IIT Bombay Course Code : EE 611 Department: Electrical Engineering - - PowerPoint PPT Presentation

IIT Bombay

Course Code : EE 611 Department: Electrical Engineering Instructor Name: Jayanta Mukherjee Email: jayanta@ee.iitb.ac.in

EE 611 Lecture 6 Jayanta Mukherjee Page 0

Lecture 6

IIT Bombay

Subtopics

• Binomial and Chebyshev matching networks

EE 611 Lecture 6 Jayanta Mukherjee Page 1

Maximally Flat (Binomial) Transformer

EE 611 Lecture 6 Jayanta Mukherjee Page 2 IIT Bombay

• The maximally flat transformer is a N section transformer
• It achieves broad-band matching by setting the reflection

coefficient and its N-1 derivatives to 0 at the center frequency (θ = π/2)

1 1 2

2

− < ≤ = Γ =       = Γ

=

N k with d d

k in k in π θ

θ π θ

0.2 0.4 0.6 0.8 1 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

θ / π Γ

in

N=1 N=2 N=3 N=4

Binomial Solution

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Many functions can be implemented For a N-section maximally flat transformer we choose

( )

( )

N j in

e A

θ

θ

2

1

−

+ = Γ

• The input reflection coefficient verifies Γin(π/2)=0
• All the derivatives of Γin(θ) upto order N-1 vanish at θ=π/2

Binomial Solution

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( )

( )

L N L L L k N j k

Z R Z R e d d Γ = + = Γ = + − = = Γ = < < + ∝ Γ

− − −

2 Z R Z

• R

2 A have we such that A2 (0) : letting by A constant the determine can We 1

• N

k 1 for 1

L L N

• N

in 2

θ θ θ

θ

Binomial Transformer Bandwidth

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( ) ( ) ( )

: and between ip relationsh following the have we , point at the t coefficien reflection for the value tolerable maximum a

• n

decide we If cos 2 in as

• f

magnitude the can write We

m m m

θ θ θ θ θ θ θ

θ θ θ

Γ = Γ = Γ Γ = + Γ = Γ Γ

− − m m N L N N j j N j L

e e e

Binomial Transformer Bandwidth

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π θ θ θ

m N L

4 2 f f by given is Bandwidh Fractional The cos gives for Solving

/ 1 m 1

• m

m

− = ∆         Γ Γ =

Fractional Bandwidth vs Number of Sections

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frequency center the around flat more becomes t coefficien reflection the and increased is sections

• f

number the as bandwidth wider have will r transforme flat maximally The cos f f by given is Bandwidh Fractional The

m 1

• N

L / 1

4 2         − = Γ Γ π ∆

0.2 0.4 0.6 0.8 1 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

θ / π Γ

in

N=1 N=2 N=3 N=4

Design of the transformer: finding the Γk

’s

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So far we have just chosen a “maximally flat” function we would like to have the reflection coefficient of our transformer to

• produce. The next step is to find how the transformer can

actually produce this maximally flat reflection coefficient. The binomial expansion formula is given by:

( ) ( )

triangle s Pascal' using calculated rapidly be also can which t" coefficien binomial " a is where, !n! n N N! C x C x

N k N k k N k N

− = = +

∑

=0

1

Design of the transformer: finding the Γk

’s

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N/k 0 1 2 3 4 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1

Pascal’s Triangle

Design of the transformer: finding the Γk

’s

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• Applying the binomial expansion to our maximally flat

function we have

• This is exactly of the same form as the input reflection

coefficient Γin of the multi-section network we analyzed in the previous lecture using the small reflection approximation

( )

( )

∑

= − − − −

= + =

N k jk N k L N N j L N in

e C e

2 2

2 1 2

θ θ

Γ Γ θ Γ ( )

L N k

• N

k N k jk k in

C Γ e Γ Γ Γ θ Γ

θ

2

2

= =∑

= −

: as identify to us permits which

k

Attribute of the Reflection Coefficients

EE 611 Lecture 6 Jayanta Mukherjee Page 11 IIT Bombay L N k N k

C Γ Γ

−

= 2

• Note that the binomial transformer will produce symmetric

Γk ‘s because Ck

N=CN N-k

• Notice that the Ck

N are always positive real numbers

• However ΓL can be positive or negative depending on whether

RL>Z0 or vice-versa

• Thus the Γk ‘s are all either positive or negative and we should

have uniformly decreasing or increasing line impedances

• Now that we know the Γk ‘s we need to determine the

characteristic impedances of the lines to obtain these Γk ‘s

Approximate Method for Finding the Zk’s

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• Since the Γk are given by Γk =(Zk+1-Zk)/(Zk+1+Zk) we could write a

system of equations and solve it to find Zk values.

• Note that in this approximate theory we have actually more

equations than variables which is a reminder that our design is approximate.

