Lecture 7: z-transform Properties, Sampling and Nyquist Sampling Theorem

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1 EE518 Digital Signal Proessing University of Washington Autumn 21 Dept. of Eletrial Engineering ure 7: z-ransform Properties, Sampling and Nyquist Sampling heorem Ot 22, 21 Prof: J. Bilmes A: Mingzhou Song 7.1 Properties of z-ransform x 1 [n] X 1 (z) x 2 [n] X 2 (z) ROC R x1 ROC R x2 Linearity ax 1 [n] + bx 2 [n] ax 1 (z) + bx 2 (z) whih follows from definition of z-transform. Ex: hen x[n] a n u[n] a n u[n N] (N ) ROC ontains R x1 R x2 x 1 [n] a n 1 u[n] X 1 (z) 1 az 1 ROC z > a x 2 [n] a n u[n N] X 2 (z) an z N 1 az 1 ROC z > a x[n] x 1 [n] x 2 [n] X(z) 1 an z N 1 az 1 However, the pole at z a of X(z) is anelled out by the zero at z a. So the ROC of X(z) is the entire z-plane when N,1. he ROC is z > when N > 1. ime Shifting y[n] x[n n ], Y (z) n x[n n ] z n X(z) x[n n ]z n k x[k]z k n z n X(z) he ROC of Y (z) is the same as X(z) exept that there are possible pole additions or deletions at z or z. 7-1

2 7-2 Multipliation by an Exponential Sequene y[n] z n x[n], Y (z) n z n x[n]z n n ( ) z n ( ) z x[n] X z z he onsequene is pole and zero loations are saled by z. If the ROC of X(z) is r R < z < r L, then the ROC of Y (z) is r R < z/z < r L, i.e., z r R < z < z r L Differentiation of X(z) X(z) z dx(z) dz n n z x[n]z n ( n)x[n]z n 1 n nx[n]z n So nx[n] z dx(z) dz ROC R x1 Conjugation of a Complex Sequene y[n] x [n], then Y (z) n x [n] X (z ) x [n]z n ( n ROC R x x[n](z ) n ) X(z ) ime Reversal y[n] x [ n], then Y (z) n x [ n]z n ( n x [ n] X (1/z ) If the ROC of X(z) is r R < z < r L, then the ROC of Y (z) is x[ n](z ) n ) ( r R < 1/z < r L i.e., When the time reversal is without onjugation, it is easy to show k 1 r L < z < 1 r R x[k](1/z ) k ) X (1/z ) x[ n] X(1/z) 1 r L < z < 1 r R

3 7-3 Convolution of Sequenes then y[n] x 1 [n] x 2 [n] Y (z) n k k k ( X 1 (z)x 2 (z) k x 1 [k]x 2 [n k]z n x 1 [k]z k n x 1 [k]z k X 2 (z) x 1 [k]x 2 [n k] x 2 [n k]z (n k) ) he ROC of Y (z) ontains R x1 R x2, beause anellation of zeros and poles may result in an ROC larger than R x1 R x2. Initial Value heorem If x[n] for n <, x[] lim z X(z) (7.1) Proof. lim z X(z) lim z n n1 x[] + x[] x[n]z n x[n] lim z z n 7.2 Sampling Sampling implements the representation of a ontinuous-time signal by a disrete-time signal, whih enables the proessing of signal by digital omputers. Q: given x(t), need we have x(t) at all times in order to represent x(t) at all times? A: No. Q: might x(t) be redundant, so we an store only a portion of x(t) without losing any information? A: Yes. Q: might this be useful when using omputers? A: Yes. Periodi Sampling An ideal ontinuous-to-disrete-time (C/D) onverter using periodi sampling is shown in ig It an be mathematially desribed as x[n] x (n ) < n < (7.2) where x (t) is the input ontinuous time signal and x[n] is the output disrete time signal. is the sampling period. f s 1 is the sampling frequeny in samples/se. s 2π is the sampling frequeny in radians/se. Question: Can we reover x (t) from x[n]?

