Ve216LectureNotesChapter2Part1

# Ve216LectureNotesChapter2Part1 - Ve216 Lecture Notes...

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Ve216 Lecture Notes Dianguang Ma Spring 2009

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2.1 Introduction In this chapter, we examine several methods for describing the relationship between the input and output signals of linear time-invariant (LTI) systems in time domain. Convolution sum/integral Linear constant-coefficient difference/differential equation Block diagram which represents the system as an interconnection of three elementary operations: scalar multiplication, addition, and time shift/integration.
2.2 The Convolution Sum An arbitrary signal can be thought of as a weighted superposition of shifted impulses. [ ] [ 1] [ 1] [0] [ ] [1] [ 1] [ ] [ ] [ ]: the entire signal [ ]: a specific value of the signal [ ] at time . k x n x n x n x n x k n k x n x k x n k δ δ δ δ =-∞ = + - + + + - + = - L L

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Figure 2.1 (p. 99) Graphical example illustrating the representation of a signal x [ n ] as a weighted sum of time-shifted impulses.
2.2 The Convolution Sum The output of an LTI system is the convolution sum of the input to the system and the impulse response of the system. [ ] [ ]* [ ] [ ] [ ] Derivation: [ ] [ ] (impulse response) [ ] [ ] (time invariance) [ ] [ ] [ ] [ ] (homogeneity) [ ] [ ] [ ] [ ] [ ] [ ] (superposition) k k k y n x n h n x k h n k n h n n k h n k x k n k x k h n k x n x k n k y n x k h n k δ δ δ δ =-∞ =-∞ =-∞ = = - - - - - = - = -

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2.2 The Convolution Sum Example 2.2 ? ] [ ] [ ] [ ] [ ] [ ] [ 4 3 ] [ = = = = n h n x n y n u n x n u n h n
2.2 The Convolution Sum Solution to Example 2.2 ] 4 3 1 [ 4 4 3 4 3 ] [ 4 3 ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ 4 3 ] [ 1 0 0 + = = - -∞ = - -∞ = - = = = - = - = = = = n n l l n k k n k k n k n k n u k u k n h k x n h n x n y n u n x n u n h

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2.3 Convolution Sum Evaluation Procedure Methods Convolution Table LTI Form Reflection and Shift Direct Form -∞ = - = = k k n h k x n h n x n y ] [ ] [ ] [ ] [ ] [
2.3 Convolution Sum Evaluation Procedure Example [ ] [ ] 2 [ 1] 3 [ 2] [ ] [ ] [ 4] [ ] [ ] [ ] ? h n n n n x n u n u n y n x n h n δ δ δ = + - + - = - - = =

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2.3 Convolution Sum Evaluation Procedure Method 1: Convolution table = + -∞ = = - = = n j i j i k j h i x k n h k x n h n x n y , ] [ ] [ ] [ ] [ ] [ ] [ ] [
2.3 Convolution Sum Evaluation Procedure Method 2: LTI form L L + - + + = - = = -∞ = ] 1 [ ] 1 [ ] [ ] 0 [ ] [ ] [ ] [ ] [ ] [ n h x n h x k n h k x n h n x n y k

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2.3 Convolution Sum Evaluation Procedure Method 3: Reflection and shift or flip and slide signal product The : ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ k n h k x k w k w k n h k x n h n x n y n k n k - = = - = = -∞ = -∞ =
2.3 Convolution Sum Evaluation Procedure Method 3: Reflection and shift or flip and slide Graph both x[k] and h[n-k] as a function of k

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