Recent content by ur5pointos2sl

Hello- I need to determine the transfer function given the pole-zero chart below. I came up with the following for the transfer function: H(s) = 1/[(s-(-2+j2)*(s-(-2-j2))] = 1/(s^2+4s+8) Could someone verify if this is inded the correct transfer function? Thanks

Thank you!

Hi I need to try and find the differential equation representing the attached circuit. My work is also being shown on the attachment. Can anyone confirm whether this is correct? If it is wrong could you please provide input as to why? Thanks. Sorry for the quality in advance.

Very helpful! Thanks!

The problem states: From the field with a radial cylindrical component only given by the following equations: E(r)= (ρ0*r3)/(4 * ε0*a2) for r<=a E(r)= (ρ0*a2)/(4*ε0*r2) for r > a obtain the corresponding charge distribution in free space in which the equation is: ρ(r) = ρ0*(r2/a2)...

Yes that makes sense, they just cancel one another. Thanks for the help!

Total charge = 0

For sin(2θ) * sin (θ) with respect to θ I got (2/3) sin(θ)^3. Evaluated from 0 to ∏ I got 0. Edit: TI89 got that answer which I have found isn't always reliable.

Alright that is good to know. So then should this be the integral to evaluate? ρs0 * r2∫∫ sin(2θ) * sin(θ) dθ dPhi 0≤θ≤∏ 0≤Phi≤2∏ After integrating that I get 2∏*ρs0*r2.

Since r is constant then can I just say dS = r2 sin θ dθ dPhi?

So then Phi from 0 to 2∏ and θ from 0 to ∏? Is this what you are saying?

You are right and trying to pull it all back together is the hardest part. Does that involve parameterization described in terms of θ and Phi?

Sorry I think you misunderstood me because that wasn't worded clearly. I originally thought that spherical coordinates were going to be used (before I posted this problem, using triple integrals) but since you stated it can be done using a double integral my thought was to use polar coordinates...

Well my first guess was a spherical coordinate system. Since it's a double integral it must be polar coordinates? What would the limits then be?

I haven't had calculus in a long time so I have forgotten most of it. ∫ ∫ ρs0 sin 2θ dA = ps0 sin 2θ * 8∏r dr dθ 0 -> ∏ 0 ->a A = 4∏r2 dA = 8∏r