Discrete Time Systems

Transcription

1 1 Discrete Time Systems {x[0], x[1], x[2], } H {y[0], y[1], y[2], } Example: y[n] = 2x[n] + 3x[n-1] + 4x[n-2]

2 2 FIR and IIR Systems FIR: Finite Impulse Response -- non-recursive y[n] = 2x[n] + 3x[n-1] + 4x[n-2] IIR: Infinite Impulse Response -- recursive y[n] = 0.5y[n-1] + 2x[n] + 3x[n-1]

3 3 Sample by Sample Processing y[n] = 0.5y[n-1] + 2x[n] + 3x[n-1] y[n] = 0.5w[n] + 3x[n] + 4v[n] w[n+1] = y[n] v[n+1] = x[n] For each new input x DO y = 0.5w + 2x + 3v w = y v = x CONTINUE

4 4 Linearity x 1 [n] a 1 x 2 [n] + x[n] H y[n] a 2 y[n] = a 1 y 1 [n]+ a 2 y 2 [n] x 1 [n] H y 1 [n] a 1 x 2 [n] H y 2 [n] a 2 + a 1 y 1 [n]+ a 2 y 2 [n]

5 5 Time-Invariance x[n] H y[n] D y[n-d] x[n] D x D [n] x[n-d] H y D [n] y D [n] = y[n-d]

6 6 Impulse Response of an LTI System δ[n] = 1, for n=0, and 0, otherwise δ[n] n H h[n] n For any input x[n], the output y[n] is given by y[n] = Σ m x[m]h[n-m] = Σ m h[m]x[n-m] (direct form)

7 7 Convolution δ[n] h[n] n δ[n-1] H h[n-1] n n δ[n-2] H h[n-2] n n H n x[n] = x[0]δ[n] + x[1]δ[n-1] + x[2]δ[n-2] + y[n] = x[0]h[n] + x[1]h[n-1] + x[2]h[n-2] +

8 8 FIR and IIR Filters An FIR filter has impulse response, h[n], that extends only over a finite time interval, say 0 n M, and is identically zero beyond that: {h[0], h[1], h[2], h[m], 0,0,0,0,0,0..} M is the filter order, and the length of the filter impulse response is M+1. An IIR filter, on the other hand, has impulse response, h[n], of infinite duration, 0 n, and is identically zero beyond that: {h[0], h[1], h[2], h[m], h[m+1],..} The convolution sum is not computationally feasible, hence it is normally represented by a constant-coefficient linear difference equation.

9 9 I/O Difference Equation Impulse Response: h[n] FIR filter: convolution sum IIR Filter: difference equation y[n] = ay[n-1] + x[n] h[n] = a n u[n]

10 10 Causality and Stability A causal sequence is a right-sided sequence, and an anti-causal sequence is a left-sided sequence. A two-sided sequence is mixed. An LTI system is causal if its impulse response, h[n], is causal. An LTI system is stable if for any bounded input sequence, the output sequence is also bounded. An LTI system is stable if and only if its impulse response is absolutely summable, i.e., Σ n h[n] <

11 11 Linear Convolution: Flip & Slide y[n] = Σ m x[m]h[n-m] If h is of order M, i.e., h = {h[0], h[1],, h[m]} is of length M+1 and x = {x[0], x[1],,x[l-1]} is of length L, then y[n] = Σ m=max(0,n-m) min(n,l-1) x[m]h[n-m] and y = {y[0], y[1],,y[l+m]} is of length (L+M)

12 12 Linear Convolution: Flip & Slide h 3 h 2 h 1 h 0 h 3 h 2 h 1 h 0 h 3 h 2 h 1 h x 0 x 1 x 2 x 3 x n-3 x n-2 x n-1 x n x L y 0 y n y L-1+M h = M+1 x = L y = h*x = L M

13 13 Transient & Steady State y[n] = Σ n m=0 h[m]x[n-m] if 0 n < M y[n] = Σ M m=0 h[m]x[n-m] if M n L-1 y[n] = Σ M m=n-l+1 h[m]x[n-m] if L n < L+M output y[n] 0 M L-1 L-1+M n

14 14 Convolution of Infinite Sequence y[n] = Σ m=max(0,n L+1) min(n,m) h[m]x[n-m] Infinite filter, finite input: M =, L < y[n] = Σ n m=max(0,n L+1) h[m]x[n-m] Finite filter, infinite input: M <, L = y[n] = Σ min(n,m) m=0 h[m]x[n-m] Infinite filter, infinite input: M =, L = y[n] = Σ n m=0 h[m]x[n-m]

