ECE 421 Introduction to Signal Processing

Transcription

1 ECE 421 Introduction to Signal Processing Dror Baron Assistant Professor Dept. of Electrical and Computer Engr. North Carolina State University, NC, USA

2 The z-transform [Reading material: Sections ] [3.4 with inverse z not covered]

3 Bigger picture Various math tools for processing discrete time signals More powerful tools support exponentially increasing signals but transform may not be defined Discrete time z transform Pro: signal can increase exponentially Con: transform not always defined Discrete time Fourier Pro: well-defined all frequencies Con: Only for well behaved signals More: subtleties for a/periodic Continuous time Laplace transform Same pros/cons Continuous time Fourier Same Fourier is special case of z/laplace LTI systems map well to these transforms Convolution becomes multiplication 3

4 But what are z transforms? Discrete time analog of Laplace transform (ECE 220, 301, 308) Similar to discrete time Fourier; covers exponential signals and not just frequencies Consider LTI system: x(n) H y(n) Time domain: convolution, y=x*h Frequency domain: multiplication (product), Y(z)=X(z)H(z) 4

5 Why does convolution become product? Exponentials are eigen-functions of LTI systems If T[v]=λv, then v and λ called e-function and e-value Eigen-function (sometimes called eigen-vector) passes through system and is merely amplified Back to ECE 421 Consider discrete time exponentials & LTI system H x(n)=(re jθ ) n y(n)=h*x=αx(n) α=h(z) (e-value) and z=re j θ 5

6 Why care about one sinusoid? Under some technical conditions linear system has Orthogonal (perpendicular) e-functions These e-functions span the entire space Signal can be uniquely expressed with coordinates Back to ECE 421 xx nn = ii αα ii ee jjθθ ii nn yy nn = h xx = ii h αα ii ee jjθθ ii nn = ii HH zz ii αα ii zz ii nn Big idea (devil is in details) Decomposed x into superposition of exponents (z i ) n Each exponent (w\coefficient α i ) amplified by H(z i ) 6

7 The z-transform Definition: XX zz = ZZ xx nn = nn= xx nn zz nn, zz C Often write xx nn zz XX(zz) This summation only defined in region of convergence (ROC) 7

8 Active learning What is the z-transform and ROC of the following signals: a) x 1 (n)={1,2,5} b) x 2 (n)={1,2,5} For finite-length x(n), ROC is entire complex plane except possibly for z=0 or z 8

9 Example with ROC Consider xx nn = 1 2 nn uu nn Will show XX zz = zz 1 Relied on infinite geometric sum, 1+A+A 2 + =1/(1-A) Requires A <1 Implies ROC = {z: 0.5z -1 <1} = {z: z -1 <2} = {z: z >0.5} Let s sketch the ROC together 9

10 More ROC Consider general case for x(n) and z=re jθ XX zz = nn= xx nn zz nn nn= xx nn rr nn ee jjθθnn Used triangle inequality XX zz 1 nn= xx nn rr nn + nn=0 Used ee jjθθnn =1 xx nn rr nn Summation over negative n requires small r to converge Summation over non-negative n requires large r ROC often comprised of concentric circles 10

11 Example revisited Earlier we discussed xx nn = 1 2 nn uu nn Showed XX zz = and ROC = {z: z >0.5} zz 1 2 Only positive indices (n>0) ROC outside some circle 11

12 Example modified Consider xx nn = nn uu nn 1 0, nn 0 = nn, nn 1 Will show XX zz = 1 1 zz 1 and ROC={z: z < } Only negative indices (n>0) ROC inside circle Note that =0.5 yields same X(z) as before; only ROC changed Specification of ROC is crucial 12

13 ROC summary Type of signal ROC Finite & causal Entire plane besides 0 Finite & anti-causal Finite & double-sided Entire plane besides Entire plane besides 0, Infinite & causal z >r 1 Infinite & anti-causal z <r 2 Infinite & double-sided r 1 < z <r 2 13

14 Inverse z-transform Recall XX zz = nn= xx nn zz nn Cauchy integral theorem, 1 2ππππ zz nn 1 kk 1, kk = nn dddd = 0, eeeeeeee Integrate over contour around origin (z=0) counter clockwise Inverse z-transform, x(n)= 1 2ππππ XX(zz)zz nn 1 dddd Computing inverse can be challenging; we don t dwell on this 14

