Bài giảng Electric circuit theory - Chapter XIII: Frequency Response - Nguyễn Công Phương

Nguy ễn Công Ph ươ ng Electric Circuit Theory Frequency Response Contents I. Basic Elements Of Electrical Circuits II. Basic Laws III. Electrical Circuit Analysis IV. Circuit Theorems V. Active Circuits VI. Capacitor And Inductor VII. First Order Circuits VIII.Second Order Circuits IX. Sinusoidal Steady State Analysis X. AC Power Analysis XI. Three-phase Circuits XII. Magnetically Coupled Circuits XIII.Frequency Response XIV.The Laplace Transform XV. Two-port Networks Frequency Response - sites.google.com/site/ncpdhbkhn 2 Frequency Response 1. Transfer Function 2. The Decibel Scale 3. Bode Plots 4. Series Resonance 5. Parallel Resonance 6. Passive Filters 7. Active Filters 8. Scaling Frequency Response - sites.google.com/site/ncpdhbkhn 3 Transfer Function (1) ω ω Iin ( ) Iout ( ) + + ω ω ω Vin ( ) H( ) Vout ( ) – – Out (ω ) H(ω ) = In (ω ) V (ω ) I (ω ) H (ω ) = out H (ω ) = out voltage ω current ω Vin ( ) Iin ( ) V (ω ) I (ω ) H (ω ) = out H (ω ) = out impedance ω admittance ω Iin ( ) Vin ( ) Frequency Response - sites.google.com/site/ncpdhbkhn 4 Transfer Function (2) ω ω Iin ( ) Iout ( ) + + ω ω ω Vin ( ) H( ) Vout ( ) – – Out (ω ) H(ω ) = In (ω ) ω = → Out ( ) 0z1 , z 2 ,... (zeros) ω = → In ( ) 0p1 , p 2 ,... (poles) Frequency Response - sites.google.com/site/ncpdhbkhn 5 Transfer Function (3) Ex. 1 5Ω ω – + + vs = 100sin t (V). Find the transfer function Vo/Vs and sketch its frequency response. v( t ) s 2 H vo V – V = j2ω s o 5+ j 2 ω V j2ω →H (ω ) =o = v + ω Vs 5j 2 j2ωω (5− j 2 ) 4 ω2 10 ω = =+=j H φ (5+−j 2ωω )(5 j 2 ) 25 + 4 ω2 25 + 4 ω 2 v v 5 – + + 4 2 16ω+ 100 ω − 5 V =φ = 1 s j2ω Vo Hv2 ; v tan 425ω+ 2 ω – Frequency Response - sites.google.com/site/ncpdhbkhn 6 Transfer Function (4) Ex. 1 5Ω ω – + + vs = 100sin t (V). Find the transfer function Vo/Vs and sketch its frequency response. v( t ) s 2 H vo 4 2 – 16ω+ 100 ω − 5 H =;φ = tan 1 v425ω2 + v 2 ω 90 1 80 ω 70 0.8 Hv ( ) 60 0.6 50 φ ω 40 v ( ) 0.4 30 20 0.2 ω ω 10 0 5 10 15 20 25 30 35 40 45 50 0 5 10 15 20 25 30 35 40 45 50 Frequency Response - sites.google.com/site/ncpdhbkhn 7 i i Ex. 2 Transfer Function (5) i o ω – + + vs = 100sin t (V). Find the transfer functions 5Ω V /V , I /I , V /I , & I /V . v( t ) o s o i o i o s s 2 H vo 1mF – Frequency Response - sites.google.com/site/ncpdhbkhn 8 Frequency Response 1. Transfer Function 2. The Decibel Scale 3. Bode Plots 4. Series Resonance 5. Parallel Resonance 6. Passive Filters 7. Active Filters 8. Scaling Frequency Response - sites.google.com/site/ncpdhbkhn 9 The Decibel Scale P = 2 G log 10 P1 P = 