4/19/2010
5_7 Chebyshev Multisection Matching Transformers
1/1
5.7 – Chebyshev Multi-section
Matching Transformer
Reading Assignment: pp. 250-255
We can also build a multisection matching network such that
the function Γ (f ) is a Chebyshev function.
Chebyshev functions maximize bandwidth, albeit at the cost
of pass-band ripple.
HO: The Chebyshev Multi-section Matching Transformer
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
1/15
The Chebyshev
Matching Transformer
An alternative to Binomial (Maximally Flat) functions (and
there are many such alternatives!) are Chebyshev polynomials.
Pafnuty
Chebyshev
1821 -1894
Chebyshev solutions can provide functions Γ (ω ) with wider
bandwidth than the Binomial case—albeit at the “expense” of
passband ripple.
It is evident from the plot below that the Chebychev
response is far from maximally flat! Instead, a Chebyshev
matching network exhibits a “ripple” in its passband. Note the
magnitude of this ripple never exceeds some maximum value
Γm (within the pass-band).
The two frequencies at which the value Γ (ω ) does increase
beyond Γm define the bandwidth of the matching network.
We denote these frequencies ωm = 2π fm (the plot above shows
the locations of the frequencies for N =4).
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
2/15
Γm
fm 2
fm 1
Figure 5.17 (p. 255)
Reflection coefficient magnitude versus frequency for the Chebyshev
multisection matching transformers of Example 5.7.
Æ Note that the bandwidth defined by fm increases as the
number of sections N is increased.
Æ Note also that the reflection coefficient in not necessarily
zero at the design frequency f0 !!!
Instead, we find:
Γ(f =f0 )
Jim Stiles
⎧⎪ 0 for odd-valued N
⎪⎪
= ⎪⎨
⎪⎪
⎪⎪⎩ Γm for even-valued N
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
3/15
Now, Chebyshev transformers are symmetric, i.e.:
Γ 0 = ΓN , Γ1 = ΓN −1 , etc.
Recall we can express the multi-section function Γ (θ ) (where
θ = ωT = β A ) in a simpler form when the transformer is
symmetric:
Γ (θ ) = 2 e − j N θ ⎡⎣Γ 0 cosN θ + Γ1 cos (N − 2 ) θ
+ " + Γn cos (N − 2n ) θ + " + G (θ ) ⎤⎦
where:
⎧1
for N even
⎪ 2 ΓN 2
⎪
G (θ ) = ⎨
⎪Γ
cos θ for N odd
⎪ (N −1) 2
⎩
Now, the reflection coefficient of a Chebyshev matching
network has the form:
Γ(θ ) = A e
− jN θ
⎛ cos θ
TN ⎜⎜
⎝ cos θm
⎞
⎟⎟
⎠
= A e − jN θ TN (cos θ sec θm )
where θm = ωmT
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
4/15
The function TN (cos θ sec θm ) is a Chebyshev polynomial of
order N.
T1 ( x ) = x
T2 ( x ) = 2x 2 − 1
T3 ( x ) = 4x 3 − 3x
T4 ( x ) = 8x 4 − 8x 2 + 1
We can determine higher-order Chebyshev polynomials using
the recursive formula:
Tn ( x ) = 2x Tn −1 ( x ) −Tn −2 ( x )
Inserting the substitution:
x = cos θ sec θm
into the Chebyshev polynomials above (and then applying a few
trig identities) gives the results shown in equations 5.60 on
page 252 of your book:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
5/15
T1 (cos θ sec θm ) = cos θ sec θm
T2 (cos θ sec θm ) = sec 2 θm (1 + cos 2θ ) − 1
= sec 2 θm cos 2θ + ( sec 2 θm − 1 )
T3 (cos θ sec θm ) = sec 3 θm (cos 3θ + 3cos θ ) − 3 sec θm cos θ
= sec 3 θm cos 3θ + (3 sec 2 θm − 3) sec θm cos θ
T4 (cos θ sec θm ) = sec 4 θm (cos 4θ + 4 cos 2θ + 3)
−4 sec 2 θm (cos 2θ + 1 ) + 1
= sec 4 θm cos 4θ
+ 4 sec 2 θm ( sec 2 θm − 1 ) cos 2θ
+ (3 sec 4 θm − 4 sec 2 θm + 1 )
Note that these polynomials have a cos N θ term, a
cos (N − 2 ) θ term, cos (N − 4 ) θ term, etc.—just like the
symmetric multi-section transformer function!
