6.003: Signals and Systems Convolution October 4, 2011 1 Mid-term - PowerPoint PPT Presentation

6.003: Signals and Systems Convolution October 4, 2011 1

Mid-term Examination #1 Wednesday, October 5, 7:30­9:30pm. No recitations on the day of the exam. Coverage: CT and DT Systems, Z and Laplace Transforms Lectures 1–7 Recitations 1–7 Homeworks 1–4 Homework 4 will not collected or graded. Solutions are posted. 1 Closed book: 1 page of notes ( 8 2 × 11 inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1­hour exam; two hours to complete. Prior term midterm exams have been posted on the 6.003 website. 2

Multiple Representations of CT and DT Systems Verbal descriptions: preserve the rationale. Difference/differential equations: mathematically compact. y [ n ] = x [ n ] + z 0 y [ n − 1] ˙( t ) = x ( t ) + s 0 y ( t ) y Block diagrams: illustrate signal flow paths. + + X Y X A Y z 0 s 0 R Operator representations: analyze systems as polynomials. Y 1 Y A = = X 1 − z 0 R X 1 − s 0 A Transforms: representing diff. equations with algebraic equations. z 1 H ( z ) = H ( s ) = z − z 0 s − s 0 3

Convolution Representing a system by a single signal. 4

Responses to arbitrary signals Although we have focused on responses to simple signals ( δ [ n ] , δ ( t ) ) we are generally interested in responses to more complicated signals. How do we compute responses to a more complicated input signals? No problem for difference equations / block diagrams. → use step­by­step analysis. 5

Check Yourself Example: Find y [3] + + X Y R R when the input is x [ n ] 1 n 1. 1 2. 2 3. 3 4. 4 5. 5 0. none of the above 6

Responses to arbitrary signals Example. + + 0 0 R R 0 0 x [ n ] y [ n ] n n 7

Responses to arbitrary signals Example. + + 1 1 R R 0 0 x [ n ] y [ n ] n n 8

Responses to arbitrary signals Example. + + 1 2 R R 1 0 x [ n ] y [ n ] n n 9

Responses to arbitrary signals Example. + + 1 3 R R 1 1 x [ n ] y [ n ] n n 10

Responses to arbitrary signals Example. + + 0 2 R R 1 1 x [ n ] y [ n ] n n 11

Responses to arbitrary signals Example. + + 0 1 R R 0 1 x [ n ] y [ n ] n n 12

Responses to arbitrary signals Example. + + 0 0 R R 0 0 x [ n ] y [ n ] n n 13

Check Yourself What is y [3] ? 2 + + 0 0 R R 0 0 x [ n ] y [ n ] 1 n n 1. 1 2. 2 3. 3 4. 4 5. 5 0. none of the above 14

Superposition Break input into additive parts and sum the responses to the parts. x [ n ] n y [ n ] = n n + + n n + + n n − 1 0 1 2 3 4 5 = n − 1 0 1 2 3 4 5 15

Linearity A system is linear if its response to a weighted sum of inputs is equal to the weighted sum of its responses to each of the inputs. Given x 1 [ n ] system y 1 [ n ] and x 2 [ n ] system y 2 [ n ] the system is linear if αx 1 [ n ] + βx 2 [ n ] system αy 1 [ n ] + βy 2 [ n ] is true for all α and β . 16

Superposition Break input into additive parts and sum the responses to the parts. x [ n ] n y [ n ] = n n + + n n + + n n − 1 0 1 2 3 4 5 = n − 1 0 1 2 3 4 5 Superposition works if the system is linear . 17

Superposition Break input into additive parts and sum the responses to the parts. x [ n ] n y [ n ] = n n + + n n + + n n − 1 0 1 2 3 4 5 = n − 1 0 1 2 3 4 5 Reponses to parts are easy to compute if system is time-invariant . 18

Time-Invariance A system is time­invariant if delaying the input to the system simply delays the output by the same amount of time. Given x [ n ] system y [ n ] the system is time invariant if x [ n − n 0 ] system y [ n − n 0 ] is true for all n 0 . 19