15 15 Finite filter, Infinite input Block Processing y 0 = h*x 0, y 1 = h*x 1, y 2 = h*x 2, x = block x 0 block x 1 block x 2 y 0 = L M y 1 = L M y 2 = L M

16 16 Finite filter, Infinite input Sample Processing y[n] = Σ m=max(0,n L+1) n h[m]x[n-m] = h[0]x[n] + h[1]x[n-1] +. h[m]x[n-m] FOR each input x DO w 0 = x y = h 0 w 0 + h 1 w 1 + h 2 w h M w M w M = w M-1 w M-1 = w M-2 w1 = w 0

17 17 Circular Buffers x h 0 w 0 y h 1 w 1 h 2 w 2 h 3 w 3 MAC: Multiply & Accumulate X Σ

18 18 Z-Transform {x[n], n = -, } X(z) = Σ n=- x[n]z -n = +x[-2]z 2 +x[-1]z 1 +x[0] +x[1]z -1 +x[2]z -2 + Z-transform of the impulse response h[n] H(z) = Σ n=- h[n]z -n is called the Transfer function of the LTI system.

19 19 Basic Properties of Z-transform Linearity: a 1 x 1 [n] + a 2 x 2 [n] Z a 1 X 1 (z) + a 2 X 2 (z) Delay: x[n-d] Z z -D X(z) Convolution: y[n] = h[n]*x[n] Z Y(z) = H(z) X(z)

20 20 Z-transform Examples δ[n] δ[n-m] u[n] a n u[n] Z Z Z Z 1 z -M 1/(1- z -1 ) for z > 1 1/(1- a z -1 ) for z > a

21 21 ROC: Region of Convergence ROC: Region of z C (complex plane) where X(z) Examples: (0.5) n u[n] X(z) = 1/(1-0.5z -1 ) if z > 0.5 -(0.5) n u[-n-1] X(z) = 1/(1-0.5z -1 ) if z < 0.5 In general a n u[n] X(z) = 1/(1-a z -1 ) if z > a -a n u[-n-1] X(z) = 1/(1-a z -1 ) if z < a

22 22 Region of Convergence: Example X(z) = 1/(1-0.8z -1 ) + 1/(1-1.25z -1 ) = (2-2.05z -1 )/(1-2.05z -1 + z -2 ) I II III Z-plane

23 23 Causal Sequence & ROC: Poles X(z) = A 1 /(1-p 1 z -1 ) + A 2 /(1-p 2 z -1 ) +.. A k /(1-p k z -1 ) + ROC: z > max p k Z-plane Causal ROC causal sequence = right-sided sequence = ROC is outside of a circle

24 24 Anti-Causal Sequence & ROC X(z) = A 1 /(1-p 1 z -1 ) + A 2 /(1-p 2 z -1 ) +.. A k /(1-p k z -1 ) + Z-plane ROC: z < min p k Anti-Causal ROC Anti-causal sequence = left-sided sequence = ROC is inside of a circle

25 25 Two-sided Sequence & ROC X(z) = A 1 /(1-p 1 z -1 ) + A 2 /(1-p 2 z -1 ) +.. A k /(1-p k z -1 ) + Z-plane Mixed ROC mixed sequence two-sided sequence = ROC is inside of a ring

26 26 Stability of a Sequence and ROC Stable Sequence = ROC contains the UNIT CIRCLE If H is stable, i.e., Σ n=- h[n] < Then on the unit circle, i.e., z =1 H(z) = Σ n=- h[n]z -n Σ n=- h[n]z -n = Σ n=- h[n] < Therefore, its ROC contain the unit circle. If H(z) s ROC contains the unit circle Then Σ n=- h[n]z -n z =1 = Σ n=- h[n] < Therefore, H is stable.

27 27 Stable and Causal Sequence Stable and Causal: All poles are inside the unit circle Z-plane Causal ROC unit circle

28 28 Stable and Anti-Causal Sequence Stable and Anti-Causal: All poles are outside the unit circle unit circle Z-plane Anti-Causal ROC

29 29 Stable and Two-sided Sequence Stable and Two-sided:Some poles are outside the unit circle Some poles are inside the unit circle unit circle Z-plane Mixed ROC

30 30 Stable System An LTI system is stable if and only if H(z) s ROC contains the unit circle.

