Microwave Circuit Design - eas.uccs.edueas.uccs.edu/~mwickert/ece5250/notes/N5250_1.pdf · Chapter 1. Review of Transmission Line Theory ... ECE 5250/4250 Microwave Circuits 1–9

Microwave Circuit Design

ECE 5250/4250 Lecture NotesFall 2009

© 1989, 1991, 1993, 1995, & 2009

Mark A. Wickert

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ECE 5250/4250 Microwave Circuits 1–1

Review of Transmission Line Theory• Transmission lines and waveguides are used to transport

electromagnetic energy at microwave frequencies from onepoint in a system to another

• The desirable features of a transmission line or waveguideare:

– Single-mode propagation over a wide band of frequencies

– Small attenuation

• The transmission line structures of primary interest for thiscourse are those for which the dominant mode of propagationis a transverse electromagnetic (TEM) wave

• Recall that for TEM waves the components of electric andmagnetic fields in the direction of wave propagation are zero

• We wish to consider transmission lines which consist of twoor more parallel conductors which have axial uniformity

• That is to say their cross-sectional shape and electrical prop-erties do not vary along the axis of propagation

Chapter

1

Review of Transmission Line Theory

1–2 ECE 5250/4250 Microwave Circuits

• The TEM wave solution for axially uniform transmissionlines can be obtained using:

– Field Analysis - (obtain electric and magnetic field wavesanalogous to uniform plane waves)

– Distributed-Circuit Analysis - (obtain voltage and currentwaves)

• Distributed circuit analysis will be at the forefront of all anal-ysis in this course, in particular consider Pozar1, “Modernmicrowave engineering involves predominantly distributedcircuit analysis and design, in contrast to the waveguide andfield theory orientation of earlier generations”

1.David Pozar, Microwave Engineering, 3rd edition, John Wiley, New York,2005.

Figure 1.1: Transmission lines with axial uniformity

Distributed Circuit Analysis



Parameters:L – series inductance per unit length due to energy storage inthe magnetic fieldC – shunt capacitance per unit length due to energy storage inthe electric fieldR – series resistance per unit length due to power loss in theconductorsG – shunt conductance per unit length due to power loss inthe dielectric. (i.e. )

• Using KCL, KVL and letting it can be shown that

(1.1)

and

(1.2)

Figure 1.2: Distributed element circuit model

! !" j!# !# 0$%–=

&z 0'

(v z t%) *–(z

--------------------- Ri z t%) * L(i z t%) *(t

-----------------+=

(– i z t%) *(z

-------------------- Gv z t%) * C(v z t%) *(t

------------------+=



• For now assume the line is lossless, that is and, so we have:

(1.3)

• Now differentiate the first equation with respect to z and thesecond with respect to time t

(1.4)

• Combine the two resulting equations to get

(voltage eqn.) (1.5)

similarly obtain

(current eqn.) (1.6)

• These are in the form of the classical one-dimensional waveequation, often seen in the form

(1.7)

where has dimension and significance of velocity

R 0=G 0=

(v(z-----– L(i

(t----=

(i(z-----– C(v

(t-----=

(2v(z2-------- L (2i

(t(z-----------–=

(2i(z(t----------- C(

2v(t2--------=

(2vz2(

-------- LC(2v(t2--------=

(2i(z2-------- LC(

2i(t2-------=

(2y(z2-------- 1

+p2-----(

2y(t2--------=

+p



• A well known solution to the wave equation is

(1.8)

– propagates in the positive z direction

– propagates in the negative z direction

• This solution can be checked by noting that

(1.9)

• A solution for the voltage wave equation is thus

(1.10)

• The current equation can be written in a similar manner, butit can also be written in terms of and since

(1.11)

or

(1.12)

Letting and using the chain rule the above equa-tion becomes

(1.13)

y y+ t z+p-----–, -

. / y– t z+p-----+, -

. /+=

y+

y–

(y 0

(z---------- (y 0

(x----------(x

(z----- 112

+p------(y 0

(x---------- x% t z

+p-----0= = =

v z t%) * v+ t z+p-----–, -

. / v– t z+p-----+, -

. /+=

V+ V–

(v–(z

--------- L(i(t----=

(v+

(z-------- (v–

(z--------+– L(i

(t----=

x t z +p30=

1–+p------(v+

(x-------- 1

+p-----(v–

(x--------+– L(i

(t----=



• This implies that

(1.14)

or

(1.15)

where

(1.16)

• At this point the general lossless line solution is incomplete.The functions v+ and v- are unknown, but must satisfy theboundary conditions imposed by a specific problem

• The time domain solution for a lossless line, in particular theanalysis of transients, can most effectively be handled byusing Laplace transforms

• If the source and load impedances are pure resistances andthe source voltage consists of step functions or rectangularpulses, then time domain analysis is most convenient

• In the following we will first consider resistive load andsource impedances, later the analysis will be extended tocomplex impedance loads using Laplace transforms

i z t%) * 1L+p--------- v+ t z

+p-----–, -

. / v– t z+p-----+, -

. /–=

i z t%) * 1Z0----- v+ t z

+p-----–, -

. / v– t z+p-----+, -

. /–=

+p1LC

----------- (velocity of propagation)%=

Z0LC---- (characteristic impedance)%=

Transient Analysis with Resistive Loads


• The Laplace transform technique will in theory allow for ageneralized time-domain analysis of transmission lines

• In Section 2 sinusoidal steady-state analysis will be intro-duced. This approach offers greatly reduced analysis com-plexity


Infinite Length Line

• Assume that the line is initially uncharged, i.e. for and

(1.17)

and

(1.18)

the above equations imply that

(1.19)

Figure 1.3: Infinite length line circuit diagram

z 04t 05

v z t%) * v+ t z+p-----–, -

. / v– t z+p-----+, -

. /+ 0= =

i z t%) * 1Z0----- v+ t z

+p-----–, -

. / v– t z+p-----+, -

. /– 0= =

v+ t z+p-----–, -

. / 0 for t z +p3– 06=



and

(1.20)

Note: For the given initial conditions only canexist on the line.

