequency of the signal.
> In the web, I learn that it is the double of the bandwith. Moreover in
> the web they state that there is a mistake in many textbooks where
> authors teach that it is the double of the highest frequency. The
> theorem in textbook is correct only when the signal is a baseband type.

You can reconstruct the signal using sinc interpolation if you sample it at
a rate more than twice its highest frequency.

You can reconstruct the signal if you sample it at a rate more than its
(double-sided) bandwidth AND your sampling rate doesn't cause aliasing
between in-band frequencies, but you will need to use different
interpolation if the rate isn't more than twice the highest frequency.

For the common case of baseband signals, these rate limits are the same.

--
Matt

```
```> You can reconstruct the signal using sinc interpolation if you sample it at
> a rate more than twice its highest frequency.
>
> You can reconstruct the signal if you sample it at a rate more than its
> (double-sided) bandwidth AND your sampling rate doesn't cause aliasing
> between in-band frequencies, but you will need to use different
> interpolation if the rate isn't more than twice the highest frequency.

Thank you very much. Now I understand something when I read further in
the textbook where they discuss about the way to sample a bandpass
signal [Bmin, Bmax]. In the case Bmax is an extremely high freq, if we
use a sinc interpolation, we need a very high sampling rate which is
not desired.
In the textbook they shift the bandpass signal back to be a low pass
signal, sample it and then use some function to reconstruct the
original bandpass signal. Is that one of the cases you want to mention?

```
```VijaKhara@gmail.com wrote:
>>You can reconstruct the signal using sinc interpolation if you sample it at
>>a rate more than twice its highest frequency.
>>
>>You can reconstruct the signal if you sample it at a rate more than its
>>(double-sided) bandwidth AND your sampling rate doesn't cause aliasing
>>between in-band frequencies, but you will need to use different
>>interpolation if the rate isn't more than twice the highest frequency.
>
>
>
> Thank you very much. Now I understand something when I read further in
> the textbook where they discuss about the way to sample a bandpass
> signal [Bmin, Bmax]. In the case Bmax is an extremely high freq, if we
> use a sinc interpolation, we need a very high sampling rate which is
> not desired.
> In the textbook they shift the bandpass signal back to be a low pass
> signal, sample it and then use some function to reconstruct the
> original bandpass signal. Is that one of the cases you want to mention?

The signal needs to be sampled at the high rate -- high enough to
reproduce the carrier -- in order to shift it without heterodyning
hardware. Once it's already sampled, what's gained by shifting it? If
it's heterodyned before being sampled, What's to talk about?

Google for Rick's clear explanation in this newsgroup. The explanation
at http://www.bitscope.com/adc/?p=3 may be useful too.

Jerry

--
Engineering is the art of making what you want from things you can get.
&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;&#2013266095;
```
```Double the highest frquency of the signal will always give you the
perfect reconstruction, but it will not always be the Nyquist
frequency. The Nyquist frequency is, in essence, the minimum frequency
required to avoid overlap in the frequency spectrum. Keeping that in
mind, in case of a bandpass signal, the Nyquist frequency is twice it's
bandwidth, and not twice its highest frquency.

Sampling results in an infinite repetition of the signal's spectrum in
the digital frequency domain, with the spacing between the spectra
equal to the sampling period. Thus sampling at twice the signal's
bandwidth is will yield the minimum spacing required for
reconstruction (maintaining the spectrum of the signal). In contrast,
sampling at the highest 2x the highest frequency, for bandpass signals,
will yield a higher sampling frequency than that required for
reconstruction (the spectra a more spaced apart than the minimum
separation).

Offcourse, these conditions become equivalent under baseband signals.

So if you have a signal at 1 kHz with bandwidth of 3 Hz. For
reconstruction, you only need to sample it at 6Hz (in practice at least
30 Hz). And not 2.6 Khz (2x hightest frequency). But, because the
digital spectrum is periodic with period of 6 Hz in the digital domain,
you have to make sure you get rid of the periodicities in the low
frequency band below the equivalent 970 Hz, and that the original
spectrum is correctly centered in the equivalent 1 KHz frequency.

```
```Ikaro wrote:
> Double the highest frquency of the signal will always give you the
> perfect reconstruction, but it will not always be the Nyquist
> frequency.

Ikaro,
for a baseband signal, you need to sample at slightly more than twice
the highest frequency.

If you sample a signal y=A*cos(2*pi*fsig + theta) at exactly fs=2fsig,
then, admittedly,  you will generate a sequence of samples that can be
used to reconstruct a sinusoidal signal component of frequency fsig.

Unfortunately, since the amplitude of the samples depends on the phase
angle theta, and since theta can not be determined from the samples,
there is no way to determine the correct amplitude of this reconstructed
component.

You can't consider a signal to have been 'perfectly' or even 'properly'
re-constructed if the amplitude of one of its components can't be
reliably determined.

Regards,
John
```