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# hw1_su10_sol - ECE 442 Summer 2010 HW#1 Solutions ...

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Unformatted text preview: ECE 442 Summer 2010 HW#1 Solutions  1.14. 1.15. 1 Now, when a resistance of 1.5kΩ is connected between 4 and ground, I= 0.77 = 0.1mA 6.15 + 1.5 1.67. Using the voltage divider rule 1 Zi = Ri || sCi 1 Yi = Ri + sCi Vi Zi 1 1 = = = = 1 Vs Z i + Rs 1 + Rs Yi 1 + Rs ( Ri + sCi ) 1+ = Rs Ri 1/(1 + RS ) 1 Ri = Rs + sCi Rs 1 + sCi 1+ Rs Ri 1 1 1 = Rs Rs Ri Ri + Rs 1 + sCi (Ri ||Rs ) 1 + Ri 1 + sCi ( Rs +Ri ) This transfer function is of the STC low-pass type with a dc gain K = and a 3-dB frequency ω = 1/C (R ||R ). For and C = 5pF, Ri /(Ri + Rs ) Rs = 20k Ω, Ri = 80k Ω, ω0 = 5× 10−12 f0 = 0 i i s i 1 = 1.25 × 107 rad/s × 20×80 × 103 20+80 ω0 1.25 × 107 = ≈ 2M Hz 2π 2π 2 Ri 1.68. Using the voltage divider rule, T (s) = R2 Vo = Vi R2 + R1 + = 1 sC R2 R1 + R2 s + which from Table 1.2 is of the high-pass type with K= and ω0 = s 1 C (R1 +R2 ) R2 R1 + R2 1 C (R1 + R2 ) As a further verication that this is a high-pass network and T (s) is a highpass transfer function we observe as at s = 0,T (s) = 0; and that as s → ∞, T (s) = R /(R + R ). Also, from the circuit observe as at s → ∞, (1/sC ) → 0 and V /V = R /(R + R ). Now, for R = 10kΩ, R = 40kΩ, and C = 0.1µF , 2 o i 1 2 2 f0 = 1 2 1 2 ω0 1 = = 31.8Hz −6 (10 + 40) × 103 2π 2π × 0.1 × 10 1 K 40 √ = 0.57V /V |T (jω0 )| = √ = 10 + 40 2 2 1.69. Using the voltage divider rule, VL RL = VS RL + RS + 1 sC = 3 RL RL + RS s + s 1 C (RL +RS ) which is of the high-pass STC type (see Table 1.2) with K= ,ω = For f ≤ 10Hz, RL RL +RS 0 1 C (RL +RS ) 0 1 2πC (RL + RS ) ⇒C≥ ≤ 10 1 2π × 10(20 + 5) × 103 Thus, the smallest value of C that will do the job is C = 0.64µF. 4 ...
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