MULTI-SECTION IMPEDANCE TRANSFORMERS - ADS...

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Dr. J. E. Rayas-Sánchez 1 Multi-Section Impedance Transformers - ADS MULTI-SECTION IMPEDANCE TRANSFORMERS - ADS Dr. J. E. Rayas-Sánchez FOUR-SECTION 3:1 BUTTERWORTH IMPEDANCE TRANSFORMER Problem Design a four-section Butterworth impedance transformer to match at 5 GHz a load impedance ZL = 150 to a transmission line with characteristic impedance Z0 = 50 . Calculate the theoretical fractional bandwidth for a maximum reflection of -40 dB. Calculate the corresponding lengths of the four sections of the impedance transformer, assuming an effective dielectric constant εe = 3.38. Solution We have N = 4, ZL = 150 , Z0 = 50 , Γm = 0.01 = -40 dB, and f0 = 5 GHz. Using Table 5.1 (Binomial Transformer Design, [1]) for N = 4 and ZL/Z0 = 3: Z1/Z0 = 1.0718 Z2/Z0 = 1.4105 Z3/Z0 = 2.1269 Z4/Z0 = 2.7990 Since Z0 = 50 , then, Z1 = 53.5900 Z2 = 70.5250 Z3 = 106.3450 Z4 = 139.9500 Since N = 4, ZL = 150 , Z0 = 50 , dB 02 . 6 0.5 0 L 0 L - = + = + - = Z Z Z Z Γ , 0.03125 2 = = - Γ N A . The expected relative bandwidth for a maximum reflection Γm = 0.01 = -40 dB is 49.0886% 2 1 cos 4 2 / 1 1 0 = - = - N m A f f Γ π Centering the impedance transformer at f0 = 5 GHz, then the expected bandwidth is f = 2.4544 GHz. Since εe = 3.38, cm 26 . 3 e 0 0 0 = = = ε λ f c f v p ; mm 8.1589 4 0 = = λ l

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Dr. J. E. Rayas-Sánchez 1 Multi-Section Impedance Transformers - ADS

MULTI-SECTION IMPEDANCE TRANSFORMERS - ADS

Dr. J. E. Rayas-Sánchez

FOUR-SECTION 3:1 BUTTERWORTH IMPEDANCE TRANSFORMER

Problem

Design a four-section Butterworth impedance transformer to match at 5 GHz a load impedance ZL = 150 Ω to a transmission line with characteristic impedance Z0 = 50 Ω. Calculate the theoretical fractional bandwidth for a maximum reflection of −40 dB. Calculate the corresponding lengths of the four sections of the impedance transformer, assuming an effective dielectric constant εe = 3.38.

Solution

We have N = 4, ZL = 150 Ω, Z0 = 50 Ω, Γm = 0.01 = −40 dB, and f0 = 5 GHz. Using Table 5.1 (Binomial Transformer Design, [1]) for N = 4 and ZL/Z0 = 3:

Z1/Z0 = 1.0718 Z2/Z0 = 1.4105 Z3/Z0 = 2.1269 Z4/Z0 = 2.7990

Since Z0 = 50 Ω, then,

Z1 = 53.5900 Ω Z2 = 70.5250 Ω

Z3 = 106.3450 Ω Z4 = 139.9500 Ω

Since N = 4, ZL = 150 Ω, Z0 = 50 Ω,

dB 02.60.50L

0L −=+=+−=

ZZ

ZZΓ , 0.031252 == − ΓNA .

The expected relative bandwidth for a maximum reflection Γm = 0.01 = −40 dB is

49.0886%21

cos4

2/1

1

0

=

−= −N

m

Af

f Γπ

Centering the impedance transformer at f0 = 5 GHz, then the expected bandwidth is ∆f = 2.4544 GHz. Since εe = 3.38,

cm 26.3e00

0 ===ε

λf

c

f

vp ; mm 8.158940 == λ

l

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Multi-Section Impedance Transformers - ADS 2

Implementation in ADS with Ideal Transmission Lines

It is confirmed a Butterworth response centered a 5 GHz, with |Γ | = 0.5 at low frequencies. Zooming-in at the |Γ | = 0.01 = −40 dB,

It is seen that the expected bandwidth ∆f (= 2.4544 GHz) is very close to the actual bandwidth for Γm = 0.01.

[1] D. M. Pozar, Microwave Engineering. Amherst, MA: Wiley, 1998.