山东大学:《生物医学信号处理 Biomedical Signal Processing》精品课程教学资源(PPT课件讲稿)Chapter 03 the Z-transform

Chapter 3 The z-Transform ◆3.0 Introduction ◆3.1z- Transform 3.2 Properties of the Region of Convergence for the z-transform 3.3 The inverse z-Transform 3. 4 z-Transform Properties 2021/1/31 Zhongguo Liu_ Biomedical Engineering_shandong Univ
2 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Chapter 3 The z-Transform ◆3.0 Introduction ◆3.1 z-Transform ◆3.2 Properties of the Region of Convergence for the z-transform ◆3.3 The inverse z-Transform ◆3.4 z-Transform Properties

3.0 Introduction Fourier transform plays a key role in analyzing and representing discrete-time signals and systems, but does not converge for all signals Continuous systems Laplace transform is a generalization of the fourier transform Discrete systems: Z-transform, generalization of ft, converges for a broader class of signals 2021/1/31 Zhongguo Liu_ Biomedical Engineering_shandong Univ
3 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. 3.0 Introduction ◆Fourier transform plays a key role in analyzing and representing discrete-time signals and systems, but does not converge for all signals. ◆Continuous systems: Laplace transform is a generalization of the Fourier transform. ◆Discrete systems : z-transform, generalization of FT, converges for a broader class of signals

3.0 Introduction Motivation of z-transform The fourier transform does not converge for all sequences and it is useful to have a generalization of the Fourier transform In analytical problems the z-Transform notation is more convenient than the Fourier transform notation 2021/1/31 Zhongguo Liu_ Biomedical Engineering_shandong Univ
4 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. 3.0 Introduction ◆Motivation of z-transform: ◆The Fourier transform does not converge for all sequences and it is useful to have a generalization of the Fourier transform. ◆In analytical problems the z-Transform notation is more convenient than the Fourier transform notation

3.1 z-Transform Z-Transform: two-sided bilateral z-transform X(2)=∑x]z=z{xm n=-00 xn Cone-sided, unilateral z-transform X(2)=∑x= n=0 CIfz=e z-transform is fourier transform X(e")=∑ enel n=-00 5 2021/1/51 zhongguo ulu_biomedIcal Engineering_shandong UniV
5 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. 3.1 z-Transform ( ) jw n jwn X e x n e =− − = ( ) [ ] n n X z x n x n z =− − = = Z ( ) = − = n 0 n X z x n z ◆one-sided, unilateral z-transform ◆z-Transform: two-sided, bilateral z-transform x n X z [ ]⎯→ ( ) Z ◆If , z-transform is Fourier transform. jw z = e

Relationship between z-transform and Fourier transform Express the complex variable Z in polar form as 2=7 x(ve)}=∑ykm 1=-00 .The Fourier transform of the product of x[n and the exponential sequence r Ifr=l, X(z)>x(en 2021/1/31 Zhongguo Liu_Biomedical Engineering_shandong Univ
6 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. ◆Express the complex variable z in polar form as Relationship between z-transform and Fourier transform jw z = re ( ) ( ) =− − − = n j w n jwn X re x n r e ◆The Fourier transform of the product of and the exponential sequence xn n r − 1, ( ) ( ) jw If r X Z X e = ⎯⎯→

Complex z plane 丌<w<丌冷 unit circle Z-plane Unit circle 4=e gRe 2021/1/31 Zhongguo Liu_Biomedical Engineering_ shandong Univ
7 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Complex z plane − w unit circle

Condition for convergence of the z-transform X()=∑xX(z)=∑小] <0 z=re"→x(ve)=∑(p"km Xx(c)s∑网小 7"<O Convergence of the z-transform for a given sequence depends only on r=z 8 2021/1/31 Zhongguo Liu_ Biomedical Engineering_shandong Univ
8 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Condition for convergence of the z-transform ( ) jw n n X re x n r − =− ( ) = − = n 0 n X z x n z ( ) 0 n n X z x n z − = = jw z = re ( ) ( ) =− − − = n j w n jwn X re x n r e ◆Convergence of the z-transform for a given sequence depends only on r z =

Region of convergence(Roc) l Fbr any given sequence, the set of values of z for which the z-transform converges is called the region Of Convergence(ROC) z-plane X()=∑” n=0 if some value of Z, say Z=Z1 is in the rocr Re then all values of z on the circle defined by z=z, Will also be in the roc if roc includes unit circle then fourier transform and all its derivatives with respect to w must be continuous functions of w
9 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. z1 Region of convergence (ROC) ◆For any given sequence, the set of values of z for which the z-transform converges is called the Region Of Convergence (ROC). if some value of z, say, z =z1 , is in the ROC, then all values of z on the circle defined by |z|=|z1 | will also be in the ROC. if ROC includes unit circle, then Fourier transform and all its derivatives with respect to w must be continuous functions of w. ( ) = − = n 0 n X z x n z

Region of convergence(Roc) SIn w n n= coS( w on),hp[n 0o<n< oo Neither of them is absolutely summable neither of them multiplied byr-n(oo<n<oo) would be absolutely summable for any value of r Thus, neither them has a z-transform that converges absolutely 2021/1/31 Zhongguo Liu_ Biomedical Engineering_shandong Univ
10 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Region of convergence (ROC) sin , lp w nc h n n x n w n [ ] cos , = ( 0 ) = ➢Neither of them is absolutely summable, neither of them multiplied by r-n (-∞<n<∞) would be absolutely summable for any value of r. Thus, neither them has a z-transform that converges absolutely. -∞<n<∞

Region of convergence(Roc) cos(wn)→∑[6(m-1+27)+6(m++27) k=-0 sin w n WThe fourier transforms are not continuous infinitely differentiable functions so they cannot result from evaluating a z-transform on the unit circle. it is not strictly correct to think of the Fourier transform as being the z-transform evaluated on the unit circle 2021/1/31 Zhongguo Liu_ Biomedical Engineering_shandong Univ
11 2021/1/31 Zhongguo Liu_Biomedical Engineering_Shandong Univ. Region of convergence (ROC) ➢The Fourier transforms are not continuous, infinitely differentiable functions, so they cannot result from evaluating a z-transform on the unit circle. it is not strictly correct to think of the Fourier transform as being the z-transform evaluated on the unit circle. cos(w n0 ) ( ) ( ) 0 0 2 2 k w w k w w k =− − + + + + ( ) = , w w , w w H e c j w c l p 0 1 sin lp w nc h n n =
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