Examples for second order filters a driven series rlc

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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...
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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.

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