Superposition Break input into additive parts and sum the responses to the parts. x [ n ] n y [ n ] = n n + + n n + + n n − 1 0 1 2 3 4 5 = n − 1 0 1 2 3 4 5 Superposition is easy if the system is linear and time-invariant . 20

Structure of Superposition If a system is linear and time­invariant (LTI) then its output is the sum of weighted and shifted unit­sample responses. δ [ n ] system h [ n ] δ [ n − k ] system h [ n − k ] x [ k ] δ [ n − k ] system x [ k ] h [ n − k ] ∞ ∞ � � x [ n ] = x [ k ] δ [ n − k ] y [ n ] = x [ k ] h [ n − k ] system k = −∞ k = −∞ 21

Convolution Response of an LTI system to an arbitrary input. x [ n ] y [ n ] LTI ∞ � y [ n ] = x [ k ] h [ n − k ] ≡ ( x ∗ h )[ n ] k = −∞ This operation is called convolution . 22

Notation Convolution is represented with an asterisk. � ∞ x [ k ] h [ n − k ] ≡ ( x ∗ h )[ n ] k = −∞ It is customary (but confusing) to abbreviate this notation: ( x ∗ h )[ n ] = x [ n ] ∗ h [ n ] 23

Notation Do not be fooled by the confusing notation. Confusing (but conventional) notation: � ∞ x [ k ] h [ n − k ] = x [ n ] ∗ h [ n ] k = −∞ x [ n ] ∗ h [ n ] looks like an operation of samples; but it is not! x [1] ∗ h [1] = ( x ∗ h )[1] Convolution operates on signals not samples. Unambiguous notation: � ∞ x [ k ] h [ n − k ] ≡ ( x ∗ h )[ n ] k = −∞ The symbols x and h represent DT signals. Convolving x with h generates a new DT signal x ∗ h . 24

Structure of Convolution � ∞ y [ n ] = x [ k ] h [ n − k ] k = −∞ x [ n ] h [ n ] ∗ n n − 2 − 1 0 1 2 3 4 5 − 2 − 1 0 1 2 3 4 5 25

Structure of Convolution � ∞ y [0] = x [ k ] h [0 − k ] k = −∞ x [ n ] h [ n ] ∗ n n − 2 − 1 0 1 2 3 4 5 − 2 − 1 0 1 2 3 4 5 26

Structure of Convolution � ∞ y [0] = x [ k ] h [0 − k ] k = −∞ x [ k ] h [ k ] ∗ k k − 2 − 1 0 1 2 3 4 5 − 2 − 1 0 1 2 3 4 5 27

Structure of Convolution � ∞ y [0] = x [ k ] h [0 − k ] k = −∞ x [ k ] h [ k ] flip ∗ k k − 2 − 1 0 1 2 3 4 5 − 2 − 1 0 1 2 3 4 5 h [ − k ] k − 2 − 1 0 1 2 3 4 5 28

Structure of Convolution � ∞ y [0] = x [ k ] h [0 − k ] k = −∞ x [ k ] h [ k ] shift ∗ k k − 2 − 1 0 1 2 3 4 5 − 2 − 1 0 1 2 3 4 5 h [0 − k ] k − 2 − 1 0 1 2 3 4 5 29

Structure of Convolution � ∞ y [0] = x [ k ] h [0 − k ] k = −∞ x [ k ] h [ k ] multiply ∗ k k − 2 − 1 0 1 2 3 4 5 − 2 − 1 0 1 2 3 4 5 h [0 − k ] h [0 − k ] k k − 2 − 1 0 1 2 3 4 5 − 2 − 1 0 1 2 3 4 5 30

Structure of Convolution � ∞ y [0] = x [ k ] h [0 − k ] k = −∞ x [ k ] h [ k ] multiply ∗ k k − 2 − 1 0 1 2 3 4 5 h [0 − k ] h [0 − k ] k k − 2 − 1 0 1 2 3 4 5 x [ k ] h [0 − k ] k − 2 − 1 0 1 2 3 4 5 31

