Active Filters: realization of complex left hand plane poles (Complex Transfer Functions regarding the frequency) using only and for the passive elements.

Xavier Gago

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Analogue Filter Characteristics

Passive Filters: L, C and R Active Filters: A.O., C and R R2 L1 C1 XL = jLω 1 R1 C1 XC = jCω 2 L2 3 Vs C2 1 Vs

Active Filters: advantages 1.- Full isolation between stages (full control of input and output impedances) - Easy design, regardless Z load - Easy multi-stage design 2.- It is possible to design complex filters without using L (expensive) 3.- Fast for a wide range of applications

Active Filters: Disadvantages

1- They require power supply VCC (but circuit Vcc is available) 2- Limited bandwidth (due to the OPAMs). 3- Low power levels (impossible to use in speakers, mains filters,..) 2

1 Analogue Filter Characteristics

Low Pass Filters types High Pass

10 10

5 5

0 0

-5 -5

-10 -10

-15 -15

-20

-20 modulo modulo

-25 -25

-30 -30

-35 -35

-40 -40

-45 -45 0 1 2 3 4 5 0 1 2 3 4 5 10 10 10 10 10 10 10 10 10 10 10 10 f(Hz) f(Hz )

Band Pass Band Reject

10 10

0 0 -10 Example: Low-Pas -10 -20

-30 -20

modulo -40 modulo -30

-50 -40 -60

-50 -70

-80 0 1 2 3 4 5 6 -60 2 3 4 5 6 10 10 10 10 10 10 10 10 10 10 10 10 f(Hz) f(Hz) 3

Analogue Filter Characteristics

Filter Order

Filter order = Vi(s) VO(s) N(s) H(s) = Pole number = H(s) D()s D(s) order

s = jω, complex variable, 10

0 VO ()s H(s) = -10 first order Vi ()s -20

-30

N order -40

-50 modulo

-60

-70 fourth order

Slope rate: 20N dB/dec -80

-90

-100 2 3 4 5 6 The higher ther order, the better the filter ! 10 10 10 10 10 f(Hz) 4

2 Filter Order and Attenuation

The Gain H(s) for a low pass filter in the attenuation zone is |H| = k / ωn

|H1| 20log H1 =20log k - 20n logω1

20log H2 =20log k - 20n logω2 |H2|

20log H1 - 20logH2 =20n log (ω2 - ω1 )

ω1 ω2 H (dB) - H (dB) 1 2 =20 n log ω2 - logω1

For a two frequencies ω 1 and ω 2 separated by a decade,

H1(dB) - H2 (dB) =20 n dB / dec Useful to determinate the filter order for a pre-fixed attenuation

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Analogue Filter Characteristics Filter Order

First Order Filters:

Low Pass High Pass ω s H(s) = G⋅ O H(s) = G⋅ s +ω O s +ω O

Second Order Filters: Transfer Function

Low Pass High Pass

ω 2 s 2 H(s) = G⋅ O H(s) = G⋅ 2 2 2 2 s + a⋅ω O ⋅s +ω O s + a⋅ω O ⋅s +ω O

Band Pass Band Reject

s s 2 +ω 2 H(s) = G⋅ H(s) = G⋅ Z s 2 + a⋅ω ⋅s +ω 2 2 2 O O s + a⋅ω O ⋅s +ω O 6

3 Analogue Filter Characteristics Filter Response

It determines the pass band characteristics

1- Butterworth 10

5 a- Moderate phase and amplitude response 0 b- Moderate oscillations for pulses -5

-10 Butterworth

2- Bessel -15 Bessel modulo -20 Chevyshev 1 dB a- Linear response -25 Chevyshev 2 dB b- Low oscillations in case of pulses -30 Chevysh3ev 1 dB c- Need high order -35

-40 2 3 4 5 10 10 10 10 3- Chebyshev f(Hz) a- Fast decrement of the gain b- Poor phase response c- Large oscillations in case of pulses d- NO monotone filter: in the bandwidth

