Chapter 2 ECE 130a - University of San Diegohome.sandiego.edu/~ekim/e171f00/lectures/tl.pdf · If we have only a going wave, the same kind of derivation we did for the going wave

Chapter 2

Transmission Lines

ECE 130a

Z l ZZ l j Z l

Z l j Z lZ

Z j Z l

Z j Z lin oL o

o Lo

L o

o L

- = ++

= ++

a f cos sincos sin

tantan

b bb b

bb

Examples of Transmission Lines: (Chapter 2)

Parallel Strip Line:

Coaxial Line:

Parallel Wire Line:

Microstrip Line:

There is a simple way to view the guided wave on a transmission line.

dw

metal conductors

radius a

radius bmetal

metal

dielectric

D

Air Line Dielectric Line

conductor radius a

ε r

w

d

Metal

Dielectric

I z t,a fV z t,a f

I z t,a fz

The potential difference (voltage) between the metal conductors with equal and opposite current flowing in them are circuit concepts, except they depend not only on time, but also on the distance z.

So we describe the wave as voltage and current waves.

Other guiding structures:

1. Waveguides -- consist of a single hollow metal tube of various cross-sectional geometry. An EM wave propagates longitudinally inside the hollow structure.

The wave propagation in waveguides is not transverse (not TEM). That is, it has longitudinal field component(s). Transverse spatial depen-dence is fairly complicated.

The propagation constant of a waveguide wave is not equal to that of plane waves, and velocity of propagation thus is not the same as light.

2. Optical Fibers -- are used at optical frequencies, at infrared, and at visi-ble wavelengths. An optical fiber consists of a very thin (50-300 mm) dielectric circular cross section cylinder. The material is usually glass or plastic. Inner and outer portions have different dielectric constants (index of refraction), as shown below.

Optical fibers do not support TEM waves, like the hollow metallic guides. Their propagation constant and mode structures are even more compli-cated than for hollow metallic guides.

generator load+

-Length

i t z,a fv t z,a f

L >> =2π β λ/

β

core radius a, index n1

cladding, index n2

Derivation of Transmission Line Equations (1-3)Let us consider a length of a transmission line at location . The circuit model is clearly a series inductance and resistance (since the whole line with a load at the end forms a loop) and a shunt capacitance and shunt leak-age conductance (between the good conductors, across the dielectric which also has a small conductance). With the above circuit parameters being

and , the model is

A straight forward application of Kirchoff’s Loop Law gives

,

and Kirchoff’s Current Law at the upper node gives

.

Dividing through by and taking the limit , we get:

Sinusoidal Analysis of Transmission Lines (using phasors)

Differentiate (1) with respect to :

z

L R C, , , G

RLi z t,a f i z z t+D ,a f

v z z t+D ,a fv z t,a f++

−−C G

v z z t v z t zi z t

tz i z t+ - = - -D D D, ,

,,a f a f a f a fL R

∂∂

i z z t i z t z v z z t zv z z t

t+ - = - + -

+D D D D

D, , ,

,a f a f a f a fG C

∂∂

∆z ∆z→0

∂∂

∂∂

vz

it

i= - -L R

∂∂

∂∂

iz

vt

v= - -C G

(1-5)

(1-6)

(1)

(2)

∂∂

wvi

zj= - +R La f

∂∂

wiv

zj= - +G Ca f

z

∂∂

w2

2

v iz

jd

dz= - +R La f

Substitute from (2) :

Similarly, we get:

The Lossless Transmission Line

(perfect conductor) (lossless dielectric)

This is not only instructive and simple, but also a good approximation of many real lines which are made of very good conductors (typically, copper) and nearly lossless dielectrics (e.g., teflon). For example, a commercial coaxial line is a nearly lossless line.

The solutions are travelling waves:

It will be shown later, when we consider transmission lines from the Electro-magnetic Field’s point of view, that these voltage and current waves corre-spond to EM waves, also. It will be shown that:

,

where v is velocity of light in the medium between the conductors.

