Derive the general solution for current in a lossless transmission line

In summary: This will satisfy the Helmholtz equation which is a special case of the field equation.In summary, the conversation discusses finding the general solution for the current I(z,t) associated with the voltage V(z,t). The steps involve substituting [1] into [2] and [3], integrating with respect to time and taking the derivative with respect to z. The solution is the well-known equation I(z,t) = (1/Z0)[f+(t-z/vp) + f-(t+z/vp)] where Z0 = √(L/C). The conversation also mentions using the value of v in terms of L and C and the necessity of equation 2 in the solution process. It is also possible to get a
  • #1
biker.josh07
5
0

Homework Statement



Find the general solution for the current I(z,t) associated with the voltage V(z,t).Do this by substituting [1] into [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 into [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
 
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  • #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
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.
 

1. What is a lossless transmission line?

A lossless transmission line is a type of electrical transmission line that has no resistance, inductance, or capacitance. This means that there is no energy loss as the current flows through the line.

2. Why is it important to derive the general solution for current in a lossless transmission line?

Deriving the general solution for current in a lossless transmission line allows us to understand and predict the behavior of the current in the line. This is important for designing and optimizing transmission systems, as well as troubleshooting any issues that may arise.

3. What factors affect the current in a lossless transmission line?

The current in a lossless transmission line is affected by the voltage, the length of the line, the characteristic impedance of the line, and the frequency of the signal being transmitted.

4. How is the general solution for current in a lossless transmission line derived?

The general solution for current in a lossless transmission line is derived using Maxwell's equations and the telegrapher's equations. These equations describe the behavior of electromagnetic waves in a transmission line.

5. Are there any limitations to the general solution for current in a lossless transmission line?

The general solution for current in a lossless transmission line assumes that the line is lossless and has a uniform cross-section. In reality, there may be some small losses and variations in the line's properties, which can affect the accuracy of the solution.

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