17: Transmission Lines

(1)

17: Transmission Lines

•

Transmission Lines

•

Transmission Line Equations

•

Solution to Transmission Line Equations

•

Forward Wave

•

Forward + Backward Waves

•

Power Flow

•

Reflections

•

Reflection Coefficients

•

Driving a line

•

Multiple Reflections

•

Transmission Line Characteristics

•

Summary

(2)

Transmission Lines

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

(3)

Transmission Lines

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Previously assume that any change in

v

S

(

t

)

appears instantly at

v

R

(

t

)

.

(4)

Transmission Lines

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

S

(

t

)

v

R

(

t

)

.

This is not true.

(5)

Transmission Lines

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

S

(

t

)

v

R

(

t

)

.

This is not true.

If fact signals travel at around half the speed of light (

c

= 30

cm/ns

)

.

Reason:

all wires have capacitance to ground and to neighbouring

conductors and also self-inductance. It takes time to change the current

through an inductor or voltage across a capacitor.

(6)

Transmission Lines

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

S

(

t

)

v

R

(

t

)

.

This is not true.

c

= 30

cm/ns

)

.

Reason:

through an inductor or voltage across a capacitor.

A

transmission line

is a wire with a uniform goemetry along its length: the

capacitance and inductance of any segment is proportional to its length.

(7)

Transmission Lines

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

S

(

t

)

v

R

(

t

)

.

This is not true.

c

= 30

cm/ns

)

.

Reason:

A

transmission line

capacitance and inductance of any segment is proportional to its length.

We represent as a large number of small inductors and capacitors spaced

along the line.

(8)

Transmission Lines

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

S

(

t

)

v

R

(

t

)

.

This is not true.

c

= 30

cm/ns

)

.

Reason:

A

transmission line

capacitance and inductance of any segment is proportional to its length.

We represent as a large number of small inductors and capacitors spaced

along the line.

(9)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Short section of line

δx

long.

v

(

x, t

)

and

i

(

x, t

) depend on both position

(10)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

δx

long.

v

(

x, t

)

and

i

(

x, t

and time.

Small

δx

⇒

ignore 2nd order derivatives:

∂v

(

x,t

)

∂t

=

∂v

(

x

+

δx,t

)

∂t

=

∂v

∂t

.

(11)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

δx

long.

v

(

x, t

)

and

i

(

x, t

and time.

Small

δx

⇒

∂v

(

x,t

)

∂t

=

∂v

(

x

+

δx,t

)

∂t

=

∂v

∂t

.

(12)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

δx

long.

v

(

x, t

)

and

i

(

x, t

and time.

Small

δx

⇒

∂v

(

x,t

)

∂t

=

∂v

(

x

+

δx,t

)

∂t

=

∂v

∂t

.

Capacitor equation:

C

∂v

∂t

=

i

(

x, t

)

−

i

(

x

+

δx, t

) =

−

∂x

∂i

δx

(13)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

δx

long.

v

(

x, t

)

and

i

(

x, t

and time.

Small

δx

⇒

∂v

(

x,t

)

∂t

=

∂v

(

x

+

δx,t

)

∂t

=

∂v

∂t

.

Capacitor equation:

C

∂v

∂t

=

i

(

x, t

)

−

i

(

x

+

δx, t

) =

−

∂x

∂i

δx

Inductor equation:

L

∂i

∂t

=

v

(

x, t

)

−

v

(

x

+

δx, t

) =

−

∂v

∂x

δx

Substitute

C

₀

=

δx

C

and

L

₀

=

δx

L

where

C

₀

and

L

₀

are capacitance

(14)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

δx

long.

v

(

x, t

)

and

i

(

x, t

and time.

Small

δx

⇒

∂v

(

x,t

)

∂t

=

∂v

(

x

+

δx,t

)

∂t

=

∂v

∂t

.