∂ ∂i / z

∂∂

w w2

2

vv

zj j= + +G C R La fa f

∂∂

w w2

2

ii

zj j= + +G C R La fa f

R = 0 G = 0

∂∂

∂∂

Vz

It

= -L ∂∂

∂∂

Iz

Vt

= -C

Also,

∂∂ w w

2

2

vv

zj j= ◊C L

∂∂

w2

2

vv

z= - 2LCc h

∂∂ b

2

22vv

z= - ∂

∂ b2

22ii

z= -

∂∂

w2

2

ii

z= - 2LCc h

β ω ω µε= =LC

e j z± β e j tω

e j t zw b-a f e j t zw b+a for or sum of both.

(with already assumed)

LC = =µε 1v

We can get the relationship of V and I by substituting back into the original transmission line equations.

Zo is called the characteristic impedance of the line. For this lossless line, Zo is real. It means that V and I are in phase. For example, power is prop-agated. None is absorbed. (Contrast this with ordinary circuits, where real impedance means power is absorbed and dissipated!)

Power Propagated on a Lossless Line

The power propagated can be calculated either from the electromagnetic wave or from the voltage-current wave. It is usual to use voltage and cur-rent for a transmission line. The power, of course, is an oscillating quantity. We are mostly interested, however, in the average power flow. If (real), then from the circuit picture:

for lossless line

If we have only a going wave, the same kind of derivation we did for the going wave gives:

( going wave)

Circuits vs. Transmission Lines

1) Q: When does the “regular” circuit (E17) method hold correct; that is, only dependence (no propagation) and when does the transmission line method need to be used?

v = -V e j t zw ba f i = -I e j t zw ba f

ddz

jv

i= − ωL ddz

ji

v= − ωC

− = −j jβ ωv iL − = −j jβ ωi vC

vi

= =ωβ

βω

LC

vi

+

+

= =L

CZo

β ω ω µε= =LC ,

+z

+z

(i.e., only going wave)

This is for going wave only.

PVZav

o

o= =Re

12

12

2

v i*

− z+ z

vi

−

−= − Zo − z

e j tω

A: Simple. When the wavelength is large compared to the dimensions, we can neglect transmission line concepts:

That is, whenever the distance scale (z) is small and/or the wavelength is long; that is, .

2) For circuits: If is real (that is, ), . This is power absorbed in the real impedance (resistance).

For a lossless transmission line: is real. If we have only + going wave:

This is not absorbed power. It is power propagating in the direction down the lossless line.

If we have - going wave:

The negative sign means that power is propagating in the negative direc-tion. Unless otherwise stated, we always will mean average power.

Transmission Line Circuits

Transmission lines are used to carry signals from one location to another. Thus, they usually have a signal generator at one end and a load at another end. Let’s now examine the behavior of a transmission line with load. We use phasors and will assume lossless transmission line.

1) Transmission line terminated in its characteristic impedance. This is called “matched.”

Let’s assume we launched a going wave on this transmission line of characteristic impedance . Then, at any point z, we know that

. At the load, . (This is definition of .) Thus, if

β πλz z= →2

0

z << λZ R= P I Rav = 1

22

Zo

PZ

Zavo

o= =++

12

12

22v

i

+ z

⇒R Zo=

Power is absorbed in the loadwhere it does “something.”to Load

PZ

Zavo

o= = − = −− −−

−Re12

12

12

22

v iv

i*

Z ZL o=Zo

+ zZo

v i+ + =/ Zo v iL L LZ/ = ZL

, then ( ) and ( ). Thus, there is no reflected wave ( going wave). Rather, the wave continues into the load, where it gets dissipated. (In practical situations, the load is a whole circuit, for example, a detector, a radio, an oscilloscope. As long as its input impedance is , the foregoing discussion is true.)

2) Transmission line terminated in other impedance (1-7, text)

Assume again we launched a signal at some frequency. Then, . But at the load, . If , cannot be

and cannot be . The boundary condition at the load can only be satisfied if there is also a reflected wave . This wave propagates backward in the direction. Then, at any point on the line:

Note the sign on the current of the reflected wave! Now we can satisfy the condition at the load . Only one value of will satisfy.

Let us define:

Then:

We solve for to get: (Problem: Solve to )

Note: is complex, in general, because may be complex.