31 31 Inverse Z-transform By inspection: First order sequence/system X(z) = 1/(1-a z -1 ) if z > a X(z) = 1/(1-a z -1 ) if z < a x[n] = a n u[n] x[n] = -a n u[-n-1] Second order sequence/system X(z) =A 1 /(1-p 1 z -1 ) + A 1* /(1-p 1 * z -1 ) if z > p 1 where A 1 = B 1 e jα 1 and p 1 = R 1 e jω 1 x[n] = 2B 1 R 1 n cos(nω 1 +α 1 )u[n] X(z) = [2B 1 cos(α 1 )-2B 1 R 1 cos(α 1 -ω 1 ) z -1 ]/[1-2R 1 cos(ω 1 ) z -1 +R 1 2 z -2 ]

32 32 Inverse Z-transform Partial Fraction Expansion: X(z) = N(z)/D(z) N(z) is of degree N & D(z) is of degree M, M<N, Real Roots = N(z)/(1-p 1 z -1 )(1-p 2 z -1 ).. /(1-p M z -1 ) = A 1 /(1-p 1 z -1 )+ A 2 /(1-p 2 z -1 )+. A M /(1-p M z -1 ) A k = (1-p k z -1 )X(z) z=pk N(z) is of degree N & D(z) is of degree M, M<N, Complex Roots X(z) = Σ k [A k /(1-p k z -1 ) + A k* /(1-p k * z -1 )] N(z) is of degree N & D(z) is of degree M, M<N, Multiple Roots N(z) is of degree N & D(z) is of degree M, M N, Complex Roots

33 33 Frequency Response & Z-Transform The DTFT (Discrete-Time Fourier Transform) X(ω) = Σ n=- x[n]e -jωn X(z) = Σ n=- x[n]z -n X(ω) = X(z) z=e jω The Inverse DTFT x[n] = (1/2π) π π X(ω)e jωn dω The Frequency Response of the LTI System H(ω) = Σ n=- h[n]e -jωn Y(ω)= H(ω)X(ω)

34 34 Parseval Theorem Σ n=- x[n] 2 = (1/2π) π π X(ω) 2 dω Digital measurement of energy of an analog signal Σ n=- x[n] 2 = (1/2π) π π X(ω) 2 dω = (1/2π)(1/T) X(Ω) 2 dω =(1/T) x(t) 2 dt Total energy = (T)Σ n=- x[n] 2

35 35 Frequency Response & Pole/Zero Locations X(z) = (1-z 1 z -1 )/(1-p 1 z -1 ) X(ω) = (e jω -z 1 )/(e jω -p 1 ) X(ω) = (e jω -z 1 ) / (e jω -p 1 ) p 1 = p 1 e jφ z 1 = z 1 e jθ X(ω) p 1 e jω z 1 θ φ ω

36 36 Transfer Function of an LTI System Impulse Response h[n] I/O Difference Equation Frequency Response Specification Transfer Function H(z) Filter Design I/O Convolution Equation Block Processing Pole/zero Pattern Block Diagram Realization Sample Processing

37 37 An Example Transfer Function: H(z) = (5+2z -1 )/(1-0.8z -1 ) Impulse Response: H(z) = /(1-0.8z -1 ) h[n] = -2.5δ[n] + 7.5(0.8) n u[n] Difference Equation: H(z) = (5+2z -1 )/(1-0.8z -1 ) = Y(z)/X(z) Y(z) 0.8z -1 Y(z) = 5X(z) +2z -1 X(z) y[n] = 0.8y[n-1] + 5x[n] + 2x[n-1] Block Diagram: Y(z)/X(z) = (5+2z -1 )/(1-0.8z -1 )

38 38 An Example - Continued Block Diagram: Y(z)/X(z) = (5+2z -1 )/(1-0.8z -1 ) x[n] 5 + y[n] z -1 z -1 x[n-1] y[n-1]

39 39 An Example - Continued Frequency Response: H(z) = (5+2z -1 )/(1-0.8z -1 ) =5(1+0.4z -1 )/(1-0.8z -1 ) H(ω) =5 (1+0.4e -jω )/(1-0.8e -jω ) Using the identity: 1-ae -jω = [1-2a cos(ω)+a 2 ] 1/2 H(ω) = 5 [1-0.8 cos(ω)+0.16] 1/2 /[1-1.6 cos(ω)+0.64] 1/ H(ω) π 5/3 ω

40 40 Direct & Canonical Forms H(z) = N(z)/D(z) = (b 0 + b 1 z -1+ b 2 z b L z -L )/(1+ a 1 z -1+ a 2 z a M z -M ) Y(z)/X(z) = (b 0 + b 1 z -1+ b 2 z b L z -L )/(1+ a 1 z -1+ a 2 z a M z -M ) Y(z)= X(z)(b 0 + b 1 z -1+ b 2 z b L z -L )/(1+ a 1 z -1+ a 2 z a M z -M ) = X(z)(b 0 + b 1 z -1+ b 2 z b L z -L ) [1/(1+ a 1 z -1+ a 2 z a M z -M )] Direct Form = X(z)/(1+ a 1 z -1+ a 2 z a M z -M ) [(b 0 + b 1 z -1+ b 2 z b L z -L )] Canonical Form

