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|V(z)|. +. V 0. +. | V0|. -. -3/4. -/2. -/4. |V(z)|. +. -. 2| V0|. V 0. |V(z)|. |V(z)|. +. 2| V0|. -. -3/4. -/2. -/4. +. 1/2. -. -3/4. -/2. -/4. = | V 0 | [ 1+ | | ² + 2| |cos(2  z +  r )]. 16.360 Lecture 9. Standing Wave. Special cases.

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slide1

|V(z)|

+

V0

+

|V0|

-

-3/4

-/2

-/4

|V(z)|

+

-

2|V0|

V0

|V(z)|

|V(z)|

+

2|V0|

-

-3/4

-/2

-/4

+

1/2

-

-3/4

-/2

-/4

= |V0| [1+ | |² + 2||cos(2z + r)]

16.360 Lecture 9

Standing Wave

Special cases

  • ZL= Z0, = 0

+

|V(z)|

= |V0|

-

ZL

Z0

 

=

+

ZL

Z0

2. ZL= 0,short circuit, = -1

+

1/2

|V(z)|

= |V0| [2 + 2cos(2z + )]

3. ZL= ,open circuit, = 1

+

1/2

|V(z)|

= |V0| [2 + 2cos(2z )]

slide2

+

V0

Z0

jz

jz

-jz

-jz

e

(e

e

(e

16.360 Lecture 9

short circuit line

B

Ii

A

Zg

Vg(t)

sc

Z0

VL

Zin

ZL = 0

l

z = - l

z = 0

ZL= 0, = -1, S = 

+

V(z) = V0 )

-

= -2jV0sin(z)

+

+

i(z) =

)

= 2V0cos(z)/Z0

V(-l)

Zin

= jZ0tan(l)

=

i(-l)

slide3

+

= 2V0cos(z)

+

V0

Z0

jz

jz

-jz

-jz

e

(e

e

(e

16.360 Lecture 9

open circuit line

B

Ii

A

Zg

Vg(t)

oc

Z0

VL

Zin

ZL = 

l

z = - l

z = 0

ZL = ,  = 1, S = 

+

V(z) = V0 )

+

-

i(z) =

)

= 2jV0sin(z)/Z0

V(-l)

oc

Zin

= -jZ0cot(l)

=

i(-l)

short circuit open circuit method

16.360 Lecture 9

Short-Circuit/Open-Circuit Method

For a line of known length l, measurements of its input impedance, one when terminated in a short and another when terminated in an open, can be used to find its characteristic impedance Z0and electrical length

slide5

16.360 Lecture 9

Line of length l = n/2

tan(l) = tan((2/)(n/2)) = 0,

Zin

= ZL

Any multiple of half-wavelength line doesn’t modify the load impedance.

slide6

(1 - )

Z0

(1 + )

-j2l

-j2l

e

e

-

+

(1

(1

)

)

Z0

16.360 Lecture 9

Quarter-wave transformer l = /4 + n/2

l = (2/)(/4 + n/2) = /2 ,

-j 

e

+

)

(1

Zin(-l)

=

=

Z0

=

-j 

e

-

(1

)

= Z0²/ZL

slide7

16.360 Lecture 9

An example:

A 50- lossless tarnsmission is to be matched to a resistive load impedance with

ZL = 100  via a quarter-wave section, thereby eliminating reflections along the feed line.

Find the characteristic impedance of the quarter-wave tarnsformer.

Z01 = 50 

ZL = 100 

/4

= Z0²/ZL

Zin

Zin = Z0²/ZL= 50 

½

½

Z0 = (ZinZL) = (50*100)

slide8

16.360 Lecture 9

Matched transmission line:

  • ZL = Z0
  •  = 0
  • All incident power is delivered to the load.
slide10

+

+

-

V0

V0

V0

Z0

Z0

Z0

-jz

jz

-jz

(e

e

e

16.360 Lecture 9

  • Instantaneous power
  • Time-average power

jz

+

e

V(z) = V0()

+

-

i(z) =

)

At load z = 0, the incident and reflected voltages and currents:

i

i

+

V = V0

i =

r

-

r

V = V0

i =

slide11

16.360 Lecture 9

  • Instantaneous power

i

i

i

P(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)]

+

+

+

+

= Re[|V0|exp(j )exp(jt)] Re[|V0|/Z0 exp(j )exp(jt)]

+

+

= (|V0|²/Z0) cos²(t +  )

r

r

r

P(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)]

-

+

-

+

= Re[|V0|exp(j )exp(jt)] Re[|V0|/Z0 exp(j )exp(jt)]

+

+

= - ||²(|V0|²/Z0) cos²(t +  + r)

slide12

1

T

+

+

(|V0|²/Z0) cos²(t +  )dt

16.360 Lecture 9

  • Time-average

Time-domain approach:

i

T

T

i

Pav =

P (t)dt

=

2

0

0

+

= (|V0|²/2Z0)

r

+

Pav

= -||² (|V0|²/2Z0)

Net average power:

i

r

Pav

+ Pav

= Pav

+

= (1-||²) (|V0|²/2Z0)

slide13

16.360 Lecture 9

  • Time-average

Phasor-domain approach

Pav

= (½)Re[V i*]

i

+

+

+

Pav = (1/2) Re[V0 V0*/Z0]

= (|V0|²/2Z0)

r

+

Pav

= -||² (|V0|²/2Z0)

+

Pav

= (1-||²) (|V0|²/2Z0)

slide14

Solution of Wave Equation

Wave Equation

TL effect

Lumped element model

TL Equation

l/>0.01

Wave (Input) Impedance

Reflection coefficient

Standing Wave

Lossless TL

+

Complete Solution

Solving for V0

Power

+

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