16.360 Lecture 2

• Transmission lines

1. Transmission line parameters, equations2. Wave propagations3. Lossless line, standing wave and reflection coefficient 4. Input impedence5. Special cases of lossless line6. Power flow7. Smith chart8. Impedence matching9. Transients on transmission lines

16.360 Lecture 2

1. Transmission line parameters, equations

Vg(t) VBB’(t)VAA’(t)

A

A’ B’

B

L

VAA’(t) = Vg(t) = V0cos(t),

VBB’(t) = VAA’(t-td) = VAA’(t-L/c) = V0cos((t-L/c)),

VBB’(t) = VAA’(t)

Low frequency circuits:

Approximate result

VBB’(t) = VAA’(t)

16.360 Lecture 2

1. Transmission line parameters, equations

Vg(t) VBB’(t)VAA’(t)

A

A’ B’

B

L

VBB’(t) = VAA’(t-td) = VAA’(t-L/c) = V0cos((t-L/c)) = V0cos(t- 2L/),

Recall: =c, and = 2

If >>L, VBB’(t) V0cos(t) = VAA’(t),

If <= L, VBB’(t) VAA’(t), the circuit theory has to be replaced.

16.360 Lecture 2

1. Transmission line parameters, equations

Vg(t) VBB’(t)VAA’(t)

A

A’ B’

B

L

= 2ft = 0.06

e. g: = 1GHz, L = 1cm

Time delay t = L/c = 1cm /3x1010 cm/s = 30 ps

Phase shift VBB’(t) = VAA’(t)

= 2ft = 0.6

= 10GHz, L = 1cm

Time delay t = L/c = 1cm /3x1010 cm/s = 30 ps

Phase shift VBB’(t) VAA’(t)

16.360 Lecture 2

• Transmission line parameters

Vg(t) VBB’(t)VAA’(t)

A

A’ B’

B

L

• time delay

VBB’(t) = VAA’(t-td) = VAA’(t-L/vp),

• Reflection: the voltage has to be treat as wave, some bounce back

• power loss: due to reflection and some other loss mechanism,

• Dispersion: in material, Vp could be different for different wavelength

16.360 Lecture 2

• Types of transmission lines

• Transverse electromagnetic (TEM) transmission lines

B

E

E

B

a) Coaxial line b) Two-wire line c) Parallel-plate line

d) Strip line e) Microstrip line

16.360 Lecture 2

• Types of transmission lines

• Higher-order transmission lines

a) Optical fiber

b) Rectangular waveguide c) Coplanar waveguide

16.360 Lecture 2

• Lumped-element Model

• Represent transmission lines as parallel-wire configuration

Vg(t) VBB’(t)VAA’(t)

A

A’ B’

B

z z z

Vg(t)

R’z L’z

G’z C’z

R’z L’z

G’z

R’z

C’z

L’z

G’z

C’z

16.360 Lecture 2

• Transmission line equations

• Represent transmission lines as parallel-wire configuration

V(z,t)R’z L’z

G’z C’z V(z+ z,t)

V(z,t) = R’z i(z,t) + L’z i(z,t)/ t + V(z+ z,t), (1)

i(z,t) i(z+z,t)

i(z,t) = G’z V(z+ z,t) + C’z V(z+ z,t)/t + i(z+z,t), (2)

16.360 Lecture 2

• Transmission line equations

V(z,t) = R’z i(z,t) + L’z i(z,t)/ t + V(z+ z,t), (1)

V(z,t)R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

-V(z+ z,t) + V(z,t) = R’z i(z,t) + L’z i(z,t)/ t

- V(z,t)/z = R’ i(z,t) + L’ i(z,t)/ t, (3)

Rewrite V(z,t) and i(z,t) as phasors, for sinusoidal V(z,t) and i(z,t):

V(z,t) = Re( V(z) ejt

), i (z,t) = Re( i (z) ejt

),

16.360 Lecture 2

• Transmission line equations

V(z,t)R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

Recall:

di(t)/dt = Re(d i e jt )/dt ), = Re(ijt

ej

- V(z,t)/z = R’ i(z,t) + L’ i(z,t)/ t, (3)

- dV(z)/dz = R’ i(z) + jL’ i(z), (4)

