Waves and Transmission Lines

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# Waves and Transmission Lines - PowerPoint PPT Presentation

Waves and Transmission Lines. Wang C. Ng. Traveling Waves. Envelop of a Standing Wave. Load. Waves in a transmission line. Electrical energy is transmitted as waves in a transmission line. Waves travel from the generator to the load (incident wave).

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Presentation Transcript

### Wavesand Transmission Lines

Wang C. Ng

Waves in a transmission line
• Electrical energy is transmitted as waves in a transmission line.
• Waves travel from the generator to the load (incident wave).
• If the resistance of the load does not match the characteristic impedance of the transmission line, part of the energy will be reflected back toward the generator. This is called the reflected wave
Reflection coefficient
• The ratio of the amplitude of the incident wave (v+ ) and the amplitude the reflective wave (v-) is called the reflection coefficient:
Reflection coefficient
• The reflection coefficient can be determine from the load impedance and the characteristic impedance of the line:
• ZL = 0
• = -1
• v - = - v + at the load
• As a result, vL = v + + v - = 0
• ZL = 
• = +1
• v - = v + at the load
• As a result, vL = v + + v - = 2 v +
• ZL = Z0
• = 0
• v - = 0 at the load
• As a result, vL = v +
• ZL = 0.5 Z0
• = - 1/3
• v - = -0.333 v + at the load
• As a result, vL = v + + v - = 0.667 v +
• ZL = 2 Z0
• = + 1/3
• v - = 0.333 v + at the load
• As a result, vL = v + + v - = 1.333 v +
• ZL = j Z0
• = + j1
• v - = v +90 at the load
• As a result, vL = v + + v - = (1 + j1) v +

= 1.414 v +45

• ZL = -j Z0
• = - j1
• v - = v +-90 at the load
• As a result, vL = v + + v - = (1 - j1) v +

= 1.414 v +-45

### Smith Chart

Transmission Line

Calculator

j1

j0.5

j2

j4

0 0.5 1 2 4

j0

-j4

ZL / Z0 = zL = 1 + j 2

-j2

-j0.5

-j1

imaginary

||

real

0 0.5 1

  0.7 45

= 0.5 + j 0.5

  0.7 45

zL= 1 + j 2

||

j1

re

j0.5

j2

im

j4

||

0 0.5 1 2 4 

j 0

-j 4

-j2

-j0.5

-j1

zL= 1 + j 2

90

135

45

j1

j2

j0.5

j4

0 0.5 1 2 4

180

0

j0

-j4

-j0.5

-j2

225

-j1

315

270

  0.7 45

90

135

45

j1

j2

j0.5

j4

0 0.5 1 2 4

180

0

j0

-j4

-j0.5

-j2

225

-j1

315

  0.45 -120

270

zL= 0.5- j 0.5

90

F

135

45

j1

j2

j0.5

j4

0 0.5 1 2 4

180

0

A

E

D C B

j0

| |

0 0.5 1

-j4

-j0.5

-j2

225

-j1

315

G

270