41 41 Direct Form Y(z)= X(z)(b 0 + b 1 z -1+ b 2 z b L z -L ) [1/(1+ a 1 z -1+ a 2 z a M z -M )] x[n] b 0 + y[n] z -1 z -1 x[n-1] z -1 b 1 -a 1 z -1 y[n-1] x[n-2] b 2 -a 2 y[n-2] x[n-l] z -1 b L -a M z -1 y[n-m]

42 42 Canonical Form Y(z)= X(z)/(1+ a 1 z -1+ a 2 z a M z -M ) [(b 0 + b 1 z -1+ b 2 z b L z -L )] x[n] + w[n] b 0 z -1 + w[n-1] y[n] -a 1 b 1 z -1 -a 2 b 2 b L z -1 -a M

43 43 Examples Example: Impulse response: h = {1, 6, 11, 6} Difference equation: y[n] = x[n] + 6x[n-1] +11x[n-2] +6x[n-3] Transfer Function: H(z) = 1+ 6z z z -3 Pole/Zeros: H(z) = (1+z -1 )(1+2z -1 )(1+3z -1 ) Frequency Response, Block Diagrams Examples: (a) Difference equation: (b) Difference equation: y[n] = 0.25 y[n-2] + x[n] y[n] = y[n-2] + x[n]

44 44 Sinusoidal Steady State Response x[n] = e jω o n DTFT: X(ω) = 2π δ(ω ω ο ) + replicas Y(ω) = H(ω)X(ω) = H(ω ο )2π δ(ω ω ο ) + replicas y[n] = H(ω ο ) e jω o n i.e., e jω o n H H(ω ο ) e jω o n = H(ω ο ) e j(ω o n+arg[h(ω o )]) cos(nω ο ) H H(ω ο ) cos(nω ο + Arg[H(ω o )] ) sin(nω ο ) H H(ω ο ) sin(nω ο + Arg[H(ω o )] ) Linearity Ae jω 1 n +Be jω 2 n H A H(ω 1 ) e jω 1 n +B H(ω 2 ) e jω 2 n

45 45 Sinusoidal Transient Response Input: x[n] = e jω o n u[n] : X(z) = 1/(1- e jω o z -1 ) LTI system: H(z) = N(z)/D(z) (assuming real poles) = N(z)/(1-p 1 z -1 )(1-p 2 z -1 )..(1-p M z -1 ) all p k < 1, k = 1, 2,., M Output: Y(z) = X(z)H(z) = N(z)/(1- e jω o z -1 )(1-p 1 z -1 )(1-p 2 z -1 )..(1-p M z -1 ) = H(ω ο ) /(1- e jω o z -1 ) + A 1 /(1-p 1 z -1 ) + A 2 /(1-p 2 z -1 ) +.. A M /(1-p M z -1 ) y[n] = {H(ω ο ) e jω o n + A 1 p 1n + A 2 p 2n +.. A M p Mn }u[n] as n y[n] H(ω ο ) e jω o n

46 46 Sinusoidal Transient Response Example: H(z) = (5+2z -1 )/(1-0.8z -1 ) Y(z) = (5+2z -1 )/(1- e jω o z -1 )(1-0.8z -1 ) Observation: 1. Stability of the Filter 2. Effective Time Constant: n eff ρ = max k p k, ρ n eff = ε n eff = ln(ε )/ln(ρ) Example 6.3.2

47 47 Unit Step Response Input: x[n] = e jω o n u[n], ω ο = 0 : X(z) = 1/(1- z -1 ) LTI system: H(z) = N(z)/D(z) (assuming real poles) = N(z)/(1-p 1 z -1 )(1-p 2 z -1 ).. /(1-p M z -1 ) all p k < 1, k = 1, 2,., M Output: Y(z) = X(z)H(z) = N(z)/(1-z -1 )(1-p 1 z -1 )(1-p 2 z -1 )..(1-p M z -1 ) = H(1) /(1- z -1 ) + A 1 /(1-p 1 z -1 ) + A 2 /(1-p 2 z -1 ) +.. A M /(1-p M z -1 ) y[n] = {H(1) + A 1 p 1n + A 2 p 2n +.. A M p Mn }u[n] as n y[n] H(1)u[n] H(1) = Σ n=0 h[n] is the DC gain

