3/6/2012 special cases of source and load present 1/22 special …jstiles/723/handouts/special cases...

3/6/2012 Special Cases of Source and Load present 1/22

Jim Stiles The Univ. of Kansas Dept. of EECS

Special Cases of Source

and Load Impedance

Let’s look at specific cases of:

1. Zg 2. and ZL,

and then determine how they affect:

1. 0V

2. and Pabs.

I z

V z

Vg LZ

z 0z

0Z gZ



The first special case: the matched load

00LZ Z j

Of course for this case—where the load is “matched” to the transmission line—we

find that 0L and thus 0in .

As a result, the complex amplitude of the plus-wave simplifies to:

0

0

0

0

0

1 1jβ

g

in g in

jβg

g

ZV V e

Z Z

ZV e

Z Z

I z

V z

Vg 0

Z

z 0z

0Z gZ



The absorbed power is just

the incident power

Likewise, the absorbed/delivered power is simply that of the incident wave:

2 2

0

0

20

0

122abs L

V

Z

VP

Z

as the matched condition causes the reflected power to be zero ( 0P )!

Now, recognizing that the input impedance of the transmission line for this case

will likewise be 0inZ Z , we can alternatively express the absorbed power

(without determining 0V !) :

2 2

0

2

02

0

2 2

0

1 1Re

2 2 2

g

g

g g

abs in

g in g

V VP Z Z

Z Z Z

V Z

Z ZZ



The 2nd special case: the conjugate match

For this case, we find the load ZL takes on whatever value

required to make in gZ Z .

The input impedance is a conjugate match

to the source impedance.

This is of course is a very important case!

First, using the fact that:

00

0 0

ginin

in g

Z ZZ ZZ Z Z Z

We can show that (trust me!) that the plus-wave amplitude simplifies to:

ginZ Z

Vg gZ

0

04Re

gjβg

g

Ze

ZV

ZV

in gZ Z



Yawn: the load absorbs the available power

Not a particularly interesting result, but now let’s look at the absorbed power.

2

2

2

2

2

2

2

2

1Re

2

1Re

2

1Re

2 2Re

1 1

2 R

8

4 e

g

abs in

g in

g

g

g g

g

g

g

g

a

g

vlg

g

VP Z

Z Z

VZ

Z

V

Z

VZ

Z

V

R

Z

P

The absorbed power is of course the available power of the source.

Since the input impedance is a conjugate match to the source, the source is

delivering energy at the highest rate it possibly can!



But, the load still reflects power!

Q: But if the power delivered is at a maximum, so too is the power absorbed by the load. I.E.:

abs del avlP P P

If the power absorbed from the load is at a maximum, then the power reflected from the load must be at a minimum. The load must be absorbing all the incident power; right?

A: You might think so—and many engineers in fact do think so.

But those engineers are incorrect!

To see why, consider the load LZ that minimizes the reflected power.

We know this load must have one specific value—the matched load 0LZ Z .



If the load is Z0 then so is Zin

But, if the load is matched to the transmission line (i.e., 0LZ Z , then the input

impedance will likewise be equal to 0

Z :

0inZ Z

And, generally speaking, an input impedance of 0inZ Z will not be equal to the

complex conjugate of the source impedance:

0 gZ Z !!!!

0,Z

ZL=Z0

0inZ Z

0in gZ Z Z

Vg gZ

del avlP P



Look closer at this result

Recall that we just determined that for 0LZ Z , the absorbed power is:

2

02

02

g

abs

g

V ZP

Z Z

It can be shown that this value is less than the available power of the source:

2 2

02

02 8

g g

avlgg

V VZP

RZ Z

Q: Huh!? This makes no sense! After all, just look at the expression for absorbed power:

2

20

0

12abs L

VP

Z

Clearly, this value is maximized when 0L (i.e., when 0LZ Z )!!!



Remember, the load affects V+ A: Let’s look closer at this result:

2

20

0

12abs L

VP

Z

Remember, we can rewrite this to show that the absorbed power is simply the

difference between the power of the incident and reflected waves:

2 2 2 2 2

20 0 0 0 0

0 0 0 0 0

12 2 2 2 2

L

abs L

V V V V VP P P

Z Z Z Z Z

ZL

Pabs

P P



Absorbed power is maximized if

the difference between incident

and reflected is maximized!

Of course, it is true that the load impedance LZ affects the minus-wave

amplitude 0V , and thus affects likewise the reflected wave power P .

Remember however, that the value of LZ likewise affects the plus-wave amplitude

0V , and thus affects likewise the incident wave power P !

* Thus, the value of LZ that minimizes P will not generally maximize P !

* Likewise the value of LZ that maximizes P will not generally minimize P .

* Instead, the value of LZ that maximizes the absorbed power absP is, by

definition, the value that maximizes the difference P P .

We find that this ideal impedance LZ is the value that results in the ideal case of

in gZ Z !