16.360 Lecture 2

• Transmission line equations

• Represent transmission lines as parallel-wire configuration

V(z,t)R’z L’z

G’z C’z V(z+ z,t)

V(z,t) = R’z i(z,t) + L’z i(z,t)/ t + V(z+ z,t), (1)

i(z,t) i(z+z,t)

i(z,t) = G’z V(z+ z,t) + C’z V(z+ z,t)/t + i(z+z,t), (2)

16.360 Lecture 4

• Transmission line equations

V(z,t)R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

- i (z+ z,t) + i (z,t) = G’z V(z + z ,t) + C’z V(z + z,t)/ t

- i(z,t)/z = G’ V(z,t) + C’ V(z,t)/ t, (5)

Rewrite V(z,t) and i(z,t) as phasors, for sinusoidal V(z,t) and i(z,t):

V(z,t) = Re( V(z) ejt

), i (z,t) = Re( i (z) ejt

),

i(z,t) = G’z V(z+ z,t) + C’z V(z+ z,t)/t + i(z+z,t), (2)

16.360 Lecture 2

• Transmission line equations

V(z,t)R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

Recall:

dV(t)/dt = Re(d V e jt )/dt ), = Re(Vjt

ej

- i(z,t)/z = G’ V(z,t) + C’ V(z,t)/ t, (6)

- d i(z)/dz = G’ V(z) + jC’ V(z), (7)

16.360 Lecture 2

V(z,t)R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

- d i(z)/dz = G’ V(z) + jC’ V(z), (7)

- dV(z)/dz = R’ i(z) + jL’ i(z), (4)

• Telegrapher’s equation in phasor domain

Take d /dz on both sides of eq. (4)

- d²V(z)/dz² = R’ di(z)/dz + jL’ di(z)/dz, (8)

16.360 Lecture 2

- d i(z)/dz = G’ V(z) + jC’ V(z), (7)

- dV(z)/dz = R’ i(z) + jL’ i(z), (4)

• Telegrapher’s equation in phasor domain

substitute (7) to (8)

d²V(z)/dz² = (R’ + jL’) (G’+ jC’)V(z),

- d²V(z)/dz² = R’ di(z)/dz + jL’ di(z)/dz, (8)

d²V(z)/dz² - (R’ + jL’) (G’+ jC’)V(z) = 0, (9)

or

d²V(z)/dz² - ²V(z) = 0, (10)

² = (R’ + jL’) (G’+ jC’), (11)

16.360 Lecture 2

V(z,t)R’z L’z

G’z C’z V(z+ z,t)

i(z,t) i(z+z,t)

- d i(z)/dz = G’ V(z) + jC’ V(z), (7)

- dV(z)/dz = R’ i(z) + jL’ i(z), (4)

• Telegrapher’s equation in phasor domain

Take d /dz on both sides of eq. (7)

- d² i(z)/dz² = G’ dV(z)/dz + jC’ dV(z)/dz, (12)

16.360 Lecture 2

- d i(z)/dz = G’ V(z) + jC’ V(z), (7)

- dV(z)/dz = R’ i(z) + jL’ i(z), (4)

• Telegrapher’s equation in phasor domain

substitute (4) to (12)

d² i(z)/dz² = (R’ + jL’) (G’+ jC’)i(z),

d² i(z)/dz² - (R’ + jL’) (G’+ jC’) i(z) = 0, (9)

or

d² i(z)/dz² - ²i(z) = 0, (13)

² = (R’ + jL’) (G’+ jC’), (11)

- d² i(z)/dz² = G’ dV(z)/dz + jC’ dV(z)/dz, (12)

16.360 Lecture 2

• Wave equations

d² i(z)/dz² - ²i(z) = 0, (13)

d²V(z)/dz² - ²V(z) = 0, (10)

= + j,

= Re (R’ + jL’) (G’+ jC’) ,

= Im (R’ + jL’) (G’+ jC’) ,

16.360 Lecture 2

• Wave equations

d² i(z)/dz² - ²i(z) = 0, (13)

d²V(z)/dz² - ²V(z) = 0, (10)

= + j,

= Re (R’ + jL’) (G’+ jC’) ,

= Im (R’ + jL’) (G’+ jC’) ,

V(z) = V0 (14)

Solving the second order differential equation

+e-z+

-V0ez

i(z) = I0 (15) + e-z+ -

I0 ez

16.360 Lecture 2

• Wave equations

V(z) = V0 (14) + e-z+

-V0ez

i(z) = I0 (15) + e-z+ -

I0 ez

where:

+V0-V0and are determined by boundary conditions.

