# Nyquist Frequency

There is a statement regarding the Nyquist frequency in one of my textbooks which I don't quite follow. I would appreciate it if someone could clarify it for me.

Now, the way I understand it - if we sample a signal with frequency $f_{s}$, then the highest possible signal frequency which we can reconstruct without any aliasing is half the sample frequency. This is called the Nyquist frequency. So we have:

$$f_{Ny} = \frac{1}{2} f_{s}$$

The statment in my textbook which confuses me is as follows:

"Recall that the Sampling Theorem approximately reconstructs a signal $f$ from samples taken uniformly at intervals of length $T$. If the signal is band-limited and its Nyquist frequency is less than $1/T$, then the reconstruction is perfect; otherwise it's an approximation"

I think this statement is somewhat confusing. After all the Nyquist frequency isn't a fixed number, but changes depending on our sample rate. So wouldn't the Nyquist frequency automatically be $\frac{1}{2T}$? In other words, why the need to to write "if the Nyquist frequency is less than $1/T$"? Wouldn't this always be the case?

I would really appreciate it if someone could clarify this statement for me!

• Can you tell what book it is – niaren May 23 '13 at 9:09
• Hi. The book is "A First Course in Wavelets with Fourier Analysis" - Second Edition, by Boggess and Narcowich. The quote is from pages 190-191. – Kristian May 23 '13 at 15:50

Nyquist rate depends only on the bandwidth of your signal and doesn't have anything to do with your sampling frequency of the signal.

You are free to sample the signal with any frequency you like, but if you want to be able to reconstruct the original signal from its samples, you have to sample with a frequency more than Nyquist rate (and the Nyquist frequency will be half of that).

All that Nyquist–Shannon sampling theorem states is that if you sample a band-limited signal with a rate more than Nyquist rate you can reconstruct it exactly.

EDIT: note that Nyquist frequency is a property of the sampling process, but Nyquist rate is a property of your original (CT) signal.Although in some textbooks like Discrete-Time Signal Processing by Oppenheim & Schafer ,Nyquist frequency is simply taken to be the maximum frequency of a bandlimited signal.

• Thanks a lot. I must admit I am still a little confused. You write: "if you want to be able to reconstruct the original from its samples, you have to sample with a frequency more than Nyquist frequency". But shouldn't it be "Twice the Nyquist frequency"? Say a signal has a maximum frequency of 3 Hz. If I sample with a frequency of 4 Hz, then the Nyquist frequency is 2 Hz. So my sample frequency is higher than the Nyquist frequency, but I still would not be able to properly reconstruct the 3 Hz frequency. Isn't that right? I'm very sorry for the confusion. This is new to me :) – user12277 May 22 '13 at 18:09
• You are right,as I have written in the last paragraph, it is Nyquist rate (=2*Nyquist frequency).corrected! – Mo's May 22 '13 at 18:17
• Great! Thanks a lot. So the book is somewhat confusing then. It's not me there's something wrong with :). Appreciate it a lot! – user12277 May 22 '13 at 18:18
• -1. You understand the concepts correctly, but you are using the term "Nyquist frequency" incorrectly. The Nyquist frequency is set by the sample rate. The signal bandwidth determines the sample rate that you need to reconstruct the signal, as you say, but it does not set the Nyquist frequency. That is done by the sample rate. – Jim Clay May 23 '13 at 12:42
• I have a copy of Oppenheim-Schafer DSP and so I looked it up. You are correct that they define the Nyquist frequency based on the signal-of-interest, and not the sample rate. I believe that they are wrong to do so, and that it is contrary to the general usage of the term. – Jim Clay May 23 '13 at 15:10

Yes, in fact a more general theorem can be proven (Shannon Theorem), which states exactly what you wrote. You can use a frequency of sampling greater than the Nyquist rate and you wil get an exact determination of the signal.

• Thank you very much for your input. Yeah, I see that this follows when using the Nyquist rate terminology. – user12277 May 22 '13 at 18:11