Lect10-Ch6-FrequencyResponse

8429 h 1000 vout 3 t 04975 cos2000t

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

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

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 logȺ1 + 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 ȹȺ Z s ( f ) = RȺ1 + jQsȹ − ȹȺ ȹ f 0 f ȺȺ Ⱥ Z s = R + j 2πfL − j € Ⱥ = RȺ1 + Ⱥ Ⱥ = RȺ1 + Ⱥ 1 2πfC ȹ L 1 ȹȺ jȹ 2πf − ȹȺ R 2πfRC ȺȺ ȹ ȹ 2πLf o f 1 f o ȹȺ jȹ − ȹȺ 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...
View Full Document

This document was uploaded on 10/01/2013.

Ask a homework question - tutors are online