Recent content by Malamala

I actually checked it carefully and in my case ##X_L = 10 \Omega##. If I connect the loop directly, and I want to compute the current through the loop, should I then divide the applied voltage by this 10 Ohm? or does the amplifier has an extra 50 ohms internally so i need to divide by 50 + 10 = 60?

Hello! I want to generate an RF magnetic field at variable frequencies (from 1 to 20 MHz) using this amplifier: https://www.minicircuits.com/WebStore/dashboard.html?model=LZY-22%2B, by passing current through a loop of current (assume the inductive resistance is negligible). How should I proceed...

I am actually a bit confused. I thought that at low frequencies, the loss would actually be higher, not smaller. For example, say I use a waveguide to send my radiation inside and I include that region in my Q factor calculation. If the waveguide size is such that its cutoff frequency is 1 GHz...

@sophiecentaur @tech99 Thank you for your answers! Basically my question was a long way of asking whether in this case I can still assume I have Lorentzian behavior at frequencies below the cutoff, or a faster decay than that. And if that is the case, how much faster is that decay of the field...

Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to...

I don't have a design yet, I am in the design phase. I am thinking of a measurement, and I did all the math, but this kind of magnetic field is the main systematic. So before deciding if I should pursue this idea, I would like to know if it is even possible at all to achieve this. At this point...

I am using python. Welch method is built it, but it has this parameter which defines how many points I have in each segment and the obtained PSD changes visually based on that. And it only looks like what I would expect, based on other papers, only when I include all the points in my data set in...

Hello! I have some time series from a self-heterodyne measurement. Basically i have a laser and I combine it with a delayed version of itself, shifted by ##\Omega = 2\pi \times 80## MHz. The delay is ##\tau = 100##ns. I remove the DC so the signal from the beat note is: $$V(t) = V_0\cos(\Omega...

Hello! My question was started from this paper. The topic of the paper is related to linewidth measurements, but my question is even before that, related to the way the noise power spectral density (PSD) spectrum shown in Fig. 1 is obtained. In Fig. 2 they show their experimental setup, which is...

What I meant was that after you suggested that commutativity might be the issue I tried that and I still didn't get ##-\sqrt{2/3}##, which I should. What I did was to apply the operator to the left which becomes: $$(<0,0|Y_1^{-1})N_+ = (<0,0|N_+)Y_1^{-1} + <0,0|(Y_1^{-1}N_+) =...

No, sorry for the confusion. I don't have independent information. I know it's not correct as the 2 answers should agree (I don't know to what value, but that value should be the same for both). So the fact that they still don't agree, even after using the non-commutativity, means that I still...

Yes, but if I assume commutativity, I still don't get the right answer (it's non-zero, but still not right).

But what should I do then? I assume that there should be a definite value for that matrix element, but I am not sure how to compute it.

Hello! I have this operator ##Y_1^1(\theta,\phi)N_-##, where ##N_-## is the lowering operator for ##|l,m>##. I want to calculate: ##<0,0|Y_1^1(\theta,\phi)N_-|1,0>## and I am getting: $$<0,0|Y_1^1(\theta,\phi)N_-|1,0> = \sqrt{2}<0,0|Y_1^1(\theta,\phi)|1,-1> = \sqrt{2}\int...

Uhh... I am still confused... if i measure the PSD experimentally using the self-heterodyne method, what would I see? Would I see 1/f + white noise, as in the first paper, or the Lorentzian shape as in the second paper? Also, it's still not clear to me how the length of the line matters i.e...