15 Properties of z-transform [Reading material: Section 3.2]

16 Properties #1 Linearity: x=αx 1 +βx 2 X(z)=αX 1 (z)+βx 2 (z) Example: x(n) = cos(ωn)u(n) = 0.5e jωn u(n)+0.5e -jωn u(n) X(z) = 0.5Z{e jωn u(n)}+0.5z{e -jωn u(n)} = 0.5{1/(1-e jω z -1 )+1/(1-e -jω z -1 )} ROC identical for both components, ROC X ={z: z >1} Time shift: if x(n) X(z), then x(n-k) z -k X(z) Scaling: if x(n) X(z), ROC={r 1 < z < r 2 } then α n x(n) X(α -1 z) and ROC={ α r 1 < z < α r 2 } 16

17 Properties #2 Time reversal: x(-n) X(z -1 ), ROC={1/r 1 < z < 1/r 2 } Proof: Z{x(-n)} = Σ n x(-n)z -n = Σ l x(l)(z -1 ) -l = X(z -1 ), where l=-n Convolution: x(n)=x 1 (n)*x 2 (n) X(z)=X 1 (z) X 2 (z) ROC X =ROC X1 ROC X2 (intersection) Can compute convolution in z domain by (i) computing X 1 (z) and X 2 (z); (ii) multiplying, X(z)= X 1 (z)x 2 (z); and (iii) x(n)=z -1 {X(z)} 17

18 Active learning Recall that correlation can be expressed, rr xxxx = xx yy Let s compute Z{r xy (l)} a) Express r xy (l) as convolution of x and y b) Express Z{x(n)} and Z{ yy(n)} c) Compute Z{r xy (l)}=r xy (z) 18

19 Properties #3 Multiplication: x(n)=x 1 (n)x 2 (n) then XX zz = 1 2ππππ XX 1 vv XX 2 zz vv vv 1 dddd Initial value theorem: if x(n) causal, then x(0)=lim z X(z) Useful table at end of Section 3.2 summarizes properties 19

20 Example (Midterm 2014; Question 4) ConsiderXX(zz) = 1 (1 1 2 zz 1 )( zz 1 ) Express X(z) expansion in the form XX zz = aa zz 1 + bb zz 1 Assuming that x(n) is causal, what is x(n)? 20

21 Rational z-transforms [Reading material: Section 3.3]

22 What s Important in Transform of Signal? Hint: Look at its poles

23 What are we trying to do here? Will consider rational form for X(z) Happens when x(n) can be written as summation of terms of form α n u(n) and/or β n u(-n-1) Poles (values of α and β) important in determining structure of signal Coming up the details! 23

24 Rational form Rational X(z) can be written as follows XX zz = BB(zz) = MM kk=0 bbkk zz kk = bb 0zz MM zzmm + AA(zz) NN aa kk zz kk aa 0 zz NN zz NN + aa 1 kk=0 bb1 bb0 zzmm bb MM bb0 aa0 zznn aa NN aa0 Can be simplified XX zz = GGzz NN MM MM kk=1 (zz zzkk ) NN kk=1 (zz pp kk ) z k zeros, X(z=z k )=0 p k poles, X(z=p k )= Intuition: components of signal aligned with zero have no energy; components aligned with poles are big 24

25 Pole zero plot Will plot locations of zeros and poles Zeros are circles o Poles are crosses x Zero/pole with multiplicity>1 denote multiplicity in number nearby Example revisited: x(n)=α n u(n) X(z)=1/(1- αz -1 )=z/(z-α), ROC Z ={z: z >α} Im(z) o x α Re(z) 25

26 Another example xx nn = nn, 0 nn MM 1 0, eeeeeeee Will show XX zz = zzmm MM zz MM 1 (zz ) Roots of numerator z k =αe j2πk/m, k {0,,M-1} Roots of denominator 0 (multiplicity M-1), α Zero and α cancel each other out Only pole z=0 ROC={z:z 0} 26

27 Locations of poles Locations of poles Inside unit circle, positive Inside unit circle, negative On unit circle, positive On unit circle, negative Outside unit circle Signal Decays to zero, same sign Decays to zero, alternates signs Constant Constant magnitude, alternates signs Blows up These results rely on single pole x(n)=α n u(n) X(z)=1/(1- αz -1 ) Double pole yields x(n)=nα n u(n) pole on unit circle blows up Locations of poles matter! 27

