17: Transmission Lines - Imperial College · PDF file17: Transmission Lines 17: ... •...

17: Transmission Lines


• Transmission Lines• Transmission LineEquations +

• Solution to TransmissionLine Equations

• Forward Wave• Forward + BackwardWaves

• Power Flow

• Reflections

• Reflection Coefficients

• Driving a line

• Multiple Reflections

• Transmission LineCharacteristics +

• Summary

E1.1 Analysis of Circuits (2017-10213) Transmission Lines: 17 – 1 / 13

Transmission Lines





• Power Flow

• Reflections


• Driving a line



• Summary


Previously assume that any change in v0(t) appears instantly at vL(t).

Transmission Lines





• Power Flow

• Reflections


• Driving a line



• Summary



This is not true.

Transmission Lines





• Power Flow

• Reflections


• Driving a line



• Summary



This is not true.

If fact signals travel at around half the speed of light (c = 30 cm/ns).

Transmission Lines





• Power Flow

• Reflections


• Driving a line



• Summary



This is not true.


Reason: all wires have capacitance to ground and to neighbouringconductors and also self-inductance. It takes time to change the currentthrough an inductor or voltage across a capacitor.

Transmission Lines





• Power Flow

• Reflections


• Driving a line



• Summary



This is not true.



A transmission line is a wire with a uniform goemetry along its length: thecapacitance and inductance of any segment is proportional to its length.

Transmission Lines





• Power Flow

• Reflections


• Driving a line



• Summary



This is not true.



A transmission line is a wire with a uniform goemetry along its length: thecapacitance and inductance of any segment is proportional to its length.We represent as a large number of small inductors and capacitors spacedalong the line.

Transmission Lines





• Power Flow

• Reflections


• Driving a line



• Summary



This is not true.



A transmission line is a wire with a uniform goemetry along its length: thecapacitance and inductance of any segment is proportional to its length.We represent as a large number of small inductors and capacitors spacedalong the line.The signal speed along a transmisison line is predictable.

Transmission Line Equations +





• Power Flow

• Reflections


• Driving a line



• Summary


A short section of line δx long:

v(x, t) and i(x, t) depend on bothposition and time.






• Power Flow

• Reflections


• Driving a line



• Summary




Small δx ⇒ ignore 2nd order derivatives:

∂v(x,t)∂t

= ∂v(x+δx,t)∂t

, ∂v∂t

.






• Power Flow

• Reflections


• Driving a line



• Summary





∂v(x,t)∂t

= ∂v(x+δx,t)∂t

, ∂v∂t

.

Basic EquationsKVL: v(x, t) = V2 + v(x+ δx, t) + V1

KCL: i(x, t) = iC + i(x+ δx, t)






• Power Flow

• Reflections


• Driving a line



• Summary





∂v(x,t)∂t

= ∂v(x+δx,t)∂t

, ∂v∂t

.


KCL: i(x, t) = iC + i(x+ δx, t)Capacitor equation: C ∂v

∂t= iC = i(x, t)− i(x+ δx, t) = − ∂i

∂xδx






• Power Flow

• Reflections


• Driving a line



• Summary





∂v(x,t)∂t

= ∂v(x+δx,t)∂t

, ∂v∂t

.



∂t= iC = i(x, t)− i(x+ δx, t) = − ∂i

∂xδx

Inductor equation (L1 and L2 have the same current):(L1 + L2)

∂i∂t

= V1 + V2 = v(x, t)− v(x+ δx, t) = − ∂v∂x

δx






• Power Flow

• Reflections


• Driving a line



• Summary





∂v(x,t)∂t

= ∂v(x+δx,t)∂t

, ∂v∂t

.



∂t= iC = i(x, t)− i(x+ δx, t) = − ∂i

∂xδx


∂i∂t

= V1 + V2 = v(x, t)− v(x+ δx, t) = − ∂v∂x

δx

Transmission Line Equations

C0∂v∂t

= − ∂i∂x

L0∂i∂t

= − ∂v∂x






• Power Flow

• Reflections


• Driving a line



• Summary





∂v(x,t)∂t

= ∂v(x+δx,t)∂t

, ∂v∂t

.



