1. - 1 - YLD 10/2/99ESINSA Tools.

2. - 2 - YLD 10/2/99ESINSA Filters Performances A filter should maintain the signal integrity. A signal does not exist alone. Signal and noise coexist. Data on a signal without a reference to its noise background are MEANINGLESS.

3. - 3 - YLD 10/2/99ESINSA Filters Performances A few performances worth to consider: Transfer function, Impulse and Step Responses, Total Signal to Noise Ratio (TSNR), Idle Channel Noise (ICN), Total Harmonic Distortion (THD), Group Delay, Gain tracking, Frequency tracking, Dynamic range, Non Linear Behavior, Power efficiency, etc..

4. - 4 - YLD 10/2/99ESINSA Tools The preferred way is to evaluate the filters performances is to simulate in the TIME DOMAIN, to analyze in TIME and FREQUENCY DOMAINS. Electrical Simulations: SPICE,.. System Simulations: C, VHDL,.. Mathematica, MatLab,.. Dedicated Simulators:SwitCap,..

5. - 5 - YLD 10/2/99ESINSA How to simulate? Digital Filters are the simplest to simulate. Analog Filters in the sampled domain are also easy to simulate, if the sampling is rigorously periodic. Jitter is causing a mathematical nightmare. Analog Filters in the continuous time domain are tricky to simulate. Master the sampling techniques. Be an expert in FFT and data windowing. Be a guru of the decimation and interpolation.

6. - 6 - YLD 10/2/99ESINSA Test Signals.

7. - 7 - YLD 10/2/99ESINSA Non Linearities In principle, linear theory does not apply. For fairly linear circuits (*), we will suppose it applies! (*) signal not ‘too’ large! No transfer function exists in a non linear circuit. Nothing is granted. The test signals are important.

8. - 8 - YLD 10/2/99ESINSA Test Signals All partners must agree on the test conditions! Several test signals are often used. Sinewave, step and impulse are the most classical and described in theorical books. We will see that other test signals are useful. A test signal should exercise an IC in such a way that the user should have no surprise in the real applications with real signals. Tough.

9. - 9 - YLD 10/2/99ESINSA A Few Test Signals (time) a b c d

10. - 10 - YLD 10/2/99ESINSA Same Test Signals (frequency) dB

11. - 11 - YLD 10/2/99ESINSA Same Test Signals (histogram) a a a a

12. - 12 - YLD 10/2/99ESINSA Same Test Signals Pseudo Random Uniformly Distributed Noise Pseudo Random Normally Distributed Noise Pseudo Random Coherent Periodic Noise Coherent Sinewave

13. - 13 - YLD 10/2/99ESINSA Sinewave. -0.5 0 0.5 1 1.5 00.00020.00040.00060.00080.0010.00120.00140.0016 current time 1.000 Vpeak -3.010 dBRMS 0 dB ref = 1.0V RMS

14. - 14 - YLD 10/2/99ESINSA (FFT) 0 dB ref = 1.0V RMS -6.030042 dB ??

15. - 15 - YLD 10/2/99ESINSA (FFT Zoom) -160 -140 -120 -100 -80 -60 -40 -20 0 12001210122012501260 freq dB 12301240 Signal leakage Window resolution? FFT resolution = 1Hz -6.030042 dB Signal = -3.010 dB 0 dB ref = 1.0V RMS

16. - 16 - YLD 10/2/99ESINSA Gaussian Noise 1.000 VRMS 0.000 dBRMS (Bandwidth 500.0 kHz) 0 dB ref = 1.0V RMS

17. - 17 - YLD 10/2/99ESINSA (FFT) 0 dB ref = 1.0V RMS Power [0..5kHz] = -20.011dB

18. - 18 - YLD 10/2/99ESINSA (FFT Zoom) -90 -85 -80 -75 -70 -65 -60 -55 -50 -45 1200121012201230124012501260 freq dB FFT resolution = 1Hz 0 dB ref = 1.0V RMS Power [0..5kHz] = -20.011dB

19. - 19 - YLD 10/2/99ESINSA Lesson learned: Document! -120 -100 -80 -60 -40 -20 0 050010001500200025003000 freq dB 0 dB reference? Windowing? FFT Resolution?

20. - 20 - YLD 10/2/99ESINSA TSNR and THD Sinewave at input, FFT on output. Evaluate at the output: the power of the fundamental the power of the harmonics the power of the noise SNR is the ratio between signal and noise TSNR is the ratio between signal and (noise + harmonics) THD is the ratio between signal and harmonics

21. - 21 - YLD 10/2/99ESINSA Method 1: Frequency Domain Analysis Method 2:Time Domain Analysis TSNR and THD

22. - 22 - YLD 10/2/99ESINSA Example 1.0 V peak Sinewave @ 1234.5 Hz 1.0 V RMS normally distributed noise, BW=500kHz F sampling = 1.0 MHz Evaluation on 1 000 000 points

23. - 23 - YLD 10/2/99ESINSA TSNR [ method 1 ] 0 dB ref = 1.0V RMS

24. - 24 - YLD 10/2/99ESINSA TSNR [ method 1 ] And the result is… Power of Fundamental:-2.999184 dB- 3 dB Power of Noise [0..5 kHz]:-20.09885 dB-20 dB TSNR [0..5 kHz]: 17.09967 dB 17 dB

