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6 003 Signals and Systems Convolution March 2 2010 Mid term Examination 1 Tomorrow Wednesday March 3 7 30 9 30pm 34 101 No recitations tomorrow Coverage Representations of CT and DT Systems Lectures 1 7 Recitations 1 8 Homeworks 1 4 Homework 4 will not collected or graded Solutions are posted Closed book 1 page of notes 8 12 11 inches front and back Designed as 1 hour exam two hours to complete Multiple Representations of CT and DT Systems Verbal descriptions preserve the rationale Difference differential equations mathematically compact y n x n z0 y n 1 y t x t s0 y t Block diagrams illustrate signal flow paths X Y z0 R X A Y s0 Operator representations analyze systems as polynomials Y 1 Y A X 1 z0 R X 1 s0 A Transforms representing diff equations with algebraic equations z 1 H z H s z z0 s s0 Convolution Representing a system by a single signal Responses to arbitrary signals Although we have focused on responses to simple signals n 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 step by step analysis Check Yourself Example Find y 3 X R Y R when the input is x n n 1 1 2 2 3 3 4 4 0 none of the above 5 5 Responses to arbitrary signals Example 0 R x n 0 R 0 0 y n n n Responses to arbitrary signals Example 1 R x n 0 R 1 0 y n n n Responses to arbitrary signals Example 1 R x n 1 R 2 0 y n n n Responses to arbitrary signals Example 1 R x n 1 R 3 1 y n n n Responses to arbitrary signals Example 0 R x n 1 R 2 1 y n n n Responses to arbitrary signals Example 0 R x n 0 R 1 1 y n n n Responses to arbitrary signals Example 0 R x n 0 R 0 0 y n n n Check Yourself What is y 3 2 0 R x n 0 R 0 0 y n n n Alternative 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 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 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 x1 n system y1 n x2 n system y2 n system y1 n y2 n and the system is linear if x1 n x2 n is true for all and 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 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 Time Invariance A system is time invariant if delaying the input to the system simply delays the output by the same amount of time Given x n system y n system y n n0 the system is time invariant if x n n0 is true for all n0 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 Structure of Superposition If a system is linear and time invariant LTI then its output is the sum of weighted and shifted unit sample responses x n X k n system h n n k system h n k x k n k system x k h n k system y n x k n k X k x k h n k Convolution Response of an LTI system to an arbitrary input x n y n X LTI y n x k h n k x h n k This operation is called convolution Notation Convolution is represented with an asterisk X 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 Notation Do not be fooled by the confusing notation Confusing but conventional notation X 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 6 x h 1 Convolution operates on signals not samples Unambiguous notation X 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 Structure of Convolution y n X x k h n k k x n h n n 2 1 0 1 2 3 4 5 n 2 1 0 1 2 3 4 5 Structure of Convolution y 0 X x k h 0 k k x n h n n 2 1 0 1 2 3 4 5 n 2 1 0 1 2 3 4 5 Structure of Convolution y 0 X x k h 0 k k x k h k 2 1 0 1 2 3 4 5 k 2 1 0 1 2 3 4 5 k Structure of Convolution y 0 X x k h 0 k k x k 2 1 0 1 2 3 4 5 flip k h k 2 1 0 1 2 3 4 5 h k 2 1 0 1 2 3 4 5 k k Structure of Convolution y 0 X x k h 0 k k x k 2 1 0 1 2 3 4 5 shift k h k 2 1 0 1 2 3 4 5 h 0 k 2 1 0 1 2 3 4 5 k k Structure of Convolution y 0 X x k h 0 k k x k 2 1 0 1 2 3 4 5 h 0 k 2 1 0 1 2 3 4 5 multiply k k h k 2 1 0 1 2 3 4 5 h 0 k 2 1 0 1 2 3 4 5 k k Structure of Convolution y 0 X x k h 0 k k x k …


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MIT 6 003 - Convolution

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Control

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QUIZ I

QUIZ I

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Modes

Modes

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