Recent content by RoyalFlush100

Okay. Just curious though, how would I perform that conversion in a less trivial case?

I believe the answer should be a multiplied by the surface area, but I a unsure how to properly evaluate the integral, converting dS to be in terms of dt and dh.

Homework Statement In the attached image. Homework Equations Gradient(x, y, z) * <f, g, h> = <fx, gy, hz> The Attempt at a Solution Because the cylinder's not capped, I know that all the flux will be in the radial direction. So, I can find a normal vector by finding the gradient of the...

Wait, so why would δ(t) delay one second? Wouldn't δ(t-1), delay a second, because that's when the impulse would occur?

So, the Laplace Transform is just e^0 = 1. If you meant when I asked if it at t=1, I was more making sure I understood the more general case. By "frequency of the signal," I meant that s would be the variable, so the magnitude of the impulse response would be dependent on s. Did I misunderstand...

2q'' + 16q' + 50q = 9(1 - U(t-2)) So, we just apply the Laplace Transform to 2q'' + 16q' + 50q = 9? How would we then solve for q(t), to account for U? Or do we only omit the unit step function when determining whether it's damped (i.e. only take the Laplace Transform of the left of the above...

Based off my formula for Q(s) poles exist at s=0 and two complex roots (-4+3i & -4-3i). The complex roots suggest underdamped. Does the simple pole at 0 impact this?

I know in the time domain, over-damped means a real, non-repeating root, under means imaginary root, and critical means real, repeating root. Underdamped will oscillate, the rest decay (same way as the exponential curve), but they will usually converge to some value. However, I'm not really...

Okay, so the reason the Laplace Transform of the impulse is 1 at zero is because it occurs at t=0, right? If it occurred at t=1, it'd be e^-s, correct? Then, the Impulse will be a function of s (dependent on what the frequency of the signal is, assumed to be at t=0, because that's when the...

I still don't get it. There's no delta function, which means there's never an instantaneous impulse. Or is the answer simply the transfer function (e.g. Laplace Transform of the above equation)?

I bolded the portions I need help with. 1. Homework Statement A series circuit consists of a resistor with a resistance of 16 ohms, an inductor with inductance of 2 H, and a capacitor with a capacitance of 0.02 F. At time t = 0 there is no charge on the capacitor and no current in the circuit...

Okay, so in that case it'd be minus then. So: i'' + 10i' - 25i = 0 s^2 + 10s - 25 = 0 s = -10/2 +/- sqrt(100-4(-25))/2 s = 5 +/- sqrt(200)/2 s = 12.071 or -2.0711 i(t) = Ae^(12.071t) + Be^(-2.0711t) i(0) = 3 = A + B B = 3 - A Now: 10 - 5 + i' = 10 i' = 5 i'(t) = 12.071Ae^(12.071t) -...

This is the way I typically do these types: 1) Take the derivative, giving a second order differential equation. Find i (the roots of which will be real and unequal, real and equal, or complex). 2) Solve for i(0), which will give an equation involving the two constants. 3) Take the first...

Homework Statement "The switch in the circuit in the figure has been closed for a long time and is opened at t = 0. Find i(50 ms)." I attached an image of the circuit below.Homework Equations KVL: Ri + q(t)/C - q(0)/C + vc(t) + Li' = vs(t) Li'' + Ri' + i/C = vs'(t) Where vc is voltage across...

How would that be done? Initially the energy in the system would be the spring energy and perhaps relative gravitational energy. However, how would that be related to period?