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

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

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

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

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

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

YLD 10/2/99ESINSA Sinewave current time Vpeak dBRMS 0 dB ref = 1.0V RMS

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

YLD 10/2/99ESINSA (FFT Zoom) freq dB Signal leakage Window resolution? FFT resolution = 1Hz dB Signal = dB 0 dB ref = 1.0V RMS

YLD 10/2/99ESINSA Gaussian Noise VRMS dBRMS (Bandwidth kHz) 0 dB ref = 1.0V RMS

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

YLD 10/2/99ESINSA (FFT Zoom) freq dB FFT resolution = 1Hz 0 dB ref = 1.0V RMS Power [0..5kHz] = dB

YLD 10/2/99ESINSA Lesson learned: Document! freq dB 0 dB reference? Windowing? FFT Resolution?

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

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

YLD 10/2/99ESINSA Example 1.0 V peak Hz 1.0 V RMS normally distributed noise, BW=500kHz F sampling = 1.0 MHz Evaluation on points

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

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

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 ) (r H ) 2 ] Noise= Power_of[Residue H+1 ]

YLD 10/2/99ESINSA TSNR [ method 2 ] time output

YLD 10/2/99ESINSA TSNR [ method 2 ]

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

YLD 10/2/99ESINSA THD [ method 2 ] time 1.0 V peak Hz 0.2 V peak Hz 1.0 V RMS normally distributed noise, BW=500kHz F sampling = 1.0 MHz Evaluation on points

YLD 10/2/99ESINSA THD [ method 2 ] time output

YLD 10/2/99ESINSA THD [ method 2 ]

YLD 10/2/99ESINSA THD [ method 2 ]

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

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.

YLD 10/2/99ESINSA Dynamic Range Amplitude (output) Analog Sigma Delta Modulator tsnr 80 dB 40 dB 60 dB 20 dB 100 dB 120 dB 140 dB

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.

YLD 10/2/99ESINSA Transfer Function e+006 freq Biquadratic Filter, statistical study, Sigma = 0.05 mag

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.

YLD 10/2/99ESINSA Impulse Response e time Nyquist FIR Filter, 365 coefficients impulse response FIR coefficients... h[174] h[175] h[176] h[177] h[178] h[179] h[180] h[181] h[182] h[183] h[184] h[185] h[186] h[187] h[188] h[189] h[190]

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

YLD 10/2/99ESINSA Step Response e time step response Nyquist FIR Filter, 365 coefficients

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 Hz a.Fs = 1.0 MHz, N = points b.Fs = 1.0 MHz, N = points c.Fs = 2.0 MHz, N = points d.Fs = 2.0 MHz, N = points

YLD 10/2/99ESINSA Remark on FFT (2) freq A = 1 MHz, pts B = 1 MHz, pts C = 2 MHz, pts D = 2 MHz, pts 0 dB ref = 1.0V RMS A,D C B

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