• We thus conclude that

(1.21)

• Suppose that at a voltage source is

applied through a source resistance , at z = 0

• Apply Ohm's law at z = 0 for t > 0 and we obtain(1.22)

or

(1.23)

which implies

(1.24)

v– t z+p-----+, -

. / 0 for all t=

v+ t z +p3–) *

v z t%) * v+ t z+p-----–, -

. /=

i z t%) * 1Z0-----v

+t z

+p-----–, -

. /=788988:

for all t z+p----- 0;–

t 0+= vg t) *

Rg

vg t) * v 0 t%) *– i 0 t%) *Rg=

vg t) * v+ t) *–RgZg------v+ t) *=

v+ t) *Z0

Z0 Rg+------------------vg t) *=



• The final result is that under the infinite line length assump-tion for any z we can write

(1.25)

(1.26)

– Note: That the infinite length of line appears as a voltagedivider to the source

– Voltages and currents along the line appear as replicas ofthe input values except for the time delay

Terminated Line

• Note: As a matter of convenience the reference point has been shifted to the load end of the line

• Suppose that a wave traveling in the direction is incidentupon the load, , at

• Thus,

(1.27)

v z t%) *Z0

Z0 Rg+------------------vg t z

+p-----–, -

. /=

i t z%) * 1Z0 Rg+------------------vg t z

+p-----–, -

. /=

z +p3

Figure 1.4: Terminated line circuit diagram

z 0=

z+

RL z 0=

v z t%) * v+ t z+p-----–, -

. / and i z t%) * 1Z0-----v+ t z

+p-----–, -

. /= =



• By applying Ohm’s law at the load, we must have(1.28)

• This condition cannot, in general, be met by the incidentwave alone since

(1.29)

• Since the line was initially discharged, it is reasonable toassume that a fraction, , of the incident wave is reflectedfrom the load resistance, i.e.,

(1.30)

• The load voltage is now

(1.31)

• Similarly the load current is

(1.32)

• To satisfy Kirchoff’s laws,

(1.33)

or the more familiar result

(1.34)

v 0 t%) * RLi 0 t%) *=

v z t%) * Z0i z t%) *=

<L

v– t) * <Lv+ t) *=

v 0 t%) * v+ t) * v– t) *+ 1 <L+) *v+ t) *= =

i 0 t%) * v+ t) *Z0

------------ v– t) *Z0

------------–1 <L+) *v+ t) *

Z0---------------------------------= =

Net load voltageNet load current--------------------------------------- RL Z0

1 <L+

1 <L–---------------= =

<LRL Z0–

RL Z0+------------------=



– Note: is real by assumption

– Note: To terminate the line without reflection use.

General Transmission Line Problem

• From earlier analysis, we know that for ,

(1.35)

• When the leading edge of has traveled to theload end of the line ( )

• Assuming a reflected wave now returns to thesource during the interval

(1.36)

<L

RL Z0=

Figure 1.5: Circuit for general transmission line problem

0 t l +p35 5 Tl=

v z t%) *Z0

Z0 Rg+------------------vg t z

+p-----–, -

. /=

i z t%) * 1Z0 Rg+------------------vg t z

+p-----–, -

. /=788988:

0 t Tl5 5

t Tl= vg t) *z l=

RL Z0$Tl t 2Tl65

v z t%) *Z0

Z0 Rg+------------------vg t z

+p-----–, -

. / Z0Z0 Rg+------------------<Lvg t 2l

+p----- z

+p-----+–, -

. /+=

= 8 8 8 > 8 8 8 ? = 8 8 8 8 > 8 8 8 8 ?

v+ t z +p3–) * v– t z +p3+) *



and

(1.37)

• When the load reflected wave arrives at thesource ( ), a portion of it will be reflected towards theload provided

• The reflection that takes place is independent of the sourcevoltage

• The wave traveling in the positive direction after seconds has elapsed, can be found by applying

Ohm's law for and :(1.38)

• Now substitute

(1.39)

and solve for

• The results is

(1.40)

i z t%) * 1Z0 Rg+------------------vg t z

+p-----–, -

. / 1Z0 Rg+------------------<Lvg t 2l

+p----- z

+p-----+–, -

. /–=

v– t z +p3+) *z 0=

Rg Z0$

Z2l +p3 2Tl=

z 0= t 2Tl=

vg 2Tl) * v 0 2Tl%) *– Rgi 0 2Tl%) *=

v 0 2Tl%) * v+ 2Tl) * v– 2Tl) *+=

i 0 2Tl%) * 1Z0----- v+ 2Tl) * v– 2Tl) *–@ A=

v+ 2Tl) *

v+ 2Tl) * vg 2Tl) *Z0

Z0 Rg+------------------ v– 2Tl) *

Rg Z0–

Rg Z0+------------------+=

vg 2Tl) *Z0

Z0 Rg+------------------ vg 0) *

Z0Z0 Rg+------------------<L<g+=

= 8 8 > 8 8 ? = 8 8 > 8 8 ?

incident part of vg t) * reflected portion of v– 2Tl) *



where is the source reflection coefficient defined as

(1.41)

Note: is real by assumption in this case

• The wave incident on the load during the interval, thus consists of the original source signal plus

a reflected component due to mismatches at both the load andsource ends of the line

(1.42)

and

(1.43)

• The process of reflections occurring at both the source andload ends continues in such a way that in general over the time interval, , the and solu-tions each require terms involving

<L

<gRg Z0–

Rg Z0+------------------=

<g

2Tl t 3Tl65

v z t%) *Z0

Z0 Rg+------------------ vg t z

+p-----–, -

. / <Lvg t z 2l–+p

-------------+, -. /+=

+ <g<Lvg t z 2l++p

-------------–, -. / 2Tl t 3Tl65%

i z t%) * 1Z0 Rg+------------------ vg t z

+p-----–, -

. / <L– vg t z 2l–+p

-------------+, -. /=

+ <g<Lvg t z 2l++p

-------------–, -. / 2Tl t 3Tl65%

nth

n 1–) *Tl t nTl5 5 v z t%) * i z t%) *n vg t) *



Example: Consider the following circuit.

