4/15/2010
The Binomial Multisection Matching Transformer.doc
1/17
The Binomial MultiSection Transformer
Recall that a multi-section matching network can be
described using the theory of small reflections as:
Γin (ω ) = Γ 0 + Γ1e − j 2ωT + Γ2e − j 4ωT + " + ΓN e − j 2N ωT
N
= ∑ Γn e − j 2n ωT
n =0
where:
T A
vp
= propagation time through 1 section
Note that for a multi-section transformer, we have N
degrees of design freedom, corresponding to the N
characteristic impedance values Zn .
Q: What should the values of Γn (i.e., Zn ) be?
A: We need to define N independent design equations, which
we can then use to solve for the N values of characteristic
impedance Zn .
First, we start with a single design frequency ω0 , where we
wish to achieve a perfect match:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
2/17
Γin (ω = ω0 ) = 0
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 Γin (ω )
maximally flat at the point ω = ω0 .
To accomplish this, we first consider the Binomial Function:
(
)
N
Γ (θ ) = A 1 + e − j 2θ
This function has the desirable properties that:
Γ (θ = π 2 ) = A (1 + e − j π )
N
N
= A (1 − 1 )
=0
and that:
d n Γ (θ )
= 0 for n = 1, 2, 3, " , N − 1
d θ n θ =π
2
In other words, this Binomial Function is maximally flat at the
point θ = π 2 , where it has a value of Γ (θ = π 2 ) = 0 .
Q: So? What does this have to do with our multi-section
matching network?
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
3/17
A: Let’s expand (multiply out the N identical product terms)
of the Binomial Function:
(
= A (C
Γ (θ ) = A 1 + e − j 2θ
N
0
)
N
+ C1N e − j 2θ + C 2N e − j 4θ + C 3N e − j 6θ + " + C NN e − j 2N θ
where:
CnN )
N!
(N − n ) ! n !
Compare this to an N-section transformer function:
Γin (ω ) = Γ 0 + Γ1e − j 2ωT + Γ2e − j 4ωT + " + ΓN e − j 2N ωT
and it is obvious the two functions have identical forms,
provided that:
Γn = A CnN
and
ωT = θ
Moreover, we find that this function is very desirable from
the standpoint of the a matching network. Recall that
Γ (θ ) = 0 at θ = π 2 --a perfect match!
Additionally, the function is maximally flat at θ = π 2 ,
therefore Γ (θ ) ≈ 0 over a wide range around θ = π 2 --a wide
bandwidth!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
4/17
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:
ω0 =
1 π vp π
=
T 2 A 2
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:
A=
vp π
1 π λ0 π λ0
=
=
=
ω0 2 β 0 2 2π 2 4
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 quarterwavelength ( λ0 4 ) at the design frequency ω0 .
Q: I see! And then we select all the values Zn such that
Γn = A CnN . But wait! What is the value of A ??
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
5/17
A: We can determine this value by evaluating a boundary
condition!
Specifically, we can easily determine the value of Γ (ω ) at
ω = 0.
Z0
Zin
Z1
Z2
A
A
"
ZN
RL
A
Note as ω approaches zero, the electrical length β A 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 (ω = 0 ) must
be:
Zin (ω = 0 ) − Z 0
Zin (ω = 0 ) + Z 0
R − Z0
= L
RL + Z 0
Γ (ω = 0 ) =
However, we likewise know that:
(
Γ ( 0 ) = A 1 + e − j 2( 0 )
)
N
N
= A (1 + 1 )
= A 2N
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
6/17
Equating the two expressions:
Γ ( 0 ) = A 2N =
And therefore:
A = 2− N
RL − Z 0
RL + Z 0
RL − Z 0
RL + Z 0
(A can be negative!)
We now have a form to calculate the required marginal
reflection coefficients Γn :
Γn = ACnN =
AN!
(N − n ) !n !
Of course, we also know that these marginal reflection
coefficients are physically related to the characteristic
impedances of each section as:
Γn =
Z n +1 − Z n
Z n +1 + Z n
Equating the two and solving, we find that that the section
characteristic impedances must satisfy:
Z n +1 = Z n
Jim Stiles
1 + Γn
1 − Γn
= Zn
1 + A CnN
1 − A CnN
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
7/17
Note this is an iterative result—we determine Z1 from Z0, Z2
from Z1, and so forth.
Q: This result appears to be our second design equation.
Is there some reason why you didn’t draw a big blue box
around it?
A: Alas, there is a big problem with this result.
