Time domain representation of signals and systems

Good to learn the basic signals and system

1. App. Phys. 602 : Digital Signal Processing Topic 2: Time domain 602 - Digital Signal Processing, Semester VIII 9/27/2022 1 Discrete-time systems Convolution Linear Constant-Coefficient Difference Equations (LCCDEs) Correlation

2. 1. Discrete-time systems A system converts input to output:  E.g. Moving Average (MA): x [ n ] DT System y [ n ] ( M = 3) x [ n ] y [ n ] + z -1 z -1 1/ M 1/ M 1/ M x [ n -1] x [ n -2] y [ n ]  1 M x [ n  k ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 2 k  M  1  { y [ n ]} = f ({ x [ n ]}) n A

3. Moving A verage (MA) x [ n ] y [ n ] + z -1 z -1 1/ M 1/ M 1/ M x [ n -1] x [ n -2] n x [ n ] 1 2 3 4 5 6 7 8 9 A n -1 x [ n -1] 1 2 3 4 5 6 7 8 9 -1 x [ n -2] -1 1 2 3 4 5 6 7 8 9 n 1 2 3 4 5 6 7 8 9 n y [ n ] -1 y [ n ]  1 M x [ n  k ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 3 k  M  1 

4. MA Smoother MA smoothes out rapid variations (e.g. “12 month moving average”) e.g. signal noise 5 k  y [ n ]  1  4 x [ n  k ] 5-pt moving average x [ n ] = s [ n ] + d [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 4

5. Accumulator Output accumulates all past inputs: x [ n ] y [ n ] + z -1 602 - Digital Signal Processing, Semester VIII 9/27/2022 5 y [ n -1]    n y [ n ]   x [  ]    n  1   x [  ]  x [ n ]  y [ n  1 ]  x [ n ]

6. Accumulator n x [ n ] -1 1 2 3 4 5 6 7 8 9 A y [ n -1] -1 1 2 3 4 5 6 7 8 9 n 1 2 3 4 5 6 7 8 9 n y [ n ] -1 x [ n ] y [ n ] + z -1 602 - Digital Signal Processing, Semester VIII 9/27/2022 6 y [ n -1]

7. Classes of DT systems Linear systems obey superposition :  if input x 1 [ n ] → output y 1 [ n ] , x 2 → y 2 ... given a linear combination  of inputs :  then output for all  , , x 1  x 2 i.e. same linear combination of outputs x [ n ] DT system y [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 7

8. Lineari t y: Example 1 Accumulator: Linear 602 - Digital Signal Processing, Semester VIII 9/27/2022 8    n y [ n ]   x [  ]    n y [ n ]     x 1 [  ]   x 2 [  ]      x 1 [  ]      x 2 [  ]     x 1 [  ]    x 2 [  ]    y 1 [ n ]    y 2 [ n ] x [ n ]    x 1 [ n ]    x 2 [ n ]

9. Lineari t y Example 2: “Teager Energy operator”: y [ n ]  x 2 [ n ]  x [ n  1 ]  x [ n  1 ] x [ n ]    x 1 [ n ]    x 2 [ n ] Nonlinear 602 - Digital Signal Processing, Semester VIII 9/27/2022 9 y [ n ]   1 2 x [ n ]   x [ n ]   2    x 1 [ n  1 ]   x 2 [ n  1 ]     x 1 [ n  1 ]   x 2 [ n  1 ]     y 1 [ n ]    y 2 [ n ]

10. Lineari t y Example 3: ‘Offset’ accumulator: but Nonlinear 602 - Digital Signal Processing, Semester VIII 9/27/2022 10 .. unless C =    n y [ n ]  C   x [  ] n  y 1 [ n ]  C   x 1 [  ]   n y [ n ]  C     x 1 [  ]   x 2 [  ]       y 1 [ n ]   y 2 [ n ]

11. Property: Shift (time) invariance 602 - Digital Signal Processing, Semester VIII 9/27/2022 11   Time-shift of input causes same shift in output i.e. if x 1 [ n ] → y 1 [ n ] the n x [ n ]  x 1 [ n  n ]  y [ n ]  y 1 [ n  n ] i.e. process doesn’t depend on absolute value of n

12. Shift-invariance counterexample Upsampler: L Not shift invariant y [ n ] = x [ n/ L ] n = , ± L, ± 2 L , . . . otherwise ( n = r · L ) y 1 [ n ] = x 1 [ n/L ] x [ n ] = x 1 [ n - n ] =} y [ n ] = x [ n/L ] = x 1 [ n/L - n ] = x 1 n - L · n L 602 - Digital Signal Processing, Semester VIII 9/27/2022 12 = y 1 [ n - L · n ] = 1 / y [ n - n ] x [ n y [ n ] {

