Chap 4. Sampling of Continuous-Time Signals

Transcription

1 Digital Signal Processing Chap 4. Sampling of Continuous-Time Signals Chang-Su Kim

2 Digital Processing of Continuous-Time Signals Digital processing of a CT signal involves three basic steps 1. Conversion of the CT signal into a DT signal 2. Processing of the DT signal 3. Conversion of the processed DT signal back into a CT signal Conceptual Block diagram CT Input Sampling DT Signal Quantization Digital Signal Digital Processor CT Output Reconstruction Filter Dequantization

3 Sampling

4 Sampling Sampling is a procedure to extract a DT signal from a CT signals (b), (c), (d) are obtained by sampling (a) Is (b) enough to represent (a)? What is the adequate sampling rate to represent a given CT signal without information loss?

5 In general, DT signal cannot represent CT signal perfectly Are these sample enough to reconstruct the original blue curve?

6 Continuous-Time Fourier Transform CTFT Formulae Forward transform X jω = Inverse transform x t = 1 2π x t e jωt dt X(jΩ)e jωt dω We will use a number of properties of CTFT without proofs They are studied in the course Signals and Systems

7 Periodic Sampling C/D (continuous-time to discrete-time) converter x n = x c nt, < n <. T : sampling period Ω s = 2π (or f T s = 1 ) : sampling frequency T

8 Periodic Sampling Conceptually, it is easier to introduce an impulse train for the C/D conversion s t = n= δ(t nt) x s t = x c t s t = n= x c nt δ t nt x s t and x n have the same information Given x s t, we can make x n. Given x n, we can make x s (t).

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10 Frequency-Domain Representation of Sampling S jω = 2π T k= δ(ω kω s ) X s jω = 1 2π X c jω S jω = = 1 T k= X c (j(ω kω s ))

11 Recovery of x c t from x s t If you can recover x c t from x s t, you can recover x c t from x n. Recovery is possible through an ideal low-pass filter when Ω s > 2Ω N.

12 Nyquist-Shannon Sampling Theorem Let x c t be a band-limited signal with X c jω = 0 for Ω Ω N. Then x c t is uniquely determined by its samples x n = x c nt, < n <, if Ω s = 2π T 2Ω N. 2Ω N is called the Nyquist rate. Under certain conditions, a CT signal can be completely represented by and recoverable from samples A low-pass signal can be reconstructed from samples, if the sampling rate is high enough. Because it is a low-pass signal, the change between two close samples is constrained (or expected).

13 Recovery of x c t from x s t h r t = sin(πt T ) πt T x s t = n= x[n]δ t nt x r t = n= x[n] sin(π(t nt) T π(t nt) T )

14 Recovery of x c t from x[n] x r t = n= nt) sin(π(t ) x[n] T π(t nt) T

15 Summary

16 s t = n= δ(t nt), x s t = x c t s t X s jω = 1 T k= X c (j(ω kω s )) X r jω = X s jω H r jω, x r t = n= x[n] sin(π(t nt) T π(t nt) T )

17 Frequency-Domain Relationship between x n and x s (t) Relationship between X e jω and X s jω X e jω = X s j ω T X s jω = X(e jωt ) Recall that X e jω is always periodic

18 Aliasing

19 Undersampling Causes Aliasing Undersampling: sampling rate is less than Nyquist rate

20 Undersampling Causes Aliasing Rotating disk 1 rotation/second To avoid aliasing, it should be motion-pictured with at least 2 frames/s. 0 1/12 2/12 3/12 4/12 5/12 6/12 7/12 8/12 9/12 10/12 11/12

21 Undersampling Causes Aliasing 12 frames/s 0 1/12 2/12 3/12 4/12 5/12 6/12 7/12 8/12 9/12 10/12 11/12

22 Undersampling Causes Aliasing 6 frames/s 0 1/12 2/12 3/12 4/12 5/12 6/12 7/12 8/12 9/12 10/12 11/12

23 Undersampling Causes Aliasing 3 frames/s 0 1/12 2/12 3/12 4/12 5/12 6/12 7/12 8/12 9/12 10/12 11/12

24 Undersampling Causes Aliasing 2 frames/s 0 1/12 2/12 3/12 4/12 5/12 6/12 7/12 8/12 9/12 10/12 11/12

25 Undersampling Causes Aliasing 12/11 = 1.09 frames/s 0 1/12 2/12 3/12 4/12 5/12 6/12 7/12 8/12 9/12 10/12 11/12

26 Examples x c t = cos 4000πt, T = 1/ πδ(Ω) cos Ω 0 t π(δ Ω Ω 0 + δ Ω + Ω 0 ) sin Ω 0 t π j (δ Ω Ω 0 δ Ω + Ω 0 )

27 Examples x c t = cos 16000πt, T = 1/6000.

28 Examples

29 DT Processing of CT Signals

30 C/D and D/C conversions C/D X e jω = 1 T k= X c j ω T 2πk T Relationship between x n and x s (t) X e jω = X s j ω T, D/C X s jω = X(e jωt ) Y r jω = H r jω Y e jωt = TY ejωt, Ω < π T 0, Otherwise

31 Overall System Effective Frequency Response Assumptions H eff jω = H ejωt, Ω < π T 0, Otherwise x c t is band-limited 2π T satisfies the Nyquist rate Relationship between x n and x s (t) X e jω = X s j ω T, X s jω = X(e jωt )

32 Example 1 H(e jω ) = 1, ω < ω c 0, ω c < ω < π

33 Example 2 y c t = d dt x c t h n = 0, n = 0 1 n nt, n 0

34 Impulse Invariance h n = Th c nt Example Ideal lowpass filter h n with cutoff frequency ω c

35 CT Processing of DT Signals

36 CT Processing of DT Signals It is rarely used, but provides a useful interpretation of some DT systems Main results H e jω = H c j ω, ω < π T H c jω = H e jωt, Ω < π T.

37 Example Fractional Delay H e jω = e jωδ, ω < π sin π n Δ h n = π n Δ

38 Changing Sampling Rate

39 Reducing Sampling Rate by an Integer Factor M sampling rate compressor Time domain x d n = x nm Frequency domain X d e jω = M 1 k=0 X(e j ω M 2πk M )

40 Reducing Sampling Rate by an Integer Factor M To avoid aliasing, we need X e jω = 0 if ω N < ω < π ω N < π M Anti-aliasing filter can be used

41 Increasing Sampling Rate by an Integer Factor L sampling rate expander Input and output x n = x c nt x i n = x c n T L Intermediate signal x e n = x[n L ] if n is a multiple of L 0 otherwise Output in terms of input x i n = k= π(n kl) sin x[k] L π(n kl) L

42 Ideal and Linear Interpolation Filters x e n = k= x k δ[n kl] x i n = x e n h i [n] = k= Ideal filter x k h i [n kl] Linear filter h i n = sin πn L πn L h lin n = 1 n L, L n L 0, otherwise

43 Changing Sampling Rate by a Noninteger Factor