∂t= iC = i(x, t)− i(x+ δx, t) = − ∂i

∂xδx


∂i∂t

= V1 + V2 = v(x, t)− v(x+ δx, t) = − ∂v∂x

δx

Transmission Line Equations

C0∂v∂t

= − ∂i∂x

L0∂i∂t

= − ∂v∂x

where C0 = Cδx

is the capacitance per unit length

(Farads/m) and L0 = L1+L2

δxis the total

inductance per unit length (Henries/m).

Solution to Transmission Line Equations





• Power Flow

• Reflections


• Driving a line



• Summary


Transmission Line Equations: C0∂v∂t

= − ∂i∂x

L0∂i∂t

= − ∂v∂x






• Power Flow

• Reflections


• Driving a line



• Summary



= − ∂i∂x

L0∂i∂t

= − ∂v∂x

General solution: v(t, x) = f(t− xu) + g(t+ x

u)

i(t, x) =f(t− x

u)−g(t+ x

u)

Z0

where u =√

1L0C0

and Z0 =√

L0

C0

.






• Power Flow

• Reflections


• Driving a line



• Summary



= − ∂i∂x

L0∂i∂t

= − ∂v∂x


u)

i(t, x) =f(t− x

u)−g(t+ x

u)

Z0

where u =√

1L0C0

and Z0 =√

L0

C0

.

u is the propagation velocity and Z0 is the characteristic impedance.






• Power Flow

• Reflections


• Driving a line



• Summary



= − ∂i∂x

L0∂i∂t

= − ∂v∂x


u)

i(t, x) =f(t− x

u)−g(t+ x

u)

Z0

where u =√

1L0C0

and Z0 =√

L0

C0

.


f() and g() can be any differentiable functions.






• Power Flow

• Reflections


• Driving a line



• Summary



= − ∂i∂x

L0∂i∂t

= − ∂v∂x


u)

i(t, x) =f(t− x

u)−g(t+ x

u)

Z0

where u =√

1L0C0

and Z0 =√

L0

C0

.



Verify by substitution:

− ∂i∂x

= −(

−f ′(t− x

u)−g′(t+ x

u)

Z0

× 1u

)






• Power Flow

• Reflections


• Driving a line



• Summary



= − ∂i∂x

L0∂i∂t

= − ∂v∂x


u)

i(t, x) =f(t− x

u)−g(t+ x

u)

Z0

where u =√

1L0C0

and Z0 =√

L0

C0

.



Verify by substitution:

− ∂i∂x

= −(

−f ′(t− x

u)−g′(t+ x

u)

Z0

× 1u

)

= C0

(

f ′(t− xu) + g′(t+ x

u))

= C0∂v∂t

Forward Wave





• Power Flow

• Reflections


• Driving a line



• Summary


Suppose:u = 15 cm/nsand g(t) ≡ 0⇒ v(x, t) = f

(

t− xu

)

Forward Wave





• Power Flow

• Reflections


• Driving a line



• Summary



(

t− xu

)

• At x = 0 cm [N],vS(t) = f(t− 0

u)

0 2 4 6 8 10Time (ns)

f(t-0/u)

Forward Wave





• Power Flow

• Reflections


• Driving a line



• Summary



(

t− xu

)

• At x = 0 cm [N],vS(t) = f(t− 0

u)

• At x = 45 cm [N],v(45, t) = f(t− 45

u)

0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u)

Forward Wave





• Power Flow

• Reflections


• Driving a line



• Summary



(

t− xu

)

• At x = 0 cm [N],vS(t) = f(t− 0

u)

• At x = 45 cm [N],v(45, t) = f(t− 45

u)

0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u)

f(t− 45u) is exactly the same as f(t) but delayed by 45

u= 3 ns.

Forward Wave





• Power Flow

• Reflections


• Driving a line



• Summary



(

t− xu

)

• At x = 0 cm [N],vS(t) = f(t− 0

u)

• At x = 45 cm [N],v(45, t) = f(t− 45

u)

0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u) f(t-90/u)


u= 3 ns.

• At x = 90 cm [N], vR(t) = f(t− 90u); now delayed by 6 ns.

Forward Wave





• Power Flow

• Reflections


• Driving a line



• Summary



(

t− xu

)

• At x = 0 cm [N],vS(t) = f(t− 0

u)

• At x = 45 cm [N],v(45, t) = f(t− 45

u)

0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u) f(t-90/u)


u= 3 ns.


