I am reading a paper about making a phasor model of an inductor. For this they use phasor transformations.
The voltage over an inductor is calculated by $$v = L \frac{di}{dt}$$ The current and voltage are in the paper defined as $$i(t) = \Re\left[\bar i(t) e^{j\int \omega_s dt}\right] $$
$$v(t) = \Re\left[\bar v(t) e^{j\int \omega_s dt}\right] $$
So now you basically fill the current and voltage in the first formula and in the paper they get the following result:
$$ \bar v(t) =L \frac{d}{dt}\bar i(t) +jL\bar i(t)\omega_s(t) $$
Now i tried to do this myself but I don't get to the same answer as they do in the paper. This is what I did:
$$\bar v(t) *\cos \left( \int\omega_s(t)dt\right) = L\frac{d\bar i(t)\cos(\int \omega_sdt)}{dt} $$
$$ \bar v(t) *\cos \left( \int\omega_s(t)dt\right)= L\left(\frac{d\bar i(t)}{dt}\cos(\int\omega_sdt)-\bar i(t)\omega_s(t)\sin(\int\omega_s(t)dt\right)$$
$$\bar v(t) = L\frac{d}{dt}\bar i(t)-\frac{\sin(\int\omega_s(t)dt)}{\cos(\int\omega_s(t)dt)}L\bar i(t)\omega_s(t) $$
$$\bar v(t) = L\frac{d}{dt}\bar i(t)-\tan(\int\omega_s(t)dt)L\bar i(t)\omega_s(t) $$
This is the point where I don't know how to go on. I have never really seen a tangent being equal to the complex number $j$. I think i am almost there but can anyone send me in the right direction on the last step?
Edit: $\omega_s$ is the instantanious frequency which is defined as $$\omega_s(t) = \frac{1}{2\pi}\frac{d\phi(t)}{dt} $$
Best Answer
You may have a typo in your answer. I suspect that the proper result is $$\bar v\left( t \right) = L\frac{d}{{dt}}\left( {\,\bar i\left( t \right)} \right) + jL\,\bar i\left( t \right){\omega _s}\left( t \right)$$ To see this, in addition to the real parts of the phasors, we may assume that their imaginary parts (and thus the total complex sums) also fulfill the constitutive relationship. Then $$\bar v\left( t \right){e^{j\int {{\omega _s}dt} }} = L\frac{d}{{dt}}\left( {\bar i\left( t \right){e^{j\int {{\omega _s}dt} }}} \right)$$ or $$\bar v\left( t \right) = L\frac{d}{{dt}}\left( {\bar i\left( t \right){e^{j\int {{\omega _s}dt} }}} \right){e^{ - j\int {{\omega _s}dt} }}$$ Expanding the derivative then leads to the modified result.