Modelling a short square-wave pulse with Telegrapher's Equation

Click For Summary
The discussion focuses on modeling a short square-wave pulse in a coaxial cable using the Telegrapher's Equation, specifically through a steady-state approximation and Fourier series. The user is exploring the voltage component of the pulse at a distance along the cable and acknowledges the limitations of their approach, particularly regarding boundary conditions. Insights are provided on the dispersive properties introduced by resistive terms in the equation, which affect the propagation speed and shape of the pulse. The conversation suggests analyzing weak damping effects and emphasizes the importance of representing spatial dependence in the differential equations. Overall, the goal is to determine the dispersed signal's wave equation and its behavior over time and distance.
bearphys
Messages
1
Reaction score
0
I am interested in looking at the voltage component of a short (possibly nanosecond width) pulse in a coaxial cable along the cable at a given distance, Z, along the line. I am "cheating" by using the steady state Telegrapher's equation [Eq. 1] and formulating a periodic pulse approximation. It may not be desirable to use this steady state equation, but some form of the Telegrapher's Eq. is advised (suggestions for alternatives are appreciated). This is unlike my "physical" setup in which I will send a single pulse of information (microwave signal). I have determined Fourier approximation [Eq. 2] of my input pulse and the Fourier Coefficient [A]. My understanding is that the infinite series of circuits, as described by the Telegrapher's Equation, is dependent on frequency. I also recognize that I am going to use he Coefficent [A] as my "weighted" distribution of frequencies to calculate my inherent, frequency dependent components.

Eq. 1
\frac{\partial}{\partial x} V(x,t) =-L \frac{\partial}{\partial t} I(x,t) - R I(x,t)
\frac{\partial}{\partial x} I(x,t) =<br /> -C \frac{\partial}{\partial t} V(x,t) - G V(x,t)​

Eq. 2
\frac{1}{2T} + \sum_{n=1}^{200} a_n \cos(\frac{2\pi n t}{T}) + b_n \sin(\frac{2\pi n t}{T})​

In my summation, 200 is the current limit as to not consume too much time approximating in Mathematica.

where an is

\frac{2}{T}\int^{\frac{T}{2}}_{0}{\cos(\frac{2 \pi n t}{T})​

and bn is

\frac{2}{T}\int^{\frac{T}{2}}_{0}{\sin(\frac{2 \pi n t}{T})​

NB: I am only using an integration for half of my period because the square-wave is 1 from 0 to \frac{T}{2} and 0 from \frac{T}{2} to T; Therefore, the second half of the period has an integration of zero.

My Fourier Coefficient is represented by b_n, since a_n is effectively zero.

Therefore, my "weighted" frequency formula, that determines the frequency dependent terms of the Telegrapher's Equation is b_n.

Now, I have to admit, I know I should solve [Eq.1] with Dirichlet approach, due to the boundary conditions, but that is the limit of my knowledge on the matter. I want to know what my pulse looks like at some t, or z.

Can anyone offer some insight into solving this initial equation and determining the dispersed signal's wave equation?

NB2: As I am using Mathematica, I can attach the existing notebook if desired.
 
Last edited:
Engineering news on Phys.org
bearphys,

As you have included resistive terms, this equation has (now) some dispersive properties.
All modes wil not be traveling at the same speed and the shape of a multimodal pulse will be modified during propagation.

Without the resistive terms, the propagation speed is v= w/k = eps 1/sqrt(LC) ,
where w is the angular frequency, k is the wavevector and v the wave velocity,
and eps means +1 or -1 .

It could be interresting to start by the analysis of weak damping.
When r and G are small enough, the wave will simply show an additional damping given by:

w/k = eps 1/sqrt(LC) - i (R/L+G/C)/(2k)

(this is a dispersion relation, approximated first order for small R and G)
(i is the imaginary constant = sqrt(-1))
(calculation to be checked)

This indicates that short wavelengths (large k) will be less attenuated than long wavelengths. Indeed, in complex notation, the wave will be represented by this kind of space-time dependance:

exp(i(k z - w t))

where w and k are related by the dispersion relation.

The time dependence can always be represented by a Fourier series, of course.
However, what you are looking for the decay of the signal along the transmission line.
Therefore, you need now to concentrate on representing the spatial dependance.
In your representation, you might assume that the an and bn are functions of z.
You need then to develop the differential equations for these functions.
 
I just realized there is a good wiki on http://en.wikipedia.org/wiki/Transmission_line" .
This wiki does even provide a compact approximate solution valid when the resistive terms are small enough:

a741e93ffe62898891323cecb2b1a7d5.png


In that case you can see that the dispersion is simply due to the decay of the wave.
Because of the decay along the line, the pulse is deformed.
Note how obvious this solution is again, except for the precise expression of the decay index.

You could use this approximation to get some insight on your method.
Indeed, by expanding this approximate solution in Fourier series, would could have a taste of what you are looking for.
Later on, you could try to get a general method valid even for large resistive terms, or for other sources of dispersion.
 
Last edited by a moderator:
I am trying to understand how transferring electric from the powerplant to my house is more effective using high voltage. The suggested explanation that the current is equal to the power supply divided by the voltage, and hence higher voltage leads to lower current and as a result to a lower power loss on the conductives is very confusing me. I know that the current is determined by the voltage and the resistance, and not by a power capability - which defines a limit to the allowable...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
Replies
5
Views
2K
Replies
13
Views
1K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 5 ·
Replies
5
Views
547
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
690