Quang_HW5

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Apr 3, 2024

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First case: Open circuit: d1 = 0.163λ , l1 = 0.339 λ Second case: Open circuit: d2 = 0.295λ, l2 = 0.161 λ
Third case: Short circuit: d1 = 0.163λ, l1 = 0.089 λ Fourth case: Short circuit: d2 = 0.295λ , l2 = 0.411 λ The results show that they do not match at exactly 6GHz (due to approximation of Smith Chart calculations). I would choose the first case as its matching point is closest to 6 GHz and has the second widest -10dB BW.
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2. (30 pts) a) Design a single-section quarter-wave matching network at 10 GHz for Z L =12.5 Ω and plot its frequency response (5-20 GHz). Z l = Z L xZ 0 = 12.5x50 = 25 Ω l = λ/4 (0.25*360 0 electrical length
b) Same, but using a 3-section Chebyshev transformer with ρ m =0.05 and 0.20. For N =3 , ρ m =0.05, Z L /Z0 =12.5/ 50 = 0.25 = > Zo/Z L = 4 From the table given with the reverse Z positions Z3 = 1.2662 * Z0 = 63.31 Ω Z2 = 2 * Z0 = 100 Ω Z1 = 3.1591 * Z0 = 157.955 Ω The design is shown as the picture below:
For N =3 , ρ m =0.20, Z L /Z0 =12.5/ 50 = 0.25 = > Zo/Z L = 4 From the table given with the reverse Z positions Z3 = 1.4333 * Z0 = 71.665 Ω Z2 = 2 * Z0 = 100 Ω Z1 = 2.7908 * Z0 = 139.545 Ω
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3.a) Design an LC matching network at 6 GHz for a 10 Ω load. Plot |S 11 | 2 in dB vs. frequency (4-10 GHz) and determine the -10 dB bandwidth.
First circuit: Serie L – Shunt C where L =0.53 nH, C = 1.061 pF
Second circuit: Serie C – Shunt L where L =0.663 nH, C = 1.326 pF
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b) Transfer the designs of (a) into t-lines with Z h =80 Ω and Z L =25 Ω, and calculate the frequency response from 4-10 GHz. Compare with (a) and comment. (you need to use equation 8.84 and 8.85 in the book) * For circuit 1: Serie L – Shunt C where L =0.53 nH, C = 1.061 pF βl = (0.53*10^-9 * 50)/80 = 3.3125 *10^-10 for inductor βl = 1.061 *10^-12 * 25 / 50 = 5.305*10^-13 for capacitor * For circuit 2: Serie C – Shunt L where L =0.663 nH, C = 1.326 pF βl = (0.663*10^-9 * 50)/80 = 4.144 *10^-10 for inductor βl = (1.326 *10^-12 * 25) / 50 = 6.63*10^-13 for capacitor