butlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutler...

8/10/2019 ButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButlerButler

http:///reader/full/butlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutl 1/11

Whites, EE 481 Lecture 26 Page 1 of 11

2013 Keith W. Whites

Lecture 26: Quadrature (90) Hybrid.

Back in Lecture 23, we began our discussion of dividers andcouplers by considering important general properties of three-and four-port networks. This was followed by an analysis ofthree types of three-port networks in Lectures 24 and 25.

We will now move on to (reciprocal) directional couplers , whichare four-port networks. As in the text, we will consider these

specific types of directional couplers:1. Quadrature (90) Hybrid,2. 180 Hybrid,3. Coupled Line, and4. Lange Coupler.

We will begin with the quadrature (90) hybrid . Fig 7.21 showsthis coupler implemented with microstrip as a 1:1 power divider:

Because of the physical symmetry, we can simplify the analysisof this circuit considerably using even-odd mode analysis . This




1

1

1

1

1

o A

4

o A

/ 8

Observe that the circuit and its boundary conditions remain thesame in both the even and odd mode configurations. It is onlythe excitation that changes. Because of this and the circuit beinglinear, by superposition the total solution is simply the sum of

the even and odd mode solutions.

Each solution (even and odd) is simpler to determine than thecomplete circuit, which is why we employ this technique.

Even mode. Because the voltages and currents must be the

same above and below the line of symmetry (LOS) in Fig7.23a, then 0 I at the LOS open circuit loads at the endsof /8 stubs, as shown.

Referring to the definition of i B ( 1, ,4i ) in Fig 7.22, wecan write from Fig 7.23a that for the even mode excitation:

1 1e e

e B A , 2 1e e

e B T A (1a)

3 2 1e e e

e B B T A , 4 1 1e e e

e B B A (1b)where 1 1 2

e A , and e and eT are the reflection andtransmission coefficients for the even mode configuration.




Odd mode. Because the voltages and currents must haveopposite values above and below the LOS in Fig 7.23b, then

0V along the LOS short circuit loads at the ends of /8stubs, as shown.

Then, 1 1o o

o B A , 2 1o o

o B T A (2a)

3 2 1o o o

o B B T A , 4 1 1o o o

o B B A (2b)where 1 1 2

o A and o and oT are reflection and transmissioncoefficients for the odd mode configuration.

Total solution. The total solution is the sum of the voltages in both circuits. From this fact, we can deduce that the total i B coefficients will be the sum of (1) and (2):

1 1 1

1 12 2

e oe o B B B (7.62a),(3)

2 2 2

1 12 2

e oe o B B B T T (7.62b),(4)

3 3 3

1 12 2

e oe o B B B T T (7.62c),(5)

4 4 4

1 12 2

e oe o B B B (7.62d),(6)

Likewise, the incident wave coefficients are

1 1 1 1 1 12 2e o A A A

4 4 4

1 10

2 2e o A A A




These match the assumed excitation in the original circuit on p. 2.

To finish the calculation of the S parameters for the quadraturehybrid, we need to determine the reflection and transmissioncoefficients for the even- and odd-mode configurations.

Your text shows that the solutions for e and eT are

0e and 1

12

eT j (7.64),(7),(8)

Here well derive solutions for o and oT .

From Fig 7.23b:o

T

o

1 21 1

1 2

1 1

We have three cascaded elements, so well use ABCD

parameters to solve for the overall S parameters of this circuit.

Elements 1 and 3. These are short circuit stubs of length 8 ,which appear as the shunt impedance

in 0 tan Z jZ l where 2 8 4l

Therefore, in0

Z j

Z , or N Y j

From the inside flap of your text:




1 0 1 0

1 1 N ABCD

Y j

(9)

Element 2. This is a /4-length of TL where2

4 2l

From the inside flap of your text:

0

0

0

0

2cos sin

02

sin cos 2 02

Z jl j l

Z ABCD

Z j l l j Z

Cascading these three ABCD matrices we find the overall ABCD matrix for odd mode excitation:

01 0 1 0 1121 1 12

2 0o

j A B j

C D j j j j

(10)

Using Table 4.2, we can convert these to S parameters (with

0 1 Z for the normalized TL):

