Recent content by TimeToShine

Try and convert all of it into an equivalent impedance first. Using this, you can find the current flowing into R2 by means of current division.

I considered this as well but from what I've heard although there are more jobs the visas are a lot more difficult to get your hands on.

That's pretty much exactly what I want to do.The only difference is that I'd love to have something lined up before I fly out, even something small and I can work my way up from there. I doubt I could afford to not have a job for the first few months. I'll do as the other person said and try the...

I thought as much. I'll start applying from September onwards and hope for the best. I'm specifically electrical energy systems and control so hopefully there is something out there for me. Thanks for your help.

There are companies here who offer 1 year work visas to the U.S. subject to an interview and background check which I think shouldn't be a problem. They're the same crowd who give students here visas to go over to the U.S. over the summer. Basically, the visa isn't what I am worried about. What...

Hi all, I'll be graduating with a masters in electrical engineering from Ireland at the end of next year. I am hoping to do a years work (maybe more) in the states immediately after this. I was just wondering if many of you who work over there have had people from abroad coming over to work...

Just use rltool, choose your specifications and then vary the zero such that the closed loop poles lie inside the area that isn't shaded. You should be able to identify an upper and lower bound for the zero.

Plot the poles and zeroes (There are no zeroes here). Use the root locus drawing rules.

The transfer functions are not the same. When you close an open loop you create a feedback network which changes the transfer function.

Anything you do to an equation you must do to both sides. If you square one side then you must square the other. As for your workings there is an error in the second line.

Find a factor first, do this by trying different values a for x. When you find a value for which f(a) = 0 then you know that x - a is a factor. Divide this by your cubic polynomial to obtain a quadratic equation and solve it in the usual way.

Find the transfer function or use voltage division to find the voltage across the output in terms of what you are given. V(o) = (V(s)*jωL)/(R+jωL) Now using ohm's law you know that the current flowing through this circuit is simply this output voltage divided (which you have all the...

Remember your transfer function can be expressed in the form H(jω) = Vo(jω)/Vi(jω), where we can denote jω by s. Here you are trying to find the input impedance Z(i), or in other words Z(jω), where Z(jω) = Z(2j). So you can conclude that ω = 2 rad/s or ω = 2*pi*f = 2 and find the...

Solve the equation and find the two complex solutions for x. Now, you know that these values are equal to ω, apply De Moivre's theorem for complex numbers to the new expression.

You could break the feedback loop in question two but this requires a knowledge of two ports which I doubt you have. In that case, my advice is to apply nodal analysis. Give the node at the positive terminal of the op amp a name such as Vx and remember that because the op amp is deal the node at...