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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 .

A finite terminated transmission line

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 + β x V - β x
and a total current function
I x I + β x I - β 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 in Z 0 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 + β L V - β L V L
and
I + β L I - β L I L

Doing kirchoff at the end of the line

Change variables!
Now, we could substitute ± V Z 0 for the two currents on the line and 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 L x

S=0 at the load and so the exponentials go away!

This then gives us for the voltage on the line (using x L s )
V s V + β L β L V - β L β L
Usually, we just fold the (constant) phase terms ± β L terms in with the V + and V - and so we have:
V s V + β s V - β 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 + β s I - β 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 β 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
V + Z 0 V - Z 0 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 Z L Z 0 V + V -
and substitute for V L in
V + V - Z L Z 0 V + V -
From which we then solve for the reflection coefficient Γ ν , the ratio of V - to V + .
V - V + Γ ν Z L Z 0 Z L Z 0
Note that since, in general, Z L will be complex, we can expect that Γ ν will also be a complex number with both a magnitude Γ ν and a phase angle θ Γ . Also, as with the case when we were looking at transients, Γ ν 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 + β s V - β 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 β s and the V - phasor now goes as β s .

We now substitute in Γ ν V + for V - in , and for reasons that will become apparent soon, factor out an β s .

V s V + β s Γ ν V + β s V + β s Γ ν β s V + β s 1 Γ ν 2 β 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 V + β s Z 0 1 Γ ν 2 β s

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
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
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
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Kristine 2*2*2=8
Bridget Reply
Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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is it 3×y ?
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J, combine like terms 7x-4y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Cied
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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AMJAD
preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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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
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Azam
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Prasenjit
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Damian
silver nanoparticles could handle the job?
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Prasenjit Reply
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.
Ali Reply
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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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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