Waveform at x = 0 completely determines the waveform everywhere else.

Forward Wave





• Power Flow

• Reflections


• Driving a line



• Summary



(

t− xu

)

• At x = 0 cm [N],vS(t) = f(t− 0

u)

• At x = 45 cm [N],v(45, t) = f(t− 45

u)

0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u) f(t-90/u)


u= 3 ns.



Snapshot at t0 = 4ns:the waveform has justarrived at the pointx = ut0 = 60 cm. 0 20 40 60 80

Position (cm)

f(4-x/u)t = 4 ns

Forward Wave





• Power Flow

• Reflections


• Driving a line



• Summary



(

t− xu

)

• At x = 0 cm [N],vS(t) = f(t− 0

u)

• At x = 45 cm [N],v(45, t) = f(t− 45

u)

0 2 4 6 8 10Time (ns)

f(t-0/u) f(t-45/u) f(t-90/u)


u= 3 ns.



Snapshot at t0 = 4ns:the waveform has justarrived at the pointx = ut0 = 60 cm. 0 20 40 60 80

Position (cm)

f(4-x/u)t = 4 ns

f(t− xu) is a wave travelling forward (i.e. towards +x) along the line.

Forward + Backward Waves





• Power Flow

• Reflections


• Driving a line



• Summary


Similarly g(t+ xu) is a wave travelling backwards, i.e. in the −x direction.






• Power Flow

• Reflections


• Driving a line



• Summary



v(x, t) =f(t− x

u) + g(t+ x

u)






• Power Flow

• Reflections


• Driving a line



• Summary



v(x, t) =f(t− x

u) + g(t+ x

u)

At x = 0 cm [N],vS(t) = f(t) + g(t)






• Power Flow

• Reflections


• Driving a line



• Summary



v(x, t) =f(t− x

u) + g(t+ x

u)

At x = 0 cm [N],vS(t) = f(t) + g(t)

At x = 90 cm [N], g starts at t = 1 and f starts at t = 6.






• Power Flow

• Reflections


• Driving a line



• Summary



v(x, t) =f(t− x

u) + g(t+ x

u)

At x = 0 cm [N],vS(t) = f(t) + g(t)

At x = 45 cm [N], g is only 1 ns behind f and they add together.At x = 90 cm [N], g starts at t = 1 and f starts at t = 6.






• Power Flow

• Reflections


• Driving a line



• Summary



v(x, t) =f(t− x

u) + g(t+ x

u)

At x = 0 cm [N],vS(t) = f(t) + g(t)


A vertical line on the diagramgives a snapshot of the entireline at a time instant t.






• Power Flow

• Reflections


• Driving a line



• Summary



v(x, t) =f(t− x

u) + g(t+ x

u)

At x = 0 cm [N],vS(t) = f(t) + g(t)



f and g first meet at t = 3.5and x = 52.5.






• Power Flow

• Reflections


• Driving a line



• Summary



v(x, t) =f(t− x

u) + g(t+ x

u)

At x = 0 cm [N],vS(t) = f(t) + g(t)



f and g first meet at t = 3.5and x = 52.5.

Magically, f and g passthrough each other entirelyunaltered.

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)

to be the forward andbackward waveforms at any point, x.

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)


i is alwaysmeasured in the+ve x direction.

Then vx(t) = fx(t) + gx(t) and ix(t) = Z−10 (fx(t)− gx(t)).

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)




Note: Knowing the waveform fx(t) or gx(t) at any position x, tells you it atall other positions: fy(t) = fx

(

t− y−xu

)

and gy(t) = gx(

t+ y−xu

)

.

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)





(

t− y−xu

)

and gy(t) = gx(

t+ y−xu

)

.

Power Flow

The power transferred into the shaded region across the boundary at x isPx(t) = vx(t)ix(t)

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)





(

t− y−xu

)

and gy(t) = gx(

t+ y−xu

)

.

Power Flow

The power transferred into the shaded region across the boundary at x isPx(t) = vx(t)ix(t) = Z−1

0 (fx(t) + gx(t)) (fx(t)− gx(t))

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)





(

t− y−xu

)

and gy(t) = gx(

t+ y−xu

)

.

