Derivative of \(arg(z(t))\) with respect to \(t\)
Context
Given
- \(x(t) \in \mathbb{R}\) is differentiable
- \(y(t) \in \mathbb{R}\) is differentiable
- \(z(t) = x(t) + jy(t)\)
- \(\theta(t) = \arg(z(t))\) is the argument (angle) of the complex signal \(z(t)\)
Problem
What is the derivativie of \(\theta(t)\) with respect to \(t\)?
Solution
\[
\theta'(t) = \frac{\partial \theta(t)}{\partial t} = \frac{x(t)y'(t) - y(t)x'(t)}{x^2(t) + y^2(t)}, \ x \neq 0 \mbox{ and } y \neq 0
\]
Derivation
The principal value of the argument of \(z(t)\) can be calculated by:
\[
Arg(z(t)) = \left\{
\begin{array}{ll}
\arctan(\frac{y(t)}{x(t)}) & \mbox{if } \ x > 0, \\
\arctan(\frac{y(t)}{x(t)}) + \pi & \mbox{if } \ x < 0 \mbox{ and } y \geq 0, \\
\arctan(\frac{y(t)}{x(t)}) - \pi & \mbox{if } \ x < 0 \mbox{ and } y < 0, \\
+\frac{\pi}{2} & \mbox{if } \ x = 0 \mbox{ and } y > 0, \\
-\frac{\pi}{2} & \mbox{if } \ x = 0 \mbox{ and } y < 0, \\
\mbox{undefined} & \mbox{if } \ x = 0 \mbox{ and } y = 0.
\end{array}
\right.
\]
Or equivalently,
\[
Arg(z(t)) = \left\{
\begin{array}{ll}
\arctan(\frac{y(t)}{x(t)}) & \mbox{if } \ x(t) > 0, \\
\arctan(\frac{y(t)}{x(t)}) + \pi & \mbox{if } \ x(t) < 0 \mbox{ and } y(t) \geq 0, \\
\arctan(\frac{y(t)}{x(t)}) - \pi & \mbox{if } \ x(t) < 0 \mbox{ and } y(t) < 0, \\
\frac{\pi}{2} - \arctan(\frac{x(t)}{y(t)}) & \mbox{if } \ y(t) > 0, \\
-\frac{\pi}{2} - \arctan(\frac{x(t)}{y(t)}) & \mbox{if } \ y(t) < 0, \\
\mbox{undefined} & \mbox{if } \ x(t) = 0 \mbox{ and } y(t) = 0.
\end{array}
\right.
\]
Note that \(Arg(z(t))\) is not differentiable w.r.t. \(t\) at \(x(t) < 0, y(t) = 0\) due to the discontinuity.
To find \(\theta'(t)\), we take the following steps:
- Define \(\theta(t)\) in a way that \(\theta(t)\) is differentiable in each of the following 4 half-planes: \(x(t) < 0\), \(x(t) > 0\), \(y(t) < 0\), \(y(t) > 0\)
- Find \(\theta'(t)\) in each of the 4 half-planes
- Confirm \(\theta'(t)\) is identical among the 4 overlapping half-planes.
To address the discontinuity in \(Arg(z(t))\), we choose to calculate \(\theta(t)\) by
\[
\theta(t) = \left\{
\begin{array}{ll}
\arctan(\frac{y(t)}{x(t)}) & \mbox{if } \ x(t) > 0, \\
\arctan(\frac{y(t)}{x(t)}) + \pi & \mbox{if } \ x(t) < 0 \mbox{ and } y(t) \geq 0, \\
\arctan(\frac{y(t)}{x(t)}) - \pi + 2\pi & \mbox{if } \ x(t) < 0 \mbox{ and } y(t) < 0, \\
\frac{\pi}{2} - \arctan(\frac{x(t)}{y(t)}) & \mbox{if } \ y(t) > 0, \\
-\frac{\pi}{2} - \arctan(\frac{x(t)}{y(t)}) & \mbox{if } \ y(t) < 0, \\
\mbox{undefined} & \mbox{if } \ x(t) = 0 \mbox{ and } y(t) = 0.