28 Complex zeros/poles Systems are typically real valued Complex poles/zeros appear in conjugate pairs Complex conjugate poles feature exponential and oscillatory characteristics 28

29 Transfer Functions

30 LTI systems and transfer functions Recall Y(z)=H(z)X(z) If we know X(z) and Y(z), then H(z)=Y(z)/X(z)=Σ n h(n)z -n H(z) called transfer function 30

31 Transfer function of difference equation Consider difference equation NN yy nn = kk=1 MM aa kk yy nn kk + Take z transform of entire equation NN YY zz = kk=1 kk=1 MM aa kk YY zz zz kk + Can simplify (assume a 0 =1) HH zz = YY(zz) XX(zz) = kk=0 MM NN kk=0 kk=1 bb kk xx nn kk bb kk XX zz zz kk bb kk zz kk aa kk zz kk LTI system described by difference equation corresponds to rational transfer function 31

32 Example Difference equation, y(n)=0.5y(n-1)+2x(n) z-transform: Y(z)=0.5Y(z)z -1 +2X(z) Rearrange terms: Y(z)[1-0.5z -1 ]=2X(z) Transfer function HH zz = YY(zz) XX(zz) = zz 1 = 2zz zz 0.5 0, 0.5 Can be shown that h(n)=2[0.5 n ]u(n) 32

33 LTI Systems in z-domain [Reading material: Section 3.5]

34 More about rational transfer functions Because difference equation corresponds to rational transfer function, let s focus more on its form Consider H(z)=Y(z)/X(z)=B(z)/A(z) and X(z)=N(z)/Q(z) B(z), A(z), N(z), Q(z) polynomials YY zz = NN zz BB(zz) QQ zz AA(zz) Where are the poles? Interested in poles, because they specify rate of decay 34

35 Impact of poles Recall H(z)=Y(z)/X(z)=B(z)/A(z) and X(z)=N(z)/Q(z) YY zz = NN zz BB(zz) = NN QQ zz AA(zz) kk=1 AA kk 1 pp kk zz 1+ LL kk=1 QQ kk 1 qq kk zz 1 response to H s poles response to X s poles Apply inverse-z: NN yy nn = kk=1 AA kk pp kk nn uu nn + LL kk=1 QQ kk qq kk nn uu nn Natural response depends on H; hopefuly transient Forced response depends on X; exponents are e-functions of LTI system, they re amplified by H 35

36 Example y(n)=0.5y(n-1)+x(n) Initially at rest (y(n=-1)=0) x(n)=e jπn/4 u(n) Let s calculate y(n) Y(z)=0.5Y(z)z -1 +X(z) H(z)=1/(1-0.5z -1 ) X(z)=1/(1-e jπn/4 z -1 ) Pole of H inside unit circle natural response decays to 0 Pole of X on unit circle forced response oscillates 36

37 Relation to stability Recall that H is BIBO stable iff Σ h(n) < Let s evaluate H(z) on unit circle: HH zz = + nn= h nn zz nn + nn= h nn zz nn Because z =1 (unit circle), H(z) + nn= h nn < LTI system BIBO stable iff ROC contains unit circle Causal system is BIBO ROC has form { z >r}, 0<r<1 Anti-causal system BIBO ROC { z <r}, r>1 37

38 Example HH zz = zz zz 1 a) What s the ROC in order for H to be BIBO stable? b) If H is causal, is it BIBO stable? Consider four options for causality (or not) of 2 components 1C component #1 causal 0.5 n u(n) finite summation 1A #1 anti-causal -0.5 n u(-n-1) infinite summation 2C yields 3 n u(n) (infinite); and 2A yields -3 n u(-n-1) (finite) Part a: H is BIBO stable finite summation 1C and 2A Part B: H is causal 1C and 2C 2C inifinite not stable 38

39 More about zeros and poles Zeros and poles can cancel out In theory effect of pole goes away In practice messy (what if zero isn t precisely on pole of finite precision computation moves them?) Multiple order poles Pole on unit circle has response polynomial * exponent Order of polynomial = multiplicity Exponent is actually oscillation Response will blow up (polynomial * oscillation) 39

40 Example (Midterm 2017; Question 3) Consider the difference equation y(n)+7y(n-1)+12y(n-2)=x(n) a) What is the transfer function, H(z)=Y(z)/X(z)? b) Given that difference equation is causal: I. Where are the poles and zeros? II. What is the ROC? III. Is the system BIBO stable? 40

