Kindly find the attached picture of a linear circuit. One differential equation from the circuit is
d(Vc(t))/dt = Vc(t)/R2C + i(t)/C + w(t)/C where w(t) is uncertainty in the input that affects state. Vc, is the voltage across capacitor.
to find the solution, i did laplace transform of the above equation and applied the inverse laplace transform to find Vc(t) and discretize to get Vc,k. However it doesn't match with the result as attached pic, which is the published answer. So i was wondering if i am using the wrong logic for getting the Vc,k?
Thank you all for your time !!
Hi, in the differential equation involving Vc, are i(t) and w(t) given or do they vary according to other differential equations?
i(t) is discharging current, its value is known, however the value changes with time. Similarly w(t) is uncertainty in the input measurement i(t). Used for modeling error in current measurement.
First, let us not consider w(t). There is a typo (?) in you equation, which should be d(Vc(t))/dt = -Vc(t)/R2C + i(t)/C (minus sign in the first term of the right hand side). Then you seem to assume that i(t) is constant on each interval [tk,tk+1). The solution in your message is then perfectly correct: indeed,
Vc(t)=exp(-(t-t0)/R2C)v(t0)+\int-t0^t exp(-(t-\tau)/R2C))i(\tau)d\tau
You find the result with t0=tk, t=tk+T=tk+1, i(t)=ik on (tk,tk+1).
As for w(t), it looks more like measurement error on Vc than on i. Measurement noise on the input should not enter the voltage/current differential equation describing the behavior of the system. Besides, this is not coherent with the solution you mention (if i and w enter the differential equation as described, the should enter the solution in the same way).
Thank you so much Dr. Martin for this explanation. It makes sense to me.
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