hw1_su08_sol - ECE 442 Summer 2008 HW#1 Solutions ...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 442 Summer 2008 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 verication 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 ...
View Full Document

This note was uploaded on 09/07/2011 for the course ECE 442 taught by Professor Schutt-aine during the Summer '08 term at University of Illinois at Urbana–Champaign.

Page1 / 4

hw1_su08_sol - ECE 442 Summer 2008 HW#1 Solutions ...

This preview shows document pages 1 - 4. Sign up to view the full document.

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