41 One-Sided z-transform [Reading material: Section 3.6]

42 One-sided z-transform Replace doubly infinite summation by summation over n 0 XX + zz = nn=0 xx(nn)zz nn Useful for dealing with initial conditions (not initially at rest) 42

43 Active learning What is the one-sided z-transform of the following signals: a) x 1 (n)={1,2,5} b) x 2 (n)={1,2,5} 43

44 Properties of one-sided z Time delay: if x(n) X + (z), then x(n-k) z -k [XX + zz + kk nn=1 xx( nn)zz nn ] Time advance: x(n+k) z k [XX + zz kk 1 nn=1 xx( nn)zz nn ] 44

45 Example Consider x 1 (n)=a n and x 2 (n)=x 1 (n-2) Will compute X 2+ (z) ZZ + xx 1 nn 2 = zz 2 XX + 1 zz + xx 1 1 zz + xx 1 2 zz 2 Transform of x 1 (n): XX + 1 zz = + nn=0 aa nn zz nn Because x 1 (n) is causal, this equals X 1 (z) XX 1 + zz = ZZ{aa nn uu(nn)} = 1 1 aazz 1 Plugging into equation: XX 2 + zz = zz 2 1 aazz 1 + aa 1 zz 1 + aa 2 45

46 Example part 2 How about the old-fashioned way to compute? x 2 (0)=x 1 (-2)=a -2 x 2 (1)=x 1 (-1)=a -1 x 2 (2)=a 0 =1 and so on XX 2 + zz = aa 2 zz 0 + aa 1 zz 1 + aa 0 zz 2 + = aa 2 [1 + aazz 1 + aa 2 zz 2 + ]= aa 2 1 aazz 1 Earlier we got XX + 2 zz = zz aazz 1 aa 1 zz 1 + aa 2 Can show expressions same 46

47 Solving difference equations Earlier we saw natural response and forced response One-sided z-transform good at incorporating initial conditions (system not initially at rest) Will use celebrated Fibonacci sequence as example Fibonacci Italian mathematician Claimed that rabbit population follows y(n)=y(n-1)+y(n-2) n y(n)

48 One-sided z-transform perspective Recall y(n)=y(n-1)+y(n-2) Need initial conditions y(1)=y(0)+y(-1) y(-1)=y(1)-y(0)=1-1=0 y(0)=y(-1)+y(-2) y(-2)=y(0)-y(-1)=1-0=1 Apply Z + to difference equation Y + (z)=z + {y(n-1)}+z + {y(n-2)} =[z -1 Y + (z)+y(-1)]+[z -2 Y + (z)+y(-1)z -1 +y(-2)] =(z -1 +z -2 )Y + (z)+1 YY + zz = 1 = 1 zz 1 zz 2 Poles at p=[1±sqrt(5)]/2 zz2 zz 2 zz 1 48

49 Expression for Fibonacci sequence Can write: YY + zz = aa + 1 pp 1 zz 1 bb 1 pp 2 zz 1 Poles p 1, p 2 seen earlier Can compute constants a, b (details in handout) YY + zz = ( )zz ( )zz 1 Inverse transform: yy nn = Will verify numerically with Matlab nn uu nn nn uu nn 49

50 Active learning example Consider y(n)=0.5y(n-1)+2 Initial condition y(-1)=0 a) Compute y(n) for n=0, 1, 2, 3 b) Can we see a pattern for y(n)? 50

51 Example via Matlab n=-1:100; % time range y=zeros(size(n)); % initialization for index=2:length(n) y(index)=0.5*y(index-1)+2; end plot(n,y); 51

52 Example via one-sided z-transform y(n)=0.5y(n-1)+2 Y + (z)=0.5z + {y del (n)}+z + {2u(n)} y del (n)=y(n-1) time delayed version Z + {2u(n)} = Σ n 2z -n =2[1+z -1 +z -2 + ] = 2/(1-z -1 ) ROC={ z >1} Z + {y del (n)} = z -1 {Y + (z)+y(-1)z} = z -1 Y + (z) Entire expression: Y + (z)= 0.5z -1 Y + (z)+2/(1-z -1 ) Y + (z)[1-0.5z -1 ]=2/(1-z -1 ) YY + 2 zz = = = 4-2 (1 zz 1 )(1 0.5zz 1 ) 1 zz zz 1 y(n)=4u(n) n u(n) 52