Derive the general solution for current in a lossless transmission line

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  • Thread starter biker.josh07
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  • #1

Homework Statement



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.

Homework 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]

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
 

Answers and Replies

  • #2
315
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.
 
  • #3
5
2
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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