Z ZL o= v vL = + Z L= i iL = + Z L=− z

Zo

Z ZL o≠Zo⇒

v i+ + =/ Zo v iL L LZ/ = Z ZL o≠ vL v+

iL i +v i− −,

− z

i i i= - = ++ - - ++ -

V

Ze

V

Ze

o

j t z

o

j t zw b w ba f a fi v

−−= −

Zo

v v v= + = ++-

-+

+ -V e V ej t z j t zw b w ba f a f

−v i/ = ZL V−

vv

−

+= Γ , the reflection coefficient

(1)

(2)

v v v+ ++ Γ =

v v v+ +−Z Z Zo o L

Γ =

Γ G

Γ = −+

Z ZZ Z

L o

L o

Γ ZL

- £ £1 1G

Fraction of the incident power reflected back:

3) Special Cases of Termination

A special case of the general load is , that is, a short circuited trans-mission line. (A transmission line may be shorted deliberately in order to achieve a certain result, or it may be shorted accidentally, something that has to be repaired.)

For this case:

We can say here that here.

If ,

Note that this time, the z dependence has two different parts. Thus, the z dependence cannot be lumped together with dependence, as it can be

Incident Power

Reflected Power

Total Power

Note: is purely imaginary.

Transmitted power (into Load)

P Z V Zi o o= LNM

OQP = L

NMOQP =+ + + +Re Re

1

2

1

2

1

22 2

v i v* / /

PT T T= LNM

OQP = + -L

NMOQP+ +Re Re

1

2

1

21 1v i v i* * *G Ga f e j

= + - -LNM

OQP+ +Re

1

21 G G GG* * *e jv i

= − ≡12Γ Pi

Pr = LNM

OQP = L

NMOQP -- - - +Re Re

1

2

1

21v i v i* * *G G a f

= -LNMOQP = -+ +Re G G2 21

2v i* Pi

1

2G G G- =*e j Im

ZL = 0

Γ = − = −ZZ

o

o1

Zo Z

Z = 0

ZL = 0

~Γ Γ= = +e ej jϕ π1

v+-= A e j t zw ba f

fv z Ae Aej z j za f= -- b b

i zZ

A e A eo

j z j za f= +-1 b b

e j twc understood

e j tω

for a wave traveling in one direction only. Now, the entire V or I, including z dependence, is the phasor:

The real voltage and current are ( for )

Both voltage and current oscillate sinusoidally in time with different maxi-mum values (amplitudes) at different locations. This is known as a standing wave--“standing,” because the amplitude remains the same at each location and the oscillating pattern is standing. Contrast this with “travelling” wave, in which a given point on the wave form progresses in distance with time. The amplitude of a travelling wave, however, is a constant value regardless of location.The instantaneous power flow is (must be calculated from real V, I ):

γv z jA za f = -2 sinb

i zA

Zz

o

a f = 2cos b

e j twc understood

A A e j= θ

I

= +2 A z tsin sinb w qa f

= +2 A

Zz t

o

cos cosb w qa fV

z

Conductor

v z t j A e z ej j t, sina f m r= -Re 2 q wb

i z tA e

Zz e

j

o

j t, cosa f =RST

UVWRe

2 qwb

= +A

Zz t

o

2

2 2sin sinb w qa f

= + +4

2A

Zz z t t

o

sin cos sin cosb b w q w qa f a f

P z t z t z t, , ,a f a f a f= v i

The average power flow (we calculate it here direction from phasors) is:

This makes physical sense, since no power flows into the short circuit. Thus, all the power must be reflected back, giving net power flow equal to zero (average). The same result, of course, could be derived from the field point of view. The short circuit is a perfectly conducting plate closing off the line. No fields penetrate into it. Thus,

The Concept of Input Impedance ( )

Characteristic impedance is the ratio of voltage to current of one wave direction.

Since transmission lines are used in conjunction with lumped circuits, we must be able to treat them as regular circuits as well. The input impedance of any circuit is at the terminals. Thus, for a transmission line, at any point on it:

,

where and are the total phasor fields at that point.

Important: Since , this is now the same as a regular circuit con-cept. So we can always replace a transmission line circuit with its input impedance!

If there is only a single wave propagating in one direction (transmission line terminated in ), then, of course, . But, in general, , and is a function of both the load and of the position z on the line.

Let’s measure as follows: at the load. Therefore, z is negative on the line .

