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?


3 Answers 3


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: \begin{equation} R = M \cos(\theta)\\ X = M \sin(\theta) \end{equation}

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.

\begin{equation} Y = \frac{R - Xj}{(R + Xj)(R - Xj)} = \frac{R - Xj}{R^2 + X^2} \end{equation}

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

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

And the phase of the admittance:

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

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:

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


If I remember correctly, the phase angle just switches the sign and that intensity decreases. So if you had impedance of 10 ohms at 45 degrees, you'd get admittance of around 0.1 siemens at -45 degrees.

We keep in mind that \$ j = \sqrt {-1}\$.

Let's see if I can derive that:

\$ Y=Z^{-1}=\dfrac{1}{R+jX}=\dfrac{1}{R+jX} \dfrac{R-jX}{R-jX}=\dfrac{R-jX}{R^2 + RjX -RjX - j^2X^2}=\dfrac{R-jX}{R^2+X^2}=\dfrac{R}{R^2+X^2}+j\dfrac{-X}{R^2+X^2}=G+jB \$

So the angle switched because the sign in front of the imaginary part switched. The intensity decreased because we got the \$R^2+X^2\$ component. There was no change in angle's absolute value because we decreased both real and imaginary parts by same amount so their ratio stayed the same. Phase angle for admittance is \$ \arctan \left(\frac{B}{G}\right)\$ and since we divided both components by same number, the absolute value of ratio remained constant.

A "quick" way to get the \$R^2+X^2\$ part would be to calculate sine and cosine of the angle multiplied by intensity. So if we have \$Z=\vert Z\vert e^{\alpha}\$ we'd use \$R^2= [\vert Z \vert \cos(\alpha)]^2, X^2=[\vert Z \vert \sin(\alpha)]^2 \$ which should be easy enough to do with a simple calculator. Then we'd add those two together, divide intensity using them and flip the angle so we get:

\$Y= \dfrac{1}{ \vert Z \vert\ e^{a}}= \dfrac { \vert Z \vert}{[\vert Z \vert \cos(\alpha)]^2+[\vert Z \vert \sin(\alpha)]^2}e^{-\alpha} =\dfrac{1}{\vert Z \vert [\cos^2(\alpha)+sin^2(\alpha)]}e^{- \alpha}=\dfrac{1}{\vert Z \vert}e^{- \alpha}\$

which should have been obvious to me from the start.

So final formula for quick conversion is: \$Y= \dfrac{1}{ \vert Z \vert\ e^{a}}=\dfrac{1}{\vert Z \vert}e^{- \alpha}\$

  • 1
    \$\begingroup\$ units of admittance is siemens, not ohms :P \$\endgroup\$ Feb 12, 2013 at 3:08
  • 1
    \$\begingroup\$ Yeah, I know that! :) That's the danger of writing answers at 4 a.m. after not sleeping for a couple of days. \$\endgroup\$
    – AndrejaKo
    Feb 12, 2013 at 3:09
  • \$\begingroup\$ Nice derivation, 4am or not \$\endgroup\$ Feb 12, 2013 at 3:21

Yes there is an easier way. If you have to have it in complex form?

(1) Convert 10/_40 to 7.6604+6.4279i.

Then take the reciprocal (1/(7.6604+6.4279i))=0.76604-0.06427i. Even a cheap Casio fx-115ES Plus will do these fast.

(Set to Complex Format) and input 10/40-->Enter, then take the reciprocal and get 0.1/-40.The Casio will convert it back to complex \$R+jX\$ if you need it.

Can also do this way(Spherical):

Given \$Z=\frac{10}{40}\$ ohms: \$R=10\cos(40)\$ and \$X=10\sin(40)\$



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