25. - 25 - YLD 10/2/99ESINSA TSNR [ method 2 ] Sinewave at input, Linear Curve Fitting on output. Output signal must be prefiltered as it is not possible to limit otherwise the bandwidth of analysis. Residue 0 = Signal with k = 0..H Fit k = r k * Sin[k  t +  k ] Residue k+1 = Residue k - Fit k Fundamental= r 1 Harmonics= Sqrt[(r 2 ) 2 +.. + (r H ) 2 ] Noise= Power_of[Residue H+1 ]

26. - 26 - YLD 10/2/99ESINSA TSNR [ method 2 ] -4 -3 -2 0 1 2 3 4 00.00050.0010.00150.0020.0025 time output

27. - 27 - YLD 10/2/99ESINSA TSNR [ method 2 ]

28. - 28 - YLD 10/2/99ESINSA TSNR [ method 2 ] OK! Data are not filtered! And the result is… TSNR [0..500 kHz]: -2.999201 dB -3 dB

29. - 29 - YLD 10/2/99ESINSA THD [ method 2 ] time 1.0 V peak Sinewave @ 1234.5 Hz 0.2 V peak Sinewave @ 2469.0 Hz 1.0 V RMS normally distributed noise, BW=500kHz F sampling = 1.0 MHz Evaluation on 100 000 points

30. - 30 - YLD 10/2/99ESINSA THD [ method 2 ] time -5 -4 -3 -2 0 1 2 3 4 5 00.00050.0010.00150.0020.0025 output

31. - 31 - YLD 10/2/99ESINSA THD [ method 2 ]

32. - 32 - YLD 10/2/99ESINSA THD [ method 2 ]

33. - 33 - YLD 10/2/99ESINSA THD [ method 2 ] And the result is… SNR: -3.082609 dB -3 dB SDR: 13.95151 dB 14 dB THD:4 %

34. - 34 - YLD 10/2/99ESINSA Dynamic Range Running the TSNR analysis for a set of sinewaves with various amplitudes allows to determine the dynamic range. An example on a Sigma-Delta Converter.

35. - 35 - YLD 10/2/99ESINSA Dynamic Range Amplitude (output) 0 20 40 60 80 100 120 -80-70-60-50-40-30-20-100 Analog Sigma Delta Modulator tsnr 80 dB 40 dB 60 dB 20 dB 100 dB 120 dB 140 dB

36. - 36 - YLD 10/2/99ESINSA Transfer Function White noise at input, FFT on input and output, RMS Average! avg[FFT( in). FFT(out)*] TF = ------------------------- avg[FFT( in). FFT( in)*] for a ‘fairly’ linear circuit.

37. - 37 - YLD 10/2/99ESINSA Transfer Function -70 -60 -50 -40 -30 -20 -10 0 10 1000100001000001e+006 freq Biquadratic Filter, statistical study, Sigma = 0.05 mag

38. - 38 - YLD 10/2/99ESINSA Impulse Response White noise at input, FFT on input and output, RMS Average! Iresp = IFFT(TF) for a ‘fairly’ linear circuit.

39. - 39 - YLD 10/2/99ESINSA Impulse Response -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 05e-0050.00010.000150.00020.000250.00030.000350.0004 time Nyquist FIR Filter, 365 coefficients impulse response FIR coefficients... h[174] 0.037855 h[175] 0.044183 h[176] 0.050129 h[177] 0.055516 h[178] 0.060176 h[179] 0.063964 h[180] 0.066760 h[181] 0.068475 h[182] 0.069052 h[183] 0.068475 h[184] 0.066760 h[185] 0.063964 h[186] 0.060176 h[187] 0.055516 h[188] 0.050129 h[189] 0.044183 h[190] 0.037855...

40. - 40 - YLD 10/2/99ESINSA Step Response Step Response is the integral of the Impulse Response. for a ‘fairly’ linear circuit.

41. - 41 - YLD 10/2/99ESINSA Step Response -0.2 0 0.2 0.4 0.6 0.8 1 1.2 05e-0050.00010.000150.00020.000250.00030.000350.0004 time step response Nyquist FIR Filter, 365 coefficients

42. - 42 - YLD 10/2/99ESINSA Remark on FFT (1) How the FFT of a sinewave behaves when the number of sampling points changes? when the sampling frequency changes? 1.0 V peak Sinewave @ 5432.1 Hz a.Fs = 1.0 MHz, N = 100 000 points b.Fs = 1.0 MHz, N = 200 000 points c.Fs = 2.0 MHz, N = 100 000 points d.Fs = 2.0 MHz, N = 200 000 points

43. - 43 - YLD 10/2/99ESINSA Remark on FFT (2) -160 -140 -120 -100 -80 -60 -40 -20 0 5200525053005350540054505500555056005650 freq A = 1 MHz, 100 000 pts B = 1 MHz, 200 000 pts C = 2 MHz, 100 000 pts D = 2 MHz, 200 000 pts 0 dB ref = 1.0V RMS A,D C B

44. - 44 - YLD 10/2/99ESINSA Remark on FFT (3) How the FFT of a sinewave behaves when the number of samples changes? the sampling frequency changes? It depends on the FREQUENCY RESOLUTION of the FFT. Sampling Frequency Frequency Resolution = ------------------ Number of Samples