Structure of Convolution � ∞ y [0] = x [ k ] h [0 − k ] k = −∞ x [ k ] h [ k ] sum ∗ k k h [0 − k ] h [0 − k ] k k − 2 − 1 0 1 2 3 4 5 x [ k ] h [0 − k ] ∞ � k − 2 − 1 0 1 2 3 4 5 k = −∞ 32

Structure of Convolution � ∞ y [0] = x [ k ] h [0 − k ] k = −∞ x [ k ] h [ k ] ∗ k k h [0 − k ] h [0 − k ] k k − 2 − 1 0 1 2 3 4 5 x [ k ] h [0 − k ] ∞ � = 1 k − 2 − 1 0 1 2 3 4 5 k = −∞ 33

Structure of Convolution � ∞ y [1] = x [ k ] h [1 − k ] k = −∞ x [ k ] h [ k ] ∗ k k h [1 − k ] h [1 − k ] k k − 2 − 1 0 1 2 3 4 5 x [ k ] h [1 − k ] ∞ � = 2 k − 2 − 1 0 1 2 3 4 5 k = −∞ 34

Structure of Convolution � ∞ y [2] = x [ k ] h [2 − k ] k = −∞ x [ k ] h [ k ] ∗ k k h [2 − k ] h [2 − k ] k k − 2 − 1 0 1 2 3 4 5 x [ k ] h [2 − k ] ∞ � = 3 k − 2 − 1 0 1 2 3 4 5 k = −∞ 35

Structure of Convolution � ∞ y [3] = x [ k ] h [3 − k ] k = −∞ x [ k ] h [ k ] ∗ k k h [3 − k ] h [3 − k ] k k − 2 − 1 0 1 2 3 4 5 x [ k ] h [3 − k ] ∞ � = 2 k − 2 − 1 0 1 2 3 4 5 k = −∞ 36

Structure of Convolution � ∞ y [4] = x [ k ] h [4 − k ] k = −∞ x [ k ] h [ k ] ∗ k k h [4 − k ] h [4 − k ] k k − 2 − 1 0 1 2 3 4 5 x [ k ] h [4 − k ] ∞ � = 1 k − 2 − 1 0 1 2 3 4 5 k = −∞ 37

Structure of Convolution � ∞ y [5] = x [ k ] h [5 − k ] k = −∞ x [ k ] h [ k ] ∗ k k h [5 − k ] h [5 − k ] k k − 2 − 1 0 1 2 3 4 5 x [ k ] h [5 − k ] ∞ � = 0 k − 2 − 1 0 1 2 3 4 5 k = −∞ 38

Check Yourself 1 1 ∗ Which plot shows the result of the convolution above? 1 1 1. 2. 1 1 3. 4. 5. none of the above 39

Check Yourself 1 1 ∗ Express mathematically: � � � n � n � k � n − k � �� 2 �� 2 ∞ �� 2 �� 2 � � = u [ k ] × u [ n − k ] u [ n ] ∗ u [ n ] 3 3 3 3 k = −∞ n � � k 2 n − k � 2 � � = × 3 3 k =0 � n � � n � n n 2 � � 2 = = 1 3 3 k =0 k =0 � � n 2 = ( n + 1) u [ n ] 3 4 4 32 80 = 1 , 3 , 3 , 27 , 81 , . . . 40

Check Yourself 1 1 ∗ Which plot shows the result of the convolution above? 3 1 1 1. 2. 1 1 3. 4. 5. none of the above 41

DT Convolution: Summary Representing an LTI system by a single signal. x [ n ] h [ n ] y [ n ] Unit­sample response h [ n ] is a complete description of an LTI system. Given h [ n ] one can compute the response y [ n ] to any arbitrary input signal x [ n ] : � ∞ y [ n ] = ( x ∗ h )[ n ] ≡ x [ k ] h [ n − k ] k = −∞ 42