Recent content by mark1234

Ok, so i solved my problem: C \frac{\partial T(t)}{\partial t} + K(T(t)-T0) = \frac {V(t)^2} {R} use boundary condition T(0)=T0=0, Laplace transform gives : T(s).(Cs-K)= \frac {V(s)^2} {R} resulting in: G(s) = \frac {T(s)} {V(s)^2} = \frac {1/R} {Cs+K} = \frac {K} {τs+1}...

Attached is the simulink model of both the time series and the transfer function. https://dl.dropboxusercontent.com/u/75646341/Knipsel.JPG and the result of the simulation: https://dl.dropboxusercontent.com/u/75646341/Knipsel2.JPG The transfer function of V(t)^2/R-K(T(t)-T0)=C*dT(t)/dt...

The laplace transform G(s) would then be (T(s)-T0)/V(s) I'll check if there's an error in my simulink model, or interpretation of it.

Buell, Thanks for your efforts. In the time function there is a squared function of V(t) present. How did you get rid of that? The transfer function, G(s) = (T(t)-T0)/V(t) Thanks,

Guess i get stuck in the Laplace transform or re-writing it to the correct form. Solving the time equation , with boundary condition T(0)=T0: T(t)-T0={\frac{V(t)^2}{KR}} (1-e^{k*t/C}) To get to the transfer function this apparently has to be rewritten in the form of...

When air resistance is ignored, the projectile will only be accelerated by gravity. I would think the velocity when hitting the water would then be exactly the same for all angles, only the distribution over the x and z component are depending on the angle.

My two cents: If the top and the bottom of the cylinder are connected by a hose, the pressure in the water column is hydro static (in case of no movement . The pressure at the bottom and the top of the piston therefore balance each other out in a static situation (whatever the position, of...

Hi All, I'm playing around with an arduino, and have build a PID controller that controls the temperature of a light bulb, measured with an NTC. All is working fine I'm looking to get a bit more theoretical on the subject and have modeled the system in a simulink like environment. I want...