# Electronic – How does the conversion between admittance and impedance affect phase angle

impedance

Wikipedia tells me that admittance \$Y\$ is the reciprocal of impedance \$Z\$:

\$Y = Z^{-1} = \dfrac{1}{R+jX}\$

which makes plain enough sense to me, except that I've forgotten how complex numbers work from high-school math. If someone gives me an impedance in polar coordinates, say "10 ohms at 40 degrees", is there a trivial way to convert this to an admittance? If it were a simple resistance, calculating the conductance is easy, \$10 \Omega = 0.1S\$. But what happens to the phase angle?

You can analyze phasor notation in a quasi-2D Cartesian fashion. The real part is the "x", and the complex part is the "y".

So given a phasor magnitude M with angle Theta,

Using trig: $$R = M \cos(\theta)\\ X = M \sin(\theta)$$

We now have the complex impedance R + Xj

To invert, you can multiply by the complex conjugate (R - Xj) to both the numerator and denominator.

$$Y = \frac{R - Xj}{(R + Xj)(R - Xj)} = \frac{R - Xj}{R^2 + X^2}$$

To compute the magnitude of the admittance, use the distance formula:

$$M_Y = \sqrt{\left(\frac{R}{R^2 + X^2}\right)^2 + \left(\frac{-X}{R^2 + X^2}\right)^2}$$

And the phase of the admittance:

$$\theta_Y = \tan^{-1}\left(\frac{-X}{R}\right)$$

Note that tangent is a bit finicky for computing the phasor angle as you have to be careful about the quadrant. If you're using a computer, they often times have an "atan2" function which takes the x and y coordinates directly and computes the CCW angle from the positive X axis.

A closer look at the phase angle mapping, and it looks like the admittance phase angle is just the reflection of the impedance phase angle about the real/X axis.

For example, an impedance phase angle of 45 degrees is equal to an admittance phase angle of -45 degrees.

And this makes sense if I had used some identities above:

$$\theta_Y = -\tan^{-1}\left(\frac{X}{R}\right) = -\theta_X$$