# Relating phase difference in a signal with its frequency

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.

Frequency is the derivative of phase over time.

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

• 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 9:57
• @Fat32 see this as I think it will answer your question: dsp.stackexchange.com/questions/40893/… – 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

$268/2\pi$ = 42.66 cycles

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

42.66 cycles over .3175 seconds = 134.4 Hz

• 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