2 GdB 10log 10 P1 V = 2 GdB 20log 10 V1 I = 2 GdB 20log 10 I1 Frequency Response - sites.google.com/site/ncpdhbkhn 10 Frequency Response 1. Transfer Function 2. The Decibel Scale 3. Bode Plots 4. Series Resonance 5. Parallel Resonance 6. Passive Filters 7. Active Filters 8. Scaling Frequency Response - sites.google.com/site/ncpdhbkhn 11 Bode Plots (1) Semilog plots of the magnitude (in decibels) and phase (in degrees) of a transfer function versus frequency 20log H H = H φ →  10 φ = = φ φ φ HHHH123... ( H 1 1)(H 2 2)(H 3 3 )... = () φ+ φ + φ + HHH1 2 3 ... 1 2 3 ... 20logHHHH= 20log + 20log + 20log + ... →  10 10 1 10 2 10 3 φ= φ + φ + φ +  1 2 3 ... Frequency Response - sites.google.com/site/ncpdhbkhn 12 Bode Plots (2) 2  ±1 jω  j2 ζω j ω  K()1 j ω +   1 +1 +    ... z ω ω  1 k  k  H(ω ) = 2  jω  j2 ζω j ω  1+   1 +2 +    ... p ω ω  1 n  n  K : gain 1 1 1 : poleat theorigin : simple pole 2 : quadratic pole jω jω j2ζ ω j ω  1 + 1+2 + p ω ω  1 n n  2 jω j2ζ ω j ω  jω : zeroat the origin 1+ : simplezero 1+1 + :quadraticzero z ω ω  1 k k  Frequency Response - sites.google.com/site/ncpdhbkhn 13 Bode Plots (3) H= 20log K H(ω ) =K →  dB 10 φ = 0 H φ 20log K 10 0 0.1 1 10 100 ω 0.1 1 10 100 ω Frequency Response - sites.google.com/site/ncpdhbkhn 14 Bode Plots (4) = − ω 1 HdB 20log 10 H(ω ) = →  jω φ = − 90 o H 20 φ 0 0o 0.1 1 10 ω 0.1 1 10 ω −20 −90 o Frequency Response - sites.google.com/site/ncpdhbkhn 15 Bode Plots (5) = ω HdB 20log 10 H(ω ) =j ω →  φ = 90 o φ H 90 o 20 0 0o 0.1 1 10 ω 0.1 1 10 ω −20 Frequency Response - sites.google.com/site/ncpdhbkhn 16 Bode Plots (6)  ω = − + j HdB 20log10 1 1  p1 H(ω ) = →  jω +  − ω  1 φ =tan 1  −  p1   p1  H 5 0 -5 -10 -15 -20 ω -25 0.1 p1 p1 10 p1 Frequency Response - sites.google.com/site/ncpdhbkhn 17 Bode Plots (7)  ω = − + j HdB 20log10 1 1  p1 H(ω ) = →  jω +  − ω  1 φ =tan 1  −  p1   p1  φ 0.1 p1 p1 10 p1 100 p1 0 ω -10 -20 -30 -40 -50 -60 -70 -80 -90 -100 Frequency Response - sites.google.com/site/ncpdhbkhn 18 Bode Plots (8)  ω = + j HdB 20log10 1 jω  z1 H(ω )= 1 + →  z ω  1 φ = −1  tan   H  z1  25 20 15 10 5 0 ω -5 0.1 z1 z1 10 z1 Frequency Response - sites.google.com/site/ncpdhbkhn 19 Bode Plots (9)  ω = + j HdB 20log10 1 jω  z1 H(ω )= 1 + →  z ω  1 φ = −1  tan   φ  z1  100 90 80 70 60 50 40 30 20 10 ω 0 0.1 z1 z1 10 z1 100 z1 Frequency Response - sites.google.com/site/ncpdhbkhn 20 Bode Plots (10)  2 j2ζ ω j ω  H =−20log 1 +2 + dB 10 ω ω  1  n n  H(ω ) = →  ζ ω ω  2 j2 2 j  − j2ζ ω / ω  1+ +   φ = − tan 1 2 n  ω ω  − ω2 ω 2  n n  1 / n  20 ζ = H 2 0.05 ζ = 10 2 0.2 ζ = 0 2 0.4 -10 ζ = 2 0.707 ζ = 1.5 -20 2 -30 ω -40 ω ω ω ω 0.1 1 1 10 1 100 1 Frequency Response - sites.google.com/site/ncpdhbkhn 21 Bode Plots (11)  2 j2ζ ω j ω  H =−20log 1 +2 + dB 10 ω ω  1  n n  H(ω ) = →  ζ ω ω  2 j2 2 j  − j2ζ ω / ω  1+ +   φ = − tan 1 2 n  ω ω  − ω2 ω 2  n n  1 / n  ω ω ω ω φ 0.1 1 1 10 1 100 1 0 ω -20 ζ = -40 2 1.5 -60 ζ = -80 2 0.707 -100 ζ = 2 0.4 -120 ζ = -140 2 0.2 -160 ζ = 2 0.05 -180 Frequency Response - sites.google.com/site/ncpdhbkhn 22 Bode Plots (12)  2 j2ζ ω j ω  H =20log 1 +2 + 2 dB 10 ω ω  j2ζ ω j ω   n n  H(ω )= 1 +2 +  →  ω ω  n n  − j2ζ ω / ω  φ = tan 1 2 n  − ω2 ω 2   1 / n  40 H ζ = 30 2 1.5 ζ = 20 2 0.707 10 ζ = 2 0.4 0 -10 ζ = ζ = 2 0.05 2 0.2 ω -20 ω ω ω ω 0.1 1 1 10 1 100 1 Frequency Response - sites.google.com/site/ncpdhbkhn 23 Bode Plots (13)  2 j2ζ ω j ω  H =20log 1 +2 + 2 dB 10 ω ω  j2ζ ω j ω   n n  H(ω )= 1 +2 +  →  ω ω  n n  − j2ζ ω / ω  φ = tan 1 2 n  − ω2 ω 2   1 / n  180 ζ = φ 2 0.05 160 ζ = 2 0.2 140 ζ = 2 0.4 120 100 80 ζ = 2 0.707 60 ζ = 2 1.5 40 20 ω 0 ω ω ω ω 0.1 1 1 10 1 100 1 Frequency Response - sites.google.com/site/ncpdhbkhn 24 Bode Plots (14) 1 N N jω  K (jω ) N 1+  (jω ) z  20N dB/decade 20log 10 K 1 ω 20N dB/decade −20N dB/decade 1 ω z ω ω 90 N o 90 N o ω 0o 0o ω ω −90 N o z z 10 z ω 10 Frequency Response - sites.google.com/site/ncpdhbkhn 25 Bode Plots (15) 1 1 N 2 N 2 N ζ ω ω   ζ ω ω   ω  j2 1 j j2 2 j + j 1+ +    1+ +    1  ω ω  ω ω  p  n n   k k   ω k p 40N dB/decade ω ω −20N dB/decade −40N dB/decade ω n ω o 180 N ω k ω ω p 10 k 10 k 10 p 10 p 0o ω o ω 0 0o ω ω 10 ω ω − o − o n n n 180 N 90 N 10 Frequency Response - sites.google.com/site/ncpdhbkhn 26 Ex. 1 Bode Plots (16) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) 10 jω 10 jω − − H(ω ) = = 90o− tan 1 (ω /5) − tan 1 ( ω /10) (1+jω /5)(1 + j ω /10) 1 ++ jj ωω /5 1 /10  1 1 H=+20log 10 20log j ω + 20log + 20log  dB 10 10 101+jω /5 10 1 + j ω /10 →   o− 11 − 1  1  φ =90 + tan + tan    ω/5  ω /10  Frequency Response - sites.google.com/site/ncpdhbkhn 27 Ex. 1 Bode Plots (17) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) 1 1 H=+20log 10 20log j ω + 20log + 20log dB 10 10 101+jω /5 10 1 + j ω /10 40 H 30 20log10 10 K 20 10 20log 10 K ω 0 ω -10 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 28 Ex. 1 Bode Plots (18) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) 1 1 H=+20log 10 20log j ω + 20log + 20log dB 10 10 101+jω /5 10 1 + j ω /10 40 H 30 20log10 10 (jω ) N 20 10 20log jω 20N dB/decade 10 ω 0 1 ω -10 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 29 Ex. 1 Bode Plots (19) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) 1 1 H=+20log 10 20log j ω + 20log + 20log dB 10 10 101+jω /5 10 1 + j ω /10 40 H 30 20log10 10 1 N 20 jω  1+  p  