Re Re1

2

1

22

2V I j A z

A

Zz

o

**

sin cosLNM

OQP = -

LNMM

OQPP

◊b b

= =Re 2 02

j A z zsin cosβ β

(purely imaginary)

Pav z ze P e P= − =+ − 0

Zin

Zo

Zo = =+

+

−

−

vi

v

i

Z V Iin = /

ZinT

T= v

i

vT iT

Zin T T= v i/

Z ZL o= Z Zin o= Z Zin o≠Zin

Z = 0z l= −

For the short circuited lossless transmission line:

The input impedance of a short circuited lossless line is purely imaginary (as it should be, since it consists of distributed capacitance and inductance.) The values of the input impedance vary with the distance if the frequency is fixed or varies with frequency if the length is constant.

Reactance can be either capacitive or inductive, and its absolute value var-ies from zero to infinity. This property can be utilized in a number of ways.

(a) Easily adjustable values of reactance.(b) Location of an accidental short on a line. (Problem)

(c) Matching of a load not equal to (will study later)

Open Circuited Lossless Line

In practice, we try to avoid open circuiting a transmission line. Reason: Theoretically, open circuit (i.e., open ends) means infinite load impedance ( ). This would mean that:

That is, all of the wave is reflected, resulting in standing waves.

However, in practice, this is not really true. The open end is not a truly infi-nite impedance. While no current flows beyond the open end, some of the electromagnetic wave goes out the open end and becomes a free space

Z l j Z l j Z l j Zf

vlin o o o- = = =a f tan tan tanb p

lp2 2

Z ll

l

j A l

lin A

Zo

- =--

=- -

-a f a f

a fa fa f

vi

22

sin

cos

b

bz l= − z = 0

z

(for a shorted line)

f

l

Zo

RL = ∞

Γ = ∞ −∞ + = +Z

Zo

o1

wave. Only a small fraction does, since the open circuited transmission line is “poorly matched” to space (i.e., it is a poor antenna). Yet, if not all the power is reflected, we obviously do not have infinite impedance.

Lossless transmission line terminated in a real load RLπ Z01

The above two cases are identical as far as T.L. 1 is concerned, since the impedance at point A is for both.

For the above, since :

is real and its value varies from to .

Line 1 has both and going wave.

Line 2 has only going wave, since it is properly terminated.

Just to the left of junction A:

Just to the right of junction A:

These are obviously equal. So we replace at A by :

A

Z01

Z01

RL

RLZ RL02 =

A

RL

We can always replace any transmission line by its equivalent input impedance. The fields up to that input point will be unchanged.

Z Z02 01≠

Γ =−+

R ZR Z

L

L

01

01

Γ −1 +1

A

Z RL02 = Z Rin A L=

1 2

+ z − z+ z

v v v= ++ −1 1

v v= +2

RL

τ = = + = + = ++

+

+ −

+

−

+

vv

v vv

vv

2

1

1 1

1

1

11 1 Γ

1

is called the transmission coefficient. Thus:

can be greater than unity, but that does not mean more than 100% of power transmitted.

Power balance:

Substituting for and and writing the sum of the reflected and transmit-ted power = incident:

Example:

Commercial coaxial line (Radio Shack) is , flat antenna line is . The above situation would happen if you connect them together without a matching transformer. (We shall learn later what a matching transformer is.) At the junction :

The fraction of power reflected is .The fraction of power transmitted to line and on to the load is

.

τ

v v2 1+ += τ τ = + = +12Γ Z

Z ZL

L o

τ

P P Pr t i+ =

PZ Z

Pro o

i= = =− +v v12

2 12

2

2 2Γ Γ

PZ Z Z

ZZt

L L o

o

L= = =+ + + ⋅v v v2

2 212

2 12

2 2 2τ τ

0 2£ £t

Γ τ

v v12 2 2

2124+ +-

+FHG

IKJ +

+FHG

IKJ

LNMM

OQPP

=Z

Z Z

Z Z

Z

Z Z

Z

Z Zo

L o

L o

L

L o

o

L ob gZ Z Z Z Z Z

Z ZL o L o L o

L o

2 2

2

2 41

+ - ++

=b g1 1= Q.E.D.