Capacitor equation:

C

∂v

∂t

=

i

(

x, t

)

−

i

(

x

+

δx, t

) =

−

∂x

∂i

δx

Inductor equation:

L

∂i

∂t

=

v

(

x, t

)

−

v

(

x

+

δx, t

) =

−

∂v

∂x

δx

Substitute

C

₀

=

δx

C

and

L

₀

=

δx

L

where

C

₀

and

L

₀

are capacitance

and inductance per unit length.

C

₀

∂v

∂t

=

−

∂x

∂i

(15)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

δx

long.

v

(

x, t

)

and

i

(

x, t

and time.

Small

δx

⇒

∂v

(

x,t

)

∂t

=

∂v

(

x

+

δx,t

)

∂t

=

∂v

∂t

.

Capacitor equation:

C

∂v

∂t

=

i

(

x, t

)

−

i

(

x

+

δx, t

) =

−

∂x

∂i

δx

Inductor equation:

L

∂i

∂t

=

v

(

x, t

)

−

v

(

x

+

δx, t

) =

−

∂v

∂x

δx

Substitute

C

₀

=

δx

C

and

L

₀

=

δx

L

where

C

₀

and

L

₀

are capacitance

and inductance per unit length.

C

₀

∂v

∂t

=

−

∂x

∂i

L

₀

∂i

∂t

=

−

∂v

∂x

General solution to two simultaneous partial differential equations will

include two arbitrary

functions

(instead of the arbitrary

constants

that you

(16)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

(17)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Transmission Line Equations:

C

₀

∂v

∂t

=

−

∂x

∂i

L

₀

∂i

∂t

=

−

∂v

∂x

General solution:

v

(

t, x

) =

f

(

t

−

x

u

) +

g

(

t

+

x

u

)

i

(

t, x

) =

f

(

t

−

x

u

)

−

g

(

t

+

x

u

)

Z

0

where

u

=

q

L

₀

1

C

₀

and

Z

₀

=

q

L

0

C

0

.

(18)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Transmission Line Equations:

C

₀

∂v

∂t

=

−

∂x

∂i

L

₀

∂i

∂t

=

−

∂v

∂x

General solution:

v

(

t, x

) =

f

(

t

−

x

u

) +

g

(

t

+

x

u

)

i

(

t, x

) =

f

(

t

−

x

u

)

−

g

(

t

+

x

u

)

Z

0

where

u

=

q

L

₀

1

C

₀

and

Z

₀

=

q

L

0

C

0

.

(19)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

C

₀

∂v

∂t

=

−

∂x

∂i

L

₀

∂i

∂t

=

−

∂v

∂x

General solution:

v

(

t, x

) =

f

(

t

−

x

u

) +

g

(

t

+

x

u

)

i

(

t, x

) =

f

(

t

−

x

u

)

−

g

(

t

+

x

u

)

Z

0

where

u

=

q

L

₀

1

C

₀

and

Z

₀

=

q

L

0

C

0

.

u

is the

propagation velocity

and

Z

₀

is the

characteristic impedance

.

(20)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

C

₀

∂v

∂t

=

−

∂x

∂i

L

₀

∂i

∂t

=

−

∂v

∂x

General solution:

v

(

t, x

) =

f

(

t

−

x

u

) +

g

(

t

+

x

u

)

i

(

t, x

) =

f

(

t

−

x

u

)

−

g

(

t

+

x

u

)

Z

0

where

u

=

q

L

₀

1

C

₀

and

Z

₀

=

q

L

0

C

0

.

u

is the

and

Z

₀

is the

characteristic impedance

.

f

()

and

g

()

can be

any

differentiable functions.

Verify by substitution:

−

∂x

∂i

=

−

f

′

(

t

−

x

u

)

−

g

′

(

t

+

x

u

)

Z

0

×

1

u

(21)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

C

₀

∂v

∂t

=

−

∂x

∂i

L

₀

∂i

∂t

=

−

∂v

∂x

General solution:

v

(

t, x

) =

f

(

t

−

x

u

) +

g

(

t

+

x

u

)

i

(

t, x

) =

f

(

t

−

x

u

)

−

g

(

t

+

x

u

)

Z

0

where

u

=

q

L

₀

1

C

₀

and

Z

₀

=

q

L

0

C

0

.

u

is the

and

Z

₀

is the

characteristic impedance

.

f

()

and

g

()

can be

any

differentiable functions.

