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Error Propagation help?

  1. May 3, 2015 #1
    1. The problem statement, all variables and given/known data
    I am given a frequency value of 95 GHz (9.5x10^10 Hz), C= 25 F, L=1.12x10^(-25) H.
    The question is to find the uncertainties in frequency by taking account of inductor being 5% accurate & capacitor being 8% accurate.

    2. Relevant equations
    I believe this is the correct formula to use--since f= 1/ 2pi * sqrt(LC) (frequency formula) is a division/fraction.

    (σf)^2= [ (df / L)^2 * σL^2 + (df / C)^2 * σC^2 ]

    3. The attempt at a solution

    Well taking the partial derivatives of f respect to L, I get: C sqrt(LC) / 2pi.
    For f respect to C I get: L sqrt(LC) / 2pi.
    So taking the partial derivatives that I had found, I plugged into the equation above Relevant equations & got:

    (σf)^2= [ (C sqrt(LC) / 2pi)^2 * σL^2 + ( L sqrt(LC) / 2pi)^2 * σC^2 ]

    I know that the percentage of accuracy should be substituted in σL^2 and σC^2 with respect to the given capacitor and inductor values, however I need a little guidance whether I am on the right track.

    Thank you!
  2. jcsd
  3. May 3, 2015 #2


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    check the formula, it is not correct! The partial derivatives are also wrong.

    You get the absolute accuracy of a quantity X if if you multiply X by the percent accuracy divided by 100.
    Last edited: May 3, 2015
  4. May 3, 2015 #3
    Sorry Ehild, but what would be a correct formula to use? My teacher hasn't really been that thorough with the subject & had us with that equation to work with, so I wasn't really sure.

    What exactly is "absolute accuracy" and after that what would I have to do?
  5. May 3, 2015 #4


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    Is this an ##LC## circuit? If so, you may want to be careful about how you are taking your partial derivatives.

    For example, here is the partial with respect to ##L##:

    Screen Shot 2015-05-03 at 9.31.49 AM.png
  6. May 3, 2015 #5
    I mean isn't
    Thank you for your response, Zondrina.
    Yes it is indeed an LC circuit.

    Ah, yeah I see where I went wrong with the derivative...but just a curiosity, would the formula that I used be incorrect?
  7. May 3, 2015 #6


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    You have the correct formula I believe, just not written properly.

    When you write:

    $$f = 1/2pi * sqrt(LC)$$

    People interpret it as:

    $$f = \frac{1}{2} \pi \sqrt{LC}$$

    When you really meant:

    $$f = \frac{1}{2 \pi \sqrt{LC}}$$

    You could have wrote it as:

    $$f = 1/(2pi * sqrt(LC))$$

    To signify everything is included in the denominator.

    You need to find the partial with respect to ##C## now, as well as the errors ##\sigma_L## and ##\sigma_C##. Then plug and chug into:

    $$\sigma_f = \sqrt{ \left(\frac{\partial f}{\partial L} \right)^2 \sigma_L^2 + \left(\frac{\partial f}{\partial C} \right)^2 \sigma_C^2 }$$
  8. May 3, 2015 #7

    Are you sure that you can just pull the sqrt C out like that? I mean if we hypothetically multipled C^(1/2) + L^(1/2), we would get LC now, not sqrt LC, right? Screen shot 2015-05-03 at 7.08.52 AM.png
  9. May 3, 2015 #8
    When I did ∂F/∂L, I get:

    -[2π (LC)^1/2 ]^(-2) * ∂/∂L[(2π (LC)^(1/2)]
    = -[(2π(LC)^1/2 ]^(-2) * (π (LC)^(-1/2) * C ]

    **the 2s cancel out on last step.
  10. May 3, 2015 #9


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    Recall from math:

    $$\sqrt{xy} = \sqrt{x} \sqrt{y}$$

    So we can write:

    $$\sqrt{LC} = \sqrt{L} \sqrt{C}$$

    Now, because the frequency ##f(L, C)## is a function of ##L## and ##C##, when we take the partial derivatives with respect to a variable, we treat all the other variables as constants. That's why I was able to pull out the ##\sqrt{C}## when taking the derivative with respect to ##L##; the variable ##C## is treated as a constant.

    Cleaning up that other partial derivative would give you:

    Screen Shot 2015-05-03 at 10.27.20 AM.png

    Notice the similarities to the other partial derivative.
  11. May 3, 2015 #10
    Ah I understand that its one of those properties to know. However I thought if you multiplied two integers/variables, you add their exponents?
    Like √(xy) = √x * √ y = xy <----
  12. May 3, 2015 #11
    Oh wait no silly me, they're two different variables so you can't add the exponents together.
  13. May 3, 2015 #12


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    You can only add the exponents if the bases are the same.