The third special case: the matched source

00gZ Z j

For this case, we find that 0V simplifies greatly:

00

0

0

0 0

1 1

1 1

1

1 1

1

2

jg

in g in

jg

in in

jg

in in

jg

ZV V e

Z Z

ZV e

Z Z

V e

V e

Look at what this says!

I z

V z

Vg LZ

z 0z

0Z 0

Z



This is the answer—regardless of the load!

It says that the incident wave in this case is independent of the load attached at

the other end (the value in

is nowhere to be found)!

Thus, for the one case 0gZ Z , we in fact can consider V z as being the source

wave.

And then the reflected wave V z is the causal result of this stimulus.

Specifically, for this case we can write directly the plus-wave—without knowing

anything about the load LZ :

0

1

2 2j βz jβ j βz

g

g jβ zV z V e V e eV

e



This blue box equation will

come in quite handy!

Therefore, the value at the load (i.e., 0z ) is:

02g jβ

VV z e

And the value of the plus-wave at the source (i.e., z ) is:

2gV

V z

This last result is very important.

Remember this result, it will come in quite handy later in the course!



All I can say is: wow!

Now consider another really important result.

Recall the power associated with the plus-wave is:

2

0

02

VP

Z

Inserting the simplified expression for 0V (i.e., when 0gZ Z ):

2 22

0

0 0 02 8 8

jg gV e VV

PZ Z Z

Wow!

Q: I don’t see why this is “wow”. Am I missing something?

A: Apparently you are. Look closer at the above result.



Do I have to explain everything?

Remember, this result is for the special case where the source impedance is a real

value, numerically equal to 0Z :

00gZ Z j

Therefore:

0Re gg

ZR Z

The source resistance is numerically equal to transmission line characteristic

impedance 0Z .

Thus, the power of the plus-wave can be alternatively written as:

2 2

08 8

g g

g

V VP

Z R Wow!

Q: This result looks vaguely familiar; haven’t we seen this before?

A: Sigh. This result is the available power of the source!!!!!!!



The incident power is the

available power (wow)!

Therefore: 2 2

08 8

g g

avlg

V VP P

Z R Wow!

For the special case 0

0gZ Z j , the power P associated with the plus-

wave (i.e., the incident wave) is equal to the available power avlP of the

source.

Q: Wow! Does this mean that the power absorbed by the load is equal to the available power avlP of the source?

A: Absolutely not!



But some of this incident

power is reflected (Doh!)

Remember, only if in gZ Z does the power absorbed by the load equal the

available power.

* Making 0

0gZ Z j causes the power of the incident wave to be equal to the

available power.

* But, the value LZ of the load impedance determines how much of that incident

power is absorbed—and how much gets reflected.

ZL

in gZ Z * abs avlP P



The really special case!

Q: Hey, that’s right! If the load is matched (i.e.,

00LZ Z j ), then wouldn’t none of the incident

power be reflected (i.e., 0P ). And so wouldn’t all the incident (i.e., available!) power be absorbed by the load?

A: That’s exactly correct!

Q: But you said earlier that minimizing the reflected power would not result in maximizing the absorbed power. Aren’t you contradicting yourself? A: Not at all.

Z0

abs avlP P

avlP P



A conjugate match!

Generally speaking, minimizing the reflected power (i.e., 0LZ Z ) will not result in

maximizing the absorbed power.

However, we are considering the special case where also 0gZ Z !

Thus, maximum power transfer will occur if there is a conjugate match in gZ Z .

Note for this special case 0gZ Z , meaning that input impedance inZ must be

equal to 0

Z for maximum power delivery.

Of course, the input impedance inZ will be equal to 0

Z when the load impedance is

matched (i.e., 0LZ Z )!

I z

V z

Vg 0

Z

z 0z

0Z 0

Z



“Zee-zeros” everywhere

Thus, in many ways, the case:

0g LZ Z Z

(i.e., both source and load impedances are numerically equal to Z0) is ideal.

Let’s summarize this “ideal case”:

* A conjugate match occurs ( in gZ Z ), and so all the available power from the

source is absorbed by the load:

del avl absP P P

I z

V z

Vg 0

Z

z 0z

0Z 0

Z



It’s just so simple; you hardly

deserve credit for this class

* The incident (plus) wave is independent of the load impedance:

2g jβ z

VV z e 0

2g jβ

VV z e

2gV

V z

* And the power associated with this incident wave is equal to the available

power of the source: 2

08

g

avl

VP P

Z

* The reflected (minus) wave is zero:

0V z

* So of course, the power associated with the reflected wave is likewise zero.

0P



Load AND source is matched—

it’s a very good thing

* This again allows us to verify that all available power is absorbed by the

load:

0abs avl avlP P P P P

* Finally, the total current and voltage simplifies nicely:

0 02 2g gjβ z jβ z

V VV z V zV z V z V z e I z e

Z Z

I z

V z

Vg 0

Z

z 0z

0Z 0

Z

3/6/2012 special cases of source and load present 1/22 special …jstiles/723/handouts/special cases...

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