4 7-4 x (t) C/D x[n]x (n) igure 7.1: An ideal C/D onverter. Answer: It depends. If x (t) has ertain harateristis and is small enough then yes. Otherwise, no. A mathematially onvenient representation of the ideal C/D onverter. ig. 7.2 shows a mathematially onvenient representation of the ideal C/D onverter. In the diagram, s(t) x (t) X x (t) s Conversion from impulse train to disrete time sequene x[n]x (n) igure 7.2: A mathematial representation of the ideal C/D onverter. s(t) n δ(t n ) x s (t) x (t)s(t) x (t) n δ(t n ) n x (n )δ(t n ) requeny Domain Representation of Sampling We now study the frequeny domain representations of the signals and sequene in ig Reall we use variable ω for disrete-time (normalized) radian frequeny. We use variable for ontinuous-time radian frequeny. x (t) X ( j) s(t) S( j) x s (t) X s ( j) x[n] X(e jω )

5 7-5 then we get X s ( j) 1 2π X ( j) S( j) [ 1 2π X 2π ( j) 1 1 k k δ( j( k s )) X ( j) δ( j( k s )) X ( j( k s )) k ] ( n δ(t n ) 2π k δ( j( k s )) ) (7.3) X s ( j) an be interpreted as a superimposed version of periodially repeated X ( j). An example is shown in ig Notie there might be overlapping among X ( j( k s )) ( < k < ). When overlapping happens, aliasing ours X (j ) N N S (j ) 2 s s s 2 s X s (j ) 2 s s s 2 s X s (j ) 2 s s s 2 s igure 7.3: requeny domain representation of sampling. in X s ( j), with respet to X ( j). If X ( j) for N, from ig. 7.3, aliasing an be avoided if s N > N. In the non-aliasing ase ( s > 2 N ), we an design a ideal lowpass filter { < H r ( j) (7.4) where is the utoff frequeny of the lowpass filter. By applying the lowpass filter on the non-aliased x s (t) and use an appropriate utoff frequeny (usually s 2 ), we an reonstrut x (t). wo onditions for reonstrution must be satisfied: 1. N (the highest frequeny of the signal) exists. 2. s > 2 N

6 7-6 In the aliasing ase ( s < 2 N ), X ( j) is no longer reoverable by LP beause of overlapping of frequeny omponents. his is a form of non-linear distortion shown as aliasing. Some frequeny omponents are aliases of those before sampling. By ertain types of filter, we an also get X ( j) suh that X s ( j) 1 X ( j( k s )) k However, is this solution uniquely determined? Apparently, X ( j) H r ( j)x s ( j) is always a solution for the above equation and X ( j) is another solution 1. But X ( j) X ( j) if there is aliasing. Nyquist Sampling heorem states that only when there is no aliasing, an x (t) be uniquely determined. heorem 7.1 (Nyquist Sampling heorem). x (t) be a bandlimited signal with X ( j) for > N hen x (t) is uniquely determined by its samples if x[n] x (n ) n,±1,±2, s 2π 2 N (7.5) where N is alled the Nyquist frequeny and 2 N is alled the Nyquist rate, whih must be exeeded by s for ideal reonstrution of x (t). ourier transform of x[n]. Sine and we get So X s ( j) { n n n x (n )δ(t n ) x (n ){δ(t n )} x (n )e jn x[n] x (n ) X s ( j) X(e jω ) ω } X(e jω ) X s ( j) ω (7.6) whih is a frequeny axis saled version of X s ( j) by 1, the sampling frequeny. Ex: Sampling a sinusoidal signal with aliasing. hen x (t) os(4πt), 1 15 X ( j) πδ( 4π) + πδ( + 4π) ( ) 2π x[n] os 3 n ig. 7.4 shows the frequeny domain representations during the sampling proess. 1 We an design ertain filter to produe X ( j), even though it may not be as trivial as the lowpass filter.

7 7-7 X (j ) 4π 4π X s (j ) 4π 2π 1π 1π 2π 4π j ω X (e ) 8π/3 4π/3 2π/3 2π/3 4π/3 8π/3 ω igure 7.4: Sampling a sinusoidal signal with aliasing.

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