Note that there are N+1 coefficients Γn (i.e., n ∈ { 0,1, ", N } )
in the Binomial series, yet there are only N design degrees of
freedom (i.e., there are only N transmission line sections!).
Thus, our design is a bit over constrained, a result that
manifests itself the finally marginal reflection coefficient
ΓN .
Note from the iterative solution above, the last transmission
line impedance Z N is selected to satisfy the mathematical
requirement of the penultimate reflection coefficient ΓN −1 :
ΓN −1 =
Z N − Z N −1
= A C NN−1
Z N + Z N −1
Thus the last impedance must be:
Z N = Z N −1
Jim Stiles
1 + AC NN−1
1 − AC NN−1
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
8/17
But there is one more mathematical requirement! The last
marginal reflection coefficient must likewise satisfy:
ΓN = A C NN = 2−N
RL − Z 0
RL + Z 0
where we have used the fact that C NN = 1 .
But, we just selected Z N to satisfy the requirement for
ΓN −1 ,—we have no physical design parameter to satisfy this
last mathematical requirement!
As a result, we find to our great consternation that the last
requirement is not satisfied:
ΓN =
RL − Z N
≠ A C NN !!!!!!
RL + Z N
Q: Yikes! Does this mean that the resulting matching network
will not have the desired Binomial frequency response?
A: That’s exactly what it means!
Q: You big #%@#$%&!!!! Why did you waste all my time by
discussing an over-constrained design problem that can’t be
built?
A: Relax; there is a solution to our dilemma—albeit an
approximate one.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
9/17
You undoubtedly have previously used the approximation:
y −x 1 ⎛y ⎞
≈ ln ⎜ ⎟
y +x 2 ⎝x ⎠
An approximation that is especially accurate when y − x is
small (i.e., when
y
x
1 ).
1 ⎛y ⎞
ln ⎜ ⎟
2 ⎝x ⎠
1.0
y −x
y +x
0.5
y
2
4
6
8
x
10
- 0.5
- 1.0
Now, we know that the values of Zn +1 and Zn in a multi-section
matching network are typically very close, such that
Zn +1 − Zn is small. Thus, we use the approximation:
Γn =
Jim Stiles
Z n +1 − Z n 1 ⎛ Z n + 1 ⎞
≈ ln ⎜
⎟
Z n +1 + Z n 2 ⎝ Z n ⎠
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
10/17
Likewise, we can also apply this approximation (although not as
accurately) to the value of A :
A = 2− N
⎛R ⎞
RL − Z 0
− ( N +1 )
ln ⎜ L ⎟
≈2
RL + Z 0
⎝ Z0 ⎠
So, let’s start over, only this time we’ll use these
approximations. First, determine A :
⎛R ⎞
− N +1
A ≈ 2 ( ) ln ⎜ L ⎟
⎝ Z0 ⎠
(A can be negative!)
Now use this result to calculate the mathematically required
marginal reflection coefficients Γn :
Γn = ACnN =
AN!
(N − n ) !n !
Of course, we also know that these marginal reflection
coefficients are physically related to the characteristic
impedances of each section as:
Γn ≈
Jim Stiles
1 ⎛ Z n +1 ⎞
ln ⎜
⎟
2 ⎝ Zn ⎠
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
11/17
Equating the two and solving, we find that that the section
characteristic impedances must satisfy:
Zn +1 = Zn exp ⎡⎣2 Γn ⎤⎦
Now this is our second design rule. Note it is an iterative
rule—we determine Z1 from Z0, Z2 from Z1, and so forth.
Q: Huh? How is this any better? How does applying
approximate math lead to a better design result??
A: Applying these approximations help resolve our overconstrained problem. Recall that the over-constraint
resulted in:
R − ZN
ΓN = L
≠ A C NN
RL + Z N
But, as it turns out, these approximations leads to the happy
situation where:
1 ⎛ RL ⎞
N
ΓN ≈ ln ⎜
⎟ = A CN
2 ⎝ ZN ⎠
Å Sanity check!!
provided that the value A is likewise the approximation given
above.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
12/17
Effectively, these approximations couple the results, such
that each value of characteristic impedance Zn approximately
satisfies both Γn and Γn +1 . Summarizing:
*
If you use the “exact” design equations to determine
the characteristic impedances Zn , the last value ΓN will
exhibit a significant numeric error, and your design will
not appear to be maximally flat.
*
If you instead use the “approximate” design equations to
determine the characteristic impedances Zn , all values Γn will
exhibit a slight error, but the resulting design will appear to
be maximally flat, Binomial reflection coefficient function
Γ (ω ) !
