math: polar form addition

Discussion in 'General Electronics Chat' started by omaroski, May 3, 2012.

  1. omaroski

    Thread Starter Active Member

    Jun 19, 2008
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    Last edited: May 3, 2012
  2. P-MONKE

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    Mar 14, 2012
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  3. omaroski

    Thread Starter Active Member

    Jun 19, 2008
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    You mean applying Pythagorean Theorem to find the hypotenuse? \sqrt{10^2+6^2}

    if i'm not wrog that would be if numbers are at 0° and 90°

    in fact i would get 11,66 in place of 14.861 as in the example.
     
  4. hasnain

    New Member

    Feb 9, 2012
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  5. crutschow

    Expert

    Mar 14, 2008
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    No. If you look at the reference listed you will see that that answer is 14.861V < 16.59°.

    You add the two by first converting from polar to rectangular form:

    10 <0° = 10 +j0
    6 <45° = 4.24 + j4.24

    Then you add the rectangular values:

    (10 +j0) +(4.24 + j4.24) = 14.24 +j4.24

    Then you convert this back to polar form:

    14.24 +j4.24 = 14.86 <16.59°
     
  6. P-MONKE

    Member

    Mar 14, 2012
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    Awww - you beat me to it

    I was just doing this longhand to show all the steps, as a few posters didn't seem to grasp the concept.

    It was good practice for me though, as I'm very rusty.

    I'm going to post it anyway because I made the effort :p
     
  7. omaroski

    Thread Starter Active Member

    Jun 19, 2008
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    I read the chapter on complex numbers more accurately and found it.

    I couldn't understand why there was multiplication, divison and inversion of polar numbers while no addition and subtraction. Then saw it was necessary convert to rectangular form using cos e sin functions, do the addition and convert to polar back again obviously.

    Thank you for your efforts!
     
  8. omaroski

    Thread Starter Active Member

    Jun 19, 2008
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    P-MONKE handwritten job made me reflect about this:

    i was calculating sin and cos with electronic calculator and with 45° i noticed that it's nothing but 1/√2 both. What's the point so?
     
  9. crutschow

    Expert

    Mar 14, 2008
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    The sin and cos of 45° are the same. Thats because a 45° right-angle triangle has two equal sides. So what's your question? :confused:
     
  10. omaroski

    Thread Starter Active Member

    Jun 19, 2008
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    The one you replied!

    It's not my field, trigonometry i mean. I couldn't tell what a sine and cosine functions would describe exactly!
     
  11. P-MONKE

    Member

    Mar 14, 2012
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    They are all about "squaring the circle" ;)

    BTW, I used "i" for the complex component, something at the back of my mind tells me that "j" (as used by crutschow) is more commonly used in electrical and electronic formulae.
    "i" tends to be used more by mathematicians, I believe.
     
  12. crutschow

    Expert

    Mar 14, 2008
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    Here's a short primer on trig.

    The reason trig is used for AC circuit calculations is that the 90° phase angle between current and voltage caused by reactive components can be viewed as two vectors with a 90° angle. Then the sum of these vectors can be solved by viewing them as the sides of a right triangle with the hypotenuse length and angle being the sum of the two vectors.

    Thus the hypotenuse length and angle is the polar form of value and the length of the sides is the rectangular form (with the "j" component being the vertical side).
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    Forgive me for pulling up a somewhat moldy thread, but this highlights a real problem with technical education (and I don't think it knows much in the way of national borders, meaning that no country is immune). We are increasingly the victims of our own technological success. We have calculators that give you sin() and cos() by punching a button and so students are taught that when you see a formula with sin() or cos() that you first punch the buttons for what's inside the parentheses, then you punch the button that says sin' or 'cos' on it. I've seen trig textbooks here that spend far more time giving illustrated examples of which buttons to press on the calculator than covering the concepts and how to work with them. The same for exponentials and logarithms. It hit me in the face when I had two extremely bright students, both of whom had been their respective high school valedictorians and both had gotten A's in AP math, come up after class because they couldn't understand the step on the board in which I had:

    e^ln(z) = z

    It took only a couple of minutes to explain the fundamental inverse relationship between the two and they understood it perfectly. But both said that, up to that point, they had only been taught that exponentials and logarithms were functions, just like the trig functions, that you would see from time to time in formulas and to evaluate the expression you push these buttons on your calculator.
     
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