# Understanding the relation between instantaneous frequency and phase of a sinusoid

Sorry for a basic question but I really wanted to understand the relation between the instantaneous frequency and instantaneous phase of a sinusoid. We know that we can get the the instantaneous frequency by differentiating the instantaneous phase and normalizing with 2*pi. Also, we could get the instantaneous phase by integrating the frequency. But i am not quite understanding how integrating a frequency gives instantaneous phase. I viewed this link - http://mriquestions.com/phase-v-frequency.html. The total area under the graph of Instantaneous frequency gives phase but how to interpret it practically?

For example, if I consider a sinusoid whose frequency is 10Hz and is constant for 10 seconds , it gives me total area as 100. Another sinusoid with 10Hz for 5 seconds and 20Hz for 5 seconds, gives a total area of 150. How these numbers (areas) can be interpreted as phase?

• it's the fundamental theorem of calculus. if you understand that differentiating instantaneous phase (w.r.t. time) yields instantaneous frequency, then you should also understand that integrating instantaneous frequency will get you back to instantaneous phase. – robert bristow-johnson Jun 15 '17 at 16:24
• Can you see that a linearly increasing phase corresponds to a constant frequency? That would be a start. – Matt L. Jun 15 '17 at 17:45
• @MattL. So, linearly increasing phase corresponds to increasing frequency... – sundar Jun 18 '17 at 16:23
• @sundar: No, that's not the case, as already mentioned in my previous comment. – Matt L. Jun 18 '17 at 16:27

An analogy: a car is at $x=0$ at $t=0$ and moves with velocity $v$ (constant). The famous formula for this is:

$$x = v \cdot t$$

By integrating velocity $v$ one gets position $x$, by derivating position $x$ one gets velocity $v$.

Now, take position to be a point in a circle (for this is convenient to use radians instead of raw meters along the circumference, but this does not really matter). Take velocity to be how fast that point moves around the circle (for this is better to choose an unit like radians per second to make it independent of the circle radius, but this does not matter neither). This would be the phasor with constant rotational speed, a.k.a sine wave. So, we can reinterpret the formula like:

$$\theta = \omega \cdot t$$

And the same logic as above applies: By integrating frequency $\omega$ one gets phase $\theta$, by derivating phase $\theta$ one gets frequency $\omega$.

You can verify by dimensional analysis that they both are the same formula actually (well, actually, radian is said to be a dimensionless unit, something like a 'dozen' or a 'mole'... that is another story):

$$[Length] = [Length/Time] \cdot [Time]$$

When you integrate with respect to time, one less $[Time]$ appear in the denominator. The derivative does the opposite.

• Thanks very much.. Now I see it.... If phasor speed is constant, we get a constant frequency. If phasor speed increases linearly, the frequency of the total sinusoid increases linearly. – sundar Jun 16 '17 at 18:37
• Yes, just remember that a phasor is actually made of a sine (projection onto the y-axis) and a cosine (projection onto the x-axis): phasor = cosine + j*sine. You can always check Euler's identities for this. – oxuf Jun 16 '17 at 18:44
• So, I just got a reply from Matt (in the questions comment section) that linearly increasing phase does not correspond to increasing frequency. And from the first comment in the answer section, the increasing phase corresponds to increasing frequency. I am confused now. – sundar Jun 19 '17 at 21:17
• linear increasing phase corresponds to constant frequency the same way that linear increasing position corresponds to constant speed – oxuf Jun 19 '17 at 21:33
• in the first comment to this answer you wrote: "phasor speed" and that is correct, "phasor speed" (frequency) is the "car speed" in the analogy – oxuf Jun 19 '17 at 21:43