# Examples for second order filters a driven series rlc

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: driven series RLC circuit can serve as a lowpass ﬁlter. i Frequency L x(t ) + − Time vL(t ) = Li￿(t ) VL(s) = sLI (s) R vR(t ) = Ri(t ) VR(s) = RI (s) C Z + 1t y(t ) vC (t ) = i(!)! C0 − EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly VC (s) = 1 I (s) sC 26 Circuit Examples for Second Order Filters A driven series RLC circuit can serve as a lowpass ﬁlter. i Frequency L x(t ) + − Time vL(t ) = Li￿(t ) VL(s) = sLI (s) R vR(t ) = Ri(t ) VR(s) = RI (s) Z + 1t y(t ) vC (t ) = i(!)! C0 − C The factors sL, R, and resistor, and capacitor. 1 sC VC (s) = 1 I (s) sC are the complex impedences of the inductor, EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 27 Transfer Function of Second Order Filter i Time L x(t ) + − R C Frequency vL(t ) = Li￿(t ) VL(s) = sLI (s) vR(t ) = Ri(t ) Z + 1t y(t ) vC (t ) = i(!)! C0 − VR(s) = RI (s) VC (s) = 1 I (s) sC In the time domain x(t) = vL(t) + vR(t) + vC (t); y (t) = vC (t) In the frequency domain (with zero initial conditions) X (s) = VL(s) + VR(s) + VC (s); EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly Y (s) = VC (s) 28 X (s) = Y (s) = ￿ 1 sL + R + sC 1 I (s) sC ￿ I (s) and the transfer function H (s) = Y (s)/X (s) H (s) = = = 1 sC I (s) ￿ 1 sL + R + sC I (s) 1 sC 1 sL + R + sC ￿ s2 1/(LC ) + (R/L)s + 1/(LC ) This is a second order lowpass ﬁlter, with DC gain 1 (0 dB). EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 29 The corner frequency ωc (assuming the underdamped case) is ωc = ￿ 1/(LC ). The total impedence of the series RLC circuit is ZT (s) = sL + R + 1/(sC ) and the impedence of the output element (the capacitor) is ZO (s) = 1/(sC ) The transfer function is a voltage divider H (s) = ZO (s)/ZT (s) = 1 sC 1 sL + R + sC which we compute as if the complex impedences were resisitors. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 30 For convenience, we’ll let C = 1, L = 1, and R = √ 2: I (s) 1/s √ s + 2 + 1/s 1 √ = s2 + 2s + 1 H (s) = s √ 2 X (s) + − 1/s + Y (s) − Bode Plot 20 logMagnitude (dB) ω)|) 10 (|H ( j 20 ! 10 1 −40 dB/decade 0 −3 dB !10 × !20 !30 !40 !1 10 1 10 ω 0 ∠H ( (deg)) Phase j ω ! × 0 10 !50 !100 −1 !150 !200 !1 10 0 10 Frequency (rad/s) ω 1 10 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 31 What if we deﬁne the output to be the voltage drop across the inductor? I (s) s √ 2 X (s) + − + Y (s) − H (s) = s √ s + 2 + 1/s = s2 √ s2 + 2s + 1 20 log10 (|H ( jω)|) Magnitude (dB) 1/s ! 1 Bode Plot 20 10 0 × −3 dB !20 !30 !40 !1 10 −1 0 10 ω 1 10 200 ! ∠H ((deg) ) Phase j ω ◦ ◦ × +40 dB/decade !10 150 100 50 0 !1 10 EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 0 10 Frequency (rad/s) ω 1 10 32 What if we dene the output to be the voltage drop across the resistor? I (s) s √ 2 X (s) + − 1/s √ 2 √ H (s) = s + 2 + 1/s √ 2s √ = 2 + 2s + 1 s + Y (s) − Bode Plot 20 logMagnitude (dB) ω)|) 10 (|H ( j 20 ! 0 dB 10 1 0 !10 × !20 !30 × ! −1 0 10 ω 1 10 100 ∠Phase jω) H ( (deg) ◦ −20 dB/decade +20 dB/decade !40 !1 10 50 0 !50 !100 !1 10 0 10 Frequency (rad/s) ω EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 1 10 33 Conclusion The frequency response of a system with a transfer function (system function) H (s) is H (j ω ), provided the j ω axis is in the region of convergence for H (s). The pole/zero placement determines the frequency response: • |H (j ω )| increases near a pole • |H (j ω )| decreases near a zero Many pole-zero patterns have been well studied, such as the Butterworth lowpass. Simple approximations accurately characterize many systems. EE102A:Signal Processing and Linear Systems I; Win 09-10, Pauly 34...
View Full Document

## This note was uploaded on 09/28/2013 for the course EE 108b taught by Professor Mukamel during the Fall '13 term at Singapore Stanford Partnership.

Ask a homework question - tutors are online