6.1 A/c line behavior

Introduction to physical Page 1 / 1

If we are going to try to use phasors on a transmission line, then we have to allow for spatial variationas well. This is simple to do, if we just let the phasor be a function of $x$ , so we have $\tilde{V} x$ . How the phasor varies in $x$ is one of the things we now have to find out.

Let's start with the Telegrapher's Equations again.

x V x t L t I x t

x I x t C t V x t

For

V x t

we can now substitute

\tilde{V} x ω t

and for

I x t

we plug in

\tilde{I} x ω t

. So we get:

x \tilde{V} x ω t L t \tilde{I} x ω t

and

x \tilde{I} x ω t C t \tilde{V} x ω t

We take the derivative with respect to time, which brings down a

ω

and then we cancel the

ω t

from both sides of each equation:

x \tilde{V} x ω L \tilde{I} x

and

x \tilde{I} x ω C \tilde{V} x

Viola ! In one simple motion, we have completely eliminated the time variable,

t

, from our equations! It is not really gone, of course, for once we figure out what

\tilde{V} x

is, we have to multiply it by

ω t

and then take the real part before we can extract once again, the actual

V x t

that we want. Nonetheless, insofar as the telegrapher's equations are concerned,

t

has disappeared from the radar screen.

To solve these we do just as we did with the transient problem. We take a derivative with respect to $x$ of , which gives us a $x \tilde{I} x$ on the right hand side, for which we can substitute , which leaves us with

x 2 2 \tilde{V} x ω 2 L C \tilde{V} x

(- times - is +, but

-1

and so we have a - in front of the

ω 2

). We then re-write as

x 2 2 \tilde{V} x ω 2 L C \tilde{V} x 0

The simplest solution to this equation is

\tilde{V} x V_{0} \pm ω L C x

from which we can then get the actual voltage signal

V x t \tilde{V} x ω t V_{0} \pm ω t ω L C x

Note that we could factor out an

ω L C

, from the exponent, which, since it is just a constant, we could include in

V_{0}

(and call it

V_{0}^{'}

, switch the order of

x

and

t

, and write as

V x t V_{0}^{'} \pm x 1 L C t

which looks a lot like the "general"

f \pm x v t

solution we were talking about earlier !

The number $ω L C$ is special. It is usually represented with a Greek letter $β$ and is called the propagation coefficient . Thus we have

V x t V_{0} \pm ω t β x

As previously,a point on the wave of constant phase requires that the argument inside the parenthesis remains constant. Thus if

V x_{1} t_{1}

is going to equal

V x_{2} t_{2}

( i.e. what was at point

x_{1}

t_{1}

is now at

x_{2}

at time

t_{2}

it must be that

\pm ω t_{1} β x_{1} \pm ω t_{2} β x_{2}

x_{2} x_{1} t_{2} t_{1} Δ x Δ t \pm ω β \pm ω ω L C \pm 1 L C v_{p}

Which one again, defines the phase velocity of the wave. Other relationships to keep in mind are

β 2 λ

λ v_{p} f ω β ω 2 2 β

The first comes from the fact that the wave varies in

x

β x

. Thus when

x γ

, the wavelength,

β γ

just increases by

2

, to get the phasor to go through one full rotation. Note also, as before, the choice of the minus sign inthe ± in represents a wave going in the

x

direction, while the choice of the + sign will give a wave going in the

x

direction. Clearly, by starting out taking the x-derivative of the equation for

I x t

we would end up with

I x t I_{0} \pm ω t β x

Let's consider the two phasors then, and define the voltage phasor associated with the positive going voltage wave as

{\tilde{V}}_{plus} x V^{+} β x

and the negative voltage phasor as

{\tilde{V}}_{minus} x V^{-} β x

We should keep in mind that both

V^{+}

and

V^{-}

can be, and probably are, complex numbers. (From now on we will drop the little ~ over the variables because its verytedious to get it to show up with this word processor. You will just have to keep in mind that any variable we do not explicitlyput inside absolute value markers ( i.e.

V^{+}

) is going to be, in general, a complex number). We will, of course, have similar expressions for the positive andnegative going current waves.

Let's consider the positive going current and voltage waves, and plug them into .

x V^{+} β x ω L I^{+} β x

The x-derivative brings down a

β

, the

β x

's cancel, and we have

V^{+} ω L β I^{+}

But, since

β ω L C

we have

V^{+} L C I^{+} Z_{0} I^{+}

as we had before.

So, what has changed? Not much from the case of transients on a line. We will now assume we have a steady state problem. This means we turned on the generator a long time ago. We assume that it has beenconnected to the line long enough so that all transient behavior has died away, and that voltages and currents are not changingany more (except oscillating at frequency $ω$ , of course).

If the line is semi-infinite (or matched with a load equal to $Z_{0}$ ) then it is pretty obvious that

V^{+} Z_{0} Z_{0} Z_{g} V_{g}

where

Z_{g}

is the source impedance, and

V_{g}

is the source voltage phasor.

A wave on a semi-infinite line

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Source: OpenStax, Introduction to physical electronics. OpenStax CNX. Sep 17, 2007 Download for free at http://cnx.org/content/col10114/1.4

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