13. Another counterexample  Hence  If then  Not shift invariant - parameters depend on n ≠ y [ n ]  n  x [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 13 scaling by time index y 1 [ n  n ]   n  n   x 1 [ n  n ] x [ n ]  x 1 [ n  n ] y [ n ]  n  x 1 [ n  n ]

14. Linear Shift Invariant ( LSI ) 602 - Digital Signal Processing, Semester VIII 9/27/2022 14   Systems which are both linear and shift invariant are easily manipulated mathematically This is still a wide and useful class of systems If discrete index corresponds to time, called Linear Time Invariant (LTI)

15. Causality  If output depends only on past and current inputs (not future), system is called causal Formally, if x 1 [ n ] → y 1 [ n ] & x 2 [ n ] → y 2 [ n ] Causa l x 1 [ n ]  x 2 [ n ]  n  N  y 1 [ n ]  y 2 [ n ]  n  N 602 - Digital Signal Processing, Semester VIII 9/27/2022 15

16. Causality example Moving average: y [ n ]  y [ n ] depends on x [ n - k ] , k ≥ → causal ‘Centered’ moving average   .. looks forward in time → noncausal .. Can make causal by delaying  1 M x [ n  k ] k  M  1  c y  n   y  n   M  1  / 2   1 M x [ n ]  602 - Digital Signal Processing, Semester VIII 9/27/2022 16 x [ n  k ]  x [ n  k ] ( M  1 ) / 2 k  1   

17. 602 - Digital Signal Processing, Semester VIII 9/27/2022 17

18. Impulse response (IR) Impulse (unit sample sequence) Given a system: if x [ n ] =  [ n ] then  “impulse response” LSI system completely specified by h [ n ] n 1 2 3 4 5 6 7  [ n ] 1 -3 -2 -1 x [ n ] DT system y [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 18 Y[n]  

19. Impulse response example Simple system: α  x [ n ] y [ n ] + z -1 z -1 α 2 α 3 n 1 2 3 4 5 6 7 x [ n ] 1 -3 -2 -1 n y [ n ] 1 2 3 4 5 6 7 α  -3 -2 -1 α  α  x [ n ]   [ n ] impulse y [ n ] = h [ n ] impulse response 602 - Digital Signal Processing, Semester VIII 9/27/2022 19

20. 2. Convolution Impulse response: Shift invariance:  + Linearity:  Can express any sequence with  s: x [ n ] = x [0]  [ n ] + x [1]  [ n -1] + x [2]  [ n -2] ..  [ n ] LSI h [ n ]  [ n-n ] LSI h [ n-n ] α ·  [ n-k ] + β ·  [ n-l ] LSI α · h [ n-k ] + β · h [ n-l ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 20

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23. Convolu t ion sum Hence, since   C on v olu t io n sum  x [ n ]   x [ k ]  [ n  k ] k  k    F or LS I , y [ n ]   x [ k ] h [ n  k ] wri tt en as y [ n ]  x [ n ] * h [ n ] Summation is symmetric in x and h i.e. l = n – k → l   x [ n ] * h [ n ]   x [ n  l ] h [ l ]  h [ n ] * x [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 23

24. Convolution properties    LSI System output y [ n ] = input x [ n ] convolved with impulse response h [ n ] → h [ n ] completely describes system Commutative: x [ n ] * h [ n ] = h [ n ] * x [ n ] Associative: ( x [ n ] * h [ n ]) * y [ n ] = x [ n ] * ( h [ n ] * y [ n ]) Distributive: h [ n ] * ( x [ n ] + y [ n ] ) = h [ n ] * x [ n ] + h [ n ] * y [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 24

25. Interpreting convolution Passing a signal through a (LSI) system is equivalent to convolving it with the system’s impulse response x [ n ] h [ n ] y [ n ] = x [ n ] ∗ h [ n ] x [ n ]={0 3 1 2 -1} h [ n ] = {3 2 1} n 1 2 3 4 5 6 7 -3 -2 -1 n 1 2 3 5 6 7 -3 -2 -1   y [ n ]   x [ k ] h [ n  k ]   h [ k ] x [ n  k ] k  k  602 - Digital Signal Processing, Semester VIII 9/27/2022 25 → →

