Finding complete response of RL circuit - Page 1
Forum section: electronics / beginners

author: Mattjd
author_rank: Regular Contributor
timestamp: October 23, 2017, 08:22:28 pm
content: Hi All,going over some signals and systems, and am having trouble solving this ODE of RC circuit. I realize you could use laplace transform but I want to know how to do it using not that. Im attempting with variation of parameters and am not having any luck.QuestionBook solutionMy attempt

author: Brutte
author_rank: Frequent Contributor
timestamp: October 23, 2017, 09:13:08 pm
content: The solution may not be a function of u(t)*I0.Take a look at the schematics - assuming u(t)=0 (short) and nonzero initial condition you will get an i(t)=I0*exp(-t) while from what 2.14 presents the output should be i(t)=0 at all times, irrespective of the initial condition. That eq seems wrong or describes a different circuit.You are looking for an algebraic sum of two solutions and not some nonlinear product.

author: Mattjd
author_rank: Regular Contributor
timestamp: October 23, 2017, 09:36:40 pm
content: I understand that. The solution should the sum of zero input response and zero state response. I can't figure out how they're getting those answers.I'm solving the homogenous form of equation 2.13 and then using that to find the nonhomogenous solution and its not working out.

author: orolo
author_rank: Frequent Contributor
timestamp: October 23, 2017, 09:48:39 pm
content: Note that: \$\int e^t \delta(t)\, dt \ = \ e^0 \ = \ 1\$, so:\$I(t) \quad = \quad B\left[u(t)e^t - 1\right]\$then your particular solution is (since t>0, and there u(t) = 1):\$i_p \quad = \quad I(t)e^{-t} \quad = \quad Bu(t) - Be^{-t} \quad = \quad B\left(1-e^{-t}\right)u(t)\$and then the general solution is (again, since t>0, and there u(t)=1):\$I(t) \quad = \quad I_0e^{-t} + B\left(1-e^{-t}\right)u(t) \quad = \quad \left[I_0e^{-t} + B\left(1-e^{-t}\right)\right]u(t)\$This said, I'd rather solve the equation for t<0 and t>0 as separate cases, and then adjust for the initial condition near t=0:For t<0, you have \$0 = I + \dot{I}\$, with general solution \$I(t) = Ce^{-t}\$. Since for t negative going to zero you want I=0, this forces \$I(t) = 0\$ for all t<0.For t>0, you have \$B =I + \dot{I}\$, with general solution \$I(t) = B + Ce^{-t}\$. Since for t positive going to zero you want \$I(0) = I_0\$, this means \$I(t) \ = \ B + (I_0 - B)e^{-t} \ =\ I_0e^{-t} + B(1-e^{-t})\$, for all t>0.Merging both solutions at t=0 gives:\$I(t) = \left[I_0e^{-t} + B(1-e^{-t})\right]u(t)\$.The factor u(t) makes the merge possible.Edit: if you want a physical interpretation, when you step the voltage at t=0, the inductor responds immediatly energizing with the impulse for an instantaneous current \$I_0e^{-t}\$, which decays with time. Simultaneously, the new voltage B starts an increasing current through the inductor \$B(1-e^{-t})\$ which stabilizes at V/R=B/1=B amps after a long time.

author: Mattjd
author_rank: Regular Contributor
timestamp: October 23, 2017, 10:16:49 pm
content: That works. Its just confusing to me because the definite integral of the impulse function is 1 not the indefinite. So, it never would have occurred to me to apply it like that, and is why I hit a wall.I also dont understand how you can just factor out these unit steps like this either\$i_p \quad = \quad I(t)e^{-t} \quad = \quad Bu(t) - Be^{-t} \quad = \quad B\left(1-e^{-t}\right)u(t)\$You factor it out but then should be e^-t over u(t) at that last step

author: orolo
author_rank: Frequent Contributor
timestamp: October 23, 2017, 10:36:05 pm
content: As long as t>0, u is constant and equal to 1, so \$-Be^{-t} = -Be^{-t}\cdot 1 = -Be^{-t}\cdot u(t)\$. Of course, you might say that the same argument holds for \$-Be^{-t} = -Be^{-t} / 1 = -Be^{-t} / u(t)\$, but when you try to extend that to negative values of t, you'd be dividing by zero for all negative t, which doesn't make sense. That's the reason why solving for t<0 and t>0 and then splicing bymultiplicationby u(t) makes things clearer, IMO.

author: rstofer
author_rank: Super Contributor
timestamp: October 23, 2017, 10:37:39 pm
content: And that's why we have Orolo do the math around here!

author: Mattjd
author_rank: Regular Contributor
timestamp: October 23, 2017, 10:40:46 pm
content: Quote from: orolo on October 23, 2017, 10:36:05 pmYes, I was just looking at graphs and what you say makes sense in accordance to thathttps://www.desmos.com/calculator/3lauhwlpo7

author: MrAl
author_rank: Super Contributor
timestamp: October 23, 2017, 10:48:34 pm
content: Hi there,This is easy...Writing the equation for the circuit:R*i+L*di/dt=E*u(t)For t<0 we have i(t)=0 but for t>0 we have to solve:R*i+L*di/dt=ESeparating the variables we get:L*di/(E-R*i)=dtthen integrating both sides:-L*ln(E-R*i)/R=t+Kand notice the K came in because of the integration.Invoking the initial condition i(0)=0 for the natural response we get K:-L/R*ln(E)=Kand so:t=(L/R)*(ln(E)-ln(E-R*i))or:(E-R*i)/E=e^(-t*R/L)or:i=E/R-E/R*e^(-t*R/L)You can plug in the values E=B and R=1 and L=1 if you like.

author: MrAl
author_rank: Super Contributor
timestamp: October 27, 2017, 06:56:36 pm
content: Hello again,I read this over and see you wanted the response when the inductor has initial energy also. This takes a little more work so i wrote the solution in the attachment. It's almost the same procedure except we first place an initial condition generator in the circuit as shown and that handles the initial current.See attachment for complete solution in the form i think you wanted to see.Take note that 'i' in the last two equations are really "iL" and that is the total current which is the solution sought after here. Also the function "log()" is really "ln()" as usual, so e^log(a)=a for example.