Here we have

(1.44)

a) Find and for

b) Find and for

• For both parts (a) and (b) the basic circuit operation is as fol-lows:

i) An 8v pulse will propagate toward the load, reaching theload in 2 ms.

ii) A -8v pulse will be reflected from the load, requiring 2 msto reach the source.

iii) A -4v pulse will be reflected at the source. It will take 2ms to reach the load.

iv) The process continues.

• The +z and -z direction propagating pulses can be displayedon a distance-time plot or bounce diagram (see Figure 1.7)

Figure 1.6: Transmission line circuit

Z0Z0 Rg+------------------ 1

4--- <L% 1 <g%– 1

2--- Tl% 2Bs= = = =

V 0 t%) * I 0 t%) * Tp 1Bs=

V 0 t%) * I 0 t%) * Tp 6Bs=



– In the bounce diagram the boundaries at z = 0 and z = 400m are represented as surfaces with reflection coefficient of1/2 and -1 respectively.

– To obtain the voltage waveform at say z = 380 m as a func-tion of time, you sum the contributions from the various +zand -z traveling waves

– For pulses that are short in comparison with the one-waydelay time of the line, only at most two wave terms need tobe included at a time.

– For long pulses (in the limit say a step function) all waveterms need to be included

Figure 1.7: Bounce diagram showing pulse propagation



a) : at we have

b) : at we have

• We can simulate this result using the Agilent AdvancedDesign System (ADS) software

• In this example we will use ideal transmission line elements

Tp 1Bs= t 4Bs=

v 0 4Bs%) * 4– 8–) *+ 12v.–= =

i 0 4Bs%) * 150------ 4– 8–) *–@ A 0.08A.= =

Tp 6Bs= t 4Bs=

v 0 4Bs%) * 8 4–) *+) * 8–) *+ 4v.–= =

i 0 4Bs%) * 150------ 8 4–) *+) * 8–) *–@ A 0.24A.= =

Source

VtPulseSRC1

Period=20 usecWidth=1 usecFall=50 nsecRise=50 nsecEdge=linearDelay=0 nsecVhigh=32 VVlow=0 V

t

TranTran1

MaxTimeStep=50 nsecStopTime=20.0 usec

TRANSIENT

TLINDTLD1

Delay=2 usecZ=50.0 Ohm

I_ProbeI_Probe1

RR2R=0 Ohm

RR1R=150 Ohm

Figure 1.8: Circuit schematic (Example1.dsn)

A short circuitCurrentmeasurementmeans

‘TLIND’ allows the lineto be spec’d in terms ofpropagation delay



• The source end voltage,

• The source end current entering the line,

• Modify the schematic by increasing the pulse width to 6

v 0 t%) *

2 4 6 8 10 12 14 16 180 20

-10

-5

0

5

-15

10

time, usec

Sou

rce

(v)

Tp 1Bs=

i 0 t%) *

2 4 6 8 10 12 14 16 180 20

0.00

0.05

0.10

0.15

-0.05

0.20

time, usec

I_P

rob

e1.i

(A)

Tp 1Bs=

Bs



• The source end voltage,

• The source end current entering the line,

v 0 t%) *

2 4 6 8 10 12 14 16 180 20

-10

-5

0

5

-15

10

time, usec

Sou

rce

(v)

Tp 6Bs=

i 0 t%) *

2 4 6 8 10 12 14 16 180 20

0.00

0.05

0.10

0.15

0.20

-0.05

0.25

time, usec

I_P

robe

1.i

(A)

Tp 6Bs=



Transient Analysis using Laplace Transforms

• In the time domain we know that for a lossless line

(1.45)

and

(1.46)

where and are determined by theboundary conditions imposed by the source and load

• Laplace transform each side of the above equations withrespect to the time variable, using the time shift theoremwhich is given by

(1.47)

where is the laplace transform of

• The result is

(1.48)

and

Figure 1.9: Lossless line with arbitrary terminations

v z t%) * v+ t z+p-----–, -

. /= v– t z+p-----+, -

. /+

i z t%) * 1Z0----- v+ t z

+p-----–, -

. / v– t z+p-----+, -

. /–=

v+ t z +p3–) * v– t z +p3+) *

L f t c–) *C D F s) *e sc–=

F s) * f t) *

v z s%) * v+ s) *esz +p3–

v– s) *esz +p3

+=



(1.49)

where and

Case 1: Matched Source

• For the special case of , the source is matched tothe transmission line which eliminates multiple reflections

• Thus, we can write

(1.50)

at , the incident wave is reflected with the reflectioncoefficient

(1.51)

so,

(1.52)

which implies that

(1.53)

• Finally, for we can write

(1.54)

i z s%) * 1Z0----- v+ s) *e

sz +p3–v– s) *e

sz +p3–@ A=

v+ s) * L v+ t) *C D= v– s) * L v– t) *C D=

Zg s) * Z0=

V+ s) * Vg s) *Z0

Z0 Z0+------------------ 1

2---Vg s) *= =

z l=

<L s) *ZL s) * Z0–

ZL s) * Z0+--------------------------=

V l s%) * V+ s) *esl +p3–

1 <L s) *+@ A=

V– s) * <L s) *es2l +p3–

V+ s) *=

0 z5 l5

V z s%) * 12---Vg s) * e

sz +p3–<L s) *e

s 2l z–) * +p3–+@ A=

12---Vg s) *e

sz +p3– 12---<L s) *e

s2l +p3–Vg s) *e

sz +p3+=



• To obtain inverse transform:

(1.55)

Example:

• Let and be a parallel RC connection

Find:

• To begin with in the s-domain we can write

(1.56)

and

(1.57)

