Incident Reflected and Absorbed Power - I2S

1/27/2009

Incident Reflected and Absorbed Power.doc

1/8

Incident, Reflected, and Absorbed Power

We have discovered that two waves propagate along a

transmission line, one in each direction (V + (z ) and V - (z ) ).

I (z ) = I +(z ) + I -(z )

z = -A

+

V (z ) =V +(z ) +V -(z )

-

A

IL

+

VL

ZL

-

z =0

The result is that electromagnetic energy flows along the transmission line at a given rate (i.e., power).

Q: How much power flows along a transmission line, and where does that power go?

A: We can answer that question by determining the power absorbed by the load!

Jim Stiles

The Univ. of Kansas

Dept. of EECS

1/27/2009

Incident Reflected and Absorbed Power.doc

2/8

You of course recall that the time-averaged power (a real value!) absorbed by a complex impedance ZL is:

{ } Pabs

=

1 2

Re

VL

IL

Of course, the load voltage and current is simply the voltage an current at the end of the transmission line (at z = 0 ). A happy result is that we can then use our transmission line theory to determine this absorbed power:

{ } Pabs

=

1 2

Re

VL

I

L

{ } = 1 Re V (z = 0)I (z = 0) 2

{( ) ( ) } = 1 Re 2 Z0

V0+ e - j 0 + 0 e + j 0

V0+ e - j 0 - 0 e + j 0

{ ( ) } = V0+ 2 Re 1 - 2 Z0

0 - 0

- 0 2

( ) = V0+ 2 2 Z0

1 - 0 2

The significance of this result can be seen by rewriting the expression as:

( ) Pabs

= V0+ 2 2 Z0

1 - 0 2

= V0+ 2 - V0+0 2 = V0+ 2 - V0- 2 2 Z0 2 Z0 2 Z0 2 Z0

Jim Stiles

The Univ. of Kansas

Dept. of EECS

1/27/2009

Incident Reflected and Absorbed Power.doc

3/8

The two terms in above expression have a very definite physical meaning. The first term is the time-averaged power of the wave propagating along the transmission line toward the load.

We say that this wave is incident on the load:

Pinc

= P+

= V0+ 2 2Z 0

Likewise, the second term of the Pabs equation describes the power of the wave moving in the other direction (away from the load). We refer to this as the wave reflected from the load:

Pref

= P-

= V0- 2 2Z0

=

LV0+ 2 2Z0

=

L

2

V+2 0

2Z0

=

L 2 Pinc

Thus, the power absorbed by the load (i.e., the power delivered to the load) is simply:

Pabs

=

V+2 0

2 Z0

- V0- 2 2 Z0

= Pinc

- Pref

or, rearranging, we find: Pinc = Pabs + Pref

This equation is simply an expression of the conservation of energy !

Jim Stiles

The Univ. of Kansas

Dept. of EECS

1/27/2009

Incident Reflected and Absorbed Power.doc

4/8

It says that power flowing toward the load (Pinc) is either

absorbed by the load (Pabs) or reflected back from the load

(Pref).

Pabs

Pinc

Pref

ZL

Now let's consider some special cases: 1. L 2 = 1 For this case, we find that the load absorbs no power!

( ) ( ) Pabs = Pinc 1 - 0 2 = Pinc 1 - 1 = 0

Likewise, we find that the reflected power is equal to the

incident:

( ) Pref

=

L

P2 inc

=

1

Pinc

= Pinc

Note these two results are completely consistent--by conservation of energy, if one is true the other must also be:

Pinc = Pabs + Pref = 0 + Pref = Pref

Jim Stiles

The Univ. of Kansas

Dept. of EECS

1/27/2009

Incident Reflected and Absorbed Power.doc

5/8

In this case, no power is absorbed by the load. All of the

incident power is reflected, so that the reflected power is equal

to that of the incident.

Pabs = 0

Pinc

Pref = Pinc

L = 1

2. L = 0

For this case, we find that there is no reflected power!

( ) Pref

=

L

P2 inc

=

0

Pinc = 0

Likewise, we find that the absorbed power is equal to the

incident:

( ) ( ) Pabs = Pinc 1 - 0 2 = Pinc 1 - 0 = Pinc

Note these two results are completely consistent--by conservation of energy, if one is true the other must also be:

Pinc = Pabs + Pref = Pabs + 0 = Pabs

In this case, all the incident power is absorbed by the load. None of the incident power is reflected, so that the absorbed power is equal to that of the incident.

Jim Stiles

The Univ. of Kansas

Dept. of EECS

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