Recent content by The Alchemist

Oh, I was too dazzled! thanks for the help. So the equation in the paper misses the sqrt...

You're right I missed the square root in the arcsin, the arctan though doesn't have one! When I solve for cosine θ \theta = \arccos{ \left( \pm \sqrt{ \frac{I_{\sigma} - I_{\pi}}{I_{\sigma} + I_{\pi}}}\right)} How can I then come to an arctan expression?

Homework Statement Given the following relation for θ: \frac{I_{\pi}}{I_{\sigma}} = \frac{\sin^2{\theta}}{1 + \cos^2{\theta}} solve for θ Homework Equations \cos^2 x + \sin^2 x = 1 The Attempt at a Solution If I solve this I get: \theta = \arcsin{\left(\pm\frac{2...

Homework Statement Find out whether particle or heat diffusion determines the width of the SOL, and whether the ions or electrons are the determining factor. (just compare all of those). Give an expression to estimate the thickness of the SOL and evaluate the expression for Te = 10 and 100...

Okay, I made my way through this. \sigma_{ij} (e_i \otimes e_j) \cdot \sigma_{kl} (e_k \otimes e_l) = \sigma_{ij} \delta_{ik} \delta_{jl} \sigma_{kl} = \sigma_{ij}^2 This is indeed a scalar, since there is no tensor space to span. The key was to create the kronecker...

Homework Statement stress tensor in cartesian components. \sigma is the stress tensor. e_i are the basis vectors Homework Equations \sigma \cdot \sigma The Attempt at a Solution I tried to write out the components with a cartesian basis: \sigma=\sigma_{ij} (e_i \otimes e_j) But then I'm...

Ok. Let me make myself more clear. I know it's about high resolution that's the reason I use these. The problem is, the molecule can only emit for about 1 second and then it is destroyed. So I take more molecules, which results indeed in less spatial resolution, to collect the needed amount of...

Hi, I'm doing some calculations on fluorescence and I'm a little stuck on statistics. Let's say I need to measure 20 seconds to collect 10000 photons from a single molecule. Then I can say well I want to measure at most 1 second and therefore I need 20 molecules in order to collect 10000...

Thank you for your efforts, I really appreciate it. And essentially it's not a 100% homework problem. It's just a problem I'm trying to solve. Your numbers are correct and your solution for the least square approximation is +- my MATLAB solution. Your last image looks very promising, with...

Thanks for your replies, I really appreciate it. I attached 3 files. one picture of the reduced A matrix, with only the 6 highest values of the matrix. one of the b vector. And the data files which include the full A matrix, the reduced and the b vector. Thanks is advance,

Thanks for your reply, I'm sorry, I kind of messed up. I meant 11 rows and 10 columns.. so that makes [11x10]. \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,10} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,10} \\ \vdots & \vdots & \ddots & \vdots \\ a_{11,1} & a_{11,2} & \cdots...

Homework Statement I have a matrix [11x10] which I want to solve (Ax=b). Matlab comes with an answer, but I would like to know if there is a possibility to give boundary conditions to this problem. like x_i can't be greater than 300 or something. is it possible to restrict the outcome?2. The...

Ok I put a capacitor parallel to the resistor and derive the new impedance which is indeed Z = \frac{R}{\left(1+(2\pi fRC)^2\right)} So \phi= 4ZkT = \frac{4RkT}{\left(1+(2\pi fRC)^2\right)} Integrating over df gives: \frac{2kT}{ \pi C}...

Well, the PSD is uniform over an infinite range of frequencies but it is proportional to the temperature of the resistor. What you want is a PSD that is cut-off at a certain frequency to limit the power. But I don't know why the PSD should have such a frequency.

Thanks, Well the power spectral density function of thermal noise is \phi = 4RkT So the power is the integral of the power spectral density function over the bandwidth. W = \int\limits_{bandwidth} \phi \mathrm{d}f = \int_{f_{1}}^{f_{2}}4RkT \mathrm{d}f = 4RkT\Delta f With \Delta f =...