[SOLVED] Nyquist rate is not true always

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. Consider the following matlab program.
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;
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 atleast 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?

In a short, the Nyquist criterion must be kept to be able reconstruct arbitrary signals.

Your example is based on assumptions. You're concluding, that the image represents the same signal due to some of it's properties. But it must not necessarily be so. There can be two signals as well. It's easy in the trivial case of a single frequency line. But if you go to complex signals, not keeping the Nyquist criterion causes complete confusion in signal analysis.

I know that it may not be possible practically. But theoretically it is indeed possible. Even for complex signals like a speech signal the above statement is possible if you know the initial phases of all. But calculation will be complex but not impossible.

But theoretically it is indeed possible.

Click to expand...

I don't see, how.

What I have shown is for two frequencies 3khz and 5khz sampled at 8khz. By Nyquist theorem you can't even find 5khz and it will be seen as 3khz. But if you make use of the angle spectrum you can see that the angle spectrum of 3khz and 5khz are different based on which we can distinguish between 3 ahd 5khz. You have to extend the same rule when you have multiple frequencies in the same signal.

I don't understand what you mean with angle spectrum. Suppose a stationary input signal of 8 kHz sampled at 5 kHz. The signal has arbitrary phase. In the complex FFT, the signal shows by a line at 3 kHz, comprised of a real and an imaginary value. How can you know if the input signal is 3 or 8 kHz without knowing it's original phase?

In your example, both sgnals can be only distinguished, because you know their phase (both or assumed as sin()). But the signal could be cos() or -sin() as well.

I never included 8khz. I said 5khz sampled at 8khz not the other way round.

---------- Post added at 16:53 ---------- Previous post was at 16:52 ----------

That's what I am saying. If you know the original phase you can distinguish between 3khz and 5khz when you sample at 8khz.

---------- Post added at 16:54 ---------- Previous post was at 16:53 ----------

I don't understand what you mean with angle spectrum

Click to expand...

I mean angle(k)

If you know the original phase you can distinguish between 3khz and 5khz when you sample at 8khz.

Click to expand...

That's it, If....It's meaningless in my opinion. Digital signal processing is more than a lab problem, where you now the signal before.

P.S.: There are in fact other examples of special signals, that are not tied to the Nyquist criterion. E.g. undersampling of a band limited signal in a digital receiver. These case are rather confirming the Nyquist theorem in my view. Nquist applies for unknown signals. If you know some signal properties, you can possibly ignore it.

Okay I get it. If I know the initial phase then I can or else I can't. I concede.