# RLC state space model

Discussion in 'Homework Help' started by MRjoker, May 8, 2011.

1. ### MRjoker Thread Starter New Member

May 8, 2011
2
0
Hello guys I need your help

There is a series RLC circuit below

And formulas that you know for this circuit

And state space model for the output voltage accross capacitor is

Now I must rewrite this state space model for the output voltage accross inductor

I'm trying but something is wrong with my trials. Can someone help me?? By the way sorry for my English.

2. ### samsmaths New Member

May 8, 2011
10
0
Hi,
This question is a very simple problem, it depends on your knowledge about circuit analysis and state equations.
In a series circuit Vs = VR + VC+ VL, so VL = Vs – (VR + VC) .....(1)
What is a state equations? It’s a set of n coupled first order ordinary differential equations. But you want to write a state equation for the voltage across the inductor.
We know that VL=L diL/dt => diL/dt= VL/L .....(2) this is our first order ordinary differential equation
Substitute equation (1) in (2)
diL/dt= 1/L [Vs – (VR + VC)]
now its simple for you to express the state space modal.
I hope this help, all the best.

3. ### MRjoker Thread Starter New Member

May 8, 2011
2
0

is it dVl/dt=dVs/dt-R*di/dt-i/c ??
dVl/dt=-R*di/dt-i/c

4. ### jegues Well-Known Member

Sep 13, 2010
735
43
Are you required to write the equation as the voltage across an inductor, or simply solve for it?

If it is the latter, write the equations in terms of $i_{L}$, and solve $V_{L}$ afterwards.

5. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
784
The overwhelming majority of worked examples on the series RLC use the series current i and Vc as the state variables.

Presumably one may use VL as a state variable - paired with what ...?

Another approach - if you know Laplace transforms is to write the Laplace form of the transfer function between VL and the source and then use the established technique of conversion from the Laplace transfer function to the State-Space form - being essentially algebraic manipulation. The only 'catch' is that one has to deal with a transfer function form in which the numerator and denominator are of the same order - in terms of the operator 's'.