1. Lecture 41 Practical sampling and reconstruction

2. Lecture 4 2 Outline F Practical sampling –Aperture effect –Non ideal filters –Non-band limited input signals F Practical reconstruction F Practical digital systems F Discrete time Fourier transform

3. Lecture 4 3 Practical sampling F Practical sampling differs from ideal in the following respects –The sample (impulse) train actually consists of pulses of duration  –Real signal are time limited, therefore cannot be band limited (the uncertainty principle of Fourier transform) –Reconstruction filters are not ideal

4. Lecture 4 4 xa(t)xa(t) n=   s(t) =   (t  nT)  x s (t) =  x a (nT)   (t  nT) n=    sample and hold filter 1 h(t)h(t)

5. Lecture 4 5 Xa(j)Xa(j) 1     s >2   1/T  /T |H s (j  )| // // H (j  ) =  sin (  /2) e -j  /2   

6. Lecture 4 6  s >2   |X s (j  )| /T/T  /T X s (j  ) = H(j  ). 1 T  X a (j  jk  s ) k= -    If  /   there is no significant distortion over signal band (otherwise equalization can compensate distortion) Aperture effect

7. Lecture 4 7 Non ideal filters F The problem of non-ideal filtering can be combatted by increasing 1/T F Effectively, we are only using the middle portion of the filter, where it is closer to perfect

8. Lecture 4 8 Non-band limited signals F We may be only interested in a low frequency portion of a wide-band signal, e.g. speech only needs upto 3-4 Khz F There may be high frequency, wideband additive noise in the input signal F So we prefilter |  c  1 0, |  H aa (j  ) = 

9. Lecture 4 9 Anti-aliasing filter F Filter prior to sampling removes higher frequency components which could have been moved into the lower frequency range by aliasing F Of course, we are distorting the signal, but this is in a frequency range in which we are not interested F If we were, we would need to use a higher sampling frequency

10. Lecture 4 10 Practical D/A conversion F We don’t have perfect interpolation F Sample and hold F The impulse response of a sample and hold filter is h(t) 1 h(t)h(t) T

11. Lecture 4 11 Frequency domain F We need to compensate by adding a compensated reconstruction filter after the sample and hold process |H s (j  )| // // H o (j  ) =  sin (  /2) e -j  /2  

12. Lecture 4 12 |H r (j  )| // e j  /2, |   sin (  /2 ) ~ Compensated reconstruction filter 0, |  ~ H r (j  ) = 

13. Lecture 4 13 Ideal system A/DD/A h(n)

14. Lecture 4 14 Practical system h(n) A/D T Antialiasing pre-filter Sample and Hold T Sample and Hold Compensated Reconstruction Filter H aa (j  ) D/A

15. Lecture 4 15 Effective frequency response H eff (j  ) = H r (j  ) H 0 (j  ) H(e j  T ) H aa (j  )

16. Lecture 4 16 xa(t)xa(t) n=   S(t) =   (t  nT)  x s (t) =  x a (nT)   (t  nT) n=   convert to discrete sequence x[n] = x a (nT)

17. Lecture 4 17 Back to sampling  Let x a (t)  aperiodic  x s (t) =  x a (nT)   (t  nT) aperiodic  X s (j  ) =   x a (nT)  (t  nT) e - j  t dt n=     

18. Lecture 4 18  X s (j  ) =  x a (nT)   (t  nT) e - j  t dt    x a (nT) e -j  nT =  x[n] e -j  n where  =  T  X( e j  ) =  x[n] e -j  n is defined as the discrete time Fourier transform of x[n] n=       

19. Lecture 4 19 F From the sampling theorem, X s (j  ) =  X a (j  kj  s ) F thus, X( e j  ) =  X a (j  j  )  In other words, on going from the continuous domain to the discrete domain, we undergo a scaling or normalization in frequency from  =  s to  = 2  F There is also a corresponding time normalization from t=T to n=1 n=    1T1T k=  1T1T TT 2  k T

20. Lecture 4 20 Discrete time Fourier transform  X(e j  ) =  x[n] e -j  n –continuous and periodic with period 2   x[n] =  X(e j  ) e j  n d  –discrete and aperiodic n=     1 

21. Lecture 4 21 Example F Let x[n] = u[n]-u[n-M] =   [n-k] Then X(e j  ) =  x[n] e -j  n =  e -j  n = = e -j  M-1)/2 k=0 M-1 n=0 M-1 n=   1-e -j  M 1-e -j  sin(  M/2) sin(  /2)

22. Lecture 4 22 sin(  M/2) sin(  /2) |X(e j  )| =  /  //  0...

23. Lecture 4 23 Reading F Discrete time signals: Sections F Sampling: Sections 8.2 F Reconstruction, quantization, coding: Sections 8.2 - p.363 F Practical sampling and reconstruction: Sections 8.2 - p.355 F Fourier transform: Chapter 4