The Binomial Multi- Section Transformer

The Binomial Multi- Section Transformer

Recall that a multi-section matching network can be described using the theory of small reflections as:

( )

2 0

j T j T j N T

in N

N j n T

n n

e e e

e

ω ω ω

ω

ω ^{− − −}

−

=

Γ = Γ + Γ + Γ + + Γ

=

∑

"

where:

propagation time through 1 section

p

T v ⁼

 A

Note that for a multi-section transformer, we have N degrees of design freedom, corresponding to the N characteristic impedance values Z_n.

Q: What should the values of ^Γ_n (i.e., Z_n) be?

A: We need to define N independent design equations, which we can then use to solve for the N values of characteristic impedance Z_n.

First, we start with a single design frequency ω₀, where we wish to achieve a perfect match:

(

)

in ω ω

Γ = =

That’s just one design equation: we need N -1 more!

These addition equations can be selected using many

criteria—one such criterion is to make the function Γ_in

( )

To accomplish this, we first consider the Binomial Function:

( )

(

)

Γ = +

This function has the desirable properties that:

( ) ( )

( )

2 1

1 1 0

j N

N

A e

A

θ π ⁻ π

Γ = = +

= −

= and that:

( )

2 n 0

n

d

d θ π

θ

θ ₌

Γ = for n ⁼1 2 3, , , ,N^{" −}1

In other words, this Binomial Function is maximally flat at the point θ π= 2, where it has a value of ^Γ

(

)

Q: So? What does this have to do with our multi-section matching network?

A: Let’s expand (multiply out the N identical product terms) of the Binomial Function:

( ) ( )

( )

2

2 4 6 2

0 1 2 3

1 ^{j N}

j j j j N

N N N N N

N

A e

A C C e C e C e C e

θ

θ θ θ θ

θ ⁻

− − − −

Γ = +

= + + + +"+

where:

( )

nN N !

C ^ N n ! n !⁻

Compare this to an N-section transformer function:

( )

in ω e^{− ω} e^{− ω} Ne^{− ω}

Γ = Γ + Γ + Γ +"+ Γ

and it is obvious the two functions have identical forms, provided that:

n A CnN

Γ = and ωT =θ

Moreover, we find that this function is very desirable from the standpoint of the a matching network. Recall that

( )

Γ = at θ π= 2--a perfect match!

Additionally, the function is maximally flat at θ π= 2,

therefore Γ

( )

Q: But how does θ π= 2 relate to frequency ω?

A: Remember that ωT =θ , so the value θ π= 2 corresponds to the frequency:

0 1

2 2

vp

T

π π

ω = =

A

This frequency (ω₀) is therefore our design frequency—the frequency where we have a perfect match.

Note that the length ^A has an interesting relationship with this frequency:

0 0

0 0

1

2 2 2 2 4

vp π π λ π λ

ω β π

= = = =

A

In other words, a Binomial Multi-section matching network will have a perfect match at the frequency where the section lengths ^A are a quarter wavelength!

Thus, we have our first design rule:

Set section lengths ^A so that they are a quarter- wavelength (λ₀ 4) at the design frequency ω₀.

Q: I see! And then we select all the values Zn such that

n A CnN

Γ = . But wait! What is the value of A ??

A: We can determine this value by evaluating a boundary condition!

Specifically, we can easily determine the value of Γ

( )

ω = .

Note as ω approaches zero, the electrical length βA of each section will likewise approach zero. Thus, the input impedance Zin will simply be equal to RL as ω → 0.

As a result, the input reflection coefficient Γ

(

)

( ) ( )

( )

0 0

0 0

0

in in L L

Z Z

Z Z

R Z R Z ω ω

ω

= −

Γ = =

= +

= − +

However, we likewise know that:

( ) (

)

( )

0 1 2 0

1 1 2

j N

N

N

A e

A A

Γ = + −

= +

=

RL

A

Z0 ^{Z1 Z2}

"

A A

Zin

Equating the two expressions:

( )

0

0 2^{N L}

L

A R Z

R Z

Γ = = −

+ And therefore:

⁰

0

2 ^{N L}

L

A R Z

R Z

− −

= + (A can be negative!)

We now have a form for the marginal reflection coefficients Γn:

( )

0 0

2 !

! !

N N L

n n

L

R Z N

AC ⁻ R Z N n n

Γ = = −

+ −

Of course, we also know that these marginal reflection coefficients are:

1 1

n n

n n n

Z Z

Z ⁺₊ Z Γ = −

+

Now, we know that the values of Z_n+₁ and Z_n are typically very close, such that Z_n+₁ −Z_n is small. It turns out for this case that we can use a helpful approximation for the marginal reflection coefficient:

1 1

1

1

n n 2 n

n n n n

Z Z ln Z

Z ⁺₊ Z Z⁺

⎛ ⎞

Γ = −+ ≈ ⎜⎝ ⎟⎠ (for ^Γ_n small)

Therefore we can conclude:

0 1

0

1 2

2 ^{n N L N}

n n

n L

R Z

ln Z C

Z^{+ −} R Z

⎛ ⎞ −

Γ = ⎜⎝ ⎟⎠ = +

Solving for Z_n+1, we find:

1 0

1

0

2 ^{N L N}

n n n

L

Z Z exp R Z C

R Z

− + +

⎡ − ⎤

= ⎢⎣ + ⎥⎦

We can further simplify this with yet another approximation:

1

0

2 ^{N L N}

n n R n

Z Z exp ln C

Z

− +

⎡ ⎛ ⎞ ⎤

≈ ⎢ ⎜ ⎟ ⎥

⎝ ⎠

⎣ ⎦

This is our second design rule. Note it is an iterative rule—

we determine Z1 from Z0, Z2 from Z1, and so forth.