V z t%) *

v z t%) * 12---vg t z

+p-----–, -

. / L1– <L s) *Vg s) *@ A

t t z 2l–) * +p3+'+=

vg t) * v0u t) *= ZL

Figure 1.10: Parallel RC circuit

v z t%) *

ZL s) *R 1

Cs------

R 1Cs------+

---------------- R1 RCs+--------------------= =

<L s) *ZL s) * Z0–

ZL s) * Z0+--------------------------

R1 RCs+-------------------- Z0–

R1 RCs+-------------------- Z0+--------------------------------= =

R Z0–

RCZ0--------------- s–

R Z0–

RCZ0--------------- s+

------------------------ b s–a s+----------- a

R Z0+

RCZ0--------------- b

R Z0–

RCZ0---------------=%=%==



• Now since

(1.58)

• To inverse transform first apply partial fractions to

(1.59)

Clearly,

(1.60)

so

(1.61)

and

(1.62)

Vg s) * v0L u t) *C D v0 s3= =

V z s%) *v02s----- e

sz +p3– b s–a s+-----------e

s 2l z–) * +p3–+=

b s–s a s+) *-------------------

K1s

------K2

s a+-----------+=

K1ba---

R Z0–

R Z0+---------------= = K2

a b+) *–a

-------------------- 2R–R Z0+---------------= =

V z s%) *v02----- 1

s---e

sz +p3–=

R Z0–

R Z0+--------------- 1

s--- 2R

R Z0+--------------- 1

s a+-----------E–E

? 7> 9= :

es 2l z–) *

+p--------------------–

+

v z t%) * L 1– V z s%) *C Dv02----- u t z

+p-----–, -

. / R Z0–

R Z0+---------------?>=

+= =

2RR Z0+---------------e

t 2l z–+p

-------------–, -. /R Z0+

RCZ0---------------–

79:u t 2l z–

+p-------------–, -

. /–



• As a special case consider

(1.63)

Example: The results of an ADS simulation when usingv, ohms, C = 50 pf, and ns is

shown below

• The ADS schematic is shown below

z l=

v z t%) *v02----- 1

R Z0–

R Z0+--------------- 2R

R Z0+---------------e

t l+p-----–, -

. /R Z0+RCZ0---------------–

–+ u t l+p-----–, -

. /=

v0R

R Z0+--------------- 1 e

t l+p-----–, -

. /R Z0+RCZ0---------------–

– u t l+p-----–, -

. /=

Figure 1.11: Sketch of the general waveformv l t%) *

v0 2= Z0 R 50= = Tl 5=

Figure 1.12: ADS Circuit diagram with nodes numbered



• Next we plot the source and load end waveforms

LoadSource

TranTran1

MaxTimeStep=0.1 nsecStopTime=20.0 nsec

TRANSIENT

CC1C=50 pF

RR2R=50 Ohm

TLINDTLD1

Delay=5 nsecZ=50.0 Ohm

VtPulseSRC1

Period=1 usecWidth=1 usecFall=0.1 nsecRise=0.1 nsecEdge=linearDelay=0 nsecVhigh=2 VVlow=0 V

t

RR1R=50 Ohm

2 4 6 8 10 12 14 16 180 20

0.2

0.4

0.6

0.8

0.0

1.0

time, nsec

So

urce

Loa

d

load end source end



Case 2: Mismatched Source Impedance

• For the general case where to reflections occur atthe source end of the line as well as at the load end

• The expression for now will consist of an infinitenumber of terms as shown below:

(1.64)

where

(1.65)

• In terms of +z and -z propagating waves we can write

(1.66)

Zg s) * Z0$

V z s%) *

V z s%) * Vg s) *Z0

Z0 Zg s) *+------------------------- e

sz +p3–<L s) *e

s 2l z–) * +p3–+@=

<L s) *<g s) *es 2l z+) * +p3–

+

<L s) *<g s) *<L s) *es 4l z–) * +p3–

+

<L s) *<g s) *<L s) *<g s) *es 4l z+) * +p3– F+ + A

<L s) *ZL s) * Z0–

ZL s) * Z0+-------------------------- and <g s) *

Zg s) * Z0–

Zg s) * Z0+-------------------------= =

V z s%) *Vg s) *Z0

Z0 Zg s) *+------------------------- e sz +p3– <L

n s) *<gn s) *e

s 2n) *l +p3–

n 0=

G

H=

esz +p3

<Ln 1+ s) *<g

n s) *es 2n 2+) *l +p3–

n 0=

G

H+



Example: Consider a circuit with v, ohms, ns, ohms, and a parallel RC circuit

with ohms and pf, as shown below in Figure1.13.

• In the s-domain the solution is of the form

(1.67)

• To inverse transform note that each series term con-sists of the product of a constant, a ratio of polynomials in ,and a time shift exponential (i.e. )

• In the time-domain each series term to within a constant is ofthe form

(1.68)

v0 2= Z0 50=Tl 5= Zg Rg 100= = ZL

R 100= C 20=

Figure 1.13: Circuit diagram with Spice nodes indicated

V z s%) * 23--- e

sz +p3– b s–) *n

s s a+) *n--------------------- 1

3---, -. / n

es 2n) *l +p3–

n 0=

G

H=

esz +p3 b s–) *n 1+

s s a+) *n 1+----------------------------- 1

3---, -. / n

es 2n 2+) *l +p3–

n 0=

G

H+

V z s%) *s

e sI–

L1– b s–) *n

s s a+) *n---------------------? 7> 9= :

t t In–'

n% 0 1 2 F% % %=



• A partial fraction expansion of the ratio of polynomials in(1.68) is

(1.69)

where

(1.70)

and

(1.71)

• To obtain a partial solution for comparison with a Spice sim-ulation we will solve (1.69) for , 1, and 2.