10 ω 20log 10 j ω p 0 ω 1 -10 20log −20N dB/decade 10 1+ jω / 5 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 30 Ex. 1 Bode Plots (20) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) 1 1 H=+20log 10 20log j ω + 20log + 20log dB 10 10 101+jω /5 10 1 + j ω /10 40 H 30 20log10 10 1 N 20 jω  1+  p  10 ω 20log 10 j ω p 0 1 1 ω 20log 10 -10 20log 10 1+ jω /10 −20N dB/decade 1+ jω / 5 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 31 Ex. 1 Bode Plots (21) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) 1 1 H=+20log 10 20log j ω + 20log + 20log dB 10 10 101+jω /5 10 1 + j ω /10 40 H Slope = 0 + 20 – 20 30 20log10 10 20 10 ω 20log 10 j 0 ω 1 1 20log -10 20log 10 1+ jω /10 10 1+ jω / 5 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 32 Ex. 1 Bode Plots (22) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) o− 11 − 1  1  φ =90 + tan + tan   ω/5  ω /10  φ 90 o (jω ) N 90 60 40 20 ω 0 -20 90 N o -40 -60 ω -90 0.1 0.5 1 5 10 20 50 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 33 Ex. 1 Bode Plots (23) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) o− 11 − 1  1  φ =90 + tan + tan   ω/5  ω /10  φ 90 o 1 90 N jω  1+  60 p  40 20 ω 0 p -20 p 10 10 p − 1  -40 tan 1 ω  0o ω -60 / 5  -90 −90 N o 0.1 0.5 1 5 10 20 50 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 34 Ex. 1 Bode Plots (24) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) o− 11 − 1  1  φ =90 + tan + tan   ω/5  ω /10  φ 90 o 1 90 N jω  1+  60 p  40 20 ω 0 p -20   p −1 1 10 10 p − 1  tan   -40 tan 1 ω /10  ω  0o ω -60 / 5  -90 −90 N o 0.1 0.5 1 5 10 20 50 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 35 Ex. 1 Bode Plots (25) 500 jω 10 jω Construct the Bode plots for H(ω ) = = (jω+ 5)( j ω + 10) (1+jω /5)(1 + j ω /10) o− 11 − 1  1  φ =90 + tan + tan   ω/5  ω /10  φ 90 o 90 60 40 20 0 ω -20 −1 1  − 1  tan   -40 tan 1   ω /10  -60 ω / 5  -90 0.1 0.5 1 5 10 20 50 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 36 Ex. 2 Bode Plots (26) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  40 H 20 20log10 8 ω 0 K -20 -40 20log 10 K -60 ω -80 -100 0.1 1 5 10 20 50 100 Frequency Response - sites.google.com/site/ncpdhbkhn 37 Ex. 2 Bode Plots (27) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  40 jω H 20log 1 + 10 20 20 20log10 8 ω 0 N jω  -20 1+  z  -40 20N dB/decade -60 z ω -80 -100 0.1 1 5 10 20 50 100 Frequency Response - sites.google.com/site/ncpdhbkhn 38 Ex. 2 Bode Plots (28) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  40 jω H 20log 1 + 10 20 20 20log10 8 ω 0 1 20log 10 1 -20 jω (jω ) N -40 -60 1 ω −20N dB/decade -80 -100 0.1 1 5 10 20 50 100 Frequency Response - sites.google.com/site/ncpdhbkhn 39 Ex. 2 Bode Plots (29) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  40 ω H 1 + j 20log 20log10 1 10 (1+ jω / 5) 2 20 20 20log10 8 ω 0 1 20log 1 10 -20 jω N jω  1+  p  -40 p -60 ω −20N dB/decade -80 -100 0.1 1 5 10 20 50 100 Frequency Response - sites.google.com/site/ncpdhbkhn 40 Ex. 2 Bode Plots (30) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  40 ω H 1 + j 20log 20log10 1 10 (1+ jω / 5) 2 20 20 20log10 8 ω 0 1 1 N 20log 2  10 ω j2ζ ω j ω  -20 j 1+2 +    ω ω  k k   -40 ω k ω -60 −40N dB/decade -80 1 20log 10 1+jω 4/10 + ( j ω /10) 2 -100 0.1 1 5 10 20 50 100 Frequency Response - sites.google.com/site/ncpdhbkhn 41 Ex. 2 Bode Plots (31) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  40 H 20 ω 0 -20 -40 -60 -80 -100 0.1 1 5 10 20 50 100 Frequency Response - sites.google.com/site/ncpdhbkhn 42 Ex. 2 Bode Plots (32) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  φ 90 50 ω 0 N jω  -50 1+  z  -90 o 90 N -180 -200 0o z z 10 z ω -270 10 -300 -360 0.1 0.2 0.5 1 2 10 50 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 43 Ex. 2 Bode Plots (33) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  φ 90 50 ω 0 1 -50 N (jω ) -90 -180 -200 ω -270 −90 No -300 -360 0.1 0.2 0.5 1 2 10 50 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 44 Ex. 2 Bode Plots (34) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  φ 90 50 ω 0 1 -50 N jω  1+  -90 p  p -180 p 10 10 p -200 0o ω -270 o −90 N -300 -360 0.1 0.2 0.5 1 2 10 50 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 45 Ex. 2 Bode Plots (35) 1000(jω + 20) 8(1+ jω / 20) Construct the Bode plots for H(ω ) = = ωω+2 ω 2 + ω +  ωω+2 + ω + ω 2  jj( 5) ( j ) 40 j 100  jj(1 /5) 1 j 4/10 ( j /10)  φ 90 50 ω 0 1 -50 2  N j2ζ ω j ω  -90 1+2 +    ω ω  k k   ω k ω ω -180 10 k 10 k -200 0o ω -270 −180 N o -300 -360 0.1 0.2 0.5 1 2 10 50 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 46 Ex. 3 Bode Plots (36) Find the transfer function from the Bode plot? 40 (jω ) N H Slope = 0 30 20 20N dB/decade 10 jω 1 ω 0 ω -10 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 47 Ex. 3 Bode Plots (37) Find the transfer function from the Bode plot? 40 Slope = 0 K H 30 20log10 10 20 20log 10 K 10 jω ω 0 ω -10 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 48 Ex. 3 Bode Plots (38) Find the transfer function from the Bode plot? 40 1 Slope = 0 N H jω  1+  30 20log10 10 p  20 p ω 10 ω −20N dB/decade j 0 ω 1 -10 20log 10 1+ jω / 5 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 49 Ex. 3 Bode Plots (39) 10 jω Find the transfer function from the Bode plot? H(ω ) = (1+jω / 5)(1 + j ω /10) 40 1 Slope = 0 N H jω  