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...

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 =...

Homework Statement Explain why the noise power across a resistor in thermal equilibrium is limited in contradiction to the formula for the noise V_{RMS} = \sqrt{4RkT\Delta f} which states that if we measure with infinite bandwidth we have infinitely large power Homework Equations...