Zin = 300Ωz = 0

A

50Ω 300Ω ZL = 300Ω

50Ω 300Ω

A z = 0a f

τ = + =1 1714Γ .

Γ = −+ = −

+ =Z ZZ Z

in o

in o

300 50300 50

0 714.

Γ 2 0 51= .300Ω

1 0 51 0 49− =. .

General Transmission Line

Now let’s return to a general case and find the total voltage and current on the line as a function of distance:

The real fields, of course, vary sinusoidally in time. and are the com-plex phasor amplitudes. Since, for a complex voltage , the maxi-mum value (or sinusoidal amplitude) is , and the sinusoidal amplitude of

along the line is:

Remember that may be positive or negative.

The minimum value of this is (if is positive):

For a general :

Similarly:

The location of consecutive minima and maxima are apart. At the load, we have either a maximum or a minimum (max. for positive). The shape of the function vs. z is not a pure sinusoid. The minima are not infinitely sharp (as they are for a short circuit).

.

v1 1 121= + = ++

-+

+ -V e V e V e ej z j zo

j z j zb b b bG Gc h

e j tω e j tωvariation understood! We have dropped

(Let V Vo1+ = )

i11 1 21= - = -+ - + + -V

Ze

V

Ze

V

Ze e

o

j z

o

j z o

o

j z j zb b b bG Gc h

v1 i1

v = Ae j tω

AV1

v121 1 2 2= + = + +-V e e V z j zo

j z j zo

b b b bG G cos sina f= + + = + +V z z V zo o1 2 2 1 2 2

2 2 2 2G G G Gcos sin cosb b ba fG(for real)

Γ

24β πλz z=

Γ

v121 2 1

min= + - = -V Vo oG G Ga f

Γ v1 1min

= -Vo Gb gv1

21 2 1max

= + + = +V Vo oG G Gb gλ /2

Γ

Load

z

An A.C. voltmeter, of course, does not measure the time variation, but only the amplitude (top solid line). (Actually the RMS value)This is called a partial standing wave. An important measurable quantity is the Standing Wave Ratio (SWR) = S.

S can range from 1 to • .

Example: For the previous example,

We note also that for positive ( , real),

and for negative ( , real),

This also agrees with our example:

If a transmission line is connected to an unknown load, the SWR can be measured and, from this, the load value found. But to fully understand how, we must consider a more general load.

Lossless Transmission Line Terminated in a General Complex Load ZL

Most practical circuits/devices (whether the device be a transmitting antenna, a TV, a receiver, a microwave amplifier, an oscilloscope, etc.) have complex input impedance. Thus, if such a device is fed by a transmission line, the transmission line sees a complex load. Without something special being done, the line is not matched to the load. There is a reflected wave.

SV= =

+−

max

minV11

ΓΓ

S = =1 7140 286

6..

Γ Z ZL o> ZL

SZ

Z

Z

Z

Z Z

Z ZZ Z

Z Z

L o

L o

L o

L o

L

o

L

o

=+

−= =

−+−+

1

1

22

Γ Z ZL o< ZL

SZ

Z

Z

Z

Z Z

Z ZZ Z

Z Z

o L

o L

o L

o L

o

L

o

L

=+

−= =

−+−+

1

1

22

6 300 50= /

Assume for simplicity that and that the load is at . There-fore, on the transmission line, . Then,

,

where l is the distance measured backwards from the load.

The input impedance at a point l distance back from the load is:

Substituting the value of and rearranging, we get the alternate forms:

When , we get .

Now returning to the voltage and current:

This is the same expression as before ( ), except now is complex . It is easiest to see how this changes the standing wave by con-

sidering phasor diagrams. The amplitude of the quantity in square brackets determines the amplitude of the standing wave, since .

Γ = −+

Z Z

Z ZL o

L o

vTj z j zV e V e= ++

−+

+β βΓ

iTo

j z

o

j zV

Ze

V

Ze= −+ − + +β βΓ

(complex, in general)

v+ = ∠ °Vo 0 Z = 0z l= −

iTo

o

j l o

o

j lV

Ze

V

Ze= − −β βΓ

vT oj l

oj lV e V e= + −β βΓ

Z Ze e

e ein o

j l j l

j l j l= +−

−

−

β β

β βΓΓ

= +−

−

−Ze

eo

j l

j l

11

2

2

ΓΓ

β

β

Γ

Z l ZZ l j Z l

Z l j Z lZ

Z j Z l

Z j Z lin oL o

o Lo

L o

o L

- = ++

= ++

a f cos sin

cos sin

tan

tan

b bb b

bb

Z ZL o= Z Zin o=

v = + −V e eoj l j lβ β1 2Γ

i = − −V

Ze eo

o

j l j lβ β1 2Γ

Z Ze

ein o

j l

j l= +−

−

−11

2

2

ΓΓ

β

β

another form

z l= − ΓΓ = ∠Γ θ r

e j lβ = 1

Definition of Generalized Reflection Coefficient

In my notation , reflection coefficient at the load (complex quantity in general) is equivalent to the book’s .

Note:

1) is real if load is real and line is lossless. is complex if load is com-plex.

2) is always complex, except at certain points. These points have spe-cial significance!

3) , that is, moving about on the line only changes the phase angle of .

It is obvious from this phasor diagram that the cases of real load and com-plex load are not fundamentally different. or for a real load, but can be anything for a complex load. The phasor rotates as l increases, making a complete cycle for ; that is, , or .

is still and , where . Obviously, the shape of the standing wave is also the same as for a real load, except the maximum or minimum no longer occurs at the load.

Location of maxima and minima: Maximum occurs at:

z l= -G GlV lV l

V eV e

ej l

j lj l( ) = ( )

( ) = ( )( )

= ( )-

+

--

++

-00

0 2b

bb

G G= 0a f ~Γr

Γ Γ

G la f

G Gla f ∫G la f

1− Γ

1+ Γ

− Γ

Γ

At

Im

Max.

Vmax.Vmin.

l = 0

Re

(moving back,away from load)

θ r = 0 180°

2 2β πl = 4 2π λ πl / = l = λ /2

Vmax 1+ Γ Vmin = −1 Γ Γ Γ=

θ β π πr l− = ± ±2 0 2 4max , , ,θ β π πr l− = ± ±2 3min , , etc.Similarly,

etc.

Example: An RF signal is sent on a parallel wire transmission line, which has , to a receiver. A VSWR of is measured on the line. The distance between two minima is and the nearest minimum to the load minimum is away. Find the frequency of the signal and the The’vnenin equivalent circuit of the receiver input.

Solution:

300Ωε r = 5 S = 3

0 67. .m0 45. .m

0 45. .m

S = 3Zo = 300Ω

ZL

0 67. .m

Γ = −+ = −

+Z ZZ Z

ZZ

L o

L o

l

L

300300

S =+−

11

ΓΓ

∴ = −+ = =Γ S

S11

24

0 5.

Γ = ∠ °= +0 5 61 2 0 24 0 44. . . .j

λ m / .2 0 67= λ m = 1 34.

λ λ ε= = =⋅m r m1 34 5 3. .

fc

MHz= = × = =λ3 10

310 100

88

β πλ

π π= = = −2 21 34

1 49 1

mm

..

− + =2β θ πl rmin ∴ + =2 98. π π θlm r

2 34 0 34 61 2. . .π θ π= = = °r

ZL 1 300 1- = +G Ga f a fZ j jL 0 76 0 44 300 1 24 0 44. . . .- = +a f a fZ

jjL = +

− = ∠ °∠− ° = ∠ °300

1 24 0 440 76 0 44

3001 32 19 50 88 30

450 49 5. .. .

. ..

.

= +292 342j Ω = +R j Lω L h=×

= × −3422 10

0 54 1086

π .

R = 292Ω L h= 0 54. µ ANS.Load is

Looking at the definition of and at the rotating phasor diagram of the pre-vious page, we also see that:

At the location of a voltage maximum:

At the location of a voltage minimum:

This is an important observation. It will be used soon for matching.

Matching a transmission line to an unmatched load

Reflections on a transmission line are undesirable for a number of reasons. For example, digital signals from PCM reflected pulses can bounce back again from the sending end if that end is also unmatched. Then false sig-nals can be detected.

Q: How do we match an unmatched load?A: There are several methods. We now study the first one.