Verify by substitution:

−

∂x

∂i

=

−

f

′

(

t

−

x

u

)

−

g

′

(

t

+

x

u

)

Z

0

×

1

u

=

C

₀

f

′

(

t

−

x

u

) +

g

′

(

t

+

x

u

)

=

C

₀

∂v

∂t

(22)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

(23)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

−

u

0

)

0

2

4

6

8

10

Time (ns)

f(t-0/u)

(24)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

−

u

0

)

At

x

= 45

cm [

N

],

v

(45

, t

) =

f

(

t

−

45

u

)

0

2

4

6

8

10

Time (ns)

f(t-0/u)

f(t-45/u)

(25)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

−

u

0

)

At

x

= 45

cm [

N

],

v

(45

, t

) =

f

(

t

−

45

u

)

0

2

4

6

8

10

Time (ns)

f(t-0/u)

f(t-45/u)

(26)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

−

u

0

)

At

x

= 45

cm [

N

],

v

(45

, t

) =

f

(

t

−

45

u

)

0

2

4

6

8

10

Time (ns)

f(t-0/u)

f(t-45/u)

f(t-90/u)

f

(

t

−

45

u

)

is the same as

f

(

t

)

but delayed by

45

u

= 3

ns.

(27)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

−

u

0

)

At

x

= 45

cm [

N

],

v

(45

, t

) =

f

(

t

−

45

u

)

0

2

4

6

8

10

Time (ns)

f(t-0/u)

f(t-45/u)

f(t-90/u)

f

(

t

−

45

u

)

is the same as

f

(

t

)

but delayed by

45

u

= 3

ns.

At

x

= 90

cm [

N

],

v

R(

t

) =

f

(

t

−

90

u

)

; now delayed by

6 ns

.

(28)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

−

u

0

)

At

x

= 45

cm [

N

],

v

(45

, t

) =

f

(

t

−

45

u

)

0

2

4

6

8

10

Time (ns)

f(t-0/u)

f(t-45/u)

f(t-90/u)

f

(

t

−

45

u

)

is the same as

f

(

t

)

but delayed by

45

u

= 3

ns.

At

x

= 90

cm [

N

],

v

R(

t

) =

f

(

t

−

90

u

)

; now delayed by

6 ns

.

Waveform at

x

= 0

completely determines the waveform everywhere else.

Snapshot at

t

₀

= 4 ns

:

the waveform has just

arrived at the point

x

=

ut

₀

= 60

cm

.

0

20

40

60

Position (cm)

80

f(4-x/u)

t = 4 ns

(29)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

−

u

0

)

At

x

= 45

cm [

N

],

v

(45

, t

) =

f

(

t

−

45

u

)

0

2

4

6

8

10

Time (ns)

f(t-0/u)

f(t-45/u)

f(t-90/u)

f

(

t

−

45

u

)

is the same as

f

(

t

)

but delayed by

45

u

= 3

ns.

At

x

= 90

cm [

N

],

v

R(

t

) =

f

(

t

−

90

u

)

; now delayed by

6 ns

.

Waveform at

x

= 0

Snapshot at

t

₀

= 4 ns

:

x

=

ut

₀

= 60

cm

.

0

20

40

60

Position (cm)

80

f(4-x/u)

t = 4 ns

(30)

Forward Wave

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

u

= 15

cm/ns

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

−

u

0

)

At

x

= 45

cm [

N

],

v

(45

, t

) =

f

(

t

−

45

u

)

0

2

4

6

8

10

Time (ns)

f(t-0/u)

f(t-45/u)

f(t-90/u)

f

(

t

−

45

u

)

is the same as

f

(

t

)

but delayed by

45

u

= 3

ns.

At

x

= 90

cm [

N

],

v

R(

t

) =

f

(

t

−

90

u

)

; now delayed by

6 ns

.

