1. Trang chủ
  2. » Kỹ Thuật - Công Nghệ

Lecture Digital signal processing: Lecture 3 - Zheng-Hua Tan

20 42 0

Đang tải... (xem toàn văn)

Tài liệu hạn chế xem trước, để xem đầy đủ mời bạn chọn Tải xuống

THÔNG TIN TÀI LIỆU

Thông tin cơ bản

Định dạng
Số trang 20
Dung lượng 1,09 MB

Nội dung

Chapter 3 introduce the z-transform. This chapter presents the following content: z-transform, properties of the ROC, inverse z-transform, properties of z-transform.

Digital Signal Processing, Fall 2006 Lecture 3: The z-transform Zheng-Hua Tan Department of Electronic Systems Aalborg University, Denmark zt@kom.aau.dk Digital Signal Processing, III, Zheng-Hua Tan, 2006 Course at a glance MM1 Discrete-time signals and systems MM2 Fourier-domain representation Sampling and reconstruction System System structures System analysis MM6 MM5 Filter MM4 z-transform MM3 DFT/FFT Filter structures MM9,MM10 MM7 Filter design MM8 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Part I: z-transform z-transform Properties of the ROC Inverse z-transform Properties of z-transform „ „ „ „ Digital Signal Processing, III, Zheng-Hua Tan, 2006 Limitation of Fourier transform Fourier transform „ ∞ ∑ x[n]e − jωn X ( e jω ) = n = −∞ x[ n] = 2π π ∫−π X (e jω )e jωn dω Condition for the convergence of the infinite sum „ | X ( e jω ) | = | ∞ ∞ ∞ n = −∞ n = −∞ n = −∞ ∑ x[n]e − jωn | ≤ ∑ | x[n] ||e − jωn | ≤ ∑ | x[n] | < ∞ If x[n] is absolutely summable, its Fourier transform exists (sufficient condition) x[n] = a n u[n] | a |< : X (e jω ) = Example − ae − jω „ „ a = : X ( e jω ) = ∞ + ∑ πδ (ω + 2πk ) − e − jω k = −∞ | a |> : Digital Signal Processing, III, Zheng-Hua Tan, 2006 z-transform Fourier transform „ X ( e jω ) = ∞ ∑ x[n] e − jω n n = −∞ z-transform „ ∞ X ( z) = ∑ x[n] z−n Z x[n] ↔ X ( z ) n = −∞ The complex variable z in polar form z = re jω „ ∞ ∞ −∞ −∞ X ( z ) = X ( re jω ) = ∑ x[ n](re jω ) − n = ∑ ( x[n]r − n )e − jωn |z| = r = 1, X ( z ) = X (e jω ) Digital Signal Processing, III, Zheng-Hua Tan, 2006 z-plane z-transform is a function of a complex variable Ỉ using the complex z-plane „ Z-transform on unit circle Fourier transform Linear frequency axis in Fourier transform ỈUnit circle in z-transform (periodicity in freq of Fourier transform) Digital Signal Processing, III, Zheng-Hua Tan, 2006 Region of convergence – ROC Fourier transform does not converge for all ∞ sequences jω − j ωn „ X (e ) = ∑ x[n]e n = −∞ z-transform does not converge for all sequences or for all values of z „ X(z) = X ( re jω ) = ∞ ∑ x[n] z −n = n = −∞ ∞ ∑ ( x[n]r − n )e − jωn n = −∞ ROC – for any given seq., the set of values of z for which the z-transform converges „ ∞ ∑ | x[n]r − n |< ∞ n = −∞ ∞ ∑ | x[n] | | z | − n < ∞ ROC is ring! n = −∞ Digital Signal Processing, III, Zheng-Hua Tan, 2006 ROC Outer boundary is a circle (may extend to infinity) Inner boundary is a circle (may extend to include the origin) If ROC includes unit circle, Fourier transform converges „ „ „ Digital Signal Processing, III, Zheng-Hua Tan, 2006 Zeros and poles The most important and useful z-transforms – rational function: „ P( z ) Q( z ) P ( z ) and Q ( z ) are polynomials in z X ( z) = Zeros: values of z for which X(z)=0 Poles : values of z for which X(z) is infinite Close relation between poles and ROC „ „ „ Digital Signal Processing, III, Zheng-Hua Tan, 2006 Right-sided exponential sequence x[n] = a n u[n] ∞ ∑ a n u[n]z − n X ( z) = n = −∞ ∞ = ∑ (az −1 ) n n=0 „ ROC ∞ ∑ | az −1 |n < ∞ n=0 „ z-transform ∞ X ( z ) = ∑ ( az −1 ) n = n=0 Z u[n] ↔ 10 , − z −1 z = , − az −1 z − a | z |>| a | | z |> Digital Signal Processing, III, Zheng-Hua Tan, 2006 Left-sided exponential sequence x[n] = − a n u[− n − 1] ∞ X ( z ) = − ∑ a n u[− n − 1]z − n n = −∞ −1 ∞ n = −∞ n=0 = − ∑ a n z − n = − ∑ (a −1 z ) n „ ROC ∞ ∑ | a −1 z |n < ∞ n=0 „ z-transform X ( z) = 11 z = , − az −1 z − a | z | and | z |> | z |> | 12 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Sum of two exponential sequence Another way to calculate: 1 x[n] = ( ) n u[n] + (− ) n u[n] Z 1 ( ) n u[n] ↔ , | z |> 2 − z −1 Z 1 , | z |> (− ) n u[n] ↔ 3 + z −1 Z 1 1 + ( ) n u[n] + (− ) n u[n] ↔ 1 − z −1 + z −1 13 | z |> Digital Signal Processing, III, Zheng-Hua Tan, 2006 Two-sided exponential sequence 1 x[n] = (− ) n u[n] − ( ) n u[−n − 1] Z 1 (− ) n u[n] ↔ , | z |> 3 + z −1 Z 1 , | z |< − ( ) n u[ − n − 1] ↔ 2 − z −1 1 1 , | z |> , | z |< X ( z) = + −1 −1 1+ z 1− z ) 12 = 1 ( z − )( z + ) 2z( z − 14 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Finite-length sequence ⎧ an , ≤ n ≤ N −1 x[n] = ⎨ otherwise ⎩0, N −1 N −1 X ( z ) = ∑ a n z − n = ∑ (az −1 ) n n =0 = − (az ) zN − aN = N −1 −1 − az z z−a N −1 ROC n =0 −1 N ∑ | az −1 n | ), − az −1 | z |>| a | x[n] = ( ) n u[n] of course, | z |< ? x[n] = −( ) n u[−n − 1] Digital Signal Processing, III, Zheng-Hua Tan, 2006 11 Partial fraction expansion „ 23 For rational function, get the format of a sum of simpler terms, and then use the inspection method Digital Signal Processing, III, Zheng-Hua Tan, 2006 Second-order z-transform X ( z) = 1− X ( z) = 1 , | z |> −1 − 2 z + z 1 −1 (1 − z )(1 − z −1 ) A1 A2 + −1 (1 − z ) (1 − z −1 ) −1 A1 = (1 − z ) X ( z ) | z =1/ = −1 A2 = (1 − z −1 ) X ( z ) | z =1/ = 2 −1 X ( z) = + −1 −1 (1 − z ) (1 − z ) X ( z) = M X ( z) = ∑b z −k ∑a z −k k =0 N k =0 k k M b X ( z) = a0 ∑ (1 − c z k =1 N k ∑ (1 − d k =1 k −1 ) z −1 ) N Ak −1 − d k =1 kz X ( z) = ∑ Ak = X ( z )(1 − d k z −1 ) | z = d k 1 x[n] = 2( ) n u[n] − ( ) n u[n] 24 Digital Signal Processing, III, Zheng-Hua Tan, 2006 12 What about M>=N? X ( z) = + z −1 + z −2 , | z |> 1 − z −1 + z −2 2 X ( z) = A1 A2 + −1 (1 − z −1 ) (1 − z ) Found by long division − −1 −2 −1 z − z +1 z + 2z +1 2 z − − z −1 + X ( z ) = B0 + B0 = − z −1 − 1 A1 = (1 − z −1 ) X ( z ) | z =1/ = −9 A2 = (1 − z −1 ) X ( z ) | z =1 = X ( z) = − 25 + −1 (1 − z −1 ) (1 − z ) (1 + z −1 ) (1 − z −1 )(1 − z −1 ) x[n] = 2δ [n] − 9( ) n u[n] + 8u[n] Digital Signal Processing, III, Zheng-Hua Tan, 2006 Power series expansion „ By long division X ( z) = , | z |>| a | − az −1 + az −1 + a z −2 + − az −1 1 − az −1 az −1 az −1 − a z − a z − = + az −1 + a z − + − az −1 x[n] = a n u[n] 26 Digital Signal Processing, III, Zheng-Hua Tan, 2006 13 Finite-length sequence −1 z )(1 + z −1 )(1 − z −1 ) 1 X ( z ) = z − z − + z −1 2 X ( z ) = z (1 − 1 x[n] = δ [n + 2] − δ [n + 1] − δ [n] + δ [n − 1] 2 27 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Part IV: Properties of z-transform „ „ „ „ 28 z-transform Properties of the ROC Inverse z-transform Properties of z-transform Digital Signal Processing, III, Zheng-Hua Tan, 2006 14 Linearity Z x1[n] ↔ X ( z ), ROC = Rx1 X ( z) = ∑ x[n] z −n n = −∞ Z x2 [n] ↔ X ( z ), ROC = Rx2 „ ∞ Linearity Z ax1[n] + bx2 [n] ↔ aX ( z ) + bX ( z ), ROC contains Rx1 ∩ Rx2 at least Z , | z |> 1 − z −1 Z z −1 u[n − 1] ↔ , | z |> 1 − z −1 Z − z −1 u[n] − u[n − 1] = δ [n] ↔ = 1, All z − z −1 u[n] ↔ 29 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Time shifting Z x[n − n0 ] ↔ z − n0 X ( z ), X ( z) = ROC = Rx1 (except for or ∞ ) „ 30 Example ∞ ∑ x[n] z −n n = −∞ , | z |> z− z −1 X ( z) = = z −1 ( ) −1 −1 1− z 1− z 4 Z n ( ) u[n] ↔ − z −1 Z n −1 ( ) u[n − 1] ↔ z −1 ( ) − z −1 X ( z) = Digital Signal Processing, III, Zheng-Hua Tan, 2006 15 Multiplication by exponential sequence Z z0n x[n] ↔ X ( z / z ), ROC = |z 0|Rx ∞ ∑ x[n] z X ( z) = −n n = −∞ F e jω0 n x[n] ↔ X (e j (ω −ω0 ) ) Examples „ Z , | z |> 1 − z −1 Z 1 a nu[n] ↔ = , | z |> a −1 − ( z / a) − az −1 1 x[n] = cos(ω0 n)u[n] = (e jω ) n u[n] + (e − jω ) n u[n] 2 1 (1 − cos ω0 z −1 ) 2 X ( z) = + = , − e jω z −1 − e − jω z −1 − cos ω0 z −1 + z − u[n] ↔ 0 31 | z |> Digital Signal Processing, III, Zheng-Hua Tan, 2006 Differentiation of X(z) dX ( z ) nx[n] ↔ − z , ROC = Rx dz Z „ Example X ( z) = ∞ ∑ x[n] z −n n = −∞ X ( z ) = log(1 + az −1 ), | z |>| a | dX ( z ) − az − = dz + az −1 Z dX ( z ) az −1 nx[n] ↔ − z = , | z |>| a | dz + az −1 nx[n] = a (− a ) n −1 u[n − 1] x[n] = 32 a (− a) n −1 u[n − 1] n Digital Signal Processing, III, Zheng-Hua Tan, 2006 16 Conjugation of a complex sequence Z x [ n] ↔ X ( z ), ROC = Rx * 33 * * X ( z) = ∞ ∑ x[n] z −n n = −∞ Digital Signal Processing, III, Zheng-Hua Tan, 2006 Time reversal Z x[ − n] ↔ X (1 / z ), ROC = / Rx „ ∞ ∑ x[n] z −n n = −∞ Example x[n] = a − nu[−n] X ( z) = 34 X ( z) = , | z || a | − az −1 Digital Signal Processing, III, Zheng-Hua Tan, 2006 17 Convolution of sequences Z x1[n] * x2 [n] ↔ X ( z ) X ( z ), X ( z) = x[n] = ( ) n u[n] h[n] = u[n] y[n] ? 35 1 , | z |> −1 1− z H ( z) = , | z |> 1 − z −1 1 Y ( z) = , | z |> −1 −1 (1 − z )(1 − z ) 1 ) = − ( −1 (1 − z −1 ) (1 − z ) 1− 2 X ( z) = Example y[n] = −n n = −∞ ROC contains Rx1 ∩ Rx2 „ ∞ ∑ x[n] z 1 (u[n] − ( ) n+1 u[n]) 1− Digital Signal Processing, III, Zheng-Hua Tan, 2006 Initial-value theorem X ( z) = x[0] = lim X ( z ) z →∞ lim X ( z ) = lim z →∞ = z →∞ ∞ ∑ x[n] lim n = −∞ z →∞ ∞ ∑ x[n] z −n n = −∞ ∞ ∑ x[n] z −n n = −∞ z− n = x[0] 36 Digital Signal Processing, III, Zheng-Hua Tan, 2006 18 Properties of z-transform 37 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Summary „ „ „ „ 38 z-transform Properties of the ROC Inverse z-transform Properties of z-transform Digital Signal Processing, III, Zheng-Hua Tan, 2006 19 Course at a glance MM1 Discrete-time signals and systems MM2 Fourier-domain representation Sampling and reconstruction System System structures System analysis MM6 MM5 Filter MM4 z-transform MM3 39 DFT/FFT Filter structures MM9,MM10 MM7 Filter design MM8 Digital Signal Processing, III, Zheng-Hua Tan, 2006 20 ... = −∞ z− n = x[0] 36 Digital Signal Processing, III, Zheng-Hua Tan, 2006 18 Properties of z-transform 37 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Summary „ „ „ „ 38 z-transform Properties... the ROC Inverse z-transform Properties of z-transform Digital Signal Processing, III, Zheng-Hua Tan, 2006 Properties of the ROC 18 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Properties... , N − Digital Signal Processing, III, Zheng-Hua Tan, 2006 Some common z-transform pairs 16 Digital Signal Processing, III, Zheng-Hua Tan, 2006 Part II: Properties of the ROC „ „ „ „ 17 z-transform

Ngày đăng: 11/02/2020, 16:30

TỪ KHÓA LIÊN QUAN