\end{array}
\right.
\]
or equivalently,
\[
\theta(t) = \left\{
\begin{array}{ll}
\arctan(\frac{y(t)}{x(t)}) & \mbox{if } \ x(t) > 0, \\
\arctan(\frac{y(t)}{x(t)}) + \pi & \mbox{if } \ x(t) < 0, \\
\frac{\pi}{2} - \arctan(\frac{x(t)}{y(t)}) & \mbox{if } \ y(t) > 0, \\
-\frac{\pi}{2} - \arctan(\frac{x(t)}{y(t)}) & \mbox{if } \ y(t) < 0, \\
\mbox{undefined} & \mbox{if } \ x(t) = 0 \mbox{ and } y(t) = 0.
\end{array}
\right.
\]
The next step is to find \(\theta'(t)\) in each of the 4 half-planes. For \(x(t)>0\), we have
\[
\tan( \theta(t) ) = \tan( \arctan(\frac{y(t)}{x(t)})) = \frac{y(t)}{x(t)}
\]
Differentiating both sides of the equation w.r.t \(t\) yields
\[
\sec^2( \theta(t) ) \theta'(t) = \frac{x(t)y'(t) - y(t)x'(t)}{x^2(t)}
\]
Given \(\sec^2 = 1 + \tan^2\) and \(\tan^2( \theta(t) ) = \frac{y^2(t)}{x^2(t)}\), the equation above can be written as
\[
\begin{align}
\left[ 1 + \frac{y^2(t)}{x^2(t)} \right] \theta'(t) &= \frac{x(t)y'(t) - y(t)x'(t)}{x^2(t)}, \\
\theta'(t) &= \frac{x(t)y'(t) - y(t)x'(t)}{x^2(t)} \left[ 1 + \frac{y^2(t)}{x^2(t)} \right]^{-1} \\
&= \frac{x(t)y'(t) - y(t)x'(t)}{x^2(t) + y^2(t)}
\end{align}
\]
For \(x(t) < 0\), we get the same result of \(\theta'(t)\) as for \(x(t) >0\) because
\[
\frac{\partial}{\partial t}\left( \arctan(\frac{y(t)}{x(t)}) + \pi \right) = \frac{\partial}{\partial t}\left( \arctan(\frac{y(t)}{x(t)}) \right)
\]
For \(y(t) < 0\), since \(\tan(\frac{\pi}{2} - \phi) = \frac{1}{\tan \phi}\) and \(\tan(- \phi)= - \tan \phi\), we have
\[
\begin{align}
\tan(\theta(t)) &= \tan \left( - \frac{\pi}{2} - \arctan(\frac{x(t)}{y(t)}) \right) \\
&= - \tan \left( \frac{\pi}{2} + \arctan(\frac{x(t)}{y(t)}) \right) \\
&= \frac{1}{\tan( \arctan(\frac{x(t)}{y(t)}))} \\
&= \frac{y(t)}{x(t)}
\end{align}
\]
which is the same as the case for \(x(t) > 0\), and therefore the solution to \(\theta'(t)\) will be the same.
YYC Note:
Can we really differentiate \(\tan(\theta(t))\) in the region of \(y(t) < 0\)? Is it differentiable at \(x(t) = 0\)?
For \(y(t) > 0\), we get the same results of \(\theta'(t)\) as for \(y(t) < 0\) because
\[
\frac{\partial}{\partial t}\left( \arctan(\frac{\pi}{2} - \frac{x(t)}{y(t)}) \right) = \frac{\partial}{\partial t}\left( - \frac{\pi}{2} - \arctan(\frac{x(t)}{y(t)}) \right)
\]
Now we conclude that the solution \(\theta'(t) =\frac{x(t)y'(t) - y(t)x'(t)}{x^2(t) + y^2(t)}\) holds for all 4 half-planes, and therefore it is the solution for all \(x(t), y(t)\) where \(x(t) \neq 0\) and \(y(t) \neq 0\).
Reference