The result is a maximally flat, Binomial reflection coefficient function ^Γ

( )

Note that as we increase the number of sections, the matching bandwidth increases.

Q: Can we determine the value of this bandwidth?

A: Sure! But we first must define what we mean by bandwidth.

As we move from the design (perfect match) frequency f0 the value ^Γ

( )

Figure 5.15 (p. 250)

Reflection coefficient magnitude versus frequency for multisection binomial matching transformers of Example 5.6 ZL = 50Ω and Z0 = 100Ω.

unacceptably high value (Γ_m, say). At that point, we no longer consider the device to be matched.

Note there are two values of frequency fm —one value less than design frequency f0, and one value greater than design frequency f0. These two values define the bandwidth ^∆f of the matching network:

( ) ( )

2 1 2 0 1 2 2 0

m m m m

f f f f f f f

∆ = − = − = −

Q: So what is the numerical value of ^Γ_m? A: I don’t know—it’s up to you to decide!

Every engineer must determine what they consider to be an acceptable match (i.e., decide Γ_m). This decision depends on the application involved, and the specifications of the overall microwave system being designed.

However, we typically set Γ_m to be 0.2 or less.

f0 fm2

fm1

Γm

( )

Γ

f f

∆

Q: OK, after we have selected ^Γ_m, can we determine the two frequencies fm ?

A: Sure! We just have to do a little algebra.

We start by rewriting the Binomial function:

( ) ( )

( )

( )

( )

1 2

2

j N

jN j j N

jN j j N

jN N

A e

Ae e e

Ae e e

Ae cos

θ

θ θ θ

θ θ θ

θ

θ

θ

−

− + −

− + −

−

Γ = +

= +

= +

=

Now, we take the magnitude of this function:

( )

jN N N

N N

A e cos A cos

θ θ θ

θ Γ = −

=

Now, we define the values θ where Γ

( )

( )

2

m m

N N

A cos m

θ θ θ Γ = Γ =

=

We can now solve for θ_m (in radians!) in terms of Γ_m:

1

1 1 1

2

m N

m cos

θ = ^{− ⎡⎢}⎢⎣ ^⎛⎜⎝^ΓA ^⎞⎟⎠ ^⎤⎥⎥⎦

1

2 1 1

2

m N

m cos

θ = ^{− ⎡⎢}⎢⎣− ⎜^⎛⎝^ΓA ^⎞⎟⎠ ^⎤⎥⎥⎦

Note that there are two solutions to the above equation (one less that 2π and one greater than 2π )!

Now, we can convert the values of θ_m into specific frequencies.

Recall thatωT =θ , therefore:

1 p

m m m

v ω =T θ = θ

A

But recall also that A = λ₀ 4, where λ₀ is the wavelength at the design frequency f₀ (not f_m!), and where λ₀ =v f_{p 0} . Thus we can conclude:

( )

4 4

p p

m m m m

v v

f

ω θ θ θ

= = λ =

A

or: 1

( )

( )

2 2

p m m

m m

v f f

f θ θ

π θ π π

= = =

A

where θ_m is expressed in radians. Therefore:

1 0 1

1 2 1

2

m N

m f

f cos

A π

− ⎡⎢ ⎛Γ ⎞ ⎤⎥

= ⎢⎣+ ⎜⎝ ⎟⎠ ⎥⎦

1 0 1

2 2 1

2

m N

m f

f cos

A π

− ⎡⎢ ⎛Γ ⎞ ⎤⎥

= ⎢⎣− ⎜⎝ ⎟⎠ ⎥⎦

Thus, the bandwidth of the binomial matching network can be determined as:

(

)

1 0 1

0

2

4 1

2 2

m

m N

f f f

f f cos

A π

−

∆ = −

⎡ ⎛Γ ⎞ ⎤

⎢ ⎥

= − ⎢⎣+ ⎜⎝ ⎟⎠ ⎥⎦

Note that this equation can be used to determine the bandwidth of a binomial matching network, given ^Γ_m and number of sections N.

However, it can likewise be used to determine the number of sections N required to meet a specific bandwidth

requirement!

Finally, we can list the design steps for a binomial matching network:

1. Determine the value N required to meet the bandwidth (^∆f and ^Γ_m) requirements.

2. Determine the impedance of each section using the iterative approximation:

1

0

2 ^{N L N}

n n R n

Z Z exp ln C

Z

− +

⎡ ⎛ ⎞ ⎤

≈ ⎢ ⎜ ⎟ ⎥

⎝ ⎠

⎣ ⎦

3. Determine section length A = λ₀ 4 for frequency f0.