Power Flow


0 (fx(t) + gx(t)) (fx(t)− gx(t))

=f2

x(t)

Z0

−g2

x(t)

Z0

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)





(

t− y−xu

)

and gy(t) = gx(

t+ y−xu

)

.

Power Flow


0 (fx(t) + gx(t)) (fx(t)− gx(t))

=f2

x(t)

Z0

−g2

x(t)

Z0

fx carries power into shaded area and gx carries power out independently.

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)





(

t− y−xu

)

and gy(t) = gx(

t+ y−xu

)

.

Power Flow


0 (fx(t) + gx(t)) (fx(t)− gx(t))

=f2

x(t)

Z0

−g2

x(t)

Z0

fx carries power into shaded area and gx carries power out independently.Power travels in the same direction as the wave.

Power Flow





• Power Flow

• Reflections


• Driving a line



• Summary


Define fx(t) = f(

t− xu

)

and gx(t) = g(

t+ xu

)





(

t− y−xu

)

and gy(t) = gx(

t+ y−xu

)

.

Power Flow


0 (fx(t) + gx(t)) (fx(t)− gx(t))

=f2

x(t)

Z0

−g2

x(t)

Z0

fx carries power into shaded area and gx carries power out independently.Power travels in the same direction as the wave.The same power as would be absorbed by a [ficticious] resistor of value Z0.

Reflections





• Power Flow

• Reflections


• Driving a line



• Summary


vx = fx + gxix = Z−1

0 (fx − gx)

From Ohm’s law at x = L, we have vL(t) = iL(t)RL

Reflections





• Power Flow

• Reflections


• Driving a line



• Summary



0 (fx − gx)


Hence (fL(t) + gL(t)) = Z−10 (fL(t)− gL(t))RL

Reflections





• Power Flow

• Reflections


• Driving a line



• Summary



0 (fx − gx)



From this: gL (t) = RL−Z0

RL+Z0

× fL (t)

Reflections





• Power Flow

• Reflections


• Driving a line



• Summary



0 (fx − gx)




RL+Z0

× fL (t)

We define the reflection coefficient : ρL = gL(t)fL(t) =

RL−Z0

RL+Z0

= +0.5

Reflections





• Power Flow

• Reflections


• Driving a line



• Summary



0 (fx − gx)




RL+Z0

× fL (t)


RL−Z0

RL+Z0

= +0.5

Substituting gL (t) = ρLfL (t) givesvL(t) = (1 + ρL) fL(t) and iL(t) = (1− ρL)Z

−10 fL(t)

Reflections





• Power Flow

• Reflections


• Driving a line



• Summary



0 (fx − gx)




RL+Z0

× fL (t)


RL−Z0

RL+Z0

= +0.5


−10 fL(t)

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

v0(t)

At source end: g0(t) = ρLf0(

t− 2Lu

)

i.e. delayed by 2Lu

= 12 ns.

Reflections





• Power Flow

• Reflections


• Driving a line



• Summary



0 (fx − gx)




RL+Z0

× fL (t)


RL−Z0

RL+Z0

= +0.5


−10 fL(t)

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

v0(t)

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

i0(t)

At source end: g0(t) = ρLf0(

t− 2Lu

)

i.e. delayed by 2Lu

= 12 ns.Note that the reflected current has been multiplied by −ρ.

Reflection Coefficients





• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1

0 1 2 3 4 5-1

0

1

RZ0-1

ρ

ρ depends on the ratio RZ0

.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment

3 +0.5






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1

vL(t)f(t) = 1 + ρ

0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment

3 +0.5 1.5






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1

vL(t)f(t) = 1 + ρiL(t)Z0

f(t) = 1− ρ 0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment

3 +0.5 1.5 0.5






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1


f(t) = 1− ρ 0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment

3 +0.5 1.5 0.5 R > Z0 ⇒ ρ > 0






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1


f(t) = 1− ρ 0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment

3 +0.5 1.5 0.5 R > Z0 ⇒ ρ > 0

13 −0.5 0.5 1.5 R < Z0 ⇒ ρ < 0






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1


f(t) = 1− ρ 0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment

3 +0.5 1.5 0.5 R > Z0 ⇒ ρ > 01 0 1 1 Matched: No reflection at all13 −0.5 0.5 1.5 R < Z0 ⇒ ρ < 0






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1


f(t) = 1− ρ 0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment

∞ +1 2 0 Open circuit: vL = 2f , iL ≡ 03 +0.5 1.5 0.5 R > Z0 ⇒ ρ > 01 0 1 1 Matched: No reflection at all13 −0.5 0.5 1.5 R < Z0 ⇒ ρ < 0






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1


f(t) = 1− ρ 0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment


0 −1 0 2 Short circuit: vL ≡ 0, iL = 2fZ0






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1


f(t) = 1− ρ 0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment



Note: Reverse mapping is R = vLiL

= 1+ρ1−ρ

× Z0






• Power Flow

• Reflections


• Driving a line



• Summary


ρ = R−Z0

R+Z0

=R

Z0−1

R

Z0+1


f(t) = 1− ρ 0 1 2 3 4 5-1

0

1

RZ0-1

ρ


.

RZ0

ρ vL(t)f(t)

iL(t)Z0

f(t) Comment



Note: Reverse mapping is R = vLiL

= 1+ρ1−ρ

× Z0

Remember: ρ ∈ −1,+1 and increases with R.

Driving a line





• Power Flow

• Reflections


• Driving a line



• Summary


From Ohm’s law at x = 0, we have v0(t) = vS(t)− i0(t)RS where RS isthe Thévenin resistance of the voltage source.

Driving a line





• Power Flow

• Reflections


• Driving a line



• Summary


vx = fx + gxix = fx−gx

Z0


Substituting v0(t) = f0 + g0 and i0(t) =f0−g0Z0

leads to:

f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t)

Driving a line





• Power Flow

• Reflections


• Driving a line



• Summary



Z0



leads to:

f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t), τ0vS(t) + ρ0g0(t)

Driving a line





• Power Flow

• Reflections


• Driving a line



• Summary



Z0



leads to:

f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t), τ0vS(t) + ρ0g0(t)

So f0(t) is the superposition of two terms:

(1) Input vS(t) multiplied by τ0 = Z0

RS+Z0

which is the same as apotential divider if you replace the line with a [ficticious] resistor Z0.

Driving a line





• Power Flow

• Reflections


• Driving a line



• Summary



Z0



leads to:

f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t), τ0vS(t) + ρ0g0(t)



RS+Z0


(2) The incoming backward wave, g0(t), multiplied by a reflectioncoefficient: ρ0 = RS−Z0

RS+Z0

.

Driving a line





• Power Flow

• Reflections


• Driving a line



• Summary



Z0



leads to:

f0(t) =Z0

RS+Z0

vS(t) +RS−Z0

RS+Z0

g0(t), τ0vS(t) + ρ0g0(t)



RS+Z0


(2) The incoming backward wave, g0(t), multiplied by a reflectioncoefficient: ρ0 = RS−Z0

RS+Z0

.

For RS = 20: τ0 = 10020+100 = 0.83 and ρ0 = 20−100

20+100 = −0.67.

Multiple Reflections





• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx

s)

s)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx

0 5 10 15 20 25 30Time (ns)

f0(t)

s)

s)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx

0 5 10 15 20 25 30Time (ns)

f0(t)

s)

0 5 10 15 20 25 30Time (ns)

gL(t)

s)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

s)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx

Each extra bit of f0 isdelayed by 2L

u(=12 ns)

and multiplied by ρLρ0 :

f0(t) =∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx


u(=12 ns)


f0(t) =∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

gL(t) = ρLf0(

t− Lu

)

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx


u(=12 ns)


f0(t) =∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

gL(t) = ρLf0(

t− Lu

)

v0(t) =f0(t) + gL

(

t− Lu

)

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx


u(=12 ns)


f0(t) =∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

gL(t) = ρLf0(

t− Lu

)

v0(t) =f0(t) + gL

(

t− Lu

)

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

0 5 10 15 20 25 30Time (ns)

v0(t)

s)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx


u(=12 ns)


f0(t) =∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

gL(t) = ρLf0(

t− Lu

)

v0(t) =f0(t) + gL

(

t− Lu

)

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

0 5 10 15 20 25 30Time (ns)

v0(t)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx


u(=12 ns)


f0(t) =∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

gL(t) = ρLf0(

t− Lu

)

v0(t) =f0(t) + gL

(

t− Lu

)

vL(t) =f0

(

t− Lu

)