26. Convolu t ion in t erpre t a t ion 1  Time-reverse h , shift by n , take inner product against fixed x k 1 2 3 5 6 7 -3 -2 -1 x [ k ] k 1 2 3 4 5 6 7 -3 -2 -1 h [0- k ] k 1 2 3 4 5 6 7 -3 -2 -1 k 1 2 3 4 5 6 7 -3 -2 -1 h [2- k ] n 1 2 3 5 7 -3 -2 -1 y [ n ] 9 9 11 2 -1 k   602 - Digital Signal Processing, Semester VIII 9/27/2022 26  y [ n ]   x [ k ] h [ n  k ] =g [ k ] h [1- k ] =g [ k -1] call h [- n ] = g [ n ]

27. Convolu t ion in t erpre t a t ion 2  Shifted x ’s weighted by points in h Conversely, weighted, delayed versions of h ...  n 1 2 3 5 7 -3 -2 -1 y [ n ] 9 9 11 2 -1 n 1 2 3 5 6 7 -3 -2 -1 x [ n ] n 1 2 3 4 6 7 -3 -2 -1 x [ n -1] n 1 2 3 4 5 7 -3 -2 -1 x [ n -2] k 1 2 3 4 5 6 7 h [ k ] -3 -2 -1  602 - Digital Signal Processing, Semester VIII 9/27/2022 27 y [ n ]   h [ k ] x [ n  k ] k 

28. Matrix interpretation 602 - Digital Signal Processing, Semester VIII 9/27/2022 28  Diagonals in X matrix are equal    

29. Convolu t ion no t es 602 - Digital Signal Processing, Semester VIII 9/27/2022 29  Total nonzero length of convolving N and M point sequences is N + M -1 Adding the indices of the terms within the summation gives n : k  i.e. summation indices move in opposite senses  y [ n ]   h [ k ] x [ n  k ] k   n  k   n

30. Convolution in MATLAB 602 - Digital Signal Processing, Semester VIII 9/27/2022 30 T he M- f ile conv implemen t s t he convolution sum of two finite-length sequences If a  [ b  [ 3 3 1 2 - 1 ] 2 1 ] the n conv(a,b) yields [ 9 9 11 2 - 1 ]

31. Connected systems Cascade connection: Impulse response h [ n ] of the cascade of two systems with impulse responses h 1 [ n ] and h 2 [ n ] is By commutativity,  * h [ n ]  h 1 [ n ] ⊛ 602 - Digital Signal Processing, Semester VIII 9/27/2022 31

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33. Inverse systems   [ n ] is identity for convolution i.e . x [ n ] *  [ n ]  x [ n ] Consider x [ n ] y [ n ] z [ n ] z [ n ]  h 2 [ n ] * y [ n ]  h 2 [ n ] * h 1 [ n ] * x [ n ]  x [ n ] i f h 2 [ n ] * h 1 [ n ]   [ n ] h 2 [ n ] is the inverse system of h 1 [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 33

34. Inverse systems   Use inverse system to recover input x [ n ] from output y [ n ] (e.g. to undo effects of transmission channel) Only sometimes possible - e.g. cannot ‘invert’ h 1 [ n ] = 0 In general, attempt to solve h 2 [ n ] * h 1 [ n ]   [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 34

35. Inverse system example Accumulator: Impulse response h 1 [ n ] = μ [ n ] Thus, ‘backwards difference’ is inverse system of accumulator. ‘Backwards difference’ .. has desired property:  [ n ]   [ n  1 ]   [ n ] n -3 -2 -1 1 2 3 4 5 6 7 n -3 -2 -1 1 2 3 4 5 6 7 602 - Digital Signal Processing, Semester VIII 9/27/2022 35

36. Parallel connection  Impulse response of two parallel systems added together is: 602 - Digital Signal Processing, Semester VIII 9/27/2022 36

37. 3. Linear Constant-Coefficient Difference Equation (LCCDE) General spec. of DT, LSI, finite-dim sys:  Rearrange for y [ n ] in causal form:  WLOG, always have d = 1 defined by { d k },{ p k } order = max( N , M ) k  k  N M  d k y [ n  k ]   p k x [ n  k ] k y [ n ]    d p k k  1 d k  d 602 - Digital Signal Processing, Semester VIII 9/27/2022 37 N M y [ n  k ]   x [ n  k ]

38. 602 - Digital Signal Processing, Semester VIII 9/27/2022 38

39. Solving LCCDEs “Total solution” y [ n ]  y c [ n ]  y p [ n ] Complementary Solution N satisfies Particular Solution for given forcing function x [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 39  d k y [ n  k ]  k 

40. Complementary Solution 602 - Digital Signal Processing, Semester VIII 9/27/2022 40 General form of unforced oscillation i.e. system’s ‘natural modes’ Characteristic polynomial of system - depends only on { d k } y c [ n ]   n  k d  n  k k  Assume y c has form N     n  N N 1 d   d  N  1  ...  d N  1 N   d     k d  N  k N  k  

41. Complementary Solution 602 - Digital Signal Processing, Semester VIII 9/27/2022 41  i factors into roots λ , i.e.  Each/any λ i satisfies eqn. Thus, complementary solution:  Any linear combination will work → α i s are free to match initial conditions k d  N  k N  k   1 2 (    ) (    )...  y [ n ]    c 1 1 2 2 3 3 n n n        ...

