1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper

    ##\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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper


    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


    User Avatar

    Staff: Mentor

    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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted