ECE 5260 — Microwave Engineering I

University of Virginia

Spring 2018

Homework 6 — Solutions

1. Slug Matching

Let’s consider the output first. The impedance looking into the slug tuner circuit from the left(see the figure below) must be Z0 for a match. From the properties of the λ/4 transformer,we have

Zin =Z2M

Z0= 12.5Ω

because ZM = 25Ω and Z0 = 50Ω. Thus, the magnitude of the reflection coefficient ontransmission line θ2 must be,

|Γin| =∣∣∣Zin − Z0

Zin + Z0

∣∣∣ = ∣∣∣Z2m − Z2

0

Z2M + Z2

0

∣∣∣ = 3

5,

in order to match at the output. Note that the magnitude of Γin does not change on theline ⇒ only the phase. Thus, the strategy is to use the first 50 Ω line (θ1) and first λ/4transformer to produce an input impedance Z ′ with reflection coefficient magnitude of 3/5.The second transmission line (θ2) is then used to move the impedance to the real axis (tonormalized value of 0.25). The final λ/4 transformer then transforms this load to 50Ω.

Consider the first two transmission lines (again, refer to the figure below):

Z =Z2M

Z ′ = Z01 + Γ

1− Γand Z ′ =

Z2M

Z0

1− Γ

1 + Γ

but

Γ′ =Z ′ − Z0

Z ′ + Z0=

Z2M (1− Γ)− Z2

0 (1 + Γ)

Z2M (1− Γ) + Z2

0 (1 + Γ)=

γ − Γ

1− γΓ

where,

γ ≡ Z2M − Z2

0

Z2M + Z2

0

.

Now, we want

|Γ′|2 =(35

)2=( γ − Γ

1− γΓ

)( γ − Γ

1− γΓ

)∗≡ ρ2 =

γ2 − γΓ− γΓ∗ + |Γ|2

1− γΓ− γΓ∗ + γ2|Γ|2

|Γ|2 − Γγ(1− ρ2)

1− γ2ρ2− Γ∗ γ(1− ρ2)

1− γ2ρ2=

ρ2 − γ2

1− ρ2γ2

completing the square gives us the equation of a circle:

∣∣∣Γ− γ(1− ρ2)

1− γ2ρ2

∣∣∣2 =ρ2(1− γ2)2

(1− γ2ρ2)2(A)

with center, Γc of,

Γc =γ(1− ρ2)

1− γ2ρ2

and radius, r of

r =ρ(1− γ2)

1− γ2ρ2

Expression (A) gives the contour on which Γ (The reflection coefficient seen on line θ1 atthe input to the first slug) must lie in order for the slug tuners to produce a match. Withρ = 3/5 and γ = −3/5, we have Γc = −0.4412 and r = 0.4412. This circle is plotted inthe figure above. The magnitude of ΓL must lie somewhere on this circle after the phase θ1is appropriately adjusted. Thus, the allowed region of the Smith Chart is found by sweepingthis circle about the Smith Chart through all possible angles (again refer to the figure above).

2. To determine how many sections we need for the uniform impedance transformer, Let’s plotthe magnitude of Γ for different numbers of sections:

|Γ(βl)| =∣∣∣RL − Z0

RL + Z0

∣∣∣ ∣∣∣ sin[(N + 1)βl]

(N + 1) sin βl

∣∣∣ (1)

This is shown below for N = 1, N + 3, and N = 4 along with |Γ| = 0.125. From the plot,we see that

we require 4 sections to meet the given specification

To determine the required transmission line impedances, we will use the “self-consistent”

method:

γn =1

N + 1

∣∣∣ZL − Z0

ZL + Z0

∣∣∣ = 2 ln(ZL

Z0

)⇒ ln

(Zn+1

Zn

)=

ln(3)

N + 1= 0.2197

Thus,

Z1 =5√3Z0 = 62.3Ω

Z2 =5√3Z1 = 77.63Ω

Z3 =5√3Z2 = 96.73Ω

Z4 =5√3Z3 = 120.53Ω

Below, we show a comparison of the analytical formula for a 4-section uniformly distributed

transformer (equation 1) with a Puff simulation using the above tline sections. The results

match quite well.

3. Let’s work this two different ways. First, we will use the iterative method demonstrated in

class. With N = 4, Z0 = 50Ω and ZL = 10Ω, we have

2NγN =RL − Z0

RL + Z0= −2

3⇒ γ4 = − 1

24

Thus,

γ4 =RL − Z4

RL + Z4= − 1

24⇒ Z4 = RL

1− γ41 + γ4

= 10.9Ω

γ3 =4!

3!(4− 3)!γ4 = 4γ4 =

Z4 − Z3

Z4 + Z3= −1

6⇒ Z3 = Z4

1− γ31 + γ3

= 15.3Ω

γ2 =4!

2!(4− 2)!γ4 = 6γ4 =

Z3 − Z2

Z3 + Z2= −1

4⇒ Z2 = Z3

1− γ21 + γ2

= 25.4Ω

γ1 =4!

