710hw2corr

710hw2corr - ftqu 2 £57m PWHQJM ’1 2015051" IQL{00.57{low/z I z 7 K” 203” 5.94632 Z" QLS/QL 21h 0‘2 52 2° 2 Z" 2 1/2 9m 2

Info iconThis preview shows pages 1–9. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ftqu 2 £57m PWHQJM ’1- 2015051" IQL... {00.57. {low/z I z, 7. K” 203” _ 5.94632 Z". .: QLS/QL 21h? ': 0‘2) 52. 2° 2 Z"- 2 1/2 9m 2 C‘DS—Dl/ Ham! / £00373 6m 51—.) 2'0 4%: z: 2 — é} QM -"—'- -7ZZ W‘b'fla/ Mm‘cflfi O A? _:.— I [03> GHZ Malta!le EEIOF - 70mm - 11/23/37 ' ISI [33- 33 " HE'FILE B MGISIU .3. HABESMJ III, _ IIIIIIIQ EIFEIIIi _ . _ I tlEiIIESIIIIIIIII EIIIIIIII I.” Problem f23>Synthesis of an impedance function (25 points) Consider the impedance we wish to implement: S3 + 453 + 75 in s3+4sz+75+12 The following circuit prototype is used: a) Using Richard's Theorem extract the unit element of impedance Z l (Hint: try two simple roots to reduce the order of Z“t ). Verify first tFuat zin(1)+zin(—1)=0. 2' (l) = I-t-Lt-t'? = L?- : l “‘ 1+ 9+7“: 2% z z. (_1) = "i‘t‘f‘? 2:- ____-L:- :1 hi- ” Ate—Tut 3 2 'l zoi “‘ '32: q- zm: L Zt‘wtfil- 5/2 a \ 9—(53'1‘r5£+75)—€(53+‘r5 +75%) 1i _5 2M3) "2'; {S'bwfgL-t— 7.912) -zs(ss+tr5?-+7S) “S‘rw253—t51—t25 ______________________l x Z #254 -753 4053—1-75 +12 l [Si-"'3 (‘59-‘13) (52...) (“231—45 ~42) I S?" + 2.5 zoutl = r 1 ash-75 +|L b) Next extract the shunt inductor L so as to reduce the impedance order. y,n 2.- _L. LS '1‘ \fon .. _ I 7guT '. \fibi 2:: Mast—r73 H2.) LS - (SzezS) Your 3 LS (SLHS) YOuT :- 5’- (25"+7S 112.)L* (5+2) 1. (St-r 25) ész—t 1.5 d: -—5 1: :: 6 6 at ’25, Fast L. =3 51123 W. 23231-5 2 C35“) __—— : $20“, 2 y -out2 = Z \S “I? ?. *2 c) Finally extract the last unit element of impedance Zo that Zin(1)+2in(-l)=0. Was is the unknown load 2L? 2. verify first Z'l‘liilct}: iii—- 7'— 2’1 : ,L 1202 2;“C‘l):_,_l_. Lt't‘L 6 z. 2. 3+2 §_ ZL3: cl‘SHL 2 _|25/+2- 251—75 2 I 8+2. "'9: W+ _ L, s l S. {35/ 2. 454:2, M _J_ ~25E+z H \ 3‘ “St—r1 z = 1L 2 section binomial transformer a B”— Term Term’l TLIN TLIN Num=1 TL1 TL2 2:50 Ohm 2:59.46 Ohm 2:84.08 Ohm E=90 E=90 F=1.5 GHz F=‘l.5 GHz ,. r A. .1 a ‘J S—PARAMETERS |t S_Param SP1 Start=.001 GHz Stop=3.0 GHz Step=.001 GHz mag(8(’l ,1)) 2 section Binomial transformer 1.5 2.0 2.5 3.0 freq. GHz Bandwidth = 2.038 - 0.957 = 1.081 GHZ Fractional Bandwidth = 0.7206 m1 freq=957.0MHz mag(8(1,1))=0.102_ q=2.038GHz mag(8(1,1))=o.1oo + Term Term2 Num=2 2:100 Ohm i— L/ 2 section Chebyshev transformer + —Term—+:|-—--—+:———MT Term Term1 TLIN TLIN Term2 Num=1 TL1 TL2 Num=2 Z=50 Ohm 2:61.81 Ohm 2:78.77 Ohm Z=1OD Ohm — E=90 E=90 _ F=1.5 GHz F=1.5 GHz S-PARAMETERS S_Param SP1 Start=.001 GHz Stop=3.0 GHz Step=.001 GHz 2 section Chebyshev transformer m1 freq=811.0MHz mag(S(1,1))=0.096 mag(8(1,1)) freq=2.186GHz mag(S(1,1))=0.095 freq, GHz Bandwidth = 2.188 - 0.811 =1.375 GHz Fractionai Bandwidth = 0.9166 4 section Binomial transformer B—E——+:i—+j . Term Term Term1 TLIN TLIN TLIN TLIN Term2 Num=1 TL1 TL2 TL3 TL4 Num=2 2:50 Ohm 2:45.22 Ohm 2:30.23 Ohm 2:16.53 Ohm 2:11.05 Ohm 2:10 Ohm E=90 E=90 E=90 E=90 — F=1.5 GHz F=1_5 GHz F=1.5 GHz F=1.5 GHz ' .4: i597] S-PARAMETERS S__Param SP1 Start=.01 GHz Stop=3.0 GHz Step=.01 GHz m89(3(1,1)) 4 section binomial transformer 0.8 —.— —|— l— 0.6—I m1 4 freq=1.02OGHz 04 _| _L m2 O-ZJ— '— mQ —J freq=1.980GHz mag(S(1,1))=0.051 0-D—lI—I—1'I—i'r'I—ir—I'r r—rr IFTIF_|_'i—f_l"f_|' 0.0 0.5 1.0 1.5 2.0 2.5 3.0 freq, GHz Bandwidth = 1.98 — 1.02 = 0.96 GHz fractional bandwidth = 0.64 4 section Chebyshev transformer + + Term Termi TLIN TLiN TLIN TLIN Term2 Num=1 TL1 TL2 TL3 TL4 Num=2 Z=40 Ohm Z=44.95 Ohm 2:47.47 Ohm Z=50.28 Ohm Z=53.09 Ohm Z=60 Ohm - E=90 E=90 E=90 E=90 — F=1.5 GHZ F=1.5 GHZ F=1.5 GHZ F=’I.5 GHZ ‘ _|_ = —_L 3 _; -: S—PARAMETERS S_Param SP1 Start=.01 GHz Stop=3.0 GHz Step=.01 GHz 4 section Chebyshev transformer m1 freq=330.0MHz mag(S(1,1))=0.094 mag(8(1,1)) m2 freq=2.670GHz mag(S(1,1))=0.094 freq, GHz bandwidth = 2.67 - 0.33 = 2.34 GHz fractional Bandwidth = 1.56 ...
View Full Document

This note was uploaded on 04/30/2011 for the course ECE 710 taught by Professor Roblin during the Spring '11 term at Ohio State.

Page1 / 9

710hw2corr - ftqu 2 £57m PWHQJM ’1 2015051" IQL{00.57{low/z I z 7 K” 203” 5.94632 Z" QLS/QL 21h 0‘2 52 2° 2 Z" 2 1/2 9m 2

This preview shows document pages 1 - 9. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online