– Case :

(1.72)

– Case n = 1

(1.73)

b s–) *n

s s a+) *n---------------------

K1s

------K12s a+-----------

K22

s a+) *2------------------- F K2n

s a+) *n-------------------+ + + +=

K1bn

an-----=

K2k1

n k–) *!------------------ d n k–) *

ds n k–) *------------------ b s–) *n

s------------------

s a–=k% 1 2 F n% % %= =

n 0=

n 0=

1s--- u t) *J

b s–s s a+) *------------------- b a3

s---------- a b+) * a3

s a+------------------------– b

a--- a b+

a------------e at–– u t) *J=



– n = 2

(1.74)

• Using Mathematica the analytical solution valid for up to25 ns was obtained

b s–) *2

s s a+) *2--------------------- b2 a23

s--------------- 1 b2 a23–) *

s a+----------------------------- b a+) *2 a3

s a+) *2--------------------------–+=

b2

a2----- 1 b2

a2-----–

, -K L. /

e at– b a+) *2

a-------------------te at––+ u t) *J

t



• This required the use of +z and -z wave solutions from theseries for and 1

• The theoretical voltage waveforms at and areshown below

• A circuit simulation using ADS was also run, the resultscompare favorably as expected

n 0=

z 0= z 1=

5 10 15 20 25

!0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

t (ns)

Load

Source

Line

Vol

tage

(v)



• Plots of and

LoadSource

VtStepSRC1

Rise=0.05 nsecDelay=0 nsecVhigh=2 VVlow=0 V

t

TranTran1


TRANSIENT

CC1C=20 pF

RR2R=100 Ohm

RR1R=100 Ohm

TLINDTLD1

Delay=5 nsecZ=50.0 Ohm

v 0 t *%) * v l t%) *

2 4 6 8 10 12 14 16 18 20 22 240 26

0.0

0.2

0.4

0.6

0.8

1.0

-0.2

1.2

time, nsec

Sou

rce

m2

Load

m1m3

m1indep(m1)=plot_vs(Load, time)=0.889

1.148E-8

m2indep(m2)=plot_vs(Source, time)=0.963

1.608E-8

m3indep(m3)=plot_vs(Load, time)=0.988

2.443E-8

Source

Load

(v)

Sinusoidal Signals on Transmission Lines



Voltage and Current Relationships

• Consider an axially uniform transmission line operating inthe TEM or Quasi-TEM mode

• Let the sinusoidal voltage and current at be given by

(1.75)

where the complex phasors V and I represent the voltage andcurrent without the time dependence

• Since a sinusoidal steady-state solution is desired, phasornotation may be used to solve the transmission line equations

• Note that with suppressed it is implied that

(1.76)

• By writing the basic line equations in phasor notation weobtain

(1.77)

Figure 1.14: Voltage and current at z on an axially uniform line

z

v t z%) * Re V z) *ejMtC D V z) * Mt V z) *N+@ Acos= =

i t z%) * Re I z) *ejMtC D I z) * Mt I z) *N+@ Acos= =

ejMt

ejMt

(nV(t n--------- jM) *nV=

(V(z------ R jML+) *I–=



and

(1.78)

• The voltage wave equation becomes

(1.79)

where R, G, L, and C are the primary transmission lineparameters defined earlier

• The general solution is

(1.80)

where

(1.81)

with being the line attenuation constant in nepers per meterand is the line phase constant in radians per meter

– is the constant associated with the wave propagating inthe +z direction and is the constant associated with thewave propagating in the -z direction

• The current can be shown to have general solution of theform

(1.82)

(I(z----- G jMC+) *V–=

(2V(z2--------- RG M2LC–) *V– jM RC LG+) *V– 0=

V z) * V+e Oz– V–eOz+=

O P jQ+=

R jML+) * G jMC+) *=

M2LC– RG jM RC LG+) *+ +=

PQ

V+

V–

I z) *

I z) * I+e Oz– I–eOz+=



• By substituting (1.77) into (1.82) we obtain

(1.83)

• The line characteristic impedance is defined as

(1.84)

which then relates the voltage and current on the line as

(1.85)

• Finally we can write as

(1.86)

• In the time domain the steady state solution voltage wave-form is

(1.87)

I z) * 1–R jML+-------------------, -. / (V z) *

(z--------------=

1–R jML+------------------- OV+e Oz–– OV–eOz+) *=

G jMC+R jML+--------------------- V+e Oz– V–eOz–) *=

Z0R jML+

O------------------- R jML+

G jMC+---------------------= =

Z0V+

I+------ V––

I–---------= =

also

I z) *

I z) * V+

Z0------e Oz– V–

Z0------eOz–=

v z t%) * V+ Mt Qz– V+N+) *e Pz–cos=

V– Mt Qz V–N+ +) *ePzcos+



Lossless Line

• Special Case: For an ideal lossless line , and reduce to

(1.88)

• Since the phase velocity, , on the line is

(1.89)

which allows to also be written as

(1.90)

• This result implies that can be obtained by knowing thevelocity of propagation in the medium and the capacitanceper unit length of the transmission line structure

• In free space

(1.91)

where and are the free space permittivity and permea-bility respectively

• For a dielectrically filled structure where it is assumedthat , and , thus

(1.92)

R G 0= = OZ0

O jQ jM LC P 0=) *= =

Z0LC----=

Q M LC= +p

+p1LC

-----------=

Z0

Z01+pC----------=

Z0

+p c 1B0!0

--------------- 3 8R10 m/sS= =

!0 B0

B B0= ! !" j!""– !0!rS= T 0=

+p c !r3=



where is the relative permittivity of the medium

• For partially filled transmission line structures such asmicrostrip, the velocity of propagation can be written as

(1.93)

where is the effective value of relative permittivity.(Note: ).

