C_Stottlemeyer_ECET350_WK1_LAB

Laboratory Report Cover Sheet DeVry University College of Engineering and Information Sciences Course Number: ECET350 Professor: Rami Abousleiman Laboratory Number: 1 Laboratory Title: Sallen-Key Active Filter Design Submittal Date: 5/12/2019 Objectives:  Design and simulate a Butterworth, low-pass Sallen-Key active filter.  Construct and test the designed Butterworth, low-pass Sallen-Key active filter. Results: Summarize your results in the context of your objectives. A.) The simulated low pass filter worked as intended with the exception of the cut off frequency being above the expected specification. Conclusions: What can you conclude about this lab based on your results? A.) I have basic understanding of how to design a Butterworth filter using the prototype equations. Student: Name Program Signature Course Number: ECET-350 Laboratory Number: 1 Page 1 of 9

Observations, Measurements, and Calculations Step 1 1. H ( s ) = ¿ 355.306 E 6 Ho s 2 + 266.479 E 3 s + 355.306 E 8 b o = ¿ 355.306 E 8 rad/sec b 1 = ¿ 266.479 E 3 rad/sec 2. R 1 = R 2 = √ 1 b o C 1 C 2 = 5.305 µΩ R 3 = 10 kΩR 4 = 5.86 kΩ 3. H o = G = 1 + R 4 R 3 = 1.586 G ( dB ) = 20log G = 4.006 dB ϵ 2 = 10 A p ( dB ) 10 − 1 = ¿ 1 M C = H o √ 1 + ϵ 2 = 2.82843 M C ( dB ) = 20 log M C = ¿ 0.903091 dB Roll-off rate: RR ≅ − 20 N = ¿ -40 dB/decade 4. Record your measured response values from the multisim simulation in Table 1. Meas. (dB) Meas. Meas. Course Number: ECET-350 Laboratory Number: 1 Page 2 of 9

Paste the steady state frequency response from the bode-plotter here in the space provided. Steady State Frequency Response—Bode Plot. Frequency Response—Wide View. Course Number: ECET-350 Laboratory Number: 1 Page 4 of 9

Step 2 Record the measured output of your filter voltage in Table 4, and then calculate and record the dB gain in the adjacent column. You may also copy your Excel data table of filter response measurements instead of completing this table. Frequency (Hz) Vin (peak to peak) Measured Filter Vout (peak to peak) Calculated Filter Gain 20 log 10(Vout/Vin) (dB) 300 1.998 3.17 4.009275567 600 1.998 3.167 4.00105159 900 1.998 3.15 3.954301398 1200 1.998 3.136 3.915611402 1500 1.998 3.141 3.92944905 1800 1.998 3.051 3.676934475 2100 1.998 2.962 3.419791406 2400 1.998 2.843 3.063627516 2700 1.998 2.684 2.563740552 3000 1.998 2.487 1.936461169 3300 1.998 2.311 1.264089232 3600 1.998 2.087 0.378539304 3900 1.998 1.941 -0.25139897 4200 1.998 1.763 -1.086863432 4500 1.998 1.564 -2.127174703 4800 1.998 1.412 -3.015215743 5100 1.998 1.279 -3.874498788 5400 1.998 1.157 -4.745242499 5700 1.998 1.057 -5.530409932 6000 1.998 0.963 -6.339383935 6300 1.998 0.878 -7.14201936 6600 1.998 0.802 -7.928422312 6900 1.998 0.751 -8.499110938 7200 1.998 0.694 -9.184720269 7500 1.998 0.643 -9.847690219 7800 1.998 0.587 -10.63914765 8100 1.998 0.543 -11.31591309 8400 1.998 0.508 -11.89463543 8700 1.998 0.475 -12.47803749 9000 1.998 0.444 -13.06425028 9300 1.998 0.416 -13.63004307 9600 1.998 0.391 -14.16837453 9900 1.998 0.368 -14.6949533 Table 4: Filter Output Measurements and Calculations 5. Paste your Excel graph of the Butterworth, low-pass filter response in the space provided. Course Number: ECET-350 Laboratory Number: 1 Page 5 of 9

2. What was the average dB per decade attenuation of your filter? This answer will need to be calculated since a full decade of frequency response measurements was not taken. A.) Average attenuation = − 3.933 dB −(− 53.341 dB ) log ( 5.128 kHz 100 kHz ) = -38.3 dB/decade 3. Compare the dB per octave attenuation measurement to the design specifications. Is it close to the expected value? A.) The dB per octave for a second-order Butterworth low-pass Sallen-Key filter is -12 dB/octave. The design produced in this experiment has a roll-off of -11.49 dB/octave, which has a margin of error of 4.25 percent. 4. Compare the dB per decade attenuation calculation to the design specifications. Is it close to the expected value? A.) The experimental design has a dB per decade attenuation of -38.3 dB, which is, again, 4.24 % off from the expected attenuation of -40 dB/decade. 5. What is the measured cutoff frequency of your filter? A.) The measured cutoff frequency is 3.37 kHz. 6. How does it compare to the design specifications? A.) The design specifies a 3.0 kHz cutoff frequency, so the measured cutoff frequency of 3.37 kHz is 12.33 % higher than the specification. 7. How closely does the overall performance of the constructed, second-order, low-pass Butterworth filter compare with the simulated version? A.) The performance of the constructed second-order low-pass Butterworth filter is, overall, close to the design specifications. The expected gain of four was achieved, and the roll-off rate is within five percent of the original specification. The largest Course Number: ECET-350 Laboratory Number: 1 Page 7 of 9

deviance is the cutoff frequency of the prototype filter, which, at 3.37 KHz, is significantly higher than the anticipated cutoff of 3.0 KHz. 8. List and explain any possible reasons for differences between the simulated and constructed filter. A.) There is a difference due to the resistors used. I picked the best available resistors that were closest to the calculations. Course Number: ECET-350 Laboratory Number: 1 Page 8 of 9