1!(4− 1)!γ4 = 4γ4 =

Z2 − Z1

Z2 + Z1= −1

6⇒ Z1 = Z2

1− γ11 + γ1

= 35.6Ω

Alternatively, we can use the “approximate” theory presented in the notes

2γn ≃ ln(Zn+1

Zn

)= 2−N

(N

n

)ln(ZL

Z0

)Applying this technique, we find:

ln(Z1

Z0

)= 2−4

(4

0

)ln(1/5) = −0.10059 → Z1 = 45.22Ω

ln(Z2

Z1

)= 2−4

(4

1

)ln(1/5) = −0.40236 → Z2 = 30.24Ω

ln(Z3

Z2

)= 2−4

(4

2

)ln(1/5) = −0.60354 → Z3 = 16.54Ω

ln(Z4

Z3

)= 2−4

(4

3

)ln(1/5) = −0.40236 → Z4 = 11.06Ω

Notice that the calculated values for Z3 and Z4 are close, but the values for Z1 and Z2

are significantly different using the two methods. What is going on .... why do the two

methods result in such different impedances? The answer is that the impedances we are

trying to match are sufficiently different that, with only four sections, the small reflection

approximation is violated. by starting at the 10Ω resistor and iteratively working our way

back to the 50Ω source impedance, the errors compound and become non-negligible. Notice

that if we calculate Z0 using the iterative technique, we get

γ0 =4!

0!(4− 0)!γ4 = γ4 =

Z1 − Z0

Z1 + Z0= − 1

24⇒ Z0 = Z1

1− γ01 + γ0

= 38.7Ω

which is quite different than what is expected (50Ω). As a result, when the partial reflections

are large enough that the small reflection approximation isn’t valid, this iterative method is

not “self-consistent.” The “approximate” technique, although not exact, has the advantage

of ensuring self-consistency.

The binomial transformer response for the iterative design method and self-consistent design

method are plotted in figure 4. Also shown is the response using the analytic formula for a

4-section binomial transformer given in class.

4. For a tapered matching section that has

d(ln(Z/Z0)

)dz

= A sin(πzL

)

with Z(0) = Z0 and Z(L) = ZL, we have

ln(Z(z)/Z0)∣∣∣Z=ZL

Z=Z0

= A

z=L∫z=0

sin(πzL

)dz ⇒ ln(ZL/Z0) = −A

(Lπ

)cos(πzL

)∣∣∣L0=

2AL

π

A =π

2Lln(ZL/Z0)

This gives us a reflection coefficient of

Γ(βL) =1

2

∫ L

0

e−j2βz d

dz

[ln(Z(z)

Z0

)]dz =

π

4Lln(ZL/Z0)

∫ L

0

e−j2βz sin(πzL

)dz

=π

4Lln(ZL/Z0)

[−j2βe−j2βz sin

(πzL

)(π/L)2 − (2β)2

−(π/L)e−j2βz cos

(πzL

)(π/L)2 − (2β)2

]L0

=

(π

2L

)2

ln(ZL/Z0)

[e−j2βL + 1

(π/L)2 − (2β)2

]=

(π

2L

)2

ln(ZL/Z0) e−jβL

[ejβL + e−jβL

(π/L)2 − (2β)2

], or

Γ(βL) =π2

2ln(ZL/Z0) e

−jβL

[cos(βL)

π2 − (2βL)2

]

Below is a plot of |Γ(βL)|/(π2

2 | ln(ZL/Z0)|) versus βL:

5. For an exponentially tapered tline, Z(z) = Z0 exp(γz), where Z0 = 50Ω and

Z(L) = Z0 exp(γL) = 100Ω ⇒ γL = ln(2)

for a line matching 100Ω to 50Ω. Using our expression for the reflection coefficient of a

tapered line,

Γ(βL) =1

2

∫ L

0

e−j2βz d

dz

[ln(Z(z)

Z0

)]dz

we have,

Γ(βL) =γ

2

∫ L

0

exp(−j2βz)dz =γ

2

1− exp(−j2βL)

j2β=

γL

2

sin(βL)

βLe−jβL

Taking the magnitude and substituting for γL, we have

|Γ(βL)| = ln(2)

2

∣∣∣ sin(βL)βL

∣∣∣

This relation is plotted in the figure below. From this plot, we see that the required (electrical)

length of a fractional bandwidth of 100% (f0 = ∆f) is, βL = 3.4π radians.

For a Chebyshev matching circuit, the reflection coefficient is given by,

Γ(θ) = Γme−jNθ TN (sec θm cos θ)

where Γm is the maximum reflection coefficient allowed in the matching band (Γm = 0.05),

N is the number of matching sections, TN is the N th-order Chebyshev polynomial, θ = βℓ,

and sec θm is given by

sec θm = cosh

[1

Ncosh−1

(1

Γm

∣∣∣∣∣ZL − Z0

ZL + Z0

∣∣∣∣∣)]

Using these relations, we find the N = 3 sections are required to yield a Chebyshev trans-

former with 100% bandwidth at the center of the matching band. The plot below shows the

magnitude of Γ for the 3-section matching circuit: