Working Examples RLC Circuit with 5V Ramp Voltage Laplace Transforms

[Cooler]

can anyone give me the working examples of current expressions of series RLC circuit using ramp voltage by Laplace transforms. thanks

R=15, L= 0.4H, C=12uF ...voltage 5v

 

[tedbradly]

You write a conservation of voltage around the loop:

-v(t) + Ri(t) + L\frac{di}{dt} + \frac{1}{c} \int i(t) \, dt=0

Take the Laplace of both sides:

-V(s) + R I(s) + LsI(s) + \frac{1}{cs}I(s)= 0

Solve for your sought after function of current:

I(s) = \frac{V(s)}{R + Ls + \frac{1}{cs}}

Remember, this solution assumes initial conditions equal to zero.

 

[Cooler]

i stuck on the 1/s^2 [r + sL + 1/sC] ...do we need multiply all inside the bracket by s^2?

 

[tedbradly]

i stuck on the 1/s^2 [r + sL + 1/sC] ...do we need multiply all inside the bracket by s^2?

Yes. Then you will most likely need to do fraction decomposition. The trick will be putting the answer in a form present in your Laplace tables.

 

[Cooler]

i don't get the fraction decomposition...

when i multiplied inside the bracket it gives me...(15s^2 + 0.4s^3 + 83.333x10^3 s) is this correct and what to do next?

 

[tedbradly]

i don't get the fraction decomposition...

when i multiplied inside the bracket it gives me...(15s^2 + 0.4s^3 + 83.333x10^3 s) is this correct and what to do next?

I(s) = \frac{1}{s(Ls^2 + Rs + \frac{1}{c})}=-\frac{c^2(R+sL)}{cLs^2+cRs+1}+\frac{c}{s}

 

[Cooler]

sorry i don't get the idea of how it end up that way...mind to explain? thnks

 

[tedbradly]

Have you googled for tutorials yet? Google "fraction decomposition" or "partial fraction decomposition"

http://www.purplemath.com/modules/partfrac.htm

Essentially, it is a procedure for reversing the addition of two fractions.

so you can say a/b + c/d = (ad + bc)/(bd)

but in partial, you start with the right hand side and find the left hand side.Effectively (though it's not the complete picture), b = s and d = (ls^2 + rs + 1/c). And the reason I chose to only distribute 1 of the s in s^2 is that 1/s and 1/(as^2+bs+c) are usually in Laplace tables.

 

[Cooler]

What answer did you get?

my answer is 7.5x10^-4[ 1 - e^-18.75t(cos 456.05t + 41.11x10^-3 sin 456.05t)