- #1
TheCanadian
- 367
- 13
Sorry if the answer is very simple, but I just had a question regarding error propagation when using a modulo operator in intermediate steps. For example, I have ## \theta = arctan(\frac {A}{B}) ## and then I do ## \theta ## % ##2\pi## (modulo ##2\pi##). This gives me an answer between ## [0,2\pi] ## instead of ##[-\pi,\pi]##. This new answer in this new range is the answer I want. Now, when I calculate ## \sigma_{\theta} ##, I get:
## \sigma_{\theta} = \frac{A/B * \sqrt{ (\sigma_{A}/A)^2 + (\sigma_{B}/B)^2}}{1 + (A/B)^2} ##
Firstly, I just want to confirm: is there anything wrong with my equation for ## \sigma_{\theta} ##?
Secondly, since I change theta itself from ##[-\pi,\pi]## to ## [0,2\pi] ##by arithmetic operations, is there any modifications to ## \sigma_{\theta} ## in this case?
Thirdly, this is just a general error propagation question (sorry, I'm a novice), but if you have constant noise (i.e. a constant error), in the computation of A/B such that ## \sigma_A = \sigma_B = 2 ##, then you have ## \sigma_{A/B} = \frac{A}{B}\sqrt{ (2/A)^2 + (2/B)^2}##, but if A = 0, then A/B = 0, but in that case, the denominator in the square root for error goes to infinity but the numerator still has 0...so how does one approach computing the error in such as case? Is ## \sigma_{A/B} = \frac{1}{B}\sqrt{ (2)^2}## valid in such a case (I just multiplied by A/A)?
Please let me know if I have any mistakes. Any advice is appreciated!
## \sigma_{\theta} = \frac{A/B * \sqrt{ (\sigma_{A}/A)^2 + (\sigma_{B}/B)^2}}{1 + (A/B)^2} ##
Firstly, I just want to confirm: is there anything wrong with my equation for ## \sigma_{\theta} ##?
Secondly, since I change theta itself from ##[-\pi,\pi]## to ## [0,2\pi] ##by arithmetic operations, is there any modifications to ## \sigma_{\theta} ## in this case?
Thirdly, this is just a general error propagation question (sorry, I'm a novice), but if you have constant noise (i.e. a constant error), in the computation of A/B such that ## \sigma_A = \sigma_B = 2 ##, then you have ## \sigma_{A/B} = \frac{A}{B}\sqrt{ (2/A)^2 + (2/B)^2}##, but if A = 0, then A/B = 0, but in that case, the denominator in the square root for error goes to infinity but the numerator still has 0...so how does one approach computing the error in such as case? Is ## \sigma_{A/B} = \frac{1}{B}\sqrt{ (2)^2}## valid in such a case (I just multiplied by A/A)?
Please let me know if I have any mistakes. Any advice is appreciated!
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