    You can do:

    $$\sqrt{x} \sqrt{x} = x^{\frac{1}{2}} x^{\frac{1}{2}} = x^{\frac{1}{2} + \frac{1}{2}} = x^1 = x$$

    You can't do:

    $$\sqrt{x} \sqrt{y} = x^{\frac{1}{2}} y^{\frac{1}{2}} = xy^{\frac{1}{2} + \frac{1}{2}} = xy^1 = xy$$
  14. May 3, 2015 #13
    Oh yeah, that's right! I really appreciate your help Zondrina!! I don't know where my head is today.

    Okay so I did get the same answer you got for partial f with respect to L----> -1/(4piLC) * sqrt(C/L)

    For partial f with respect to C I got answer of ----> -1/(4piLC) * sqrt(L/C)

    In terms of this equation, partial derivatives are determined.
    Screen shot 2015-05-03 at 7.53.23 AM.png
    So the absolute errors we know is that inductor is 5 percent and capacitor is 8 percent...& know that C= 25 F and L=L=1.12x10^(-25) H.

    Then taking C=25 F...I would take that number and multiply it with 8 percent and divide by 100 correct?
  15. May 3, 2015 #14


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    ##\sigma_L## and ##\sigma_C## represent the errors associated with measuring the inductance and capacitance of each component.

    For example, we know if we measure the inductance, we may write it as:

    $$L = (1.12 \times 10^{-25} \pm \sigma_L) \space H$$

    If we measure the capacitance we may write it as:

    $$C = (25 \pm \sigma_C) \space F$$

    You are told the inductor is ##5 \%## accurate and the capacitor is ##8 \%## accurate. By accurate, they mean how much the inductance/capacitance may vary according to the nominal value.

    So if the capacitor is ##8 \%## accurate, then ##\sigma_C = 25 \times 0.08 = 2## means we can write the capacitance as:

    $$C = (25 \pm 2) \space F$$
  16. May 3, 2015 #15
    Ah I see.
    I understand that the measurement includes the best value and error and also that the +/- indicates a range of the possible correct value...but silly question...when you take C=(25±2)F and plug it into the absolute error of the equation...how do I go on about inputting that in my calculator? Would I just use the "2" ?
  17. May 3, 2015 #16


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    I think you are referring to this equation from prior:

    $$\sigma_f = \sqrt{ \left(\frac{\partial f}{\partial L} \right)^2 \sigma_L^2 + \left(\frac{\partial f}{\partial C} \right)^2 \sigma_C^2 }$$

    Perhaps it would be insightful if I included some extra information:

    $$\sigma_f = \sqrt{ \left(\frac{\partial f}{\partial L} (1.12 \times 10^{-25}, 25) \right)^2 \sigma_L^2 + \left(\frac{\partial f}{\partial C} (1.12 \times 10^{-25}, 25) \right)^2 \sigma_C^2 }$$

    You need to evaluate the partial derivatives at the point ##(L, C) = (1.12 \times 10^{-25}, 25)##. You also need to plug in ##\sigma_C = 2## and ##\sigma_L = ?##. The answer that comes out is in ##Hz##, so make sure to convert to ##GHz##.
  18. May 3, 2015 #17
    Here is what I did:

    Does this seem correct to you? Screen shot 2015-05-03 at 9.35.29 AM.png
    ***EDIT: Excuse my error in typing from the picture I attached. I forgot to multiply (2)^2 at the far end of the square root. And I meant to have the " ]^2 " part at the very end to be INSIDE the square root.

    Just to be 100% clear, the σL and σC still contain its units of H and F respectively right?
    Last edited: May 3, 2015
  19. May 3, 2015 #18


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    I suggest just converting everything to standard S.I. units for calculations, and omitting the units when you do the actual calculation. This makes it a bit easier to look at, and it will give the answer in standard S.I. units.
  20. May 3, 2015 #19
    Thank you so much for your help Zondrina! This site needs more of you :smile:
  21. May 4, 2015 #20


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    Are your sure you needed to go the partial derivatives route?

    Because if you just needed to know the result, you could have calculated the upper and lower extremes of f using the extreme values of the elements, e.g., for the upper

    ##\dfrac 1{2\pi\sqrt{L*0.95*C*0.92}}\\\\=\ \dfrac 1{2\pi\sqrt{LC}}*\dfrac1{\sqrt{0.95*0.92}}\\\\\\=\ \dfrac 1{2\pi\sqrt{LC}}*1.07\\\\\\##

    The other extreme of f is less percent, so I'd keep the error as ± 7%

    How does this compare with your calculation using partial derivatives?
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