42. Complementary Solution Repeated roots in chr. poly:  Complex λ i s → sinusoidal y c [ n ]   i  i n  y [ n ]    c 1 1   n  n   n 2  n 2 1 3 1  ...   n L  1  n  ... L 1 1 2 n (    ) L (    )...  602 - Digital Signal Processing, Semester VIII 9/27/2022 42

43. Particular Solution 602 - Digital Signal Processing, Semester VIII 9/27/2022 43 Recall: Total solution  Particular solution reflects input  ‘Modes’ usually decay away for large n leaving just y p [ n ] Assume ‘form’ of x [ n ] , scaled by β : e.g. x [ n ] constant → y p [ n ] = β  x [ n ] = λ → y [ n ] = β · λ p n n ( λ ∉ λ i ) or = β n L λ n ( λ ∈ λ i ) y [ n ]  y c [ n ]  y p [ n ]

44. LCCDE example  Need input : x [ n ] = 8 μ [ n ] Need initial conditions : y [-1] = 1, y [-2] = -1  y [ n ]  y [ n  1 ]  6 y [ n  2 ]  x [ n ] x [ n ] + y [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 44

45. LCCDE example 602 - Digital Signal Processing, Semester VIII 9/27/2022 45 Complementary solution: y [ n ]  y [ n  1 ]  6 y [ n  2 ]  ;      3    2   → roots λ = -3, λ = 2 1 2  y c [ n ]   1   3    2  2  n n α 1 , α 2 are unknown at this point y [ n ]   n   n  2   2    6  

46. LCCDE example 602 - Digital Signal Processing, Semester VIII 9/27/2022 46  Particular solution: Input x [ n ] is constant = 8 μ [ n ] assume y p [ n ] = β , substitute in: y [ n ]  y [ n  1 ]  6 y [ n  2 ]  x [ n ]      6   8  [ n ]   4   8     2 (‘large’ n )

47.  Solve for unknown α i s by substituting  n = LCCDE example from ICs initial conditions into DE at n = 0, 1, ... y [ n ]  y [ n  1 ]  6 y [ n  2 ]  x [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 47    1 2     n n 3   2   T o t al solu t ion y [ n ]  y c [ n ]  y p [ n ] y [ ]  y [  1 ]  6 y [  2 ]  x [ ]        1  6  8 1 2      3 1 2

48. LCCDE example n = 1    Don’t find α i s by solving with ICs at n = -1,-2 (ICs may not reflect natural modes; Mitra3 ex 2.37-8 (4.22-3) is wrong ) 602 - Digital Signal Processing, Semester VIII 9/27/2022 48 y [ 1 ]  y [ ]  6 y [  1 ]  x [ 1 ]   (  3 )   ( 2 )          6  8 1 2 1 2 1 2   2   3   18 solve: α 1 = -1.8 , α 2 = 4.8 Hence, system output: y [ n ]   1.8(  3) n  4.8(2) n  2 n 

49. LCCDE solving summary 602 - Digital Signal Processing, Semester VIII 9/27/2022 49 Difference Equation (DE): Ay [ n ] + By [ n -1] + ... = Cx [ n ] + Dx [ n -1] + ... Initial Conditions (ICs): y [-1] = ... DE RHS = with y [ n ]= λ n → roots { λ i } gives complementary soln Particular soln : y p [ n ] ~ x [ n ]  solve for βλ n “at large n ” α i s by substituting DE at n = 0, 1, ... ICs for y [-1], y [-2]; y t = y c + y p for y [0], y [1]  c y [ n ]    i i n 

50. LCCDEs: zero input/zero state 602 - Digital Signal Processing, Semester VIII 9/27/2022 50 Alternative approach to solving LCCDEs is to solve two subproblems:  y zi [ n ] , response with zero input (just ICs) y zs [ n ] , response with zero state (just x [ n ] )  Because of linearity, y [ n ] = y zi [ n ]+ y zs [ n ] Both subproblems are ‘fully realized’ But, have to solve for α i s twice (then sum them)  

