chapter 17 - Chapter 17- Two-Port and Three Port Networks...

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Chapter 17- Two-Port and Three Port Networks Exercises Ex. 17.4-1 1 25(100) 10 250 ac abc RR R RRR = == ++ 2 (125)(125) 12.5 250 bc R = 3 100(125) 50 250 ab R = Ex. 17.5-1 12 21 1 21 YY =− = ( ) ( ) 11 12 11 22 21 22 11 1 21 42 42 42 1 1 10.5 1/ 7 21 10.5 Y Y += = − − = −= 3 41 21 21 7 = Y 2 1 1 1 11 0 1 2 22 0 2 1 12 21 0 2 42 (21 10.5) 1 8 42 31.5 10.5(63) 9 73.5 6 I I I V Z I V Z I V ZZ I = = = + = + == = = 21 2 2 18 6 10.5 42(10.5) Since , then 6 Z = 73.5 73.5 6 9 II V I I = 17-1
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Ex. 17.6-1 1 11 1 2 21 1 1 6 1 .167 6 I Y V I Y V == −= 1 12 2 2 22 2 0.0567 0.944 I Y V I Y V Ex. 17.7-1 22 1 6, (9 1 ) 10 , 1 I iV i i = =+ = = 2 22 2 1 12 2 6 0 . 6 S 10 0.1 10 Ii h Vi h == = 1 1 1 1 2 1 10 99 44 5 V I ii V I = =+ = =−= Therefore () 1 11 1 2 21 1 0.9 10 9 44 9 4.4 10 9 h I i i I h I i = = 17-2
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Ex. 17.8-1 21 2 1 12 6 411 15 5 5 5 and S 30 1 2 75 50 30 1 2 3 4 10 5 10 15   =∆ = = =  YYZ = Ex. 17.8-2 () 2/5 1 1/10 4 10 1/30 2/15 1/3 4/3 −− == T Ex. 17.9-1 ab c 1 12 1 0 1 3 = , = and = 0 1 1/6 1 0 1       TT T 1 12 0 3 12 3 3 21 0 1 1/ 6 1 1/ 6 1 0 1 1/ 6 3/ 2 abc c    =       TTT T Problems Section 17-4: T-to-T1 Transformations P17.4-1 17-3
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P17.4-2 P17.4-3 21 1 2 21 2i 22 L 1 22 L (forward current gain) zI I z IA zR I −− =⇒ = = ++ () 12 i 1 11 1 1 1 2 2 1 2 2 1 in 11 11 1 22 L (input resistance) zA I Vz I z I z z Rz z II I + == =− + iL 2 22 Li L 1 1 i n 1 v 1i n and (forward voltage gain) AR V VI RA R I VR I A = = = = L 2 pi vi in R AA R 17-4
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P17.4-4 First, simplify the circuit using a -Y transformation: eq 1 || 5|| 20 4 3 R RR = == Mesh equations: 2 2 30 18 10 50 10 20 I I I I = =− Solving for the required current: 30 10 50 20 100 0.385 A 18( 20) ( 10)10 260 I = −− P17.4-5 17-5
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Section 17-5: Equations of Two-Port Networks P17-5-1 12 11 12 11 22 21 22 6 12 18 3 9 Z ZZ Z Z = =Ω⇒ =Ω = Ω 2 1 1 1 11 1 0 12 12 21 22 0 22 0 1 S 14 1 S (6 12) 21 /7 1 S 7 V V V I Y V II Y VV IV Y = = = == = = + === Υ 11 14 21 = 21 7    Υ P17.5-2 24 4 4 2 j j j j −− = −+ Z 17-6
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P17.5-3 2 2 12 11 21 11 0 0 and V V II YY VV = = == 12 2 2 and II I Vb GG G V + + =+ = 1 so 2 1 1 2 2 1 ( ( 1) ) 1 and ( 3 I Gb GV V I b G V V =− = = Finally 11 21 Y 1 S and Y 3 S = −= Next 22 32 and 2 I + 1 1 1 12 2 2 0 2 22 2 3 2 0 1 S 4 S V V I YG V I G V = = = + = P17.5-4 Using Fig. 17.5-2 as shown: 12 21 12 21 11 12 22 21 0.1 S or 0.1 S 0.2 0.3 S 0.05 0.15 S = = =−= = P17.5-5 12 21 11 12 11 10 S 13.33 S 23.33 S Y µ = += = 22 21 22 20 S 30 Y S + =⇒ = 17-7
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