Low Loss Line

• Special Case: For most practical transmission line structuresthe losses are small, which is to say, and

• This low-loss assumption allows the expression for to besimplified

• To begin with write

(1.94)

using the binomial expansion

• Thus

(1.95)

!r

+pc!eff

------------=

!eff!eff !r5

R ML« G MC«

O

O jM LC 1 RG jM RC LG+) *+

M2LC–------------------------------------------------+

1 23=

1 x+) *1 23 1 12---x for x 1«+S

O jM LC 1 12--- RG–

M2LC-------------- j

M---- R

L--- G

C----+, -

. /–+? 7> 9= :

S

= > ? 0U



or

(1.96)

Note that did not change from the lossless case

• Using a similar approach on results in

(1.97)

Note also that under the low-loss assumption is stillapproximately real

O P jQ 12--- LC R

L--- G

C----+, -

. / jM LC+S+=

Q

Z0

Z0R jML+G jMC+--------------------- R jML+) * G jMC–) *

G2 M2C2+--------------------------------------------------= =

RG jM GL RC–) * M2LC+ +

M2C2---------------------------------------------------------------------S

LC---- 1 RG jM LG RC–) *+

M2LC------------------------------------------------+=

LC---- 1 1

2---j G

MC-------- R

ML-------–, -

. /+ LC----US

Z0



Terminated Lossless LineFirst consider the case where the generator or source driving theline is matched to the line characteristic impedance as shownbelow in Figure 1.15.

• For an arbitrary load impedance, , the boundary conditionswill require both forward (+z) and backward (-z) propagatingwaves to exist

• At , the voltage and current equations with are

(1.98)

and

(1.99)

• By convention, the voltage reflection coefficient at the load isdefined as

(1.100)

Z0

Figure 1.15: Lossless line terminated with impedance Z0

ZL

z 0= P jQ=

V 0) * V+e jQ 0) *– V–ejQ 0) *+ VL= =

I 0) * 1Z0-----V+e jQ 0) *– 1

Z0-----V–ejQ 0) *– IL= =

< V–

V+------=



• From the voltage and current equations we can now solve for in terms of and

(1.101)

or

(1.102)

• The voltage transmitted to the load due to the incident volt-age wave can be defined in terms of the voltage transmissioncoefficient

(1.103)

so

(1.104)

• At a discontinuity in a transmission line system, such as load terminating the line at , the ratio of power incident

to the power reflected is a quantity of interest

• A common scalar network analyzer measurement is returnloss (RL) which is 10log10 of this power ratio

• The return loss is related to in the following way

(1.105)

where

< ZL Z0

V 0) *I 0) *----------- ZL

V+ 1 <+) *

V+ 1 <–) * Z03---------------------------------- Z0

1 <+1 <–-------------= = =

<ZL Z0–

ZL Z0+------------------=

T

VL TV+ 1 <+) *V+= =

T 1 <+V

ZL z 0=

<

RL 10log10PincidentPreflected--------------------, -. /=



(1.106)

• Now

(1.107)

so

(1.108)

• If then and the magnitude of the voltagealong the line is just that of the incident wave which is

• In general so

(1.109)

and

(1.110)

by writing (polar form) then we can write

(1.111)

which inspired the vector diagram of shown below

Pincident12---Re V+ I+) **C D 1

2--- V+ 2

Z0------------==

Preflected12---Re V– I–) **C D 1

2--- V– 2

Z0-----------==

PincidentPreflected-------------------- V+ 2

V– 2------------ 1

< 2---------= =

RL 20log10 < dB–=

ZL Z0= < 0=V+

< 0$

V z) * V+e jQz– <V+ejQz+=

V z) * V+ 1 <ej2Qz+=

< WejX=

V z) * V+ 1 Wej X 2Qz+) *+=

V z) *



• With the aid of the vector diagram shown above, it is clearthat takes on maximum and minimum values of

(1.112)

where is an integer

• The variation in is sinusoidal with the distancebetween maxima and between minima each being

• Recall that where is the wavelength of TEMwaves in the medium

• The incident and reflected voltage waves interfere to producethe voltage standing-wave pattern shown below in Figure1.17

Figure 1.16: Vector diagram showing V z) *

V z) *

V z) * max V+ 1 W+ X 2Qz+% 2nY= =

V z) * min V+ 1 W– X 2Qz+% 2nY Y+= =

n

V z) *

d Y Q3= Z 23=

Q 2Y Z3= Z



• The ratio of to is defined as the voltagestanding-wave ratio (VSWR), or simply SWR

(1.113)

• Comparison of termination characterization parameters

RL VSWR

0.0 1.0

0.1 +20 dB 1.2222

0.3162 +10 dB 1.9520

0.5012 +6 dB 3.0095

0.8913 +1 dB 17.3910

0.9441 +0.5 dB 34.7532

Figure 1.17: Voltage wave along the line for various real loads.

V z) * max V z) * min

VSWR 1 W+1 W–------------ 1 <+

1 <–---------------= =

<

+G



The analysis of a terminated transmission line up to this point hasassumed that the source impedance and line characteristicimpedance are equal. This simplifies the analysis of voltage andcurrent along the line since under this assumption the wavereflected at the load is completely absorbed when it arrives at thesource.

• To consider the mismatched source case, the impedance seenlooking into an arbitrarily terminated transmission line ishelpful

Figure 1.25: Circuit for finding the input impedance of a line terminated in ZL.

• To find , simply form the ratio with

(1.114)

Recall that the load reflection coefficient is given by

(1.115)

Figure 1.18: Circuit for finding the input impedance of a line ter-minated in ZL.

Zin V z) * I z) *3 z l–=

ZinV z) *I z) *-----------

z l–=

Z0V+ejQl V–e j– Ql+

V+ejQl V–e j– Ql–---------------------------------------= =

<LV–

V+------

ZL Z0–

ZL Z0+------------------= =



so

(1.116)

Finally after rearranging using Euler’s identity for tan( ) reduces to

(1.117)

• Two special cases of interest that will be considered laterresult when the operating frequency is such that or

• The input impedances for these cases are

(1.118)

• The input impedance of open and short circuit terminatedtransmission lines will also be of interest

(1.119)

Zin Z0

ejQl ZL Z0–

ZL Z0+------------------e

jQl–+

ejQl ZL Z0–

ZL Z0+------------------e

jQl––

---------------------------------------------=

Zin

Zin Z0ZL jZ0 Qltan+

Z0 jZL Qltan+----------------------------------=

l Z 23=l Z 43=

Zin l Z 23=) * ZL=

Zin l Z 43=) *Z0

2

ZL------=

Zin ZL 0'jZ0 Qltan=

Zin ZL G'jZ0 Qlcot–=



• The reflection coefficient at any can also be obtainedby noting that

(1.120)

so

(1.121)

• The above result will appear later during the discussion ofscattering parameters.

