Recent content by lorenb

as long as the elements are connected to each "node" the same way. It does not matter where you put them. Make sure you have a good understanding of what a node is. In bother figures there are only two nodes

i'm pretty sure you need to sample at a frequency of over two times the original frequency to prevent "aliasing"

it all sounds correct so far. You may want to make a graph of velocity vs time. You also may want to look up some equations for terminal velocity, etc.

make sure you pay attention to your y-axis units. Looks like multisim is in magnitude while orcad is in volts. if you are using .1 volts they both seem to be the same. . .

well, i don't know how to vary the voltage at the same time as the frequency. but i attached a graph of 10-5k frequency with 1V

ya, that looks right. Just think that the 2 amps is making a loop through the 10 ohm resistor. If you made a circle on the left loop with an arrow pointing in the same direction as the current source. The current would go up through the 10 ohm resistor. So put your voltage source so the + side...

ya, i guess the gain and decibel values don't having anything to do with the capacitor. The capacitor will block high frequency sources.

your gain and decibel value are right for the given the capacitor value you found. I've never used that method or formula to find the capacitor value though. I know Zc = 1/jwc... where'd you get the (omega = 1/R2C1)?

well, when you use the "for" part and say for x = -1:.1:3 . MATLAB only uses this for the loop. it does not make a vector out of it. well, i think it just keeps changing the value of x so that the value of x is the value for your last loop try putting x = -1:.1:3 after your loop and keeping the...

well, I've noticed most people here won't give a lot of help unless you make some sort of effort. i googled miller effect and found an equation for it. You may want to start by writing equations like that and other relevant equations.

F=pi^2 EI/(KL)^2} where F = maximum or critical force (vertical load on column), E = modulus of elasticity, I = area moment of inertia, L = unsupported length of column, K = column effective length factor, whose value depends on the conditions of end support of the column...

if the resistors are on the same "line" (no branches between them) they are in series. -the equivalent resistance of the two resistors is the Sum (i.e. R1+R2) if the resistors are not on the same "line" yet they share the same nodes. (i.e. they are connected to each other though a node on...

there's no way you'll get a real number from your equations. considering the only values given are resistances. The is probably why we are trying to get a solution that has some constants.(a and b). not really sure how to solve for a and b though

well, if we make up our own are size... we'll get a resistance per unit area... so use like 1 m^2 for your area and R will be 1/4 ohm per meter^2 Then if we are actually given an area like 1/2 m^2 we could use the above equation or just times it by 1/4 to get 1/8 ohms. not sure what else you...

this doesn't sound right... Electricty is always trying to get to a lower potential energy (not necessarily back to the source)? So if your generator is producting a voltage (potential energy) higher than that of ground (around 0V). Then you connected these two points there would be a current...