Comments on: How do I convert impedance to this weird phasor-form?
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform/
Comments on Ask MetaFilter post How do I convert impedance to this weird phasor-form?Mon, 13 Apr 2009 15:22:23 -0800Mon, 13 Apr 2009 15:22:23 -0800en-ushttp://blogs.law.harvard.edu/tech/rss60Question: How do I convert impedance to this weird phasor-form?
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform
ElectricalEngineeringFilter: How do I add electrical impedances together, and convert the result to what looks to be a phasor? <br /><br /> So I'm in an electric circuits class, we have a test in three days and one simple concept is really confusing me. <br>
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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.<br>
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<small>(If MeFi formatting messes up the above example, that's 90 + j120 turns into 150, angle sign, 53.13)</small><br>
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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?post:ask.metafilter.com,2009:site.119418Mon, 13 Apr 2009 15:16:30 -0800DManelectricalengineeringelectricalphasorBy: fatllama
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709592
90 forms the real and 120 the imaginary part of a complex number, z. |z| = sqrt(x^2 + y^2) = 150 is the magnitude. 53.13 is the angle this vector makes in the complex plane, i.e. arctan(90/120) = 53.13 degrees.comment:ask.metafilter.com,2009:site.119418-1709592Mon, 13 Apr 2009 15:22:23 -0800fatllamaBy: fatllama
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709596
Er, arctan(120/90), that is.comment:ask.metafilter.com,2009:site.119418-1709596Mon, 13 Apr 2009 15:25:28 -0800fatllamaBy: muddgirl
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709597
Imagine the impedance value plotted on a <a href="http://en.wikipedia.org/wiki/File:Complex_conjugate_picture.svg">complex plane</a>, 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 <i>x</i> and <i>y</i> and are trying to find <em>r</em> and <em>phi</em>. fatllama gave you the formula but it's pretty easy to derive from trigonometry if you recognize that the length <em>r</em> is the hypotenuse of a right triangle with sides <em>x</em> and <em>y</em>.comment:ask.metafilter.com,2009:site.119418-1709597Mon, 13 Apr 2009 15:26:53 -0800muddgirlBy: DMan
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709600
Wow, typical AskMe, takes 6 minutes to get the exact right answer.<br>
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Thank you so much fatllama, that makes complete sense!comment:ask.metafilter.com,2009:site.119418-1709600Mon, 13 Apr 2009 15:27:58 -0800DManBy: Luddite
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709602
All complex numbers have an equivalent angle / amplitude representation. This is most clearly seen if you look at an <a href="http://en.wikipedia.org/wiki/Argand_Diagram">Argand Diagram</a>:<br>
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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.comment:ask.metafilter.com,2009:site.119418-1709602Mon, 13 Apr 2009 15:29:30 -0800LudditeBy: muddgirl
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709603
Furthermore, you can read more about phasors at <a href="http://mathworld.wolfram.com/Phasor.html">this MathWorld page</a>. It's important to become familiar with phasor notation; later on, you'll see them expressed like in Eq. (1) from that page:<pre>|z|=e^(j*phi)</pre> where <em>z</em> is the magnitude and <em>phi</em> is the angle.comment:ask.metafilter.com,2009:site.119418-1709603Mon, 13 Apr 2009 15:29:33 -0800muddgirlBy: DMan
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709604
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.comment:ask.metafilter.com,2009:site.119418-1709604Mon, 13 Apr 2009 15:30:20 -0800DManBy: muddgirl
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709605
Argh, messed up that equation. It should be <pre>x + yj= |z|e^(j*phi)</pre>comment:ask.metafilter.com,2009:site.119418-1709605Mon, 13 Apr 2009 15:30:32 -0800muddgirlBy: d13t_p3ps1
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709606
d13t_p3ps1's boyfriend, graduating senior in EE here:<br>
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:<br>
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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. <br>
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I hope that helps! <br>
If this isn't in depth enough, send MeFi mail to us and I'll see if I can help better.comment:ask.metafilter.com,2009:site.119418-1709606Mon, 13 Apr 2009 15:30:34 -0800d13t_p3ps1By: Chocolate Pickle
http://ask.metafilter.com/119418/How-do-I-convert-impedance-to-this-weird-phasorform#1709612
The keyword you need is "<a href="http://en.wikipedia.org/wiki/Polar_coordinate_system">polar coordinates</a>". (As opposed to <a href="http://en.wikipedia.org/wiki/Cartesian_coordinate_system">Cartesian coordinates</a>.)comment:ask.metafilter.com,2009:site.119418-1709612Mon, 13 Apr 2009 15:40:15 -0800Chocolate Pickle