Lesson_13

# Lesson_13 - Lesson 13 Lesson 13 Challenge 12 Lesson 13...

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Unformatted text preview: Lesson 13 Lesson 13 Challenge 12 Lesson 13: Frequency response function Challenge 13: Lesson 13 Challenge 12 A linear time-invariant system is assembled by cascading two MA filters. The 1 st stage is a 5 th order MA filter. The 2 nd stage is a 7 th order MA filter. What is the overall impulse response of the cascade system. Let h 1 =(1/5)[1,1,1,1,1] Let h 2 =(1/7)[1,1,1,1,1,1,1] h=h 1 *h 2 = (1/35)[1 2 3 4 5 5 5 4 3 2 1] The length of h is (5+7-1)=11 samples, with a length 4 sample transition build-up, length 4 sample transition build-down, and length 3 steady-state interval. Lesson 13 Challenge 12 |H 1 (e j ω )| |H 2 (e j ω )| Lesson 13 Challenge 12 |H 1 (e j ω )| |H 2 (e j ω )| Combined (Cascaded) Notice location of the nulls. Lesson 13 Lesson 13 Consider an LTI system having a known impulse response. The output is defined by the convolution sum: Suppose the input is with a signal of magnitude A, normalize frequency ω , and phase offset φ . [ ] [ ] k n x h n y N k k- = ∑ = [ ] ( 29 π ω π ϖ ω φ ω < ≤- = = + : / ; s n j T Ae n x Lesson 13 Lesson 13 Convolution: (N+1)-order FIR and x[k] Define: frequency response function or simply frequency response . for inputs x[k] (complex exponentials) [ ] ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 n j j n j j k j N k k k j n j j N k k k n j N k k e Ae e Ae e h e e Ae h Ae h n y ω φ ω φ ω ω ω φ φ ω ω ℑ = = = =- =- = +- = ∑ ∑ ∑ ( 29 ( 29 ( 29 C ∈ = = ℑ- = ∑ k j N k k j e h e H ω ω ω Lesson 13 Lesson 13 Outcome: ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 C C C ∈ + =...
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## This note was uploaded on 08/21/2010 for the course EEL 3135 taught by Professor ? during the Spring '08 term at University of Florida.

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Lesson_13 - Lesson 13 Lesson 13 Challenge 12 Lesson 13...

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