# Nyquist rate is not true always

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I have come up with something. I don't know if it is true or not. I just want your opinion. Nyquist said that the sampling frequency should be at least twice than that of the maximum frequency. But that is not always true if we make use of the phase of the signal.

For eg:

take Sampling frequency fs=8000;

So t=0:1/8000:1;

Consider a signal:

y=sin(2*pi*3000*t);

Consider another signal:

y2=sin(2*pi*5000*t);

Now if we find the fft

k=fft(y);

k2=fft(y2);

a1=abs(k);

a2=abs(k2);

b1=angle(k);

b2=angle(k2);

figure,plot(b1),figure,plot(b2)

The magnitude spectrum of the two waves y and y2 will be same but not the angle spectrum. So take the signal together

z=y+y2;

z2=y;

k=fft(z);

k2=fft(z2);

If you find the magnitude spectrum for both they will be same which states that there is only one frequency component in both of the signals z and z2. So we say there is aliasing. But we can avoid this by taking the angle / phase spectrum into consideration which is

ang1=angle(k);

ang2=angle(k2);

If you plot both they will be different. Using this phase spectrum we can indeed differentiate two signals and say that the signal z consists of 3000Hz and 5000hz waves though the sampling frequency is less than twice that of the maximum frequency of the signal which is 5000Hz. So I believe Nyquist rate is not always true. But anyway the sampling frequency should be at least be equal to the maximum frequency of the wave in the above case or else we will have aliasing for sure. What do you say?

##### 0 Comments

### Accepted Answer

Daniel Shub
on 5 Nov 2011

The Nyquist theorem basically says that if you sample at twice the highest frequency you can reconstruct the signal exactly without any additional information.

What you are asking is: if the phase, as a function of frequency, of the signal is known, what sample rate is required to reconstruct the signal. This is a very different question from the one Nyquist address. I believe the answer is that you need to sample at same same frequency as the highest frequency component (i.e., half the Nyquist rate).

EDIT As Wayne points out in one of his answers, the above belief is ridiculously wrong.

Amazingly, if the amplitude, as a function of frequency, of the signal is known, you also only need to sample at 1/2 the Nyquist rate.

As Wayne points out, if the amplitude and phase, as a function of frequency, of the signal is known you don't even need to sample your signal to reconstruct it.

### More Answers (11)

Wayne King
on 5 Nov 2011

@iVenky, @Daniel "I need to know only phase. I don't need the amplitude. I assumed that the initial phase is zero. Then I can find or else I can't."

I still do not agree with this but that's fine.

t = 0:0.01:1-0.01;

x = cos(2*pi*10*t);

x1 = cos(2*pi*90*t);

xdft = fft(x);

xdft1 = fft(x1);

Look at the infinitesimal difference between these time domain signals.

max(abs(x-x1))

% and

norm(x-x1)

The DFT of these signals should only have two nonzero DFT coefficients, at [-10, 10] Hz. Which means the phase is zero at all DFT bins except those two.

Of course, the fft() output does contain nonzero DFT coefficients at other bins, but you cannot count on these numerical imprecisions in the computer implementation of the DFT on different architectures and different FFT algorithms as the basis to differentiate between x1 and x above.

The maximum difference between these signals in the time domain is on the order of 10^{-14}. In the Fourier domain,

max(abs(xdft-xdft1))

10^{-13}.

Are you going to assert this is robust enough to tell whether you have x or x1?

And by the way, look at the DFT coefficients at the only frequency that matters:

xdft1(11)

xdft(11)

identical.

##### 4 Comments

Wayne King
on 5 Nov 2011

Hi Venky, That is not correct. There are a few flaws in your reasoning:

How do you presume to tell that the frequency is actually 5 kHz from the phase and not a component at 3 kHz with a different phase? Or an alias from a different frequency? With a sampling rate of 8 kHz, 13 kHz also gets aliased to 3 kHz, as well as an infinite number of other frequencies.

t=0:1/8000:1;

x = cos(2*pi*13000*t);

y2 = cos(2*pi*5000*t);

How will you reproduce the correct analog signal from these samples? HOw do you discern that x came from a 13 kHz signal with a phase of 0, while y2 came from a 5 kHz signal with a phase of 0?

Your general statement, however, can be true. There are instances where you do not need to sample the data at least at the Nyquist rate to be able to reconstruct the original signal (or image).

Read about compressed sensing, or compressive sensing. It's a very important developing area of research, which exactly addresses that question: when is the Nyquist rate not necessary?

##### 0 Comments

Wayne King
on 5 Nov 2011

##### 0 Comments

Wayne King
on 5 Nov 2011

Venky, with all due respect, you're missing the point. How are you going to tell that it was 5 kHz and tell its correct phase? That is what you need to reconstruct the signal.

How do you know that you didn't take 13 kHz? That is the problem faced by undersampling.

By your reasoning, if I know exactly what my input was, then I can reconstruct the output exactly even if it is not sampled adequately.

That is trivially true: Here is the my new theorem:

"If I know exactly what the continuous-time input is, then I can reconstruct the signal exactly based on 1 sample taken at any time." (actually better than that I can reconstruct it based on 0 samples).

The Nyquist theorem is not there to guide us when we know what the input signal contains exactly.

Finally, you still have not shown how you can tell 3 kHz from 5 kHz in this situation. If I created files for you with signals consisting of either 3 or 5 kHz (where 5 kHz is alised to 3), you have no way of telling me which frequency it contains. You have no technique that would perform better than chance.

##### 0 Comments

Wayne King
on 5 Nov 2011

OK, here is a 30-point signal

0.9239

0.5225

-0.0785

-0.6494

-0.9724

-0.9239

-0.5225

0.0785

0.6494

0.9724

0.9239

0.5225

-0.0785

-0.6494

-0.9724

-0.9239

-0.5225

0.0785

0.6494

0.9724

0.9239

0.5225

-0.0785

-0.6494

-0.9724

-0.9239

-0.5225

0.0785

0.6494

0.9724

I'll tell you that the sensor only allowed inputs [0,100] Hz. The data is sampled at 100 Hz. Tell me the correct frequency and phase. You need to be able to tell me the correct frequency and phase, or I can't reconstruct the signal.

##### 2 Comments

Wayne King
on 5 Nov 2011

and realize this is much simplified since you have a certain probability of getting it by chance.

Wayne King
on 5 Nov 2011

Here is an even better test for you (same scenario):

1.6310

1.6795

1.3566

0.7856

0.1846

-0.2168

-0.2653

0.0577

0.6286

1.2296

1.6310

1.6795

1.3566

0.7856

0.1846

-0.2168

-0.2653

0.0577

0.6286

1.2296

1.6310

1.6795

1.3566

0.7856

0.1846

-0.2168

-0.2653

0.0577

0.6286

1.2296

How you take this data and show me exactly what frequencies and phases are present so I can reconstruct the analog signal.

Mohammed Sanad
on 22 Oct 2018

Can some one tell me what is the defferent between nyquist frequency and Same as Max frequency?

##### 0 Comments

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