# 5.5 Terminated lines

 Page 1 / 1
This module looks at terminated transmission lines.

If, on the other hand, we have a finite line, terminated with some load impedance, we have a somewhat morecomplicated problem to deal with .

There are several things we should note before we head off into equation-land again. First of all, unlike the transient problems we looked atin a previous chapter , there can be no more than two voltage and current signals on the line, just ${V}^{+}$ and ${V}^{-}$ , (and ${I}^{+}$ and ${I}^{-}$ ). We no longer have the luxury of having ${V}_{1}^{+}$ , ${V}_{2}^{+}$ , etc. , because we are talking now about a steady state system . All of the transient solutions which built up when the generator was first connectedto the line have been summed together into just two waves.

Thus, on the line we have a single total voltage function , which is just the sum of the positive and negative going voltage waves

$V(x)={V}^{+}e^{-(i\beta x)}+{V}^{-}e^{i\beta x}$
and a total current function
$I(x)={I}^{+}e^{-(i\beta x)}+{I}^{-}e^{i\beta x}$
Note also that until we have solved for ${V}^{+}$ and ${V}^{-}$ , we do not know $V(x)$ or $I(x)$ anywhere on the line. In particular, we do not know $V(0)$ and $I(0)$ which would tell us what the apparent impedance is looking into the line.
${Z}_{\mathrm{in}}=Z(0)=\frac{{V}^{+}+{V}^{-}}{{I}^{+}+{I}^{-}}$
Until we know what kind of impedance the generator is seeing, we can not figure out how much of the generator's voltage will becoupled to the line! The input impedance looking into the line is now a function of the load impedance, the length of the line,and the phase velocity on the line. We have to solve this before we can figure out how the line and generator will interact.

The approach we shall have to take is the following. We will start at the load end of the line, and in a manner similar to the one we usedpreviously, find a relationship between ${V}^{+}$ and ${V}^{-}$ , leaving their actual magnitude and phase as something to be determined later. We can then propagate the two voltages(and currents) back down to the input, determine what the input impedance is by finding the ratio of ( ${V}^{+}+{V}^{-}$ ) to ( ${I}^{+}+{I}^{-}$ ), and from this, and knowledge of properties of the generator and its impedance, determine what the actual voltagesand currents are.

Let's take a look at the load. We again use KVL and KCL ( ) to match voltages and currents in the line and voltages and currents in the load:

${V}^{+}e^{-(i\beta L)}+{V}^{-}e^{i\beta L}={V}_{L}$
and
${I}^{+}e^{-(i\beta L)}+{I}^{-}e^{i\beta L}={I}_{L}$
Now, we could substitute $\frac{±(V)}{{Z}_{0}}$ for the two currents on the line and $\frac{{V}_{L}}{{Z}_{L}}$ for ${I}_{L}$ , and then try to solve for ${V}^{-}$ in terms of ${V}^{+}$ using and but we can be a little clever at the outset, and make our (complex) algebra a good bit cleaner . Let's make a change of variable and let
$s\equiv L-x$
This then gives us for the voltage on the line (using $x=L-s$ )
$V(s)={V}^{+}e^{-(i\beta L)}e^{i\beta L}+{V}^{-}e^{i\beta L}e^{-(i\beta L)}$
Usually, we just fold the (constant) phase terms $e^{±(i\beta L)}$ terms in with the ${V}^{+}$ and ${V}^{-}$ and so we have:
$V(s)={V}^{+}e^{i\beta s}+{V}^{-}e^{-(i\beta s)}$
Note that when we do this, we now have a positive exponential in the first term associated with ${V}^{+}$ and a negative exponential associated with the ${V}^{-}$ term. Of course, we also get for $I(s)$ :
$I(s)={I}^{+}e^{i\beta s}+{I}^{-}e^{-(i\beta s)}$
This change now moves our origin to the load end of the line, and reverses the direction of positive motion. But , now when we plug into $e^{i\beta s}$ the value for $s$ at the load ( $s=0$ ), the equations simplify to:
${V}^{+}+{V}^{-}={V}_{L}$
and
${I}^{+}+{I}^{-}={I}_{L}$
which we then re-write as
$\frac{{V}^{+}}{{Z}_{0}}-\frac{{V}^{-}}{{Z}_{0}}=\frac{{V}_{L}}{{Z}_{L}}$
This is beginning to look almost exactly like a previous chapter . As a reminder, we solve for ${V}_{L}$
${V}_{L}=\frac{{Z}_{L}}{{Z}_{0}}({V}^{+}-{V}^{-})$
and substitute for ${V}_{L}$ in
${V}^{+}+{V}^{-}=\frac{{Z}_{L}}{{Z}_{0}}({V}^{+}-{V}^{-})$
From which we then solve for the reflection coefficient ${\Gamma }_{\nu }$ , the ratio of ${V}^{-}$ to ${V}^{+}$ .
$\frac{{V}^{-}}{{V}^{+}}\equiv {\Gamma }_{\nu }=\frac{{Z}_{L}-{Z}_{0}}{{Z}_{L}+{Z}_{0}}$
Note that since, in general, ${Z}_{L}$ will be complex, we can expect that ${\Gamma }_{\nu }$ will also be a complex number with both a magnitude $\left|{\Gamma }_{\nu }\right|$ and a phase angle ${\theta }_{\Gamma }$ . Also, as with the case when we were looking at transients, $\left|{\Gamma }_{\nu }\right|< 1$ .

Since we now know ${V}^{-}$ in terms of ${V}^{+}$ , we can now write an expression for $V(s)$ the voltage anywhere on the line.

$V(s)={V}^{+}e^{i\beta s}+{V}^{-}e^{-(i\beta s)}$
Note again the change in signs in the two exponentials. Since our spatial variable $s$ is going in the opposite direction from $x$ , the ${V}^{+}$ phasor now goes as $i\beta s$ and the ${V}^{-}$ phasor now goes as $-(i\beta s)$ .

We now substitute in ${\Gamma }_{\nu }{V}^{+}$ for ${V}^{-}$ in , and for reasons that will become apparent soon, factor out an $e^{i\beta s}$ .

$V(s)={V}^{+}e^{i\beta s}+{\Gamma }_{\nu }{V}^{+}e^{-(i\beta s)}={V}^{+}(e^{i\beta s}+{\Gamma }_{\nu }e^{-(i\beta s)})={V}^{+}e^{i\beta s}(1+{\Gamma }_{\nu }e^{-(2i\beta s)})$
We could have also written down an equation for $I(s)$ , the current along the line. It will be a good test of your understanding of the basic equations we are developinghere to show yourself that indeed
$I(s)=\frac{{V}^{+}e^{i\beta s}}{{Z}_{0}}(1-{\Gamma }_{\nu }e^{-(2i\beta s)})$

how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!