Analog Circuits and Systems Prof. K Radhakrishna Rao Lecture 27: State Space Filters 1
Review Q enhancement of passive RC using negative and positive feedback Effect of finite GB of the active device on filter parameters LP Passive RC second order ω = ω + K 1 + 1 0 p Qa Qp K where ω p is the passive filter normalizing frequency where is the passive filter Q; Q p K is the gain of the inverting amplifier used in the negative feedback loop 2
Review (contd.,) HP passive RC second order ω ω p,q = Q 1 + K 0 a p 1 + K LP + HP (notch) passive ω ω = + ( ),Q Q 1 KQ 0 p a p p BP passive filter (RC) passive feedback Q ω ω,q = p 0 p a 1 KQ p 3
Review (contd.,) Positive feedback because of high sensitivity to K used for only low Q a Negative feedback because of low sensitivity to K but sensitivity to GB in case of 3 and 4 Independent Q adjustment and w 0 adjustment not possible Q = ( GB) 1 Q ( total phase lag error in the loop) a Q a fq 0 a product is the criteria for low sensitivity to GB of Q a Q a ( φ) V = 1 4
State Variable Filters Are also known as Biquad filters (use two integrators) KHN filters (Kervin, Heulessman and Newcomb of Burr-Brown) Universal Active filters (UAF) 5
Active filter design as solution of differential equation n th order linear differential equation n th order filter design n n 1 n 1 o n 1 L 0 o i i dv d V o + L K + K V = KV dt n dt n n 1 n 1 o n 1 L 0 o i i dv d V o = K + K V KV dt n dt th ( 1) from n state variable derive n- state variables using integrators Then sumup these with input. Connect the summer output to input results in the solution of nth order differential equation. 6
First-Order Filter Is represented by a first order differential equation dv dt o + KV = 0 o KV i i rewritten as dv dt o = -KV + 0 KV o i i 7
Simulation of LP and HP filters using ideal integrators 8
First-order filter using Op Amps 9
Second-order filter can be represented by a second order differential equation 2 dv dt o 2 dv + K o + K V = KV 1 dt 0 o i i rewritten as 2 dv dt o 2 ω dv = - 0 o - ω 2 V + H ω 2 V Q dt 0 o 0 0 i 10
Simulation of LP, HP and BP filters using ideal integrators; f 0 =1.59kHz, Q=5 11
Phase Plot using ideal integrators; f 0 =1.59kHz, Q=5 12
Transient Plot using ideal integrators; f 0 =1.59kHz, Q=5 13
Second-order filter using Op Amps 14
Simulation Second-order filter with Op Amps (where effect of GB is minimal and f 0 is 1.59 khz; Q=5) 15
Simulation Phase Plot (where effect of GB is minimal and f 0 is 1.59 khz; Q=5 changed from 1, 5 and 9) 16
Simulation Transient (where effect of GB is minimal and f 0 is 1.59 khz; Q=5) 17
Outputs of UAF for a square-wave input at f 0 18
Simulation of Second-order filter with Op Amps (where effect of GB is significant and f 0 is 15.9 khz; Q=5) The effect of finite GB is on the peak and notch 19
Simulation Transient (where effect of GB is significant and f 0 is 15.9 khz; Q=5) 20
Third-order filter 3 2 o o o 3 2 2 1 0 dv dv dv + K + K + K V = KV dt o i i dt dt rewritten as 3 2 o o o 3 2 2 1 0 dv dv dv = -K - K - K V + K V dt o i i dt dt 21
Third-order filter using Ideal Integrators 1 3 0 and 2 3 2 3 s s s s s s 1+ 2 + 2 + 1+ 2 + 2 + ω ω ω ω ω ω 2 3 2 3 0 0 0 0 0 0 s ω 3 22
Third-order filter using Ideal Integrators 23
Third-order Butterworth using LF353 or TL082 (where the effect of GB is minimal and f 0 is 1.59 khz ) 24
Third-order Butterworth using LF353 or TL082 (where the effect of GB is significant and f 0 is 15.9 khz) 25
Butterworth Low-Pass Filter Synthesis as third-order filter (Using LM741) 26
Butterworth Low-Pass Filter Synthesis as second-order filter followed by first-order filter (Using LM 741) 27
Observations Higher even-order filters can be realized by cascading second order filters functions. Higher odd-order filters is can be realized cascading one first-order filter with required number of second order filters. Direct realization of higher order (> 3) using any of the Op Amps will lead to inferior performance due to cumulative phase error in the feedback loop 28
Outputs at different points in a second-order filter Output can be taken at several points in the circuit: V, V, V and V o1 o2 o3 o4 Input, output relationships ( H ω 2) s 2 0 0 ( 2 2) ( ) Vo1 = V s ω + s ω Q + 1 i 0 0 - High Pass Filter 29
Simulation Gain at ω? ω is H and at ω=ω is H Q; Q=5; H = 1; f = 1.59kHz 0 0 0 0 0 0 30
Outputs at different points in a second-order filter ( H ω ) V s o2 = 0 0 V s ω + s ω Q + 1 ( 2 2) ( ) i 0 0 - Band Pass Filter Gain at ω=ω 0 is H Q 0 31
Simulation Q=5; H0 0 = 1; f = 1.59kHz The image cannot be displayed. Your computer may not have enough memory to open the image, or the image may have been corrupted. Restart your computer, and then open the file again. If the red x still appears, you may have to delete the image and then insert it again. 32
Outputs at different points in a second-order filter V H o3 = 0 V s ω + s ω Q + 1 ( 2 2) ( ) i 0 0 - Low Pass Filter Gain at ω= ω is H and at ω=ω is H Q 0 0 0 0 33
Simulation Q=5; H0 0 = 1; f = 1.59kHz 34
Outputs at different points in a second-order filter ( 2 2) H V 1+ s ω o4 0 0 = V s ω + s ω Q + 1 ( 2 2) ( ) i 0 0 - Band Stop Filter Gain at ω= ω and ω? ω is H, and at ω=ω is zero 0 0 0 0 35
Simulation Q=5; H0 0 = 1; f = 1.59kHz 36
Adding V o1, V o2 and V o3 It is possible to realize any second order filter function α + α + α 2 + + 1 o1 2 o2 3 o3 = V V V as bs c where a 1, a 2 and a 3 can be negative, positive or zero and a, b and c can be positive or negative and of zero or any non-zero value ( 2 2) ( ) V s ω + s ω Q + 1 i 0 0 37
All pass filter design α + α + α 2 + + 1 o1 2 o2 3 o3 = V V V as bs c ( 2 2) ( ) V s ω + s ω Q + 1 i 0 0 = ( 2 2) ( 0 0 ) ( 2 2) ( 0 0 ) s ω s ω Q + 1 H s ω + s ω Q + 1 0 ω φ= 2tan ω 1 0 1 Q ω ω 0 2 38
Simulation - All pass filter design H 0 =1;Q=1;f 0 =1.59kHz; a 1 = a 2 = a 3 =1 39
Conclusion 40