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Response filter comparisons

Butterworth Filter • Flat Response and Steep Attenuation • OK for Step Function

Chebyshev Filter • Steeper than Butterworth • More Non-Linear than Butterworth 8

4 Analogue Filter Characteristics

Chebyshev Filters

Larger ripple Faster gain decrement

10 10

0 5 -10 0 -20

-5 -30

-40 -10 3dB Orden 3 -50 modulo modulo 1dB -15 -60 Orden 6

-70 -20 -80 -25 -90

-30 -100 2 3 4 5 2 3 4 5 6 10 10 10 10 10 10 10 10 10 f(Hz) f(Hz )

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First Order Active Filters

First order low pass filter

XBP1 1 1 VO ()s = ⋅ Vi ()s in out C ⋅s R ⋅C ⋅s +1 V ()s = 1 ⋅V ()s 1 1 O 1 i 2 +R1 ω O 1 R1 3 C1 ⋅s H(s) = ω O = 1 s +ω R C 1kohm O 1 1 1 C1 1uF C ⋅R 1 Vs H()s = 1 1 ω = ω = 1 C 3dB O s + R1C1 C1 ⋅R1 ω0 = pole frequency

10 0

0 -20

-10 -40 -20

-60 -30 fase

modulo -40 -80

-50 -100 -60

-120 -70

-80 -140 0 1 2 3 4 5 6 0 1 2 3 4 5 6 10 10 10 10 10 10 10 10 10 10 10 10 10 10 f(Hz ) f(Hz) 10

5 n- order Active Filters

Second order low pass filter

10 Type a kO Bessel 1.732 1.274 5 Butterworth 1.414 1 0 Chebyshev 1dB 1.045 0.863 -5 -10 Bessel Chebyshev 2dB 0.895 0.852 -15 Butterworth

Chebyshev 3dB 0.767 0.841 modulo -20 Chevyshev 1 dB -25 Chevyshev 2 dB 2 -30 ω O Chevysh3ev 1 dB H(s) = G⋅ -35 2 2 s + a⋅ω O ⋅s +ω O -40 2 3 4 5 10 10 10 10 f(Hz) ω = ω ⋅k O C 3dB O

Un-dumped natural frequency

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n- order Active Filters

Second order high pass filter

Type a kO Bessel 1.732 1.274 10 Butterworth 1.414 1 5 0 Chebyshev 1dB 1.045 0.863 -5

Chebyshev 2dB 0.895 0.852 -10

Chebyshev 3dB 0.767 0.841 -15 Bessel modulo -20 Butterworth -25 Chevyshev 1 dB 2 s -30 H(s) = G⋅ Chevyshev 2 dB s 2 a s 2 -35 + ⋅ω O ⋅ +ω O -40 Chevysh3ev 1 dB 2 3 4 5 6 10 10 10 10 10 f(Hz) ω = ω k O C 3dB O

Un-dumped natural frequency

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6 n- order Active Filters

Third order filter

Second order section First order section

Type a kO Type kO Bessel 1.447 1.274 Bessel 1.328 Butterworth 1.414 1 Butterworth 1 Chebyshev 1dB 0.496 0.911 Chebyshev 1dB 0.452 Chebyshev 2dB 0.402 0.913 Chebyshev 2dB 0.322 Chebyshev 3dB 0.326 0.916 Chebyshev 3dB 0.299

Fourth order filter

Second order section Second order section

Type a kO Type a kO Bessel 1.916 1.436 Bessel 1.241 1.610 Butterworth 1.848 1 Butterworth 0.765 1 Chebyshev 1dB 1.275 0.502 Chebyshev 1dB 0.281 0.943 Chebyshev 2dB 1.088 0.466 Chebyshev 2dB 0.224 0.946 Chebyshev 3dB 0.929 0.443 Chebyshev 3dB 0.179 0.950 13

n- order Active Filters

Fifth order filter

2º order 2º order 1º order

Type a kO akO kO Bessel 1.775 1.613 1.091 1.819 1.557 Butterworth 1.618 1 0.618 1 1 Chebyshev 1dB 0.714 0.634 0.180 0.961 0.280 Chebyshev 2dB 0.578 0.624 0.142 0.964 0.223 Chebyshev 3dB 0.468 0.614 0.113 0.967 0.178

