# Phase of the chirp signal

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 ?

• What do you expect you will see? – Dan Boschen Dec 13 '19 at 0:29
• I wasn't able to understand the quadratic nature of time and how does that affect the phase. – Amu Dec 13 '19 at 3:30
• I see, Did Envidia answer your question or is there still confusion? – Dan Boschen Dec 13 '19 at 3:30
• I'm still confused. I'm really sorry. – Amu Dec 13 '19 at 3:32
• 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 3:34

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.

• Thank you @envidia for taking the time to answer my question. Maybe I'm overthinking it. – Amu Dec 13 '19 at 3:34
• @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 4:57
• Yes. Thank you @Envidia for your time again. – Amu Dec 13 '19 at 5:49

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.

• 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 4:05
• You mean what $e^{j\theta}$ represents, simply? – Dan Boschen Dec 13 '19 at 4:06
• *last paragraph. Sorry. – Amu Dec 13 '19 at 4:07
• by integral of frequency versus time, do you mean instantaneous frequency ? – Amu Dec 13 '19 at 5:26
• 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 5:27