Terminated Lossless Line with Arbitrary Source ImpedanceThe circuit of interest is shown below in Figure 1.26

• One approach to this problem is to consider all the reflectionsand re-reflections and then sum the infinite series

• Since the infinite series is geometric and , it is summa-ble in closed form

z l–=

V+z l–= V+ejQl=

V–z l–= V–e j– Ql=

< l) * V–e j– Ql

V+ejQl----------------- <Le j2Ql–= =

Figure 1.19: Terminated line with arbitrary source impedance.

< 15



• A second approach is to use the boundary conditions at thesource and load terminations to obtain the steady-state solu-tion directly

• Using (1.117) for we can write

(1.122)

• Now since

(1.123)

so

(1.124)

• Now substitute

(1.125)

into (1.124) to obtain

(1.126)

Zin

V z = l) * VgZin

Zin Zg+------------------- V+ejQl V–e jQl–+= =

V– V+<L=

VgZin

Zin Zg+------------------- V+ ejQl <Le jQl–+@ A=

V+ VgZin

Zin Zg+------------------- e jQl–

1 <Le j2Ql–+-----------------------------=

Zin Z01 <L e j2Ql––+

1 <Le j2Ql––-----------------------------------=

V+ VgZ0e jQl–

Z0 1 <Le j2Ql–+) * Zg 1 <Le j2Ql––) *+@ A--------------------------------------------------------------------------------------------=

VgZ0

Z0 Zg+------------------ e jQl–

1 <g<Le j2Ql––------------------------------------=



where is the reflection coefficient looking towards thegenerator

• Finally, we can write for any

(1.127)

An equation for may be obtained in a similar manner

Source to Load Power Transfer Considerations

• The power delivered to the load via the lossless line is justthe input power which is given by

(1.128)

• Let and , then

(1.129)

– As a special case suppose that

– Clearly , , and

(1.130)

<g

l– z 05 5

V z) * VgZ0

Z0 Zg+------------------

1 <Lej2Qz+

1 <g<Le j2Ql––------------------------------------e jQ z l+) *–=

I z) *

P 12---Re V l–) *I* l–) *C D 1

2--- V l–) *V

* l–) *Zin

----------------? 7> 9= :

==

12--- Vg

2 ZinZin Zg+-------------------

2Re 1

Zin*

-------? 7> 9= :

=

Zin Rin jXin+= Zg Rg jXg+=

P 12--- Vg

2 Rin

Rin Rg+) *2 Xin Xg+) *2+-------------------------------------------------------------=

ZL Z0=

<L 0= Zin Z0=

P 12--- Vg

2 Z0

Z0 Rg+) *2 Xg2+

--------------------------------------=



– As another special case suppose that

– Note that the condition can be satisfied throughadjustment of and

(1.131)

– How do we obtain maximum power transfer?

– For fixed we know from circuit theory that for maxi-mum power transfer we choose

– The resulting power delivered is as expected

(1.132)

– Note that if is real then (1.130) and (1.131) yield themaximum power transfer result

Terminated Lossy Line

• The analysis of a terminated lossy line is very similar to thelossless case except now must be replaced by

• It will be assumed that is small enough to imply that can still be considered real

Zin Zg=

Zin Zg=Ql Z0

P 12--- Vg

2 Rg

4 Rg2 Xg

2+) *---------------------------=

ZgZin Zg

*=

P 12--- Vg

2 14Rg---------=

Zg

jQ O P jQ+=

P Z0

Figure 1.20: Terminated lossy line.



• To begin with we can immediately write

(1.133)

• Additionally and respectively become

(1.134)

and

(1.135)

• The power delivered to the line input is

(1.136)

• The power actually delivered to the load is

(1.137)

V z) * V+ e Oz– <LeOz+@ A=

I z) * V+

Z0------ e Oz– <LeOz–@ A=

<L Zin

< l) * <Le j2Ol– <Le j2Ql–@ Ae 2Pl–= =

Zin Z0ZL Z0 Oltanh+

Z0 ZL Oltanh+----------------------------------=

Pin12---Re V l–) *I* l–) *C D=

V+ 2

2Z0------------ e2Pl <L

2e 2Pl– <Le j2Ql– <Le j2Ql–) **–+–? 7> 9= :

=

V+ 2

2Z0------------ 1 < l) * 2–@ Ae2Pl=

= 8 > 8 ?

< l) * 2e2Pl

PL12---Re V 0) *I* 0) *C D V+ 2

2Z0------------ 1 <L

2–@ A= =



• The power lost in the line is

(1.138)

Line Attenuation CalculationThe perturbation method of loss analysis assumes that the fieldsassociated with the lossy line are very similar to the fields of thelossless line.

• The power flow along the line is given by

(1.139)

where is the power input at , and is the attenua-tion parameter of interest

• The power loss per unit length is

(1.140)

• Rearranging we obtain

(1.141)

• Note that this analysis assumes that is known and canbe found from the fields of the lossless line

Ploss Pin PL– V+ 2

2Z0------------ e2Pl 1–) * <L

2 1 e 2Pl––) *+@ A= =

P z) * P0e 2Pz–=

P0 z 0= P

Pl(P–(z

---------- 2PP0e 2Pz– 2PP z) *= = =

PPl z) *2P z) *--------------

Pl z = 0) *2P0

----------------------= =

P0 Pl



Dispersion

• In the first order approximation to given by (1.96) wefound that

(1.142)

• If is not of the form , then the phase velocity

• If is a function of frequency then the frequency compo-nents associated with a broadband signal will arrive at theload end of the line at different times

• This phenomenon is known as dispersion.

