4/25/2016

Lesson 7: Anti- Aliasing Filtering

ET 438b Sequential Control and Data Acquisition

Department of Technology

Lesson 7_et438b.pptx 1

Learning Objectives

After this presentation you will be able to:

 Explain why anti-aliasing filters are used.

 Design a first order anti-aliasing filter using an OP AMP

 Design a 2nd order anti-aliasing filter using an OP AMP

 Verify the performance of an anti-aliasing filter using simulation software.

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Anti-Aliasing Filters Dealing with Aliasing in practical systems

Exact frequency components of a sampled signal are unknown.

Can not determine if signal component is alias or real

Sampling rate limited by hardware selection.

Lab DAQ cards rate is 200 kHz-250 kHz.

fnyquist = 100 kHz – 125 kHz sets input frequency limits.

Bandwidth limit input signals using anti-aliasing filters to eliminate

frequencies above fnyquist.

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Anti-Aliasing Filters

Goal- reduce amplitude of all frequencies above fnyquist to zero level Ideal low pass filter Bode plot

Cut-off

Frequency, fc

Set fc = fnyquist for perfect

Amplitude Stop band signal elimination

Frequency

Pass band Cut-off Frequency, fc Practical (Butterworth) filters

have sloping characteristics Amplitude 1st Order: -20 dB/decade nd 2 Order: -40 dB /decade Frequency 3rd Order: -60 db/decade Lesson 7_et438b.pptx 4

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Anti-Aliasing Filters: OP AMP Circuits

First order Butterworth filter

R2 Design formulas

C1  R2 Av  Voltage gain in the pass band -12V R1 U1 R1 UA741 1 Vin Vo1 fc  Cut - off frequency (Hz) + 2R2C1 To DAQ Channel 12V

Design Procedure

1.) Determine the acceptable level of signal gain, Av, at sampling frequency fs 2.) Use the formula below to determine the value of fc based of the designed signal gain f f  s for A 1 c 2 v 1 Av 2 2 Av 3. Select a value of C1 from standard values and compute value of R2

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Anti-Aliasing Filters: OP AMP Circuits Design Procedure (continued) 4.) Set R1=R2 to give pass band gain of -1 (o dB). Amplify signal after filtering to reduce noise and unwanted signal components. (e.g. 60 Hz)

Design Example: A data acquisition systems samples an analog signal at st Ts=0.0004 seconds. Design a 1 order anti-aliasing filter that will reduce the voltage level of all signal above the Nyquist frequency to 0.4

Solution: Determine the sampling frequency from Ts then find the fnyquist

1 1 fs 2500 Hz fs    2500 Hz fnyquist   1250 Hz Ts 0.0004 s 2 2

f 2500 Hz 2500 Hz f  s    546 Hz Find the value of fc for the c 2 2 1 Av 1 0.4 2 5.25 given level of Av above 2 2 2 2 Av 0.4 fnyquist Lesson 7_et438b.pptx 6

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Design Example (continued) Select a capacitor value of 0.1 mF. Use larger values for lower f’s to keep 2700  R2 values of resistors in range of 1k to 820k C1 0.1 mF 1 f  1 -12V c 2R2C1 R2  R1 U1 UA741 2C1fc Vin Vo1 2700  + 1 To DAQ Channel R2   2916  12V 20.1106 546 Set R1=R2 to give gain of -1 Select standard value of 2700 ohms and Design complete. Check result with compute value of fc 1 circuit simulation. C1  590 Hz 20.1106 2700

@ 1.254 kHz Av=-7.4 dB (dB/20) dB=20Log(Av) 10 =Av (-7.4/20) 10 =Av=0.43

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Anti-Aliasing Filters- 2nd Order Filters Second order Butterworth filter C3 Unity gain is fixed by negative feedback loop in this design. 12V U2 R4 R5 UA741 + Vin Vo2 Use same design procedure as To DAQ Channel C2 -12V 1st order filter but use following formula to find cut-off frequency. f Design formulas f  s for A 1 c 2 v 1 Av 4 2 2 Av 1 Voltage gain in the pass band Av 1 f  Cut - off frequency (Hz) c 2 R4R5C3C2 2nd order filter produces -40 With R4  R5 and C3  2C2 dB/decade rolloff in stop band

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Anti-Aliasing 2nd Order Filter Design

Design Example 2: Repeat the design of the 1st order filter example using a 2nd order filter nd Ts=0.0004 seconds sampling. Design a 2 order anti-aliasing filter that will reduce the voltage level of all signal above the Nyquist frequency to 0.4

Same fs and fnyquist as previous case

1 1 fs 2500 Hz fs    2500 Hz fnyquist   1250 Hz Ts 0.0004 s 2 2

f 2500 Hz 2500 Hz Find the value of fc for the f  s    826 Hz nd c 2 2 4 given level of A using 2 1 Av 1 0.4 2 5.25 v 4 4 2 2 2 2 order filter formula Av 0.4

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Design Example (continued) C3 Select value of C2 and compute C3. Let C2=0.1 mF so….. 0.2 mF 12V U2 R4 R5 UA741 1.5 k + m m Vin C3= 2∙C2=2∙(0.1 F)= 0.2 F Vo2 1.5 k To DAQ Channel C2 -12V 0.1 mF

1 R4  Use design formulas to find value of R4=R5 2 2C2fc 1 1 R4  6 fc  With R4  R5 and C3  2C2 2 20.110 F826 Hz 2 R4R5C3C2 1 1 fc   R4 1363   R5 2 R42 2C22 2 2 R4C2 Use standard value of 1.5k. Solve for R4 given C2 and f Response very sensitive to R c values 1 1 R4  fc  2 2 R4C2 2 2C2fc

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Design Example (continued) Design complete. Check result with circuit simulation.

@ 1.249 kHz Av=-9.4 dB (dB/20) dB=20Log(Av) 10 =Av (-9.4/20) 10 =Av=0.34

Flatter response in pass band sharper roll off.

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End Lesson 7: Anti- Aliasing Filtering ET 438b Sequential Control and Data Acquisition

Department of Technology

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