I. Quarter Wave Matching ( “Transformer”)

Suppose we have a transmission line terminated in a real load

If the length of this line is exactly , then , since

Zin

Z Z Z Sin o o=+−

=11

ΓΓ

Z ZZSin o

o= -+ =1

1GG

Z Z Sin o=

ZZSin

o=

at voltage maximum

at voltage minimum

λ /4

Z RL L∫b gZin ⇒

ZL

Z02Z02 is intrinsic impedance.

l = λ /4 ZZZin

L

= 022

cosβl = 0

sin sin sinβ πλ

λ πl = ⋅ = =2

4 21

and

This then offers a method to match this load to any lossless transmission line , intrinsic impendance to be matched to . real

We insert a transmission line of , length ( ), and having .

Then, at ( ):

!

Since , we are matched. No reflection, since at point A, we could replace the whole combination by .

How is this possible physically? We won’t go into a detailed analysis of what really happens to the right of point A. But briefly, we do have (in actual physical fact) reflection of waves at point A and at ZL. The two reflections, however, cancel each other out on the line .

Q: This obviously works if the load is real, since is also real, resulting in real . But if the load is complex, then what to do?

A: Recall that the input impedance at a voltage minimum or maximum is real and . Thus, these are the steps for matching a complex load

to a line .

1) Add a section of line to the load. The length should be lmax or lmin of the standing wave that results. Say, we use :

This is equivalent to a load of .

Z01 Z0ZL ZL RLb g

Z02

λ4

Z01 ZL1

A2

λ /4 2 Z Z ZL02 01= ⋅A

ZZ

Z

Z Z

ZZin

L

L

L

= = =022

01

01

Z Zin A = 01

Zin A

Z 01Z Zin = 01

1

Z01

Z02

Z So Z So /ZL Z01

Z01

lmax

lmax

ZL

Z S ZL = 01

2) Now use , matching on this equivalent load. For example, insert a line:

The impedance at point B then is:

Thus we are matched. No reflection going back on line 1. Same com-ment applies as before. Actually, there are reflections at points A and B, but they are 180∞ out of phase and cancel out.

The 50W - 300W “transformer” supplied for TV and video components is a lumped circuit simulation of a transformer!

Important Note: means

The matching is only valid at one frequency! Since , if f is

changed, the wavelength is changed and for the same l.

Example: (a) Match a 50 coaxial cable to a 300W real input impedance device at a frequency of 180 MHz. Assume all transmission lines used have . (b) Find the reflection coefficient of the matched circuit at 90 MHz.

λ /4

λ /4

λ /4Z Z Z S Z Z S ZL02 01 01 01 01= = = , length

Z01Z01

lmax

AB Zo

ZS ZZ

S ZSZ

Zin BL

= = =012

1

012

0101

λ /4

l = λ /4 β πλ

λ πl = =⋅2

4 2

β πν

πλ

= =2 2f

β πl ≠ /2

ε r = 6

(a) At 180 MHz:

We cannot readily buy a 122.5W transmission line. However, any transmis-sion line at these R.F. frequencies can be simulated by a proper lumped cir-cuit network. (In this class, we won’t study how.)(b) If the frequency is changed to 90 mHz:

We have:

Reflection coefficient at A is:

Fraction of power reflected back at A is .

length needed = =0 684

017.

. . .m ans

λ o

c

fm= = ×

×=3 10

18 10167

8

8.. .

λ λ λm

o m= =6

0 68. .

Z ans02 50 300 122 5= =⋅ . .ΩT.L. impedance needed

on the T.L.,

λ4 017= . .m

50Ω 122 5. Ω ZL = 300Ω

λ m = × =0 68 2 1 36. . β π πl = × =2

1 360 17

4..

l = λ8

β πl =

4tan βl = 1

Zj

jin A = + ×+ × = ∠ °

∠ −122 5300 122 5 1122 5 300 1

122 5324 22 2

324 90 22.

..

..

= ∠ − °= −122 5 45 6 85 8 87 5. . . .j

300Ω122 5. Ω50Ω

Γ = − −− + = −

−85 8 87 5 5085 8 87 5 50

33 5 87 5135 8 87 5

. .

. .. .. .

jj

jj

= ∠− °∠− ° = ∠− °937 69

161 5 32 80 58 36 2

.. .

. . .ans

Γ 2 0 34= .