0 0

110 0

1 2 1 10

1 11 2

A B Z CZ D j jS

A B Z CZ D j j

(11)




210 0

2 2 11 11 2

2 2 22 2 1

S A B Z CZ D j j

j j

(12)

Since the ports are matched, then:11 0o S (7.66a),(13)

and 212 1

11

121

o

jS

j jT j (7.66b),(14)

Finally, using (7), (8), (13), and (14) in (3)-(6) we find:

1 0 B (7.67a),(15)

21 1

1 12 2 2 2 2

j B j j

(7.67b),(16)

31 1 1 1

1 12 2 2 2 2

B j j (7.67c),(17)

4 1 10 0 02 2 B (7.67d),(18)

When properly interpreted, these results tell us much about thecircuit. In particular, when port 1 is excited and all other portsterminated in matched loads, then:

1 0 B port 1 is matched.

2 2 B j 90 phase shift from port 1 to port 2, and

one half of the time average input power is delivered to port 2.




3 1 2 B 180 phase shift from port 1 to port 3

(90 phase shift between ports 3 and 2), and one half of theinput power is delivered to port 3.

4 0 B no power output to port 4.

Because of the high degree of symmetry, we can treat any portas the input port. Then, the isolation is the other port on thesame side as the input and the outputs are the two ports on theother side of the circuit.

Employing this concept and the results above, we can constructthe other three columns in the full S matrix for the quadrature(90) hybrid by simply transposing rows of the first column:

0 1 0

0 0 11

1 0 020 1 0

j

jS

j j

(7.61),(19)

That is, the first column in (19) is the results from (15)-(18)when the input was assumed at port 1. In the second column, wecan directly deduce that the outputs are at ports 1 and 4, theinput is at port 2 and the isolation is at port 4. Furthertransposition of the rows in column 1 produces columns 3 and 4.




Example N26.1. Design a branch line hybrid coupler using 100- microstrip on 32-mil RO4003C for a center frequency of 2.5

GHz. Include the effects of copper and substrate losses.

Because there are two different characteristic impedancesneeded for the 90 hybrid device, two different widths ofmicrostrip must be computed (because W /d depends on Z 0) andtwo different /4-lengths must be determined (because ,r e depends on W /d ).

0 100 Z sections . Using LineCalc, 18.02W mil and, 2.424r e . The guide wavelength at this frequency is then

8

9

2.998 107.702

2.5 10 2.424

c f

cm

Hence, this branch line coupler should have 100- lines withlength 4 1.93 cm.

0 2 70.71 Z sections . Using LineCalc, 39.62W mil

and , 2.545r e . The guide wavelength is then8

9

2.998 107.517

2.5 10 2.545

c f

cm

Hence, this branch line coupler should have 70.71- lineswith length 4 1.88 cm.

The following S parameter results were obtained for this designusing ADS.


http:///reader/full/butlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbut 10/11


1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.41.4 3.6

0.2

0.4

0.6

0.0

0.8

freq, GHz

m a g ( S ( 1

, 1 ) )

m1 m2 m a g ( S ( 2

, 1 ) )

m3

m a g ( S ( 3

, 1 ) )

m a g ( S ( 4

, 1 ) )

m1freq=mag(S(1,1))=0.102

2.370GHz

m2freq=mag(S(1,1))=0.102

2.640GHz

m3freq=mag(S(2,1))=0.703

2.510GHz

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.41.4 3.6

-100

0

100

-200

200

freq, GHz

p h a s e ( S ( 1

, 1 ) )

p h a s e ( S ( 2

, 1 ) )

m4 p h a s e ( S ( 3

, 1 ) )

m5

p h a s e ( S ( 4

, 1 ) )

m4freq=phase(S(2,1))=-91.709

2.510GHz

m5freq=phase(S(3,1))=178.357

2.510GHz


http:///reader/full/butlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbut 11/11


1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.41.4 3.6

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.0

0.8

freq, GHz

m a g ( S ( 4

, 3 ) )

m a g ( S ( 3

, 2 ) )

1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.0 3.2 3.41.4 3.6

-150

-100

-50

-200

0

freq, GHz

p h a s e ( S ( 3

, 2 ) )

butlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutlerbutler...

Documents