Waveform at

x

= 0

Snapshot at

t

₀

= 4 ns

:

x

=

ut

₀

= 60

cm

.

0

20

40

60

Position (cm)

80

f(4-x/u)

t = 4 ns

(31)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

(32)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Similarly

g

(

t

+

x

u

)

is a wave travelling backwards, i.e. in the

−

x

direction.

v

(

x, t

) =

f

(

t

−

u

x

) +

g

(

t

+

x

u

)

0

2

4

6

8

Time (ns)

10

f(t-0/u)

g(t+0/u)

(33)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Similarly

g

(

t

+

x

u

)

−

x

direction.

v

(

x, t

) =

f

(

t

−

u

x

) +

g

(

t

+

x

u

)

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

) +

g

(

t

)

0

2

4

6

8

Time (ns)

10

f(t-0/u)

g(t+0/u)

x=0

(34)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Similarly

g

(

t

+

x

u

)

−

x

direction.

v

(

x, t

) =

f

(

t

−

u

x

) +

g

(

t

+

x

u

)

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

) +

g

(

t

)

0

2

4

6

8

Time (ns)

10

f(t-0/u)

g(t+0/u)

x=0

f(t-90/u)

g(t+90/u)

x=90

(35)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Similarly

g

(

t

+

x

u

)

−

x

direction.

v

(

x, t

) =

f

(

t

−

u

x

) +

g

(

t

+

x

u

)

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

) +

g

(

t

)

0

2

4

6

8

Time (ns)

10

f(t-0/u)

g(t+0/u)

x=0

f(t-90/u)

g(t+90/u)

x=90

f(t-45/u)+g(t+45/u)

x=45

At

x

= 45

cm [

N

],

g

is only

1

ns behind

f

and they add together.

(36)

Forward + Backward Waves

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Similarly

g

(

t

+

x

u

)

−

x

direction.

v

(

x, t

) =

f

(

t

−

u

x

) +

g

(

t

+

x

u

)

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

) +

g

(

t

)

0

2

4

6

8

Time (ns)

10

f(t-0/u)

g(t+0/u)

x=0

f(t-90/u)

g(t+90/u)

x=90

f(t-45/u)+g(t+45/u)

x=45

At

x

= 45

cm [

N

],

g

is only

1

ns behind

f

At

x

= 90

cm [

N

],

g

starts at

t

= 1

and

f

starts at

t

= 6

.

Snapshots:

At

t

= 2

ns

f

and

g

have not

yet met.

0

20

40

60

80

Position (cm)

f(2-x/u)+g(2-x/u)

(37)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Similarly

g

(

t

+

x

u

)

−

x

direction.

v

(

x, t

) =

f

(

t

−

u

x

) +

g

(

t

+

x

u

)

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

) +

g

(

t

)

0

2

4

6

8

Time (ns)

10

f(t-0/u)

g(t+0/u)

x=0

f(t-90/u)

g(t+90/u)

x=90

f(t-45/u)+g(t+45/u)

x=45

At

x

= 45

cm [

N

],

g

is only

1

ns behind

f

At

x

= 90

cm [

N

],

g

starts at

t

= 1

and

f

starts at

t

= 6

.

Snapshots:

At

t

= 2

ns

f

and

g

have not

yet met.

By

t

= 5

ns,

f

and

g

have

just crossed.

0

20

40

60

80

Position (cm)

f(2-x/u)+g(2-x/u)

t = 2 ns

0

20

40

60

80

Position (cm)

f(5-x/u)+g(5-x/u)

t = 5 ns

(38)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Similarly

g

(

t

+

x

u

)

−

x

direction.

v

(

x, t

) =

f

(

t

−

u

x

) +

g

(

t

+

x

u

)

At

x

= 0

cm [

N

],

v

S

(

t

) =

f

(

t

) +

g

(

t

)

0

2

4

6

8

Time (ns)

10

f(t-0/u)

g(t+0/u)

x=0

f(t-90/u)

g(t+90/u)

x=90

f(t-45/u)+g(t+45/u)

x=45

At

x

= 45

cm [

N

],

g

is only

1

ns behind

f

At

x

= 90

cm [

N

],

g

starts at

t

= 1

and

f

starts at

t

= 6

.

