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Unformatted text preview: terminated on both sides with Zo , it follows thatS11 = S22 = 0.
b. We know that when terminated with ZS = ZL = 50Ω the 2port is conjugately matched at both ports. In that case,
G = GT = GA = GA,max = 20 dB. When the load impedance is changed to 1000Ω, the available gain is still 20 dB (because
Pout
available gain does not depend on the load termination), so we can use GT = GA Pavo , i.e., in dB, the transducer gain will be equal
4·50·1000
to the available gain plus the output mismatch factor, which is 10 log( (50+1000)2 ) = −7.4 dB. Thus GT = 20 − 7.4 = 12.6dB.
S21 ΓL
c. ΓIN  = S11 + S12S22 ΓL  = S12 S21 ΓL . Worst case is when ΓL =1. We know S21  = 1020/10 = 10 and S12  = 10−30/10 =
1−
1
√
= 0.0316. Thus, ΓIN  ≤ 10 · 0.0316 = 0.316.
1000 d. With S11 = S22 = 0, GT = S21 2 (1−ΓS 2 )(1−ΓL 2 )
.
1−S12 S21 ΓS ΓL 2 With ZS = ZL = 200, ΓS = ΓL = 0.6. We do not know the phase of either S12 or S21 , so only upper and lower bounds on GT can be calculated. Upper bound is GT =
Lower bound is GT = 102 (1−0.62 )(1−0.62 )
1+(10)(0.0316)(0.6)(0.6)2 ⇒ 15.2 dB. 102 (1−0.62 )(1−0.62 )
1−(10)(0.0316)(0.6)(0.6)2 ⇒ 17.1 dB. 826 Solution
A passive, lossless, reciprocal 2port will have S12 = S21 . Such a 2port is fully described by the 3 complex parameters S11 , S22 ,
and S21 . These parameters must also satisfy:
S21 2 + S22 2 = 1
(1)
S21 2 + S11 2 = 1 ∗
∗
S11 S21 + S21 S22 = 0 (2)
(3) c. In this case, we can determine S22 directly. Since this network transforms a 50 Ω source impedance to Zout = (10 + j 200) Ω:
S22 = 10 + j 200 − 50
= 0.862 + j 0.459
10 + j 200 + 50 a. From (1):
S21  =
b. From (1) and (2):
1 − S22 2 = 0.214 S11  = S22  = 0.977...
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This homework help was uploaded on 03/13/2014 for the course ECE 453 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff

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