1+  30 20log10 10 p  20 p ω 10 1 ω 20log 10 −20N dB/decade j 1+ jω /10 0 ω 1 -10 20log 10 1+ jω / 5 -20 0.1 1 5 10 20 100 200 500 Frequency Response - sites.google.com/site/ncpdhbkhn 50 Frequency Response 1. Transfer Function 2. The Decibel Scale 3. Bode Plots 4. Series Resonance 5. Parallel Resonance 6. Passive Filters 7. Active Filters 8. Scaling Frequency Response - sites.google.com/site/ncpdhbkhn 51 Series Resonance (1) 1 E = E θ R jω L m jω C – + I 1 1  Z=+RjLω + =+ RjL ω −  = H ( ω ) jω C ω C  1 Resonance : Im(Z )=→ 0ωL − = 0 ωC ω =1 = 1 Resonant frequency :0 rad/s;f 0 Hz LC2π LC Frequency Response - sites.google.com/site/ncpdhbkhn 52 Series Resonance (2) E I = m R2 +()ω L − 1/ ω C 2 ω=1 = ω ω 0 1 2 Em LC R I 2 ω =−R + R  + 1 E 1   0.707 m 2L 2 L  LC R B R R  2 1 ω = +  + 2 2L 2 L  LC R B =ω − ω = 2 1 L ω 0 0 ω ω ω ω Q = 1 0 2 B Frequency Response - sites.google.com/site/ncpdhbkhn 53 Ex. Series Resonance (3) ω ω ω 10sin ωt V 20 Ω 6H 0.02F Find 0, 1, 2, B, Q? – + ω =1 = 1 = 0 2.89 rad/s LC 6× 0.02 i R R 21 20  20 2 1 ω =−+ +=−+  + = 1.67 rad/s 1 2L 2 L LC 26×  26 × 60.02 × R R 21 20  20 2 1 ω =+ +=+  + = 5.00 rad/s 2 2L 2 L LC 26×  26 × 60.02 × =−=ω ω − = B 2 1 5.00 1.67 3.33 rad/s ω 2.89 Q =0 = = 0.87 B 3.33 Frequency Response - sites.google.com/site/ncpdhbkhn 54 Frequency Response 1. Transfer Function 2. The Decibel Scale 3. Bode Plots 4. Series Resonance 5. Parallel Resonance 6. Passive Filters 7. Active Filters 8. Scaling Frequency Response - sites.google.com/site/ncpdhbkhn 55 Parallel Resonance (1) 1 J = J θ m + jω C V R – jω L 11 1 1  Y =+ +jCω =+ jC ω −  = H ( ω ) RjLω R ω L  1 Resonance : Im(Y )=→ 0ωC − = 0 ωL ω =1 = 1 Resonant frequency :0 rad/s;f 0 Hz LC2π LC Frequency Response - sites.google.com/site/ncpdhbkhn 56 Parallel Resonance (2) J V = m R2 +()ω C − 1/ ω L 2 ω=1 = ω ω 0 1 2 LC Jm R V 2 ω =−1 + 1  + 1 1   2RC 2 RC  LC 0.707 Jm R B 1 1  2 1 ω = +  + 2 2RC 2 RC  LC 1 B =ω − ω = 2 1 RC ω 0 0 ω ω ω ω Q = 1 0 2 B Frequency Response - sites.google.com/site/ncpdhbkhn 57 Ex. Parallel Resonance (3) Find the resonant frequency? 2Ω 4Ω + – 0.01H ω 1mF Em sin t Frequency Response - sites.google.com/site/ncpdhbkhn 58 Frequency Response 1. Transfer Function 2. The Decibel Scale 3. Bode Plots 4. Series Resonance 5. Parallel Resonance 6. Passive Filters 7. Active Filters 8. Scaling Frequency Response - sites.google.com/site/ncpdhbkhn 59 Passive Filters (1) H H 1 1 ω ω ω 0 c ω 0 c H(0)= 1; H ( ∞ ) = 0 H(0)= 0; H ( ∞ ) = 1 Lowpass Highpass H Bandpass Bandstop H 1 1 ω ω ω ω ω ω 0 1 2 0 1 2 H(0)= 0; H ( ∞ ) = 0 H(0)= 1; H ( ∞ ) = 1 Frequency Response - sites.google.com/site/ncpdhbkhn 60 Passive Filters (2) H H 1 1 H 1 ω ω 0 1 ω 0 2 ω ω H H 0 c ω 1 1 ω ω ω ω 0 2 0 1 H H H 1 1 1 ω ω 0 c ω ω ω ω ω ω 0 1 2 0 1 2 Frequency Response - sites.google.com/site/ncpdhbkhn 61 H Passive Filters (3) 1 =1 × Vi Vo ω jCω R+1/ jC ω 0 c ω ω Vo 1/j C 1 →==H(ω ) = – V R+1/ jω C 1 + j ω RC + i R + v( t ) i vo ( t ) Presonant =1 →ω = 1 C H(c ) – Paverage 2 2 ω=1 → ω = 1 – + H()c H () c + 1+ jω RC + ω2 2 2 R c 1 c R C V i 1 Vo jω C – →ω = 1 c RC Frequency Response - sites.google.com/site/ncpdhbkhn 62 H Passive Filters (4) 1 ω ω V 0 c V = R i o R+1/ jω C – + C + v( t ) i v( t ) V R jω RC R o →==H(ω ) o = +ω + ω – Vi R1/ j C 1 j RC – + 1 + 1 ω = Vi c jω C V RC R o – Frequency Response - sites.google.com/site/ncpdhbkhn 63 H Passive Filters (5) 1 ω ω ω V 0 1 2 V = R i o R+ jLω + 1/ jC ω – + C + v( t ) V R i v( t ) →H(ω ) =o = R o +ω + ω L Vi R jL1/ jC – = R RjL+(ω − 1/ ω C ) – + 1 + V i jω C V ω R o j L – Frequency Response - sites.google.com/site/ncpdhbkhn 64 H Passive Filters (6) 1 ω ω ω ω+ ω 0 1 2 = (jL 1/ jC ) Vi Vo R+ jLω + 1/ jC ω – + R + vi ( t ) C V jLω+1/ jC ω vo ( t ) →H(ω ) =o = +ω + ω L Vi R jL1/ jC ω− ω – = j( L 1/ C ) +ω − ω RjL( 1/ C ) – + R + Vi 1 jω C Vo jω L – Frequency Response - sites.google.com/site/ncpdhbkhn 65 Ex. 1 Passive Filters (7) What is the type of this filter? – + + 1k Ω ω = ω Vs →Vo =j L = ω vs ( t ) v Vo j L H( ) 2 H o + ω + ω µ – R j L Vs R j L 1 F jω2 j ω 2/1000 j ω →=H(ω ) = = 1000+jω 2 1 + j ω 2/1000 500(1 + j ω /500) 60 40 20 0 -20 -40 -60 1 5 10 20 50 100 200 500 1,000 10,000 Frequency Response - sites.google.com/site/ncpdhbkhn 66 Ex. 2 Passive Filters (8) + – vo What is the type of this filter? – + 1k Ω v( t ) s 2 H 1µ F Frequency Response - sites.google.com/site/ncpdhbkhn 67 Ex. 3 Passive Filters (9) Find the cutoff frequency? – + + 1k Ω ω = ω Vs →Vo =j L = ω vs ( t ) v Vo j L H( ) 2 H o + ω + ω µ – R j L Vs R j L 1 F jω2 (1000− j ωωω 2)2 j 42 2000 ω →=H(ω ) = = + j 1000+ jω 2 100022+ 4ω 1000 22 + 4 ω 1000 22 + 4 ω 16ω4+ 4.10 6 ω 2 →H(ω ) = 4ω2+ 10 6 ω4+ 6 ω 2 ω=→1 16 4.10 =→= 1 ω H()c c 500 rad/s 24ω2+ 10 6 2 Frequency Response - sites.google.com/site/ncpdhbkhn 68 Ex. 4 Passive Filters (10) ω + Find L if c = 400 rad/s? – + 1k Ω v( t ) s C L vo – Frequency Response - sites.google.com/site/ncpdhbkhn 69 Ex. 5 Passive Filters (11) The filter is to reject a 200-Hz sinusoid while – + 100 Ω + passing other frequencies, its bandwith is 100 Hz. v( t ) i C Find L and C? vo ( t ) L =π =× π = π B2 f 2 100 200 rad/s – R R 100 B=→== L = 0.1592 H L B 200 π ωππ= =× = π 02f 0 2 200 400 rad/s 1 1 1 ω =→==C = 3.9777µ F 0 ω2 π 2 × LC 0 L (400 ) 0.1592 Frequency Response - sites.google.com/site/ncpdhbkhn 