Snapshots:

At

t

= 2

ns

f

and

g

have not

yet met.

By

t

= 5

ns,

f

and

g

have

just crossed.

0

20

40

60

80

Position (cm)

f(2-x/u)+g(2-x/u)

t = 2 ns

0

20

40

60

80

Position (cm)

f(5-x/u)+g(5-x/u)

t = 5 ns

(39)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

to be the forward and

(40)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

to be the forward and

backward waveforms at any point,

x

.

i

is

always

measured in the

+ve

x

direction.

Then

v

x(

t

) =

f

x(

t

) +

g

x

(

t

)

and

i

x(

t

) =

Z

−

1

0

(

f

x(

t

)

−

g

x(

t

))

.

(41)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

.

i

is

always

measured in the

+ve

x

direction.

Then

v

x(

t

) =

f

x(

t

) +

g

x

(

t

)

and

i

x(

t

) =

Z

−

1

0

(

f

x(

t

)

−

g

x(

t

))

.

Note:

Knowing the waveform

f

x(

t

)

or

g

x(

t

)

at any position

x

, tells you it at

all other positions:

f

y

(

t

) =

f

x

t

−

y

−

x

u

and

g

y

(

t

) =

g

x

t

+

y

−

x

u

.

(42)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

.

i

is

always

measured in the

+ve

x

direction.

Then

v

x(

t

) =

f

x(

t

) +

g

x

(

t

)

and

i

x(

t

) =

Z

−

1

0

(

f

x(

t

)

−

g

x(

t

))

.

Note:

f

x(

t

)

or

g

x(

t

)

at any position

x

, tells you it at

all other positions:

f

y

(

t

) =

f

x

t

−

y

−

x

u

and

g

y

(

t

) =

g

x

t

+

y

−

x

u

.

Power Flow

The power transferred into the shaded region across the boundary at

x

is

(43)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

.

i

is

always

measured in the

+ve

x

direction.

Then

v

x(

t

) =

f

x(

t

) +

g

x

(

t

)

and

i

x(

t

) =

Z

−

1

0

(

f

x(

t

)

−

g

x(

t

))

.

Note:

f

x(

t

)

or

g

x(

t

)

at any position

x

, tells you it at

f

y

(

t

) =

f

x

t

−

y

−

x

u

and

g

y

(

t

) =

g

x

t

+

y

−

x

u

.

Power Flow

x

is

(44)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

.

i

is

always

measured in the

+ve

x

direction.

Then

v

x(

t

) =

f

x(

t

) +

g

x

(

t

)

and

i

x(

t

) =

Z

−

1

0

(

f

x(

t

)

−

g

x(

t

))

.

Note:

Knowing the waveform

f

x(

t

)

or

g

x(

t

)

at any position

x

, tells you it at

f

y

(

t

) =

f

x

t

−

y

−

x

u

and

g

y

(

t

) =

g

x

t

+

y

−

x

u

.

Power Flow

x

is

P

x(

t

) =

v

x(

t

)

i

x(

t

) =

Z

₀

−

1

(

f

x

(

t

) +

g

x(

t

)) (

f

x(

t

)

−

g

x(

t

))

=

f

2

x

(

t

)

Z

0

−

g

x

2

(

t

)

Z

0

(45)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

.

i

is

always

measured in the

+ve

x

direction.

Then

v

x(

t

) =

f

x(

t

) +

g

x

(

t

)

and

i

x(

t

) =

Z

−

1

0

(

f

x(

t

)

−

g

x(

t

))

.

Note:

f

x(

t

)

or

g

x(

t

)

at any position

x

, tells you it at

f

y

(

t

) =

f

x

t

−

y

−

x

u

and

g

y

(

t

) =

g

x

t

+

y

−

x

u

.