+I0-I0and are related to -V0

+V0 and by characteristic impedance Z0.

16.360 Lecture 2

recall:

- dV(z)/dz = R’ i(z) + jL’ i(z), (4)

V(z) = V0 (14) ++

-V0ez

i(z) = I0 (15) + e-z+ -

I0 ez

e-z

e-zV0+

ezV0-- = (R’ + jL’) i(z), (16)

i(z) = (R’ + jL’)

e-z(V0

+ ezV0-

- )

I0+

=(R’ + jL’)

V0

+I0-

=(R’ + jL’)

-V0

-(17) (18)

• Characteristic impedance Z0

16.360 Lecture 5

• Characteristic impedance Z0

I0+

=(R’ + jL’)

V0

+

I0-

=(R’ + jL’)

-V0

-

(17)

(18)

=(R’ + jL’)

+Z0

Define characteristic impedance Z0

I0+

V0

=

= (R’ + jL’) (G’+ jC’)

(R’ + jL’)(G’+j C’)

recall:

16.360 Lecture 5

• Summary:

= (R’ + jL’) (G’+ jC’)

+Z0

I0+

V0 =(R’ + jL’)(G’+j C’)

V(z) = V0 (14) ++

-V0ez

i(z) = I0 (15) + e-z+ -

I0 ez

e-z

(19)

(20)

16.360 Lecture 5

• Example, an air line :

solution:

R’ = 0 , G’ = 0 /, Z0 = 50, = 20 rad/m, f = 700 MHz

L’ = ? and C’ = ?

Z0 =(R’ + jL’)(G’+j C’) =

jL’j C’ = 50

= (R’ + jL’) (G’+ jC’) = j L’C’

= + j,

= L’C’ = 20 rad/m

16.360 Lecture 5

• lossless transmission line :

= (R’ + jL’) (G’+ jC’)

= + j,

If R’<< j L’ and G’ << jC’,

= (R’ + j L’ ) (G’+ jC’)

= j L’C’

= 0

= L’C’lossless line

16.360 Lecture 2

• lossless transmission line :

= 0

= L’C’

lossless line

Z0 =(R’ + jL’)(G’+j C’) =

jL’j C’

Z0 =L’C’

= 2/ = 2/ = L’C’

1

Vp = / =L’C’

1

16.360 Lecture 2

• For TEM transmission line :

Vp =L’C’

1

L’C’ =

=

1

= L’C’ =

=rr

c

Z0 =L’C’

• summary :

V(z) = V0 +

+-

V0ejz

i(z) = I0 + e-jz+ -

I0 ejz

e-jz

=rr

cVp =

L’C’

1

= L’C’ =

16.360 Lecture 5

• Voltage reflection coefficient :

Vg(t) VL

A

z = 0

B

l

ZL

z = - l

Z0Vi

i(z) =

V(z) = V0 +

+-

V0ejz

-

e-jz

e-jz+

V0

Z0ejz

-V0

Z0

VL = += -V0

+V0V(z)

z = 0

-+

V0

Z0

-V0

Z0i(z)

z = 0iL = =

ZL = VL

iL= +

-V0

+V0

-+

V0

Z0

-V0

Z0

+V0

-V0

=ZL Z0-

ZL Z0+

16.360 Lecture 5

• Voltage reflection coefficient :

-V0

+V0=

ZL Z0-

ZL Z0+

• Current reflection coefficient :

-i0

+i0= - i

-V0

+V0= -

• Notes :

1. || 1, how to prove it?