• In linear system theory we would say the system has nonlin-ear phase

• Fortunately since is typically much less than ,we see that for low loss lines is very nearly linear in

Example: Consider the transmission line circuit shown below inFigure 1.21.

O

Q M LC RG2M2 LC----------------------–=

Q Q aM=+p M Q constant$3=

+p

RG 2M2 LCQ M

Figure 1.21: Match terminated low loss line.



• We can immediately write that

(1.143)

• The transfer function denoted is thus givenby

(1.144)

• The magnitude and phase are

(1.145)

• For we see that the phase function is composed oflinear and nonlinear components

Distortionless Line

• It is possible for a lossy line to have a linear phase factor ifthe line parameters satisfy

(1.146)

• To demonstrate this insert (1.146) into (1.96)

vL M) * 12---vg M) *e Pl– e

jM LC RG2M2 LC---------------------––

=

H M) * H 2Yf) *=

H M) *vL M) *vg M) *-------------- 1

2---e Pl– e

jM LC RG2M2 LC---------------------––

= =

H M) * 12---e Ml–=

H M) *N Ml LC– RGl2M LC-------------------+=

RG 0$

RL--- G

C----=



(1.147)

• Clearly, is now a linear function of frequency and theattenuation, , is nearly constant with frequency

• Note that the line resistance is usually a weak function of fre-quency

Example: RG-58/U Coax

• To better illustrate the impact of dispersion, consider the spe-cial case of RG-58/U coaxial cable

• In the Pozar text (and Collin) the transmission parameters ofcoax are shown to be

where is the surface resistivity of the conductor, and isgiven by

, (1.148)

Coax

O jM LC 1 R2

M2L2------------ 2j R

ML-------–+

1 23=

jM LC 1 j RML-------–=

R CL---- jM LC+ P jQ+==

QP

b

a

L B2Y------ b

a---ln= C 2Y!"

b a3ln---------------=

RRs2Y------ 1

a--- 1

b---+, -

. /= G 2YM!""b a3ln

-----------------=

Rs

RsMB2T-------=



where here with

• For RG-58/U the dielectric diameter is in and thedielectric material is polyethylene having @ 10GHz and

• Since the characteristic impedance is nominally 50 ohms wecan determine the inner radius, , by setting

,

thus in

• The transmission line parameters are given by:

(1.149)

• To obtain the frequency response of a one meter section ofproperly terminated RG-58/U cable, we simply insert theabove transmission line parameters into

B B0Br B0= = B0 4Y 10 8–R=

2b 0.116=!r 2.25=

!"" 0.004!0!r=

a

Z0LC----S 60

!r

-------- ba---ln 50= =

2a 0.033=

L B2Y------ b

a---ln

!rZ0c

--------------- 250 nH/m= = =

C 2Y!"b a3ln

---------------!r

cZ0-------- 100 pf/m= = =

RRs2Y------ 1

a--- 1

b---+, -

. / 1.273 10 4– f ohms/mRR= =

G 2YM!""b a3ln

----------------- MC!""!"

--------------- 2.513 10 13– f s/mRR= = =



(1.150)

making sure to explicitly include the frequency dependenceon and , i.e., and

• The equivalent impulse response of the system can beobtained by inverse Fourier transformation, i.e.,

(1.151)

• The response of the system to an arbitrary pulse waveform,, can be obtained in the transform domain, or by

direct convolution

(1.152)

where

• The following results were obtained using Mathematica

– Magnitude and phase frequency response for m

– Impulse response computed using a 1024 point inverseFourier transform (IFFT) for m

– Simulated pulse response to a 500 Mb/s return-to-zero(RZ) data pattern of the form ...000101001000... (eachpulse is 2ns long)

H f) *vL f) *vg f) *------------ 1

2---e Ol–= =

12--- j2Yf LC 1 RG j2Yf RC LG+) *+

4Y2f2LC–-----------------------------------------------------+ l–exp=

R G R R f) *= G G f) *=

h t) * F1– H f) *C D H f) *ej2Yft fd

G–

G

[= =

vg t) * p t) *=

vL t) * p t) **h t) * F1– P f) *H f) *C D= =

P f) * F p t) *C D=

l 2=

l 2 10 and 50% %=



• Mathematica modeling

Z0

Z0

p t) * l

vL t) * y y) *=

+

-

t (ns)

22 nsp t) *

6 10 16



• Frequency response

107 108 109 1010 1011!10

!8

!6

!4

!2

0

Gai

n (d

B)

f (Hz)

Gain in dB lessthe 1/2 loss factorof 6 dB



• The phase response and pulse response

– The input is a real signal, but the frequency response is notconjugate symmetric, so the impulse response is complex

– This makes the pulse response contain real and imaginaryparts as well

108 109 1010 1011

0.005

0.010

0.015

0.020

0.025

0.030

0.035

10 20 30 40 50

!0.4

!0.2

0.2

0.4

0.6

Pha

se (d

eg)

f (Hz)

t (ns)

Linear phaseterm removed

Real

Imaginary?

The model makes itimpossible to eliminatethe imaginary part



• ADS schematic when using the lossy coax model

• Input and output waveforms with ADS are more realistic, buta warning regarding a complex impulse response is posted

Input Output

COAX_MDSTL1

TanD=0.004Er=2.25Mur=1.0Cond2=5.8E+7Cond1=5.8E+7T=2.0 milL=2 meterRo=60 milRi=58 milA=16.5 mil

VtPulseSRC3

Width=2 nsecDelay=16 nsec

t

VtPulseSRC2


t

VtPulseSRC1


t

RR1R=50 Ohm

TranTran1


TRANSIENT

RR2R=50 Ohm

5 10 15 20 25 30 350 40

-0.0

0.2

0.4

0.6

0.8

-0.2

1.0

time, nsec

Inpu

tO

utp

ut

Input Output

Documents

Microwave Circuit Design - eas.uccs.edueas.uccs.edu/~mwickert/ece5250/notes/N5250_1.pdf · Chapter 1. Review of Transmission Line Theory ... ECE 5250/4250 Microwave Circuits 1–9