• However, if Zk+1 is not too different from Zk (required to keep our

theory of small reflections valid), we have seen that we can approximate Γk by :

• This is the recommended approach as it yields self-consistent

solution

k k k k k k k

Z Z Z Z Z Z

1 1 1

ln 2 1

+ + +

≈ + − = Γ

Approximate Method for Finding the Zk’s

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1

ln 2 2 2 2 ln Z R C C Z R Z R Z Z

L N k N N k L L N k k k − − +

≈         + − = = Γ

• Since we know the Γk’s we can obtain the following expression
• where the last expression is valid provided RL and Z0 are within

around a factor of 2 or ½.

• We can step through the above equation starting at k=0 to find

all the impedance values in the transformer.

2 Section Example

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1 ln ln 2 1 ln ln ln 4 1 ln ln 2 2 2 2 ln

2 / 1 1 2 4 / 1 1 1

=         = = =         = = ≈         + − = =

− − +

k for Z R Z R Z Z for k Z R Z R Z Z Z R C C Z R Z R Z Z

L L L L L N k N N k L L N k k k

: have we Γ

Let us consider a two section example N=2

• Starting from

2 Section Example

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4 / 1 4 / 3 2 4 / 1 4 / 1 1

, Z R Z Z R Z

L L

= =

The resulting characteristic impedances are then

• Note that for k=2 solution using this method is not accurate.

This is due to the approximate nature of the design equations we have derived. Pozar (pg 281) provides a table with more accurate values.

Design Example

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Design a three section binomial transformer to match a load RL=100 ohms to a 50 ohm source. If a reflection coefficient of 0.1 is considered tolerable, what is the relative bandwidth obtained? Since Γm=0.1,

2 1 3 i 3 3 2 3 1 3 m

3(0.0417) 0.0417 thus are s ' The 1 C 3 C 3 C C are cubic) a g 3(expandin N for ts coefficien binomial The 93.39%

• r

0.9339 / 4

• 2

then is bandwidth relative The Γ Γ Γ Γ Γ π θ Γ θ = = = = = = = = = = =         − + =

− 3 3 / 1 1

1 8374 . cos rads Z R Z R

L L m m

Design Example

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2599 . ln 2 ln 2599 . ln 2 ln 0866 . ln 2 ln ln 2 2 2 2

3 2 3 2 3 3 1 3 1 2 3 3

= = = = = = ≈         + − = =

− − − − − +

Z R C Z Z Z R C Z Z Z R C Z R C C Z R Z R

L L L L N k N N k L L N k 1 k 1 k k

Z Z from s Z' Find Z Z ln : use we exact the using

• f

Instead Γ Γ

Design Example

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Pozar. in table the using

• btained

those to close very are values These

• hm.

100 and

• hm

50 between values increasing uniformly are These 91.6909 Z 70.7055, Z 54.523, Z gives Solving

3 2 1

= = =

Result

EE 611 Lecture 6 Jayanta Mukherjee Page 19 IIT Bombay

0.2 0.4 0.6 0.8 1 0.05 0.1 0.15 0.2 0.25 0.3 0.35

θ / π Γ

in

Exact Binomial Approximate Design

Chebyshev Transformer

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• To improve on the maximally flat transformer we consider

next the Chebyshev transformer.

• We will get more bandwidth for a given Γin if we allowed some

ripples inside the passband (i.e. between θ = θm and θ=π-θm)

0.2 0.4 0.6 0.8 1 0.02 0.04 0.06 0.08 0.1 0.12 0.14 θ/π Γ(in) N=1 N=2 N=3 N=4

• No perfect match at center

Frequency for N even

• Γin has N zeros

Chebyshev Polynomials

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To derive a coefficient Γ(θ) with equal ripples we shall rely on the use

• f Chebyshev polynomials TN(x)
• 2
• 1

1 2

• 2
• 1.5
• 1
• 0.5

0.5 1 1.5 2

θ Γ(θ)

N=1 N=2 N=3 N=4

Chebyshev Polynomials

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These polynomials have the following properties.

• The first three Chebyshev polynomials are given by:

x x x T x x T x x T 3 4 ) ( 1 2 ) ( ) (

3 3 2 2 1

− = − = =

• Higher order polynomials can be found from
• The Chebyshev polynomial verifies the property:
• As a result we have |TN(x)|<1 for |x|<1 and in this region the

Polynomials oscillate between +/- 1

) ( ) ( 2 ) (

2 1

x T x xT x T

n n n − −

− =

) cos( ) (cos θ θ n Tn =

Application of Chebyshev Polynomials

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• If we map the range |x|<1 to the passband of our device, we can

design Γ to be less than Γm in the passband

• For |x|>1,|TN(x)|>1 and the reflection will be larger than Γm
• utside the passband
• The properties of these polynomials will give our transformer an

“equal-ripple” response in the passband

• For this purpose we need to define x in terms of θ such that |x|=1

is mapped to the passband endpoints [θm,π-θm]

• One approach is to select:

m m

x θ θ θ θ sec cos cos cos = =

Application of Chebyshev Polynomials

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• The resulting reflection coefficient is