Figure 5.15 (p. 250)
Reflection coefficient magnitude versus frequency for multisection
binomial matching transformers of Example 5.6 ZL = 50Ω and Z0 = 100Ω.
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
13/17
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 (f ) will increase. At some frequency (fm, say) the
magnitude of the reflection coefficient will increase to some
unacceptably high value ( Γm , say). At that point, we no longer
consider the device to be matched.
Γ (f )
Δf
Γm
f
fm1
f0
fm2
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 Δf of
the matching network:
Δf = fm 2 − fm 1 = 2 (f0 − fm 1 ) = 2 (fm 2 − f0 )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
14/17
Q: So what is the numerical value of Γm ?
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 Γm ). This decision depends on
the application involved, and the specifications of the overall
microwave system being designed.
However, we typically set Γm to be 0.2 or less.
Q: OK, after we have selected Γm , can we determine the two
frequencies fm ?
A: Sure! We just have to do a little algebra.
We start by rewriting the Binomial function:
(
Γ (θ ) = A 1 + e − j 2θ
)
N
= Ae − jN θ (e + j θ + e − j θ )
N
= Ae − jN θ (e + j θ + e − j θ )
N
N
= Ae − jN θ (2 cos θ )
Now, we take the magnitude of this function:
Γ (θ ) = 2N A e − jN θ cos θ
= 2N A cos θ
Jim Stiles
N
N
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
15/17
Now, we define the values θ where Γ (θ ) = Γm as θm . I.E., :
Γm = Γ (θ = θm )
= 2N A cos θm
N
We can now solve for θm (in radians!) in terms of Γm :
θm 1
⎡ ⎛ Γ ⎞ 1N ⎤
1
= cos −1 ⎢ ⎜ m ⎟ ⎥
⎢2 ⎝ A ⎠ ⎥
⎣
⎦
θm 2
1
⎡
⎛ Γm ⎞ N ⎤
−1 ⎢ 1
= cos − ⎜
⎟ ⎥
⎢ 2⎝ A ⎠ ⎥
⎣
⎦
Note that there are two solutions to the above equation (one
less that π 2 and one greater than π 2 )!
Now, we can convert the values of θm into specific
frequencies.
Recall that ωT = θ , therefore:
ωm =
1
T
θm =
vp
A
θm
But recall also that A = λ0 4 , where λ0 is the wavelength at
the design frequency f0 (not fm !), and where λ0 = v p f0 .
Thus we can conclude:
ωm =
Jim Stiles
vp
A
θm =
4v p
λ0
θm = ( 4f0 ) θm
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
or:
fm =
16/17
( 4f0 ) θm = (2f0 ) θm
1 vp
θm =
2π A
2π
π
where θm is expressed in radians. Therefore:
fm 1
1
⎡
⎤
⎛
⎞
2f0
1 Γm N ⎥
−1 ⎢
=
cos + ⎜
⎟
⎢ 2⎝ A ⎠ ⎥
π
⎣
⎦
fm 2
1
⎡
⎛
⎞
Γm N ⎤⎥
2f0
−1 ⎢ 1
=
cos − ⎜
⎟
⎢ 2⎝ A ⎠ ⎥
π
⎣
⎦
Thus, the bandwidth of the binomial matching network can be
determined as:
Δf = 2 (f0 − fm 1 )
1
⎡
⎤
4f0
1 ⎛ Γm ⎞ N ⎥
−1 ⎢
= 2f0 −
cos + ⎜
⎟
⎢ 2⎝ A ⎠ ⎥
π
⎣
⎦
Note that this equation can be used to determine the
bandwidth of a binomial matching network, given Γm 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!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/15/2010
The Binomial Multisection Matching Transformer.doc
17/17
Finally, we can list the design steps for a binomial matching
network:
1. Determine the value N required to meet the
bandwidth ( Δf and Γm ) requirements.
2. Determine the approximate value A from Z 0 , RL and N.
3. Determine the marginal reflection coefficients
Γn = A CnN required by the binomial function.
4. Determine the characteristic impedance of each
section using the iterative approximation:
Zn +1 = Zn exp ⎡⎣2 Γn ⎤⎦
5. Perform the sanity check:
ΓN ≈
1 ⎛ RL ⎞
N
ln ⎜
⎟ = A CN
2 ⎝ ZN ⎠
6. Determine section length A = λ0 4 for design
frequency f0.
Jim Stiles
The Univ. of Kansas
Dept. of EECS