For example, a 4-section Chebyshev matching network will
have the form:
Γ(θ ) = A e − j 4θ T4 (cos θ sec θm )
= A e − j 4θ ⎡⎣sec 4 θm cos 4θ
+ 4 sec 2 θm sec 2 θm − 1 cos 2θ
(
(
)
)
+ 3 sec 4 θm − 4 sec 2 θm + 1 ⎤
⎦
While the general form of a 4-section matching transformer
is a polynomial with these same terms:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
6/15
Γ 4 (θ ) = 2 e − j N θ ⎡⎣Γ 0 cos N θ + Γ1 cos (N − 2 ) θ +
= 2 e − j 4θ ⎡⎣Γ 0 cos4θ + Γ1 cos ( 4 − 2 ) θ +
= 2 e − j 4θ ⎡⎣Γ 0 cos 4θ + Γ1 cos2θ +
1
Γ2 ]
2
1
Γ2 ]
2
N =4
1
Γ2 ]
2
Thus, we can determine the values of marginal reflection
coefficients Γ 0 , Γ1 , Γ2 by simply equating the 3 terms of the
two previous expressions:
2 e − j 4θ Γ 0 cos 4θ = A e − j 4θ sec 4 θm cos 4θ
⇒ Γ0 =
1
A sec 4 θm
2
(
(sec
)
− 1)
2 e − j 4θ Γ1 cos 2θ = A e − j 4θ 4 sec 2 θm sec 2 θm − 1 cos 2θ
⇒ Γ1 = A 2 sec 2 θm
2 e − j 4θ
2
θm
1
Γ2 = A e − j 4θ 3 sec 4 θm − 4 sec 2 θm + 1
2
⇒ Γ2 = A 3 sec 4 θm − 4 sec 2 θm + 1
(
(
)
)
And because it’s symmetric, we also know that Γ3 = Γ1 and
Γ4 = Γ0 .
Now, we can again (i.e., as we did in the binomial matching
network) determine the values of characteristic impedance
Zn, using the iterative (approximate) relationship:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
7/15
Zn +1 = Zn exp ⎡⎣2 Γn ⎤⎦
Q: But what about the value of A ?
A: Also using the same boundary condition analysis that we
used for the binomial function, we find from our transmission
line knowledge that for any multi-section matching network,
at θ = 0 :
R − Z0
Γ (θ = 0 ) = L
RL + Z 0
Likewise, a Chebyshev matching network will have the specific
value at θ = 0 of:
Γ (θ = 0 ) = A e − jN ( 0 ) TN ( sec θm cos ( 0 ) )
= ATN ( sec θm )
These two results must of course be equal, and equating them
allows us to solve for A :
A=
RL − Z 0
1
RL + Z 0 TN ( sec θm )
Here again (just like the binomial case) we will find it
advantageous to use the approximation:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
8/15
RL − Z 0 1 ⎛ RL ⎞
≈ ln ⎜
⎟
RL + Z 0 2 ⎝ Z 0 ⎠
So that the value A is approximately:
A≈
1 ln (RL Z 0 )
2TN ( sec θm )
(A can be negative!)
Q: Gosh, both the values of marginal reflection coefficients
Γn and value A depend explicitly on sec θm . Just what is this
value, and how do we determine it?
A: Recall that θm = ωmT , where T is the propagation time
through one section (i.e., T =
A
vp
) and ωm = 2π fm defines the
bandwidth associated with ripple value Γm . Thus, for a given
ripple Γm and value N , we can find sec θm by solving this
equations
Γm = Γ (θ = θm )
= A e − jN θm TN ( sec θm cos θm )
= A e − jN θm TN ( sec θm cos θm )
= A TN (1 )
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
9/15
Q: Yikes! The value sec θm disappeared from this equation;
how can we use this to determine sec θm ?
A: Don’t forget the value A !! Inserting this into the
expression:
Γm = Γ (θ = θm )
= A TN (1 )
1 ln (RL Z 0 )
=
T (1 )
2 TN ( sec θm ) N
One property of Chebyshev polynomials is that:
TN (1 ) = 1 for all values N
Therefore, we conclude:
Γm = Γ (θ = θm )
1 ln (RL Z 0 )
=
2 TN ( sec θm )
And so rearranging:
TN ( sec θm ) =
1
ln (RL Z 0 )
2 Γm
Note that RL , Z 0 and Γm are design parameters, thus we can
use the above to determine TN ( sec θm ) , and thus sec θm !!!
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
10/15
Q: Um, I’m not at all clear about how one determines sec θm
from the value TN ( sec θm ) . Can you be more specific??
A: Perhaps I should.
Recall that the Chebyshev polynomials we use in the design of
this matching transformer were of the form TN (cos θ sec θm ) ,
i.e.:
Γ(θ ) = A e − jN θ TN (cos θ sec θm )
Note the value TN ( sec θm ) is just the magnitude of the
Chebyshev TN (cos θ sec θm ) when evaluated at cos θ = 1 , which
is the case when θ = 0 !!
TN (cos θ sec θm )
θ =0
= TN (cos 0 sec θm ) = TN ( sec θm )
Now, it can be shown that (Å a phrase professors
use while in hand-waving mode!) for all values θ
outside the passband of the matching network, the
general form of the Chebyshev polynomial can also be
written as:
TN ( sec θm cos θ ) = cosh ⎡⎣N cosh −1 ( sec θm cos θ ) ⎤⎦
Note the value θ = 0 = ωT means that the frequency ω = 0 .