51. Impulse response of LCCDEs Impulse response: i.e. solve with x [ n ] = δ [ n ] → y [ n ] = h [ n ] (zero ICs)  With x [ n ] = δ [ n ] , ‘form’ of y p [ n ] = βδ [ n ] → solve y [ n ] for n = 0,1, 2... to find α i s δ [ n ] LCCDE h [ n ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 51

52. LCCDE I R example    t hus 1 602 - Digital Signal Processing, Semester VIII 9/27/2022 52 n ≥ Infinite length e.g. y [ n ]  y [ n  1 ]  6 y [ n  2 ]  x [ n ] (from before); x [ n ] = δ [ n ]; y [ n ] = 0 for n <0 y [ n ]    c 1 2 3   2     n n h [ n ]  0. 6   3  n  0. 4  2  n y p [ n ] = βδ [ n ] n = 0: y [ ]  y [  1 ]  6 y [  2 ]  x [ ]  α 1 + α 2 + β = 1 n = 1: α 1 (–3) + α 2 (2) + 1 = n = 2: α 1 (9) + α 2 (4) – 1 – 6 =  α 1 = 0.6, α 2 = 0.4, β =

53. System property: Stability Output grows without limit in some conditions z -1 2 n Certain systems can be unstable e.g. x [ n ] + y [ n ] y [ n ] -1 1 2 3 4 ... 602 - Digital Signal Processing, Semester VIII 9/27/2022 53

54. Stability Several definitions for stability; we use Bounded-input, bounded-output (BIBO) stable For every bounded input x [ n ]  B x  n output is also subject to a finite bound, y [ n ]  B y  n 602 - Digital Signal Processing, Semester VIII 9/27/2022 54

55. Stability example MA f il t er: → BIBO Stable M k  y [ n ]  1  M  1 x [ n  k ] M k  y [ n ]  1  M  1 x [ n  k ] M k   1  M  1 x [ n  k ] M  1 M  B  B 602 - Digital Signal Processing, Semester VIII 9/27/2022 55 x y

56. S t abili t y & LCCDEs 602 - Digital Signal Processing, Semester VIII 9/27/2022 56 LCCDE output is of form: y [ n ]    n    n  ...   n 1 1 2 2 0  α s and β s depend on input & ICs, but to be bounded for any input we need |λ     ...

57. 4. Correlation Correlation ~ identifies similarity between sequences:  Cross of x against y “lag” correla t ion 602 - Digital Signal Processing, Semester VIII 9/27/2022 57   Note:   Set m = n-l  

58. Correlation and convolution 602 - Digital Signal Processing, Semester VIII 9/27/2022 58 Correlation: Convolu t ion: Hence: Correlation may be calculated by convolving with time-reversed sequence      

59. Autocorrelation 602 - Digital Signal Processing, Semester VIII 9/27/2022 59 Energy of sequence x [ n ] xx Note : r [ ]  2  n    x [ n ]   x Autocorrelation (AC) is correlation of signal with itself:  

60. Correla t ion maxima Note:   From geometry,  when x // y , cos θ = 1 , else cos θ < 1 angle between x and y r x x [  ]  r x x [ ]  r x x [  ] r x x [ ]  1 xy Similarly: r [  ]  x y    r x y [  ] r x x [ ] r y y [ ]  1 i xy   x i y i  x y cos  x i 2  602 - Digital Signal Processing, Semester VIII 9/27/2022 60

61. AC of a periodic sequence Sequence of period N : x ˜ [ n ]  x ˜ [ n  N ] Calculate AC over a finite window: if M >> N r x ˜ x ˜ [  ]  1 2 M  1 n   M M  x ˜ [ n ] x ˜ [ n   ]  1 N n  N  1 602 - Digital Signal Processing, Semester VIII 9/27/2022 61  x ˜ [ n ] x ˜ [ n   ]

62. AC of a periodic sequence  i.e AC of periodic sequence is periodic Average energy per sample or Power of x x ˜ x ˜ r [ ]  1 N 2 x ˜ [ n ]  P N  1  n  x ˜ r x ˜ x ˜ [   N ]  1 N n  N  1 602 - Digital Signal Processing, Semester VIII 9/27/2022 62  x ˜ [ n ] x ˜ [ n    N ]  r x ˜ x ˜ [  ]

63. Cross correlation r x y [  ] What correlations look like AC of any x [ n ]  AC of periodic r x x [  ] 602 - Digital Signal Processing, Semester VIII 9/27/2022 63  r x ˜ x ˜ [  ]  

64. What correlation looks like 602 - Digital Signal Processing, Semester VIII 9/27/2022 64