Recent content by ijmbarr

Mike_In_Plano, thank you. It became obvious as soon as I connected the scope that oscillations were occurring. Lesson learned: sometimes its worth getting the oscilloscope out rather than struggling along with a multimeter. Thanks again. Iain

I'm currently in the process of building a constant current source using an op amp. Circuit diagram: http://upload.wikimedia.org/wikipedia/en/9/97/Op-amp_current_source_with_pass_transistor.png I'm using a voltage reference instead of a zener diode. Currently the load resistance is 0...

I mean how does the physics of the situation determine which you choose?

Why do you choose the k+ to be higher and the k- pole to be lower? It seems like you could just as easily choose it the other way around and change the result?

Doesn't jordan's lemma only apply as the radius tends to infinity?

Thanks for the replies. The section I was referring to is page 864 RHB third edition. I'm treating the small indents around the poles as semicircles A, so their contribution to the integral is: \int_{A} f(z) dz Where f(z) is the function being integrated. Expanding it as a...

Any ideas? Is the integral Multivalued?

So why would you choose the contour to go around the pole at +k rather than both poles or none of them? does this mean the integral has several possible values? Also, while the pole at -k isn't include in the contour, doesn't the half circle that goes around it, whose radius tends to zero...

Sorry, that was a typo, there should be no z there. The residues should read: R(k) = e^{irk}/2 R(-k) = e^{-irk}/2 They were worked out using res(f,c) = \frac{1}{(n-1)!} \lim_{z \to c} \frac{d^{n-1}}{dz^{n-1}}\left( f(z)(z-c)^{n}...

Homework Statement I am trying to evaluate the integral: I = \int^{\infty}_{-\infty} \frac{z e^{irz}\ }{(z-k)(z+k)} dz The way I attepted it was to use contour integration around a semicircle in the top half of the argan diagram, with two small indents above the poles. This means that the...