Power Flow

x

is

P

x(

t

) =

v

x(

t

)

i

x(

t

) =

Z

₀

−

1

(

f

x

(

t

) +

g

x(

t

)) (

f

x(

t

)

−

g

x(

t

))

=

f

2

x

(

t

)

Z

0

−

g

x

2

(

t

)

Z

0

(46)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

backward waveforms at any point,

x

.

i

is

always

measured in the

+ve

x

direction.

Then

v

x(

t

) =

f

x(

t

) +

g

x

(

t

)

and

i

x(

t

) =

Z

−

1

0

(

f

x(

t

)

−

g

x(

t

))

.

Note:

f

x(

t

)

or

g

x(

t

)

at any position

x

, tells you it at

f

y

(

t

) =

f

x

t

−

y

−

x

u

and

g

y

(

t

) =

g

x

t

+

y

−

x

u

.

Power Flow

x

is

P

x(

t

) =

v

x(

t

)

i

x(

t

) =

Z

₀

−

1

(

f

x

(

t

) +

g

x(

t

)) (

f

x(

t

)

−

g

x(

t

))

=

f

2

x

(

t

)

Z

0

−

g

x

2

(

t

)

Z

0

f

x

carries power

into

shaded area and

g

x

carries power

out

independently.

(47)

Power Flow

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

Define

f

x(

t

) =

f t

−

x

u

and

g

x(

t

) =

g t

+

u

x

.

i

is

always

measured in the

+ve

x

direction.

Then

v

x(

t

) =

f

x(

t

) +

g

x

(

t

)

and

i

x(

t

) =

Z

−

1

0

(

f

x(

t

)

−

g

x(

t

))

.

Note:

f

x(

t

)

or

g

x(

t

)

at any position

x

, tells you it at

f

y

(

t

) =

f

x

t

−

y

−

x

u

and

g

y

(

t

) =

g

x

t

+

y

−

x

u

.

Power Flow

x

is

P

x(

t

) =

v

x(

t

)

i

x(

t

) =

Z

₀

−

1

(

f

x

(

t

) +

g

x(

t

)) (

f

x(

t

)

−

g

x(

t

))

=

f

2

x

(

t

)

Z

0

−

g

x

2

(

t

)

Z

0

f

x

carries power

into

shaded area and

g

x

carries power

out

independently.

Power travels in the

same direction as the wave

.

(48)

Reflections

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

x

=

f

x

+

g

x

i

x

=

Z

−

1

0

(

f

x

−

g

x)

(49)

Reflections

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

x

=

f

x

+

g

x

i

x

=

Z

−

1

0

(

f

x

−

g

x)

From Ohm’s law at

x

=

L

, we have

v

L

(

t

) =

i

L

(

t

)

R

L

(50)

Reflections

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

x

=

f

x

+

g

x

i

x

=

Z

−

1

0

(

f

x

−

g

x)

From Ohm’s law at

x

=

L

, we have

v

L

(

t

) =

i

L

(

t

)

R

L

Hence

(

f

L

(

t

) +

g

L(

t

)) =

Z

₀

−

1

(

f

L

(

t

)

−

g

L(

t

))

R

L

From this:

g

L

(

t

) =

RL

−

Z

0

(51)

Reflections

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

x

=

f

x

+

g

x

i

x

=

Z

−

1

0

(

f

x

−

g

x)

From Ohm’s law at

x

=

L

, we have

v

L

(

t

) =

i

L

(

t

)

R

L

Hence

(

f

L

(

t

) +

g

L(

t

)) =

Z

₀

−

1

(

f

L

(

t

)

−

g

L(

t

))

R

L

From this:

g

L

(

t

) =

RL

−

Z

0

RL

+

Z

0

×

f

L

(

t

)

We define the

reflection coefficient

:

ρ

L

=

gL

fL

(

t

)

=

RL

−

Z

0

(52)

Reflections

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

x

=

f

x

+

g

x

i

x

=

Z

−

1

0

(

f

x

−

g

x)

From Ohm’s law at

x

=

L

, we have

v

L

(

t

) =

i

L

(

t

)

R

L

Hence

(

f

L

(

t

) +

g

L(

t

)) =

Z

₀

−

1

(

f

L

(

t

)

−

g

L(

t

))

R

L

From this:

g

L

(

t

) =

RL

−

Z

0

RL

+

Z

0

×

f

L

(

t

)

We define the

:

ρ

L

=

gL

fL

(

t

)

=

RL

−

Z

0

RL

+

Z

0

= +0

.