2. If ZL = Z0, = 0. Impedance match, no reflection from the load ZL.

16.360 Lecture 2

• An example :

A’

z = 0

A

Z0 = 100RL = 50

CL = 10pF

f = 100MHz

ZL = RL + j/CL = 50 –j159

=ZL Z0-

ZL Z0+

16.360 Lecture 2

• Standing wave • Input impedance

i(z) =

V(z) = V0 +

+-

V0ejz

-

e-jz

e-jz+

V0

Z0ejz

-V0

Z0

-V0

+V0with =

i(z) =

V(z) = V0 ( )++e jz

-

e-jz

(e-jz+

V0

Z0 ejz )

|V(z)| = |V0| | |+ e-jz || e jz+ e jr

= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2

16.360 Lecture 2

• Standing wave

i(z) =

V(z) = V0 +

+-

V0ejz

-

e-jz

e-jz+

V0

Z0ejz

-V0

Z0

-V0

+V0with =

i(z) =

V(z) = V0 ( )++e jz

-

e-jz

(e-jz+

V0

Z0 ejz )

|V(z)| = |V0| | |+ e-jz || e jz+ e jr

= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2

16.360 Lecture 2

• Standing wave

i(z) =

V(z) = V0 ( )

-

++e jz

e-jz

(e-jz+

V0

Z0 ejz )

|i(z)| = |V0| /|Z0|| |+ e-jz || e jz- e jr

= |V0|/|Z0| [1+ | |² - 2||cos(2z + r)] + 1/2

= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2

|V(z)|

16.360 Lecture 2

Special cases

= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2

|V(z)|

1. ZL= Z0, = 0

= |V0| +|V(z)|

2. ZL= 0, short circuit, = -1

|V(z)|

|V0| +

- -3/4 -/2 -/4

= |V0| [2 + 2cos(2z + )] + 1/2

|V(z)||V(z)|

2|V0| +

- -3/4 -/2 -/4

16.360 Lecture 2

Special cases

= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2

|V(z)|

3. ZL= , open circuit, = 1

= |V0| [2 + 2cos(2z )] + 1/2

|V(z)|

|V(z)|

2|V0| +

- -3/4 -/2 -/4

16.360 Lecture 2

• Voltage maximum

= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2

|V(z)|

|V0| [1+ | |],+|V(z)|max = when 2z + r = 2n.

–z = r/4 + n/2

n = 1, 2, 3, …, if r <0

n = 0, 1, 2, 3, …, if r >= 0

16.360 Lecture 2

• Voltage minimum

= |V0| [1+ | |² + 2||cos(2z + r)] + 1/2

|V(z)|

|V0| [1 - | |],+|V(z)|min = when 2z + r = (2n+1).

–z = r/4 + n/2 + /4

Note:

voltage minimums occur /4 away from voltage maximum, because of the 2z, the special frequency doubled.

16.360 Lecture 2

• Voltage standing-wave ratio VSWR or SWR

1 - | ||V(z)|min

S |V(z)|max =

1 + | |

S = 1, when = 0,

S = , when || = 1,

16.360 Lecture 2

• An example

Vg(t) VL

A

z = 0

B

l

ZL

z = - l

Z0Vi

Voltage probe

S = 3, Z0 = 50, lmin = 30cm, lmin = 12cm, ZL=?

lmin = 30cm, = 0.6m,

S = 3, || = 0.5,

Solution:

-2lmin + r = -, r = -36º,

, and ZL.

16.360 Lecture 2

• Input impudence

Vg(t) VL

A

z = 0

B

l

ZL

z = - l

Z0Vi

ZgIi

Zin(z) = V(z)

I(z)

=

++e jz

e-jzV0( )

+- e jz

e-jzV0( )

Z0 =+(1 e j2z

)

-(1 e j2z )

Z0

+(1 e-j2l )

-(1 e-j2l )

Z0Zin(-l) =

16.360 Lecture 2

An example

A 1.05-GHz generator circuit with series impedance Zg = 10- and voltage source given by Vg(t) = 10 sin(t +30º) is connected to a load ZL = 100 +j5-through a 50-, 67-cm long lossless transmission line. The phase velocity is 0.7c.Find V(z,t) and i(z,t) on the line.

Solution:

Since, Vp = ƒ, = Vp/f = 0.7c/1.05GHz = 0.2m.

= 2/, = 10 .

