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I have a 134 point complex signal. Using MATLAB, I have observed that the phase difference between any two adjacent samples of this signal is 2 radians i.e. the phase increases linearly from 0 to 268 radians over 134 samples.

How can I relate this phase difference between adjacent samples with the frequency of the same signal?

PS I am sampling at 422 Hz.

Chaitanya Patil
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Frequency is the derivative of phase over time.

So, what's the time between two adjacent samples? That directly gives you the frequency.

Marcus Müller
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    Hi ! I understood that by the phase, OP means the complex number $z[n] = a[n] + jb[n]$ 's phase $\phi = \tan^{-1}(b/a)$..? So what's the relation to time derivative of phase as it's in the argument of a sinusoidal signal $\cos(\theta(t)) = \cos(\omega t + \phi)$ whose derivate yields $\theta(t)' = \omega$ as the frequency ? Am I missing something ? Need some more coffe may be :-) ? – Fat32 Aug 31 '18 at 09:57
  • @Fat32 see this as I think it will answer your question: https://dsp.stackexchange.com/questions/40893/what-is-an-intuitive-explanation-of-the-phase-of-a-signal/40894#40894 – Dan Boschen Aug 31 '18 at 11:31
  • @DanBoschen I still need more coffee :-). Eventhough the IQ de-modulation complex signal representation makes sense, if I take a real signal (zero-phase) as a special case of the general complex signal, then its frequency is zero? So probably I don't get the meaning of freuqency here? is it the modulation information on the signal or the signal spectrum of the signal? I assumed the latter. But the IQ modulation assumption assumes the former... So I guess OP wanted to say complex signal from IQ demodulation... – Fat32 Aug 31 '18 at 11:40
  • The real signal has a positive and negative frequency ($e^{j\omega t}$ is a phasor spinning counter-clockwise on the complex IQ plane representing a positive frequency). Look at Euler's identity and how it represents both a positive and negative frequency! $Ke^{j\theta}$ is the same as writing magnitude K and angle $\theta$ in case you didn't know that relationship. – Dan Boschen Aug 31 '18 at 11:43
  • With this view, individual frequencies are all just one spinning complex phasor in time on the complex IQ plane - and that is what becomes a single impulse in the frequency domain, while something like a cosine or sine must be represented with two impulses in frequency (positive and negative), as you must have two spinning at the same rate and magnitude in opposite directions for the imaginary component to cancel and result in a real signal. – Dan Boschen Aug 31 '18 at 11:46
  • This is not specific to IQ modulation, it just decomposes signals into real and imaginary components. $e^{j\theta} = cos(\theta)+j sin(\theta)$ and $cos(\theta)=\frac{1}{2}e^{j\theta}+\frac{1}{2}e^{-j\theta}$ and thus we see how this can be represented and viewed. Once we do start modulating and manipulating frequencies and signals, having this view is very helpful. – Dan Boschen Aug 31 '18 at 11:50
  • Further--- Yes the real signal as the sum of all the phasor components has a phase that only changes between 0° and 180° (as for a cosine or sine), as you observed when you said "zero phase". But to understand the frequency content of any signal, we decompose it first into individual complex spinning phasors that each spin at a constant rate with constant magnitude. At this point we can see the individual frequency components and measure their phase versus time to determine their frequency. This in fact is the Fourier Transform. – Dan Boschen Aug 31 '18 at 11:54
  • @DanBoschen I don't get yor point now. When you say frequency of a signal I assume its Fourier spectrum, but if you model the signal as IQ de-moulation then it's the modulating message signal's frequency? So $x[n] = (1 + j0) u[n] $ is a real signal whose frequency (spectrum) is $X(w) = \frac{1}{1 - e^{jw}} + \pi \delta(w)$, so what's the derivative of the complex phasor thing for this $x[n]$ ? it seems zero, since number is real for all n? So as I said in my reply, may be OP implicitly wants to say IQ demodulated complex data which supports the derivative of phase = frequency argument. – Fat32 Aug 31 '18 at 12:02
  • Let us continue this discussion in chat. – Dan Boschen Aug 31 '18 at 12:03
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$268/2\pi$ = 42.66 cycles

$1/422$ $\times$ 134 samples = .3175 seconds

42.66 cycles over .3175 seconds = 134.4 Hz

If I had enough coffee

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    I look at it as $\omega = 2\pi f = 2 radians$, which means a normalized frequency of $f = 2/(2\pi)$, to get the actual frequency we scale by $F_s$ and get the same result. – learner Aug 31 '18 at 16:48