# DISTRIBUTED ELEMENT MODEL IN TRANSMISSION LINES (PART II)

|In our previous post in this section, we stated what we will study next to characterize the Distributed Model:

- The transmission lines equations
- Solve and exercise applying the equations
- Propagation constant and characteristic impedance of the line
- Solve and exercise calculating this parameters on a real line

In this post, we will see points 1 and 2.

## Distributed Element Model

Recall the representation and meaning of the distributed elements in a transmission line:

These parameters vary according to the type of line. For example:

**Note: click on the images to see a better resolution. Then, click to go back in your browser to come back to the post ðŸ˜‰**

## The transmission lines equations

When using the distributed element model, we can apply theÂ Kirchhoff laws:

## Exercise Explained:

TheÂ distributed coefficients of a transmission line withÂ w =10^{4} rad/sec are:

R = 0.053 â„¦ /m L = 0.62 mH/m G = 950 pS/m C = 39.5 pF/m.

In the z-coordinate over the line, the instantaneous current is given by:

*i(t)* = 75 cos 10 ^{4}t mA

a) Obtain the expression for the voltage gradient along the line, in the point z.

b) What is the maximum value of the voltage gradient?

## Solution:

**a)**

In the time domain, the voltage gradient is given by:

Substituting values, we have:

*= – 0.053 ( 0.075 cos 10 ^{4} t) + ( 0.62 x 10 ^{-6} ) (0.075 x10 ^{4} sin 10 ^{4} t) =*

*= – 3.98 x10 ^{-3} cos 10 ^{4} t + 0.465 x 10 ^{-3} sin 10^{4} t =*

*…*

*= 4.006 x10 ^{-3} cos( 10 ^{4} t – 3.03) Volts/meter.*

= 4.006 x10 ^{-3} cos( 10 ^{4} t – 173.4 Â° ) Volts/meter.

**b)**

The maximum voltage gradient is equal to the amplitude, 4 mV. This happens when:

*cos(10 ^{4 }t – 3.03) = 1*

This implies that,

*10 ^{4 }t – 3.03 = 0 , 2*

*Ï€*

*, 4*

*Ï€*

*… Radians.*

This occurs in the following instants of time:

*t _{0} = 3.03 / 10^{4} = 3.03 x 10^{-4} sec , *

Then

*Â t _{0} = 303 ms *

*t _{1} = (2*

*Ï€ + 3.03) / 10*

^{4}= 9.31 x 10^{-4}sec , t_{1}= 931 ms, …*t _{n} = (n*

*Ï€ + 3.03) / 10*

^{4}sec, with n = 0, 2, 4 …## What’s next?

We hope this post was useful ðŸ™‚ In the next post in this section, we will cover the points:

3. Propagation constant and characteristic impedance of the line

4. Solve and exercise calculating this parameters on a real line

After that, we will move to theÂ the parameters when the line is in a transient stateÂ which analyses what happen when there is a sudden change in the conditions of the line.

Does this sound interesting? ðŸ˜€