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.
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:
Now, we
could substitute
$\frac{\pm (V)}{{Z}_{0}}$ 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
$x=L-s$ )
Usually, we just fold the (constant) phase terms
$e^{\pm (i\beta L)}$ 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:
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
$e^{i\beta s}$ .
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
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.