Lect10-Ch6-FrequencyResponse

# 8429 h 1000 vout 3 t 04975 cos2000t

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Unformatted text preview: .71°) + 3.535 cos(200πt − 45°) + 0.4975 cos(2000πt − 84.29°) H(f) in Decibels | H(f) | Logarithmic Decibel (dB) scale compresses range of numbers, for plots 100 40 10 20 2 H ( f ) dB = 20 log H ( f ) | H(f) |dB 6 1 0 0.5 -6 0.1 -20 0.01 -40 Logarithmic Scale Decade = range of frequencies wherein highest is ten times lowest (10:1) ȹ f2 ȹ number of decades = log10 ȹ ȹ ȹ f1 Ⱥ Octave = range of frequencies wherein highest is twice lowest (2:1) ȹ f2 ȹ number of octaves = log2 ȹ ȹ ȹ f1 Ⱥ 1 ȹ f ȹ 1 + jȹ ȹ ȹ f B Ⱥ 1 H( f ) = ȹ f ȹ 2 1 + ȹ ȹ ȹ f B Ⱥ H( f ) = Transfer Function in Decibels ȹ ȹ ȹ ȹ ȹ ȹ 1 H ( f ) dB = 20Log H ( f ) = 20Logȹ ȹ 2 ȹ f ȹ ȹ ȹ ȹ 1 + ȹ f ȹ ȹ ȹ B Ⱥ Ⱥ ȹ ȹ f ȹ 2 ȹ f ȹ 2 H ( f ) dB = 20Log(1) - 20Log 1 + ȹ ȹ = −20Log 1 + ȹ ȹ ȹ f B Ⱥ ȹ f B Ⱥ ȹ ȹ ȹ 2 ȹ f H ( f ) dB = -10Logȹ 1 + ȹ ȹ ȹ ȹ ȹ f Ⱥ ȹ B ȹ Ⱥ Bode Plot f<<fB fB = Half Power Freq., a.k.a. ‘’Break Frequency’’ f>>fB Bode Magnitude Plot: |H(f)|dB vs. log f Exercise 6.1: Find Bode Plot 1 fB = = 1000 2πRC ȹ f ȹ ∠H ( f ) = − arctanȹ ȹ ȹ f B Ⱥ € asymptotes Bode Magnitude Plot: |H(f)|dB vs. log f asymptotes Bode Phase Plot:∠H(f) vs. log f First-Order Highpass Filter ȹ f ȹ jȹ ȹ Vout ȹ f B Ⱥ H( f ) = = ȹ f ȹ Vin 1 + jȹ ȹ ȹ f B Ⱥ fB = 1 2πRC H( f ) = Vout = IR = Vin R ( j 2 πfRC )Vin = 1 + R 1 + j 2 πfRC j 2πfC € ȹ f ȹ ȹ ȹ ȹ f B Ⱥ ȹ f ȹ 2 1 + ȹ ȹ ȹ f B Ⱥ Ⱥ ȹ ȹ 2 Ⱥ ȹ f ȹ f H ( f ) dB = 20 logȹ ȹ − 10 logȺ1 + ȹ ȹ Ⱥ Ⱥ ȹ f B Ⱥ Ⱥ ȹ f B Ⱥ Ⱥ Ⱥ Frequency Response of Highpass Filter Actual Bode First-Order Highpass Filter with Inductor Vin = I ( R + j 2πfL) Vin ( R + j 2πfL) Vout = I ( j 2πfL) Vin Vout = ( j 2πfL) ( R + j 2πfL) Vout ȹ j 2πfL ȹ H( f ) = = ȹ ȹ Vin ȹ R + j 2πfL Ⱥ ȹ j 2πfL ȹ ȹ jf ȹ ȹ ȹ ȹ f B ȹ R ȹ , H ( f ) = ȹ ȹ = ȹ j 2πfL ȹ jf ȹ ȹ 1 + ȹ ȹ 1 + ȹ ȹ R Ⱥ ȹ f B Ⱥ I= Hambley, Exercise 6.13 R fB = 2πL Series Resonant Circuit (RLC) KVL : j 2 πfLI + RI + 1 I = Vs j 2πfC Series impedance : Z s = j 2πfL + R + 1 j 2πfC 1 Z s ( f ) = R + j 2πfL − j = ℜ( f ) + jX ( f ) 2πfC @ resonance freq. f0, reactance X(f) = Im Zs(f) → 0 1 2πf0 L = , resonant freq : 2πf 0C f 0= 1 2π LC Quality Factor (Qs) Defines narrowness of ‘’peak’’ in transfer function |H(f)|, about resonance freq. fo Quality factor: 2πf 0 L 1 Qs = = , R 2πf 0 RC Series Zs Ⱥ ȹ f f 0 ȹȺ Z s ( f ) = RȺ1 + jQsȹ − ȹȺ ȹ f 0 f ȺȺ Ⱥ Z s = R + j 2πfL − j € Ⱥ = RȺ1 + Ⱥ Ⱥ = RȺ1 + Ⱥ 1 2πfC ȹ L 1 ȹȺ jȹ 2πf − ȹȺ R 2πfRC ȺȺ ȹ ȹ 2πLf o f 1 f o ȹȺ jȹ − ȹȺ R f o 2πRCf o f ȺȺ ȹ Series Resonant Bandpass Filter Vs I= Zs( f ) Vs R ȹ f f 0 ȹ 1 + jQsȹ − ȹ ȹ f 0 f Ⱥ Vs VR = IR = ȹ f f 0 ȹ 1 + jQsȹ − ȹ ȹ f 0 f Ⱥ Vout VR 1 = = ȹ f Vin Vs f 0 ȹ 1 + jQsȹ − ȹ ȹ f 0 f Ⱥ I= Bandpass Response Bandpass Bandwidth Bandwidth: High & low pass freq: f0 B = f H − fL = Qs B f H ≅ f0 + & 2 B fL ≅ f0 − 2 Parallel Resonant Bandpass Filter Ideal Filters Passes lows (f < fH) Stops highs Passes mids (fL< f < fH) Stops lows & highs Passes highs (f > fL) Stops lows Passes lows & highs Stops mids Second-Order Lowpass Filter Second-Order Highpass Filter Second-Order Notch Filter Actuator Transfer Function •  Force Actuator •  Magnetostriction –  Magnetic field ⇒ strain –  Current ⇒ displacement...
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## This document was uploaded on 10/01/2013.

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