+ gL(t)

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

0 5 10 15 20 25 30Time (ns)

v0(t)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx


u(=12 ns)


f0(t) =∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

gL(t) = ρLf0(

t− Lu

)

v0(t) =f0(t) + gL

(

t− Lu

)

vL(t) =f0

(

t− Lu

)

+ gL(t)

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

0 5 10 15 20 25 30Time (ns)

v0(t)

0 5 10 15 20 25 30Time (ns)

vL(t)

s)






• Power Flow

• Reflections


• Driving a line



• Summary


ρ0 = − 23

ρL = 12

vx = fx + gx


u(=12 ns)


f0(t) =∑

∞

i=0 τ0ρiLρ

i0vS

(

t− 2Liu

)

gL(t) = ρLf0(

t− Lu

)

v0(t) =f0(t) + gL

(

t− Lu

)

vL(t) =f0

(

t− Lu

)

+ gL(t)

0 5 10 15 20 25 30Time (ns)

f0(t)

0 5 10 15 20 25 30Time (ns)

gL(t)

0 5 10 15 20 25 30Time (ns)

v0(t)

0 5 10 15 20 25 30Time (ns)

vL(t)

Transmission Line Characteristics +





• Power Flow

• Reflections


• Driving a line



• Summary


Integrated circuits & Printed circuit boardsHigh speed digital or high frequency analoginterconnectionsZ0 ≈ 100Ω, u ≈ 15 cm/ns.

Long CablesCoaxial cable (“coax”): unaffacted by external fields;use for antennae and instrumentation.Z0 = 50 or 75Ω, u ≈ 25 cm/ns.Twisted Pairs: cheaper and thinner than coax andresistant to magnetic fields; use for computer networkand telephone cabling. Z0 ≈ 100Ω, u ≈ 19 cm/ns.

When do you have to bother?Answer: long cables or high frequencies. You can completely ignoretransmission line effects if length ≪ u

frequency = wavelength.• Audio (< 20 kHz) never matters.• Computers (1GHz) usually matters.• Radio/TV usually matters.

Summary





• Power Flow

• Reflections


• Driving a line



• Summary


• Signals travel at around u ≈ 12c = 15 cm/ns.

Only matters for high frequencies or long cables.

Summary





• Power Flow

• Reflections


• Driving a line



• Summary




• Forward and backward waves travel along the line:

fx(t) = f0(

t− xu

)

and gx(t) = g0(

t+ xu

)

Summary





• Power Flow

• Reflections


• Driving a line



• Summary





fx(t) = f0(

t− xu

)

and gx(t) = g0(

t+ xu

)

Knowing fx and gx at any single x position tells you everything

Summary





• Power Flow

• Reflections


• Driving a line



• Summary





fx(t) = f0(

t− xu

)

and gx(t) = g0(

t+ xu

)


• Voltage and current are: vx = fx + gx and ix = fx−gxZ0

Summary





• Power Flow

• Reflections


• Driving a line



• Summary





fx(t) = f0(

t− xu

)

and gx(t) = g0(

t+ xu

)



• Terminating line with R at x = L links the forward and backward waves:

backward wave is gL = ρLfL where ρL = R−Z0

R+Z0

Summary





• Power Flow

• Reflections


• Driving a line



• Summary





fx(t) = f0(

t− xu

)

and gx(t) = g0(

t+ xu

)





R+Z0

the reflection coefficient, ρL ∈ −1,+1 and increases with R

Summary





• Power Flow

• Reflections


• Driving a line



• Summary





fx(t) = f0(

t− xu

)

and gx(t) = g0(

t+ xu

)





R+Z0


R = Z0 avoids reflections: matched termination.

Summary





• Power Flow

• Reflections


• Driving a line



• Summary





fx(t) = f0(

t− xu

)

and gx(t) = g0(

t+ xu

)





R+Z0



Reflections go on for ever unless one or both ends are matched.

Summary





• Power Flow

• Reflections


• Driving a line



• Summary





fx(t) = f0(

t− xu

)

and gx(t) = g0(

t+ xu

)





R+Z0



Reflections go on for ever unless one or both ends are matched.

f is infinite sum of copies of the input signal delayed successivelyby the round-trip delay, 2L

u, and multiplied by ρLρ0.

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