= (ZL-Z0)/(ZL+Z0), = 0.45exp(j26.6º)

+(1 e-j2l )

-(1 e-j2l )

Z0Zin(-l) = = 21.9 + j17.4

V0[exp(-jl)+ exp(jl)] +

=Zin(-l) + Zg

Zin(-l)Vg

16.360 Lecture 2

short circuit line

ZL= 0, = -1, S =

Vg(t) VL

A

z = 0

B

l

ZL = 0

z = - l

Z0

ZgIi

Zinsc

i(z) =

V(z) = V0 )- ejz

+

(e-jz

(e-jz+

V0

Z0 ejz )

= -2jV0sin(z)+

= 2V0cos(z)/Z0+

Zin =V(-l)

i(-l)= jZ0tan(l)

16.360 Lecture 2

short circuit line

Zin =V(-l)

i(-l)= jZ0tan(l)

• If tan(l) >= 0, the line appears inductive, jLeq = jZ0tan(l),

• If tan(l) <= 0, the line appears capacitive, 1/jCeq = jZ0tan(l),

l = 1/[- tan (1/CeqZ0)], -1

• The minimum length results in transmission line as a capacitor:

16.360 Lecture 2

An example:

tan (l) = - 1/CeqZ0 = -0.354,

Choose the length of a shorted 50- lossless line such that its input impedance at2.25 GHz is equivalent to the reactance of a capacitor with capacitance Ceq = 4pF. The wave phase velocity on the line is 0.75c.

Solution:

Vp = ƒ, = 2/ = 2ƒ/Vp = 62.8 (rad/m)

l = tan (-0.354) + n, = -0.34 + n,

-1

16.360 Lecture 2

open circuit line

ZL= 0, = 1, S =

Vg(t) VL

A

z = 0

B

l

ZL =

z = - l

Z0

ZgIi

Zinoc

i(z) =

V(z) = V0 )+ ejz

-

(e-jz

(e-jz+

V0

Z0 ejz )

= 2V0cos(z)+

= 2jV0sin(z)/Z0+

Zin =V(-l)

i(-l)= -jZ0cot(l)

oc

16.360 Lecture 2

Application for short-circuit and open-circuit

• Network analyzer

• Measure ZinocZin

scand

• Calculate Z0

• Measure S paremeters

Zin = -jZ0cot(l)ocZin = jZ0tan(l)sc

=Z0 ZinscZin

oc

• Calculate l

= -jZ0Zin

sc

Zinoc

16.360 Lecture 2

Line of length l = n/2

= ZL

tan(l) = tan((2/)(n/2)) = 0,

+(1 e-j2l )

-(1 e-j2l )

Z0Zin(-l) =

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

16.360 Lecture 2

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

= Z0²/ZL

l = (2/)(/4 + n/2) = /2 ,

+(1 e-j2l )

-(1 e-j2l )

Z0Zin(-l) =+(1 e

-j )

-(1 e-j

)Z0=

(1 + ) Z0(1 - )

=

16.360 Lecture 2

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.

ZL = 100

/4

Z01 = 50

Zin = Z0²/ZL= 50

Z0 = (ZinZL) = (50*100)½ ½

16.360 Lecture 2

Matched transmission line:

1. ZL = Z0

2. = 03. All incident power is delivered to the load.

16.360 Lecture 8

• Instantaneous power• Time-average power

i(z) =

V(z) = V0 ( )++e jz

-

e-jz

(e-jz+

V0

Z0 ejz )

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

V = V0+ i =

i +V0

Z0

i

V = V0-r

i = -

V0

Z0

r

16.360 Lecture 2

• Instantaneous power

+

iP(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)] i i

= Re[|V0|exp(j )exp(jt)] Re[|V0|/Z0 exp(j )exp(jt)] + + +

= (|V0|²/Z0) cos²(t + ) + +

-

rP(t) = v(t) i(t) = Re[V exp(jt)] Re[ i exp(jt)] r r

= Re[|V0|exp(j )exp(jt)] Re[|V0|/Z0 exp(j )exp(jt)] + - +

= - ||²(|V0|²/Z0) cos²(t + + r) + +

16.360 Lecture 2

• Time-average

(|V0|²/Z0) cos²(t + )dt + +

Time-domain approach:

Pav =i

T1

0

TP (t)dt

i= 2

0

T

= (|V0|²/2Z0)+

Pavr

= -||² (|V0|²/2Z0)+

= PaviPav + Pav

r

= (1-||²) (|V0|²/2Z0)+

Net average power:

16.360 Lecture 2

• Time-average

Phasor-domain approach

Pavr

= -||² (|V0|²/2Z0)+

= (1-||²) (|V0|²/2Z0)+

= (½)Re[V i*]Pav

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

= (|V0|²/2Z0)

Pav