step response of RLC circuit derivation

Discussion in 'Math' started by PG1995, Nov 28, 2011.

  1. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi

    Please have a look on the following attachments to see my questions. Please help me with them.

    The following two attachments have the queries:
    1: http://forum.allaboutcircuits.com/attachment.php?attachmentid=37034&stc=1&d=1322527789
    2: http://forum.allaboutcircuits.com/attachment.php?attachmentid=37035&stc=1&d=1322527789

    Here, this is the standard linear homogeneous differential equation we are dealing dealing with along with its solution. Please also have a look here.

    Here this is complete treatment of Eq. 8.4 the author is talking about. You might the highlights in "Eq. 8.4 #2" useful.
    Eq. 8.4 #1: http://forum.allaboutcircuits.com/attachment.php?attachmentid=37044&stc=1&d=1322565823
    Eq. 8.4 #2: http://forum.allaboutcircuits.com/attachment.php?attachmentid=37045&stc=1&d=1322565823
    Eq. 8.4 #3: http://forum.allaboutcircuits.com/attachment.php?attachmentid=37042&stc=1&d=1322565132


    Regards
    PG
     
    Last edited: Nov 29, 2011
  2. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I'm tied up on an urgent project, so I'll only handle Q1 for now. I'll look at the others later.

    For Q1, the author is correct. The Vs is different from the system variable v. The Vs is a forcing function that goes from zero to Vs. So it's constant once the system is turned on. The homogeneous equation assumes the forcing function is zero and that solution gives you the natural response of the system. The system poles are not affected by the forcing function.
     
    PG1995 likes this.
  3. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    For Q2, you can always calculate alternative initial conditions if you need them.

    For example, you know that i=C dv/dt, hence if you know i(0), then you know dv(0)/dt.

    By the way, the notation dv(0)/dt is not good. v(0) is a constant, hence the derivative of v(0) is zero. The context makes it clear that the notation is supposed to mean that you first take the derivative of v(t) with respect to t and then you substitute in t=0. However, the accepted notation for this is ...

     {{{\rm d}v(t)}\over{{\rm d}t}}|_{t=0}
     
    PG1995 likes this.
  4. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    For Q3, the author is saying that any variable in the system has that form for the complete solution. The idea is that the homogeneous solution always decays to zero as time goes to infinity. This means that the final value is a valid particular solution for that variable. Just add the homogeneous solution (for that variable) to the final value of the system solution and you have your complete solution.
     
    PG1995 likes this.
  5. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Thanks a lot, Steve.

    The Q1 is confusing me so much and I'm still striving to get hold of it.

    Let's start from the start. This is the standard linear homogeneous ordinary differential equation (LHODE); notice the "by" term. I can understand the case when there is no source in RCL circuit; I mean source free RLC circuit because we get normal and straightforward LHODE. But when we have a step response RLC circuit then we get a constant term on the right side. Please look here. When we try to get the equation in standard form by taking the constant term on the right side to the left side, we don't even have a standard LHODE. Do we? If we have, then how do we get "b" for the term 'by'? Even if you did (I mean you were able to get "by" term), I don't think you get the same solution as for the equation for source free RLC circuit because the "b's" of the two equations would be different. But as you see here the author is saying so. Please help me with it. Thank you very much.

    Best wishes
    PG
     
  6. steveb

    Senior Member

    Jul 3, 2008
    2,433
    469
    I'm heading out the door for an appointment, and need to come back to this later today. But, in the meantime, consider that you should not be trying to bring the constant term over to the left had side. The constant term is the input function, or the forcing function. The homogeneous solution is the solution assuming that the forcing function is zero, hence the poles (or roots of the characteristic equation) are unaffected by the forcing function. The forcing function determines the particular solution.

    EDIT: I found a wireless hotspot while waiting at my appointment, so I can comment further.

    Note that you can't just add the Vs term to the v term. The variable "v" is a variable, while Vs is a constant. They are two different things. It's not clear to me why you feel they need to be added together.
     
    Last edited: Nov 30, 2011
    PG1995 likes this.
  7. PG1995

    Thread Starter Active Member

    Apr 15, 2011
    753
    5
    Hi Steve

    Good Morning.You are so nice. Rarely these days we come across such nice good people. :) I hope your appointment goes well.

    With best wishes
    PG
     
    steveb likes this.
Loading...