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I need a closed-form expression for the time derivative of the arctangent of a ratio of two time-dependent variables. I need advice because a credible reference I found doesn't produce the expected result.

After decomposing a real-valued signal into its in-phase ( I(t) ) and quadrature ( Q(t) ) components, I know that arctan(Q/I) is the instantaneous phase angle of the signal. I need to calculate the time derivative of arctan(Q/I) to obtain the instantaneous frequency of the signal.

A paper from MIT OpenCourseWare uses a straightforward derivation to reduce this problem to a simple result:

$$\frac{d}{dt}\arctan(x) = \frac{1}{1+x^2}$$

For a test signal comprising two tones of equal amplitude, simulation demonstrates that arctan(Q/I) increases linearly with time, implying a constant time derivative. The "MIT" expression for the time derivative produces a sine wave whose frequency is the sum of the frequencies of the two tones comprising the test signal.

I applied the result of each expression to the input terminal of a voltage-controlled oscillator. The constant time derivative produces the desired response: a sinusoid with "phase flips" at the zeros of the envelope and a spectrum comprising two "tones" at the frequencies of input signals. The "MIT" derivative produces a sinusoid without the "phase flips" and a spectrum that appears to be an FM signal with peaks corresponding to the sum of the input signal's component tones.

Can anyone explain what I have done wrong?

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  • $\begingroup$ As you provide no details of your functions $I$ and $Q$, it's not possible to determine what's gone wrong. But, did you use the chain rule? $\endgroup$ Aug 24, 2018 at 15:28

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I think the problem might actually be not with the differentiation of $\arctan\frac{Q}{I}$, but your claim that this is the phase. It's a common mistake, but one easily identified with the following observation: multiplying both of $Q,\,I$ by $-1$ should change the phase by $\pi$. The phase is denoted $\operatorname{atan2}(Q,\,I):=2\arctan\frac{Q}{\sqrt{Q^2+I^2}+I}$. The partial derivatives are given here. One way to obtain these results is to note that the phase $\theta$ satisfies $\tan\theta=\frac{Q}{I}$. Differentiating both sides, then dividing by $\sec^2\theta=1+\frac{Q^2}{I^2}=\frac{Q^2+I^2}{I^2}$, gets the results I've linked to.

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  • $\begingroup$ Having been cast into doubt, I searched the internet for, "i and q components of an ssb signal." Page 8 of tinyurl.com/y8wlzhjo both confirms my interpretation of "phase" and follows your train of thought re: partial derivatives. Using these partials produces the value of instantaneous frequency I seek. Thanks. $\endgroup$
    – Brian K1LI
    Aug 24, 2018 at 19:31
  • $\begingroup$ @BrianK1LI Glad I could help. Please consider selecting my answer or a later one if you feel inclined to do so. $\endgroup$
    – J.G.
    Aug 24, 2018 at 19:43
  • $\begingroup$ I would select your answer, but I do not yet have that privilege on this board. Perhaps another viewer can do the honors. $\endgroup$
    – Brian K1LI
    Aug 24, 2018 at 21:01
  • $\begingroup$ @BrianK1LI Ah, my confusion; I thought new users could select answers but couldn't upvote them, but perhaps they've changed that. $\endgroup$
    – J.G.
    Aug 24, 2018 at 21:17
  • $\begingroup$ Actually, I had to apply the chain rule twice to get the answer I needed. All set now! $\endgroup$
    – Brian K1LI
    Aug 25, 2018 at 19:43

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