70 Frequency Response 1. Transfer Function 2. The Decibel Scale 3. Bode Plots 4. Series Resonance 5. Parallel Resonance 6. Passive Filters 7. Active Filters 8. Scaling Frequency Response - sites.google.com/site/ncpdhbkhn 71 R f Active Filters (1) V Z R ω =o = − f i C f H( ) + – + Vi Z i + V i Vo = – – ZiR i Z 1 f R f ω Z j Cf R f i = = + Z f – + 1 1+ jω R C + R + f f V f ω i Vo j C f – – R 1 →H(ω ) =−f × R R + ω 1 2 Ri1 jRC f f vi ii 1 – ω = + vo c R2 Rf C f v= − v o i io R1 Frequency Response - sites.google.com/site/ncpdhbkhn 72 Active Filters (2) V Z R f H(ω ) =o = − f Vi Z i Ri Ci + – + + 1 V Z =R + i Vo i i ω – – j C i = Z fR f R jRCω →H(ω ) =−f =− f i 1 1+ jω R C R + i i i ω j C i ω = 1 c Ri C i Frequency Response - sites.google.com/site/ncpdhbkhn 73 H Active Filters (3) 1 Lowpass Highpass vi Gain v filter filter o ω 0 2 ω H R R R f 1 R C2 + C R – 1 Ri + – + + – v + ω ω i v 0 1 – o – H R jω R C R −f × 1 − f i − f 1 + ω + ω Ri1 jRC f f 1 j Ri C i Ri V 1 jω RC  R  H(ω ) ==−o −2 − f 0 ω ω ω +ω  + ω    1 ω 2 Vi 1jRC1  1 jRC 2   R i  0 R 1 jω RC 1 1 =−×f × 2 ω=; ω = ; ωωω = +ω + ω 2 1 012 Ri 1 j RC1 1 j RC 2 RC1 RC 2 Frequency Response - sites.google.com/site/ncpdhbkhn 74 H Lowpass Active Filters (4) 1 v filter i Summing vo amplifier Highpass ω 0 1 ω filter H 1 R R C1 ω ω – R f 0 2 + Ri H 1 – + C2 + R R R + i – v + o 0 ω ω ω vi 1 2 – – V R 1 jω RC  H(ω ) ==−o f − − 2 +ω + ω  ViR i 1 jRC1 1 jRC 2  Frequency Response - sites.google.com/site/ncpdhbkhn 75 Frequency Response 1. Transfer Function 2. The Decibel Scale 3. Bode Plots 4. Series Resonance 5. Parallel Resonance 6. Passive Filters 7. Active Filters 8. Scaling Frequency Response - sites.google.com/site/ncpdhbkhn 76 Scaling (1)  L  K L' = K L L' = L' = m L ω= ω m ω= ω  ω= ω  '   ' K   K f ' K f   K f →  C f →  →  = = = = R' Km R  C ' R' R  C R' K R  C  K C ' = m C ' =  m    K f  Km K f R' = K R ω' = K ω  R' = R ω' = K ω  R' = K R ω' = ω   m f   f   m →  C L →  C K →  C = = = C ' = = m C ' = L' Km L  C ' L'   L' L    Km K f   K f K f   Km K f ω' = ω  ω' = K ω  R' = R ω' = K ω  R' = K R  R' = K R f   f   m C →  m C →  L C →  K = = = L' = = L' = m L C '  L' Km L C '   C '   Km  K f   K f Km K f   K f Frequency Response - sites.google.com/site/ncpdhbkhn 77 Scaling (2) Ex. – + + The cutoff frequency is 500 rad/s. Scale the circuit for 1k Ω a cutoff frequency of 25 kHz using a 10-kΩ resistor? v( t ) s 2 H vo – ω'= 2 π ×× 25 103 = 5 π × 10 4 rad/s 4 ω π × + =' = 5 10 = π – + K f 100 10k Ω ω 500 v( t ) s vo R ' 10 – K = = = 10 0.064 H m R 1 ω= ω ' K f  K 10  →=L'm L = 2 = 0.064 H = π R' Km R  K f 100 Frequency Response - sites.google.com/site/ncpdhbkhn 78