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Homework Help: Complex equation

  1. Dec 11, 2011 #1
    1. The problem statement, all variables and given/known data

    I've set up this equation to find the integral time in a PI-regulator.

    [tex] \frac{\sqrt{0.02^2 - 4 \times \frac{0.02}{X}}}{2} = \frac{\pi}{100} [/tex]

    This is not solvable by normal means because the X has to be postive, thus rendering the square root negative.

    I have tried by substistuting values for X, and found that if X = approx. 18,4 it will all add up.

    The problem is that when solving this equation the "normal" way, it will no longer be complex when squaring the root. The answer results in approx -22.54.

    What do I have to do to solve this equation and get an exact answer for X in the imaginary plane?
  2. jcsd
  3. Dec 11, 2011 #2

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    Hi again Twinflower! :smile:

    Your equation has 1 solution for X, but that solution is negative.
    I'm afraid he result that you found is not a solution since the right hand side of your equation is not imaginary.

    So... what is it that you want?
  4. Dec 11, 2011 #3
    Hm, when I come to think about it, the right side is indeed imaginary.
    But it doesnt seem like my calculator's solver function will accept that (Casio CFX-9860GII)
  5. Dec 11, 2011 #4

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    If your calculator is the only problem, just negate the argument of your square root and everything will be real.
  6. Dec 11, 2011 #5
    Yes, I did try that but when negating the argument the solution is not identical-but-negated.

    I'll try to explain the whole problem:

    Determine Ti so that the cycle for the regulation is 200 seconds.

    The cycle is determined by the imaginary part of the augmented equation of a differential equation.

    This is the augmentet equation:
    [tex]\lambda^2 + 0.02 \lambda + \frac{0.02}{Ti} = 0[/tex]

    This yields something like this:
    [tex]\alpha +/- j \beta[/tex]

    And the period of the cycle is defined like this:

    Beta equals rad pr second for the sine wave, and the cycle is defined as 2pi/beta.
    That means that if the cycle has to be 200, then beta has to be j pi/100.

    Because beta is the imaginary part, and Ti is the only unknown in the standard polynomial equation for 2 unknowns, I tried to solve it the way I described in my first post.

    Failing miserably ;)
  7. Dec 11, 2011 #6
    I tried this as well, but it can't be done because:

    [tex]\frac{\sqrt{0.02^2 - 4 \times \frac{0.02}{18.4}}}{2} = 0.0314158[/tex]

    [tex]\frac{\sqrt{0.02^2 - 4 \times \frac{0.02}{-18.4}}}{2} = 0.0344522[/tex]
  8. Dec 11, 2011 #7

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    Looks like you did it right in your opening post, except that you forgot to include an [itex]i[/itex] on the RHS of your equation.

    Perhaps you can solve this?
    [tex]\frac{\sqrt{-(0.02^2 - 4 \times \frac{0.02}{X})}}{2} = \frac{\pi}{100}[/tex]
  9. Dec 11, 2011 #8
    It worked perfecly!

    X = 18.399933
    (my estimate was pretty close)

    Thanks, for the nth time :)
    You really deserve your homework helper badge. And you should know that it's guys like you that made me donate to this forum :)
  10. Dec 11, 2011 #9

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    TBH, it's people like you that make me spend so much time on this forum. ;)
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