Recent content by Bromio

When I say that there isn't an "at rest" situation I'm talking about the frictionless case. If there isn't friction, the spring will experiment an non-stop simple harmonic motion, won't it? Thank you! I've solved that integral and I've got that result. Amazing!

No misinterpretation. Now I see all you're trying to explain. For some reason I thought there wasn't friction. It's obvious that, if there isn't it, I can't apply energy equations "at rest" situation, simply because there isn't that situation. Am I right? So, when I study energies at the...

Okay. So the dissipated energy has a value of \frac{1}{2}k\Delta x^2, hasn't it? What I don't understand is why, if we are in a conservative field, we can't just consider initial and final states. Could you explain me the reason? Thank you.

Intuitively I understand what you say, but I don't see how to write it analytically. I mean, as we are in a conservative field, we only have to consider initial and final states. How should I have written the main equation (mgh_0 = mgh + \frac{1}{2}k\Delta x^2) to include what you say? Thanks!

Homework Statement A particle is connected to a spring at rest. Because of weight, the mass moves a distance \Delta x. Calculate the value of the elastic constant k Homework Equations U_{PE} = \frac{1}{2}k\Delta x^2 U = mgh F = ma (in general) F = kx (Hooke's Law) The Attempt at a...

I don't understand it. Isn't the formula G_P (dB) = G_V (dB) + G_I (dB) always valid?

Thank you for answering. They come from G_V (dB) = 20log(G_V). If G_V(dB)= 20 dB, then G_V = 10^{G_V(dB)/20} = 10^{20/20} = 10, isn't it?

Hi. I know G_P (dB) = 10log(G_P) and G_V (dB) = 20log(G_V) and G_I (dB) = 20log(G_I). I also know that G_P (dB) = G_V (dB) + G_I (dB). So, if I have a common base amplifier whose current gain is G_I (dB) = 0, then G_P (dB) = G_V(dB). Suppose G_V (dB) = 20 dB. So G_V = 10^{20/20} = 10 G_I...

I rolled two wires around the coil: one coming from the sinusoidal signal generator and another one going to the spectrum analyzer, which also showed "power" (dissipated power, I suppose). So, I was measuring the voltage drop in the coil (which can be modeled as a RLC circuit). Sweeping...

Hi. When using spectrum analyzer to measure the response of a coil (a RLC circuit), I see that there is a peak at one frequency (resonance frequency). This is logical because we can model a coil as a RLC circuit. If I change the frequency of the sinusoidal source, the peak reduces its value...

Hi. I've been working with a bolometer and a directional coupler for calculating the module of the reflection coefficient of a load. I used several loads (matched and unmatched) and found out that the incident power (I measured a sample at the coupled port) was the same each case. Why? I...

Thank you. I really wanted to find the Y-parameter matrix of a Schottky diode working as mixer, whose model is this: Should I conclude that it isn't posible to get it by adding capacitor Cj and resistor Rj Y-matrices, going to Z-parameters, adding the result with Rs Z-matrix, and finally...

In that case I have the same problem as before. I know that Y_{11}=\frac{Z_{22}}{\Delta_Z}, where \Delta_Z = Z_{11}Z_{22}-Z_{12}Z_{21}. Because Z_{11} = Z_{12} = Z_{21} = Z_{22} = \frac{1}{j\omega C}, \Delta_Z = 0, so Y_{11}\to\infty. Where's the mistake? Thank you.

OK. Thanks. So, V_1 = Z_{11}I_1 + Z_{12}I_2 V_2 = Z_{21}I_1 + Z_{22}I_2 I have no problem finding Z_{11}: Z_{11} = \left.\frac{V_1}{I_1}\right|_{I_2=0} = \frac{1}{j\omega C} However, when finding Z_{12} I'm not sure if I draw the circuit properly in order to calculate the parameters...

Homework Statement Calculate the Y parameters of a capacitor in parallel (see the attached figure). Homework Equations Y-Parameters: I_1=Y_{11}V_1+Y_{12}V_2 I_2=Y_{21}V_1+Y_{22}V_2 Y_{11} = \frac{I_1}{V_1} when V_2=0 Y_{12} = \frac{I_1}{V_2} when V_1=0 Y_{21} = \frac{I_2}{V_1}...