5

Substituting

g

L

(

t

) =

ρ

L

f

L

(

t

)

gives

v

L

(

t

) = (1 +

ρ

L

)

f

L

(

t

)

and

i

L

(

t

) = (1

−

ρ

L)

Z

−

1

(53)

Reflections

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

x

=

f

x

+

g

x

i

x

=

Z

−

1

0

(

f

x

−

g

x)

From Ohm’s law at

x

=

L

, we have

v

L

(

t

) =

i

L

(

t

)

R

L

Hence

(

f

L

(

t

) +

g

L(

t

)) =

Z

₀

−

1

(

f

L

(

t

)

−

g

L(

t

))

R

L

From this:

g

L

(

t

) =

RL

−

Z

0

RL

+

Z

0

×

f

L

(

t

)

We define the

:

ρ

L

=

gL

fL

(

t

)

=

RL

−

Z

0

RL

+

Z

0

= +0

.

5

Substituting

g

L

(

t

) =

ρ

L

f

L

(

t

)

gives

v

L

(

t

) = (1 +

ρ

L

)

f

L

(

t

)

and

i

L

(

t

) = (1

−

ρ

L)

Z

−

1

0

f

L

(

t

)

0 2 4 6 8 10 12 14 16 18 Time (ns) v 0(t)

(54)

Reflections

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

v

x

=

f

x

+

g

x

i

x

=

Z

−

1

0

(

f

x

−

g

x)

From Ohm’s law at

x

=

L

, we have

v

L

(

t

) =

i

L

(

t

)

R

L

Hence

(

f

L

(

t

) +

g

L(

t

)) =

Z

₀

−

1

(

f

L

(

t

)

−

g

L(

t

))

R

L

From this:

g

L

(

t

) =

RL

−

Z

0

RL

+

Z

0

×

f

L

(

t

)

We define the

:

ρ

L

=

gL

fL

(

t

)

=

RL

−

Z

0

RL

+

Z

0

= +0

.

5

Substituting

g

L

(

t

) =

ρ

L

f

L

(

t

)

gives

v

L

(

t

) = (1 +

ρ

L

)

f

L

(

t

)

and

i

L

(

t

) = (1

−

ρ

L)

Z

−

1

0

f

L

(

t

)

0 2 4 6 8 10 12 14 16 18 Time (ns) v 0(t) 0 2 4 6 8 10 12 14 16 18 Time (ns) i 0(t)

At source end:

g

₀

(

t

) =

ρ

L

f

₀

t

−

2

u

L

i.e. delayed by

2

u

L

= 12

ns.

(55)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

ρ

=

R

−

Z

0

R

+

Z

0

=

R

Z

₀

−

1

R

Z

₀

+1

0

1

2

3

4

5

-1

0

1

RZ

₀-1 ρ

ρ

depends on the ratio

Z

R

0

.

R

Z

0

ρ

vL

(

t

)

f

(

t

)

iL

(

t

)

Z

0

f

(

t

)

Comment

3

+0

.

5

(56)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

ρ

=

R

−

Z

0

R

+

Z

0

=

R

Z

₀

−

1

R

Z

₀

+1

vL

(

t

)

f

(

t

)

= 1 +

ρ

0

1

2

3

4

5

-1

0

1

RZ

₀-1 ρ

ρ

depends on the ratio

Z

R

0

.

R

Z

0

ρ

vL

(

t

)

f

(

t

)

iL

(

t

)

Z

0

f

(

t

)

Comment

3

+0

.

5

1

.

5

(57)

•

Transmission Lines

•

Forward Wave

•

Power Flow

•

Reflections

•

Driving a line

•

Summary

ρ

=

R

−

Z

0

R

+

Z