( )

) cos (sec θ θ θ Γ

θ m N jN in

T Ae− =

Some Simplifications

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• We can determine A by taking θ=0:

( )

) (sec ) (sec 1 ) (sec

m N L L L m N m N L L L in

T Z R Z R T A AT Z R Z R θ Γ θ θ Γ Γ =         + − = ⇒ = + − = =

• We can then rewrite the reflection coefficient as:
• The maximum reflection coefficient Γm is then related to θm by:

( )

) cos (sec ) (sec 1 θ θ θ Γ θ Γ

θ m N jN m N L in

T e T

−

=

( )

) (sec

m N L m in m

T θ Γ θ Γ Γ = =

Some Simplifications

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• Note that TN (x = secθm) is always positive as we have

x = sec θm>1

• Since we have Γm=|A|, we can rewrite in terms of Γm:

( )

) cos (sec Z Z for and Z Z for T e

L L m N jN m in

< − > + ± =

−

θ θ Γ θ Γ

θ

BandWidth of Chebyshev Transformer

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• Equation

establishes a relationship between θm and Γm given the order N and the load reflection coefficient ΓL

• Using an alternative definition for Chebyshev polynomials
• Once θm is known, the fractional bandwidth is given by

) (sec ) (

m N L m in m

T θ Γ θ Γ Γ = =

                + − = =

− − 1 1

1 cosh 1 cosh sec ) cosh cosh( ) ( Z R Z R N x n x T

L L m m N

Γ θ θ get to sec for solve can we

m

π θ ∆

m

f f 4 2 − =

Design Procedure

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• We can use our earlier procedure of small reflections to find the

Zk values from the Γk values

• Γk values are determined by comparing our cosine series for

Γin(θ) with the corresponding Chebyshev polynomial.

• For the design of 3rd and 4th order Chebyshev transformer we

can make use of the following identity

k k k

Z Z

1

ln 2 1

+

= Γ

( ) ( ) ( )

1 1 2 cos sec 4 3 2 cos 4 4 cos sec ) cos (sec cos sec 3 cos 3 3 cos sec ) cos (sec

2 4 4 3 3

+ + − + + = − + = θ θ θ θ θ θ θ θ θ θ θ θ θ θ

m m m m m m

T T

3rd Order Example

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• With N=3, from our small reflection theory for symmetrical

transformers we have (Eqn 5.46 Pozar) and from Chebyshev theory

• Thus for this third order transformer we have
• Note the symmetry of the reflection coefficients of the

Chebyshev transformer

( )

{ }

θ Γ θ Γ θ Γ

θ

cos 3 cos 2

1 3

+ =

− j in

e

( )

( )

[ ]

( )

m m m j m j in

Ae T Ae θ θ θ θ θ θ θ θ Γ

θ θ

sec sec cos 3 3 cos sec cos sec

3 3 3 3 3

− + = =

− −

[ ]

m m m

A A θ θ Γ Γ θ Γ Γ sec sec 2 3 sec 2 1

3 2 1 3 3

− = = = =

Design Example

EE 611 Lecture 6 Jayanta Mukherjee Page 30 IIT Bombay

Design a three section Chebyshev transformer to match a load RL=100 ohms to a 50 ohm source. If a reflection coefficient of 0.1 Is considered tolerable, what is the relative bandwidth? Since Γ =0.1,

( )

from s Z' and calculate can we results section 3 derived previously the Using . A have we n Z larger tha is Z Since result. binomial section 3

• ur

n better tha is that this Note percent. 125.19

• r

/ 4

• 2

then is bandwidth Relative

k L m

Γ Γ π θ θ θ 1 . 2519 . 1 5876 . 2015 . 1 / 1 cos 2015 . 1 1 . 1 cosh 3 1 cosh sec

1 1

= + = = = = =                 + − =

− − m m L L m

Z R Z R

Design Example(2)

EE 611 Lecture 6 Jayanta Mukherjee Page 31 IIT Bombay

[ ]

81.86 Z 69.77, Z 59.47, Z find to 2 1

3 2 1

= = = = = = = − = = = = = 0799 . ln 2 1 0799 . sec sec 2 3 ln 2 1 0867 . sec 2 1 50 ln

1 2 3 2 3 1 2 1 3 1

Γ Γ θ θ Γ θ Γ Z Z A Z Z A Z

m m m

Comparison of Approximate and Desired Frequency Response

EE 611 Lecture 6 Jayanta Mukherjee Page 32 IIT Bombay

0.2 0.4 0.6 0.8 1 0.05 0.1 0.15 0.2 0.25 0.3

θ/π Γin Exact Binomial Approximate Design

Exact results (including multiple reflections) from our approximate design (neglecting multiple reflections) are compared below with the desired Chebyshev response. Pozar has a table pg 286 to assist in designing Chebyshev transformers for Γm=0.05 and Γm=0.2