48 48 Pole/Zero Designs First Order Filter H(z) = G(1+bz -1 )/(1-az -1 ) H(0), H(π), and speed of response n eff H(0) =G(1+b)/(1-a), H(π) = G(1-b)/(1+a) H(π) / H(0) = (1-b)(1-a)/(1+b)(1+a) a = ε 1/n eff Example H(z) = (5+2z -1 )/(1-0.8z -1 ) = 5(1+0.4z -1 )/(1-0.8z -1 )

49 49 Second-Order Filter & Resonator Z-Plane ω o Re jω o H(z) = G/(1- Re jω o z -1 )(1- Re -jω oz -1 ) = G/(1-2Rcos(ω o )z -1 + R 2 z -2 ) Unit circle Re -jω o H(ω) = G/(1- Re jω o e -jω )(1- Re -jω oe -jω ) G: normalize H(ω o ) = 1

50 50 Second-Order Filter & Resonator H(ω) 2 = G/(1-2Rcos(ω ω o ) + R 2 )(1-2Rcos(ω+ω o ) + R 2 ) 1 1/2 H(ω) 2 ω o ω (3 db bandwidth) π ω ω 2(1-R)

51 51 Second-Order Filter & Resonator Transfer Function: H(z) = G/(1-2Rcos(ω o ) z -1 + R 2 z -2 ) Impulse Response: h[n] = [G/sin(ω o )] R n sin(nω o ) Difference Equation: y[n] = 2Rcos(ω o )y[n-1] - R 2 y[n-2] Gx[n] x[n] G + y[n] z -1 a 1 = -2Rcos(ω o ) a 2 = R 2 -a 1 -a 2 z -1 y[n-1] y[n-2]

52 52 Second-Order Filter & Resonator Example: Design a 2-pole resonator filter with peak frequency f o = 500 Hz, and 3-dB bandwidth f = 32 Hz. The sampling frequency is f s = 10KHz. R = 0.99, G = , a 1 = and a 2 = H(z) = /( z z -2 )

53 53 Parametric Equalizer Unit circle ω o Re jω o p 1 =R e jω o z 1 =r e jω o p 1* =R e -jω o z 1* =r e -jω o Z-Plane Re -jω o H(z) = (1- re jω o z -1 )(1- re -jω oz -1 ) /(1- Re jω o z -1 )(1- Re -jω oz -1 ) = (1 + b 1 z -1 + b 2 z -2 )/(1 + a 1 z -1 + a 2 z -2 ) a 1 = -2Rcos(ω o ), a 2 = R 2 b 1 = -2rcos(ω o ), b 2 = r 2 H(z) = (1 + b 1 z -1 + b 2 z -2 )/(1 + (R/r)b 1 z -1 + (R/r) 2 b 2 z -2 )

54 54 Parametric Equalizer H(z) = (1 + b 1 z -1 + b 2 z -2 )/(1 + (R/r)b 1 z -1 + (R/r) 2 b 2 z -2 ) b 1 = -2rcos(ω o ), b 2 = r 2 H(ω) 2 = (1-2rcos(ω ω o ) + r 2 )(1-2rcos(ω+ω o ) + r 2 ) /(1-2Rcos(ω ω o ) + R 2 )(1-2Rcos(ω+ω o ) + R 2 ) H(ω) 2 ω o π ω

55 55 Notch Filter Unit circle Z-Plane ω o Re jω o Re -jω o Notch Filter: Zeros on the unit circle r = 1 H(z) = (1 + b 1 z -1 + z -2 )/(1 + (R)b 1 z -1 + R 2 z -2 ) b 1 = -2cos(ω o ),

56 56 Notch Filter Notch polynomial N(z) = Π k (1-e jω k z -1 ) Real Coefficient N(z) = Π k (1-2cosω k z -1 +z -2 ) Notch Filter = N(z)/N(ρ -1 z) = (1 + b 1 z -1 + b 2 z b 2 z -(M-1) + b 1 z -M )/ (1 + ρb 1 z -1 + ρ 2 b 2 z ρ -(M-1) b 2 z -(M-1) + ρ M b 1 z -M )

57 57 Comb Filter From notch filter moves the zeros inside the poles H(z) = (1 + rb 1 z -1 + r 2 b 2 z r -(M-1) b 2 z -(M-1) + r M b 1 z -M )/ (1 + ρb 1 z -1 + ρ 2 b 2 z ρ -(M-1) b 2 z -(M-1) + ρ M b 1 z -M ) r ρ and r < ρ