I'm having a tough time understanding the phase increments which are quadratic in nature in the chirp signal. When I see consecutive samples, what will I see in terms of phase and if I have to measure the phase difference between the two samples, what will I see ?
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What do you expect you will see? – Dan Boschen Dec 13 '19 at 00:29
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I wasn't able to understand the quadratic nature of time and how does that affect the phase. – Amu Dec 13 '19 at 03:30
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I see, Did Envidia answer your question or is there still confusion? – Dan Boschen Dec 13 '19 at 03:30
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I'm still confused. I'm really sorry. – Amu Dec 13 '19 at 03:32
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No worries don't apologize! Welcome to Signal Processing Stack Exchange as well. It's possible to root of your problem isn't signal processing but math- but I can try to help you figure out where your stuck and where to go for help. – Dan Boschen Dec 13 '19 at 03:34
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Many get confused with "phase" and how we use it - so this answer may help you: https://dsp.stackexchange.com/questions/40893/what-is-an-intuitive-explanation-of-the-phase-of-a-signal/40894#40894 and maybe this one: https://dsp.stackexchange.com/questions/41998/regarding-phase-and-frequency-components-in-the-signal/41999#41999 – Dan Boschen Dec 13 '19 at 03:36
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Great posts ! So what I've gathered from them is that since the phase isn't changing at a constant rate i.e. changing quadratically (if that is a word), it is changing frequencies ? – Amu Dec 13 '19 at 03:52
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See if my answer below is helpful or too simplistic. I am guessing this is where you were having trouble. My answer is to introduction so that you can then understand Envidia's good answer. – Dan Boschen Dec 13 '19 at 03:53
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@Amu To your question, a phase function that is linear will yield a constant instantaneous frequency as you identified via the derivative. If you take the derivative of $\phi(t)$ in my answer, you will see that the frequency is now a linear function of time, hence the relationship between a chirp's linear frequency and its quadratic phase. – Envidia Dec 13 '19 at 17:30
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Take the chirp signal
$$ x(t) = e^{j\pi\alpha t^2} = e^{j\phi(t)} $$
Where $\alpha$ is the chirp rate of the signal. You can see that the function describing the phase, $\phi(t)$, is in the form of a quadratic equation. So you can plot
$$ \phi(t) = \pi\alpha t^2 $$
and see the quadratic nature of the phase at any point in time. The actual phase-change between each sample depends on what your sample rate is.
Envidia
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Thank you @envidia for taking the time to answer my question. Maybe I'm overthinking it. – Amu Dec 13 '19 at 03:34
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@Amu I think you were! Also chew on Dan's answer below, which goes into how the derivative of $\phi(t)$ yields the instantaneous frequency of the signal. – Envidia Dec 13 '19 at 04:57
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It may help to know that frequency is the derivative of phase. A change in phase versus a change in time is frequency by definition. A phase that keeps growing in the positive direction linearly represents a positive frequency. If you looked at this on the complex plane you would see a phasor rotating counter clockwise: constant magnitude and linearly increasing phase. If the phase rotates $2\pi$ it will have completed 1 cycle. Thus we get to cycles/sec and Hz. To complete this visual picture, a similar phasor that rotated clockwise would be a negative frequency: the phase is increasing in the negative direction with time. Add the two of them together and you get a result that stays on the real axis (so actually has a phase that is only 0 and $\pi$) which is a real cosine given by Euler's Identity:
$$cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2}$$
Note that $e^{j\theta}$ is the same thing as $1\angle \theta$, so when you have the expression $e^{j\omega t}$ you have the spinning phasor I described with a magnitude one and a phase that is growing linearly with time at rate $\omega$.
With that in mind hopefully it is much clearer to you now how the phase of a chirp signal relates to its frequency: Phase versus time is the integral of frequency versus time. If the frequency was constant (positive) with time, the phase would be increasing linearly with time. Similarly if you have a frequency ramp (frequency increasing linearly with time which is a chirp), the phase would be quadratically increasing with time.
Dan Boschen
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Oh wow ! The last paragraph especially made things very clear. I will have to go through it in my head a couple of times so that I fully understand it. Thanks a lot. – Amu Dec 13 '19 at 04:05
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by integral of frequency versus time, do you mean instantaneous frequency ? – Amu Dec 13 '19 at 05:26
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Yes the frequency at any instance in time. Are you able to visualize the spinning phasor on the complex plane with regards to frequency? – Dan Boschen Dec 13 '19 at 05:27
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Thanks. So if I take conjugate product of two consecutive samples of a chirp, I would be able to measure the increment in phase which is quadratic. And if I need to measure the rate of change of frequency i.e. chirp rate, I would need one more measurement, so that I can take the difference between the two measurements right ? – Amu Dec 13 '19 at 05:42
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Yes, and the angle of the conjugate product is the phase. Also see this: https://dsp.stackexchange.com/questions/31578/simulation-of-a-frequency-ramp – Dan Boschen Dec 13 '19 at 05:46
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