This frequency is most definitely outside the passband, and
thus according to the above expression:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
11/15
TN ( sec θm cos (θ = 0 ) ) =TN ( sec θm )
= cosh ⎡⎣N cosh −1 ( sec θm ) ⎤⎦
Likewise,
Q: I know I should remember exactly what function cosh is,
but I don’t. Can you help refresh my memory?
A: The function cosh is the hyperbolic cosine. Recall that
cosine (the “regular” kind) can be expressed as:
cos θ =
e j θ + e −j θ
2
Similarly, the hyperbolic cosine is:
cosh x =
e x + e −x
2
Plotting coshx from x =− 2 to x =+ 2 :
3.0
2.5
2.0
1.5
1.0
0.5
2
Jim Stiles
1
0
The Univ. of Kansas
1
2
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
12/15
Hopefully it is apparent to you, from the above expression for
cosh x , that it results in a real value that is always greater
than or equal to one (provided x is real):
cosh x ≥ 1
for − ∞ < x < ∞
Thus cosh x = cosh x .
Q: What about cosh−1y ?
A: The function cosh−1y is the inverse hyperbolic cosine, aka
the hyperbolic arccosine (arcosh y). Note that this value is
defined only for y ≥1 , and is specifically:
(
cosh−1 y = ± ln y + y 2 − 1
)
for y ≥ 1
Note that there is always two solutions (positive and
negative) for the inverse hyperbolic cosine!
You will find that most scientific calculators support the cosh
and cosh-1 functions as well.
Anyway, combining the previous results, we find:
TN ( sec θm ) =
R
1
ln L = cosh ⎡⎣N cosh −1 ( sec θm ) ⎤⎦
2Γm
Z0
Therefore:
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
13/15
R
1
ln L = cosh ⎡⎣N cosh −1 ( sec θm ) ⎤⎦
2Γm
Z0
Now (finally!) we have a form that can be manipulated
algebraically. Solving the above equation for sec θm :
⎡1
⎛ 1
RL ⎞ ⎤
−1
sec θm = ± cosh ⎢ cosh ⎜
ln
⎟⎟ ⎥
⎜
N
Z
Γ
2
⎢⎣
0 ⎠⎥
⎝ m
⎦
Of course, the specific values of θm can be determined with
the sec−1 (i.e., arcsecant) function.
Note there are two solutions for sec−1 (one for the plus sign
and one for the minus); The two solutions are related as:
θ1 = π − θ 2
Now, we can convert the values of θm into specific bandwidth
frequencies fm 1 and fm 2 .
Since θ = ωT , we find:
ωm 1 =
And similarly:
ωm 2 =
Jim Stiles
1
T
1
T
θm =
θm =
vp
A
vp
A
θ1
θ2
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
14/15
Note that there are two solutions θm for this equation—one
value of θm (θ1 )will be less than π 2 (defining the lower
passband frequency), while the other (θ2 ) will be greater than
π 2 (defining the upper passband frequency).
Moreover, we find that the two values of θm will be symmetric
about the value π 2 ! For example, if the lower value θ1 is
π 2 − π 10 , then the upper value θ2 will be π 2 + π 10 .
Q: So??
A: This means that the center of the passband will be defined
by the value θ = π 2 —and the center of the passband is our
design frequency ω0 ! In other words, since ω0T = π 2 :
ω0 =
π 1 π vp
=
2T
2 A
Thus, we set the center (i.e., design) frequency by selecting
the proper value of section length A . Note the above
expression is precisely the same result obtained for the
Binomial matching network, and thus we have precisely the
same design rule!
That design rule is, set the section lengths A such that they
are a quarter wavelength at the design frequency ω0 :
Jim Stiles
The Univ. of Kansas
Dept. of EECS
4/19/2010
The Chebyshev Matching Transformer
A=
15/15
λ0
4
where λ0 = v p ω0 .
Summarizing, the Chebyshev matching network design
procedure is:
1. Determine the value N required to meet the
bandwidth and ripple Γm requirements.
2. Determine the Chebychev function
Γ (θ ) = A e − jN θ TN (cos θ sec θm ) .
3. Determine all Γn by equating terms with the
symmetric multisection transformer expression:
Γ (θ ) = 2 e − j N θ ⎡⎣Γ 0 cosN θ + Γ1 cos (N − 2 ) θ
+ " + Γn cos (N − 2n ) θ + " + G (θ ) ⎤⎦
4. Calculate all Zn using the approximation:
Γn =
1 Z n +1
ln
2
Zn
5. Determine section length A = λ0 4 .
Jim Stiles
The Univ. of Kansas
Dept. of EECS