1. Prof. Brian L. Evans Dept. of Electrical and Computer Engineering The University of Texas at Austin Lecture 4 http://courses.utexas.edu/ EE 345S Real-Time Digital Signal Processing Lab Fall 2006 Sampling and Aliasing

2. 4 - 2 Sampling: Time Domain Many signals originate as continuous-time signals, e.g. conventional music or voice By sampling a continuous-time signal at isolated, equally-spaced points in time, we obtain a sequence of numbers k  {…, -2, -1, 0, 1, 2,…} T s is the sampling period. Sampled analog waveform impulse train s(t)s(t) t TsTs TsTs Review

3. 4 - 3 Sampling: Frequency Domain Sampling replicates spectrum of continuous-time signal at integer multiples of sampling frequency Fourier series of impulse train where  s = 2  f s  G()G() ss  s  s  s  F()F() 2  f max -2  f max Modulation by cos(2  s t) Modulation by cos(  s t) Review

4. 4 - 4 Amplitude Modulation by Cosine Multiplication in time: convolution in Fourier domain Sifting property of Dirac delta functional Fourier transform property for modulation by a cosine Review

5. 4 - 5 Amplitude Modulation by Cosine Example: y(t) = f(t) cos(  0 t) Assume f(t) is an ideal lowpass signal with bandwidth  1 Assume  1 <<  0 Y(  ) is real-valued if F(  ) is real-valued Demodulation: modulation then lowpass filtering Similar derivation for modulation with sin(  0 t) which is available on slides 9-7 and 9-8  0 1     F()F()  0 Y()Y() ½         +           +    ½F    ½F    Review

6. 4 - 6 Shannon Sampling Theorem Continuous-time signal x(t) with frequencies no higher than f max can be reconstructed from its samples x(k T s ) if samples taken at rate f s > 2 f max Nyquist rate = 2 f max Nyquist frequency = f s / 2 Example: Sampling audio signals Human hearing is from about 20 Hz to 20 kHz Apply lowpass filter before sampling to pass frequencies up to 20 kHz and reject high frequencies Lowpass filter needs 10% of maximum passband frequency to roll off to zero (2 kHz rolloff in this case) What happens if f s = 2 f max ? Review

7. 4 - 7 Sampling Theorem Assumption Continuous-time signal has no frequency content above f max Sampling time is exactly the same between any two samples Sequence of numbers obtained by sampling is represented in exact precision Conversion of sequence to continuous time is ideal In Practice

8. 4 - 8 Bandwidth Bandwidth is defined as non-zero extent of spectrum in positive frequencies Lowpass spectrum on right: bandwidth is f max Bandpass spectrum on right: bandwidth is f 2 – f 1 Definition applies to both continuous-time and discrete-time signals Alternatives to “non-zero extent”? Bandpass Spectrum f1f1 f2f2 f –f2–f2 –f1–f1 Lowpass Spectrum f max -f max f

9. 4 - 9 Bandpass Sampling Bandwidth: f 2 – f 1 Sampling rate f s must greater than analog bandwidth f s > f 2 – f 1 For replicas of bands to be centered at origin after sampling f center = ½ (f 1 + f 2 ) = k f s Lowpass filter to extract baseband Bandpass Spectrum f1f1 f2f2 f –f2–f2 –f1–f1 Sample at f s Sampled Bandpass Spectrum f1f1 f2f2 f –f2–f2 –f1–f1

10. 4 - 10 Sampling and Oversampling As sampling rate increases, sampled waveform looks more like original In some applications, e.g. touchtone decoding, frequency content matters not waveform shape Zero crossings: frequency content of a sinusoid –Distance between two zero crossings: one half period. –With sampling theorem satisfied, a sampled sinusoid crosses zero the right number of times even though its waveform shape may be difficult to recognize DSP First, Ch. 4, Sampling & interpolation demo http://users.ece.gatech.edu/~dspfirst

11. 4 - 11 Aliasing Analog sinusoid x(t) = A cos(2  f 0 t +  ) Sample at T s = 1/f s x[n] = x(T s n) = A cos(2  f 0 T s n +  ) Keeping the sampling period same, sample y(t) = A cos(2  (f 0 + l f s ) t +  ) where l is an integer y[n]= y(T s n) = A cos(2  (f 0 + lf s )T s n +  ) = A cos(2  f 0 T s n + 2  lf s T s n +  ) = A cos(2  f 0 T s n + 2  ln +  ) = A cos(2  f 0 T s n +  ) = x[n] Here, f s T s = 1 Since l is an integer, cos(x + 2  l) = cos(x) y[n] indistinguishable from x[n]

12. 4 - 12 Aliasing Since l is any integer, a countable but infinite number of sinusoids will give same sequence of samples Frequencies f 0 + l f s for l  0 are called aliases of frequency f 0 with respect to f s All aliased frequencies appear to be the same as f 0 when sampled by f s

13. 4 - 13 Folding Second source of aliasing frequencies From negative frequency component of a sinusoid, -f 0 + l f s, where l is any integer f s is the sampling rate f 0 is sinusoid frequency Sampling w(t) with a sampling period of T s = 1/f s So w[n] = x[n] = x(T s n) x(t) = A cos(2  f 0 t +  )

14. 4 - 14 Aliasing and Folding Aliasing and folding of a sinusoid sin(2  f input t) sampled at f s = 2000 samples/s with f input varied Mirror image effect about f input = ½ f s gives rise to name of folding Apparent frequency (Hz) Input frequency, f input (Hz) 1000 200030004000 f s = 2000 samples/s

15. 4 - 15 DSP First Demonstrations Web site: http://users.ece.gatech.edu/~dspfirst Aliasing and folding (Chapter 4) Strobe demonstrations (Chapter 4) Disk attached to a shaft rotating at 750 rpm Keep strobe light flash rate F s the same Increase rotation rate F m (positive means counter-clockwise) Case I: Flash rate equal to rotation rate Vector appears to stand still When else does this phenomenon occur? F m = l F s For F m = 750 rpm, occurs at F s = {375, 250, 187.5, …} rpm

16. 4 - 16 Strobe Demonstrations Tip of vector on wheel: r is radius of disk  is initial angle (phase) of vector F m is initial rotation rate in rotations per second t is time in seconds For F m = 720 rpm and r = 6 in, with vector initially vertical Sample at F s = 2 Hz (or 120 rpm), so T s = ½ s, vector stands still:

17. 4 - 17 Strobe Demonstrations Sampling and aliasing –Sample p(t) at t = T s n = n / F s : –No aliasing will occur if F s > 2 | F m | –Consider F m = -0.95 F s which could occur for any countably infinite number of F m and F s values: –Rotation will occur at rate of -1.9  rad/flash, which appears to go counterclockwise at rate of 0.1  rad/flash

18. 4 - 18 Strobe Movies Fixes the strobe flash rate Increases rotation rate of shaft linearly with time Strobe initial keeps up with the increasing rotation rate until F m = ½ F s Then, disk appears to slow down (folding) Then, disk stops and appears to rotate in the other direction at an increasing rate (aliasing) Then, disk appears to slow down (folding) and stop And so forth