Sixth order filter

2º order 2º order 2º order

Type a kO akO akO Bessel 1.959 1.609 1.636 1.694 0.77 1.910 Butterworth 1.932 1 1.414 1 0.518 1 Chebyshev 1dB 1.314 0.347 0.455 0.733 0.125 0.977 Chebyshev 2dB 1.121 0.321 0.363 0.727 0.0989 0.976 Chebyshev 3dB 0.958 0.298 0.289 0.722 0.0782 0.975 14

7 Sallen – Key structure: low pass filter

C1 1 2 R ⋅C1 ⋅C 2 R R U1 H()s = 2 1 1 s 2 + ⋅s + 3 2 Vs R⋅C R ⋅C ⋅C 2 1 1 2

2 C2 ω H(s) = K ⋅ O 2 2 s + a⋅ω O ⋅s +ω O In a Sallen-Key structure, K=1 2 1 a and ω from a⋅ω = ω 2 = O O R⋅C O 2 Filters Response data tables 1 R ⋅C1 ⋅C 2

1- kO is chosen as desired (ωO = kO ωC), then a is chosen in the table

2- C1 value is chosen 2 1 3- Others components are calculated as R = C = a C 2 2 2 ⋅ωO ⋅ 1 C1 ⋅R ⋅ωO 15

Sallen – Key structure: low pass filter

Example: Second order low pass filter, Butterworth response, fC = 1 kHz and Gain=1

3 2 k0=1; ωO = 2 π 10 R = = 225 k By design, C1 = 1 nF 3 −9 a = 1.414 1.414 ⋅2⋅π ⋅10 ⋅10 Attention: it may be very high value

C1 = 100 nF R = 2.25 kΩ

1 C = = 5⋅10 −8 = 50 nF 2 2 C = 50 nF 10 −7 ⋅()2.25 ⋅10 3 ⋅ 4⋅π 2 ⋅10 6 2

100nF 0 C1

XBP1 -20

2.25kohm 2.25kohm U1 in out R R -40 1 3 Vs -60

2 modulo 50nF C2 -80

-100

-120

2 3 4 5 6 10 10 10 10 10 f(Hz) 16

8 Sallen – Key structure: high pass filter

R1 s 2 H()s = 2 1 C C U1 s 2 + ⋅s + 1 2 R 2 ⋅C C ⋅R1 ⋅R 2 3 Vs 2 R2 s 2 H(s) = K ⋅ 2 2 s + a⋅ω O ⋅s +ω O

In a Sallen-Key structure, K=1 2 1 a and ω0 from a⋅ω = ω 2 = Filters Response data tables O R ⋅C O 2 2 C ⋅R1 ⋅R 2

1- kO is chosen as desired (ωO = kO ωC), then, a is chosen in the table

2- C value is chosen 2 1 R = R = 3- Components calculated as 2 1 2 2 a⋅ω O ⋅C R 2 ⋅C ⋅ω O 17

Sallen – Key structure: high pass filter

Example: Second order high pass filter, Bessel response, fC = 5 kHz and Gain=1

4 kO = 1.274 ωO = ωC / kO = 2.466 10 2 a = 1.732 By design C = 10 nF R = = 4.68 k 2 1.732⋅2.466⋅10 4 ⋅10 −8

1 R = = 3.51k 1 2 2 4.68 ⋅10 3 ⋅()(10 -8 ⋅ 2.466 ⋅10 4 )

10

3.51kohm 0 R1 -10

XBP1 10nF 10nF -20 U1 in out C C 1 -30

3 modulo Vs 2 -40 4.68kohm R2 -50

-60

-70 2 3 4 5 6 10 10 10 10 10 f(Hz) 18

9 Signal Conditioning: Summary

Signal conditioning circuits amplify low-level signals and then isolate and filter them for more accurate measurements.

Operational amplifier circuits can be used to provide the following signal conditioning: - increase the amplitude of the signal, and filter the signal, - decrease the signal output impedance, or - provide a variable gain and offset control (to calibrate transducer's output).

The dynamic range of a signal from an input transducer may be too large to process through the DAQ system (eg. the ADC is often the limiting factor). To design the right signal conditioning approach for your application, there are several steps: 1.- Know all about your sensors and what kind of signals they are supposed to produce: May need amplifiers and filters !!: 2.- Consider grounding and shielding. 3.- List your specifications and by the right data acquisition (DAQ) hardware.

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