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Homework Help: Archived Derive the general solution for current in a lossless transmission line

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the general solution for the current I(z,t) associated with the voltage V(z,t).Do this by substituting [1] in to [2] and [3], integrate with respect to time, and then take the derivative with respect to z.

    2. Relevant equations

    V(z,t)= f+(t-z/vp) + f-(t+z/vp) [1] where f is some arbitrary function.

    ∂V/∂z= -L(∂I/∂t) [2]

    ∂I/∂z= -C(∂V/∂t) [3]

    3. The attempt at a solution

    Ok,so I tried starting with [3] and integrating with respect to time: ∫(∂I/∂z)∂t= ∫(-C(∂V/∂t))∂t
    which gives (I think) V=-1/C ∫(∂I/∂z)∂t

    now differentiating this with respect to z: ∂V/∂z= ∂/∂z[-1/C ∫(∂I/∂z)∂t]

    and substituting the RHS of this in to [2]:∂/∂z[-1/C ∫(∂I/∂z)∂t]=-L(∂I/∂t)

    now I'm stuck and not sure where to go from here. The solution is the well known equation
    I(z,t)= (1/Z0)[f+(t-z/vp) + f-(t+z/vp)] where Z0=√(L/C)

    I would appreciate knowing the steps to get there.

    Thanks very much
  2. jcsd
  3. Apr 30, 2017 #2
    1. Substitute [1] into [3]. (Note: how do you know that equation 1 is a valid solution?)
    2. Take derivatives. You will need to use the chain rule.
    3. Integrate this equation with respect to t (you need to use a u-substitution but it is trivial).

    This will give you the equation for I, but your above answer for I has something wrong about it. Also, you will need to use the value of v in terms of L and C.

    Finally, it seems like we ignored equation 2. Actually, we didn't. Do you see why it is necessary?

    Edit: apologies to all, just saw this was from 2013. I am not sure why it popped up in my feed.
  4. May 17, 2017 #3
    It is also possible to get a solution in terms of Maxwell equations to get the second order diff. eq. for electric field.known as the field equation.
    Make the resistivity zero so as to get a pure sinusoidal Tesla solution for the electric field on the transmission line.
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