# How do I convert impedance to this weird phasor-form?

April 13, 2009 3:16 PM Subscribe

ElectricalEngineeringFilter: How do I add electrical impedances together, and convert the result to what looks to be a phasor?

So I'm in an electric circuits class, we have a test in three days and one simple concept is really confusing me.

I understand that electrical impedance (Z) can be in a complex format, like, for example, 90+j120. But then the book somehow converts that into what looks like a phasor. So 90+j120 turns into 150 /_ 53.13.

(If MeFi formatting messes up the above example, that's 90 + j120 turns into 150, angle sign, 53.13)

I have no idea where the 150 and 53.13 came from, nor do I understand how to calculate those values. I can't find a clear explanation of it in the book anywhere. Can anyone help me?

So I'm in an electric circuits class, we have a test in three days and one simple concept is really confusing me.

I understand that electrical impedance (Z) can be in a complex format, like, for example, 90+j120. But then the book somehow converts that into what looks like a phasor. So 90+j120 turns into 150 /_ 53.13.

(If MeFi formatting messes up the above example, that's 90 + j120 turns into 150, angle sign, 53.13)

I have no idea where the 150 and 53.13 came from, nor do I understand how to calculate those values. I can't find a clear explanation of it in the book anywhere. Can anyone help me?

Imagine the impedance value plotted on a complex plane, where the real part (90) is on the horizontal axis and the imaginary (120) is on the vertical axis. If you plot that point and draw a line to the origin, that's a vector with a certain length and angle from the horizontal axis. You have

posted by muddgirl at 3:26 PM on April 13, 2009

*x*and*y*and are trying to find*r*and*phi*. fatllama gave you the formula but it's pretty easy to derive from trigonometry if you recognize that the length*r*is the hypotenuse of a right triangle with sides*x*and*y*.posted by muddgirl at 3:26 PM on April 13, 2009

Response by poster: Wow, typical AskMe, takes 6 minutes to get the exact right answer.

Thank you so much fatllama, that makes complete sense!

posted by DMan at 3:27 PM on April 13, 2009

Thank you so much fatllama, that makes complete sense!

posted by DMan at 3:27 PM on April 13, 2009

All complex numbers have an equivalent angle / amplitude representation. This is most clearly seen if you look at an Argand Diagram:

If the X axis is the Real part of the complex number, and the Y axis is the Imaginary part then if you have a complex number A+Bj, then you draw a point at the point X=a, Y=b, and draw a line connecting that point to the origin. The length of that line is the amplitude, and the angle it forms with the X axis is the angle.

posted by Luddite at 3:29 PM on April 13, 2009

If the X axis is the Real part of the complex number, and the Y axis is the Imaginary part then if you have a complex number A+Bj, then you draw a point at the point X=a, Y=b, and draw a line connecting that point to the origin. The length of that line is the amplitude, and the angle it forms with the X axis is the angle.

posted by Luddite at 3:29 PM on April 13, 2009

Furthermore, you can read more about phasors at this MathWorld page. It's important to become familiar with phasor notation; later on, you'll see them expressed like in Eq. (1) from that page:

posted by muddgirl at 3:29 PM on April 13, 2009

|z|=e^(j*phi)where

*z*is the magnitude and*phi*is the angle.posted by muddgirl at 3:29 PM on April 13, 2009

Response by poster: Thanks to you too muddgirl, yeah, at one point I had thought that maybe it had something to do with that, but just couldn't think of how to derive it.

posted by DMan at 3:30 PM on April 13, 2009

posted by DMan at 3:30 PM on April 13, 2009

Argh, messed up that equation. It should be

posted by muddgirl at 3:30 PM on April 13, 2009

x + yj= |z|e^(j*phi)

posted by muddgirl at 3:30 PM on April 13, 2009

d13t_p3ps1's boyfriend, graduating senior in EE here:

90+j120 are the respective real axis and imaginary axis values of the impedance. They are the magnitudes of the legs of a triangle. The phasor magnitude is the length of the hypotenuse of that triangle. From there you can do a sin or cos inverse to get the angle for the phasor. For example:

sqrt(90^2+120^2)=150, cos^-1(90/150)=53.13 is what it should be. I checked my work with an online calculator, so they should be correct.

I hope that helps!

If this isn't in depth enough, send MeFi mail to us and I'll see if I can help better.

posted by d13t_p3ps1 at 3:30 PM on April 13, 2009

90+j120 are the respective real axis and imaginary axis values of the impedance. They are the magnitudes of the legs of a triangle. The phasor magnitude is the length of the hypotenuse of that triangle. From there you can do a sin or cos inverse to get the angle for the phasor. For example:

sqrt(90^2+120^2)=150, cos^-1(90/150)=53.13 is what it should be. I checked my work with an online calculator, so they should be correct.

I hope that helps!

If this isn't in depth enough, send MeFi mail to us and I'll see if I can help better.

posted by d13t_p3ps1 at 3:30 PM on April 13, 2009

The keyword you need is "polar coordinates". (As opposed to Cartesian coordinates.)

posted by Chocolate Pickle at 3:40 PM on April 13, 2009

posted by Chocolate Pickle at 3:40 PM on April 13, 2009

« Older What more can I do to help the environment? | Will a ADHD diagnosis or treatment make me lose my... Newer »

This thread is closed to new comments.

posted by fatllama at 3:22 PM on April 13, 2009