Hi Folks,
Got a quick question regarding an R-L-C natural overdamped response. I am basing my equation on the voltae across the capacitor (vC).

vC = A*exp(m1*t) + B*exp(m2*t)

Now to obtain the values for A and B. I have assumed two intial conditions which is the only way to do it. The two I have chosen are:

1) vC at time 0 is 2 V

2) the derivative of vC (dvC/dt) at time 0 is 0.

I'm a bit concerened about my second condition, the rate of change of voltage with respect to time at time 0. It shouldnt be 0 should it, it should be a non-zero value. Are there any other conditions I can use.

Any help will be appreciated.

 

In your second equation, if dt = 0, how do you divide by 0?

i = C*dv/dt

dv = i / C*dt (if I did my algebra correctly?)

A current must be injected into the capacitor over some period of time, to create a voltage across the capacitor.

 

I may be a bit off, but you do not need initial conditions to solve your equation. You should be aware of the difference between a particular solution and a general solution.

Steve

 

Hi Folks,
Thanks for the reply, I dont understand the first reply I am not dividing by 0. if vC has an intial value of 2 V at time 0:

2 = A*exp(m1*t) + B*exp(m2*t)

then dvC/dt

0 = m1*A*exp(m1*t) + m2*B*exp(m2*t)

I know it doesnt really matter what value I choose I will still get the familar shape of decay. However I wanted to relate it to a real case. I was concerned that if the voltage is already 2 V across the capacitor then at time 0 the voltage would decay to reach 0 and the derivative of vC cant be zero, any ideas of what it should be?.
Regarding the second reply to my original question, I havnt given all the information again. The transient response I was looking at was the overdamped where vC = A*exp(m1*t) + B*exp(m2*t) because there are two unknown constants there must be two intial values. At least thats my understanding which could be wrong.

Jag.

 

Hi Folks,
Sorry more information I left out.

a= LC = 0.5
b= RC = 1.5
c = 1

The component values I chose were R = 3R , L = 1H and C= 0.5 F.

I know they are not very realistic values, however it makes the maths easier for me!

m1 = (-b+ sqrt(b^2 - 4*a*c))/2*a

m2 = (-b - sqrt(b^2 - 4*a*c))/2*a

m1 = -1 and m2 = -2
at time 0 e(m1*t) will equal 1 therefor equations will look as follows:

2 = 1A + 1B
0 = -1A -2B

Simmulatanous equation of two unkowns.