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Lec12-handout-6up - 6.003 Signals and Systems Lecture 12 1 6.003 Signals and Systems CT Frequency Response and Bode Plots Mid-term Examination#2

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Unformatted text preview: 6.003: Signals and Systems Lecture 12 October 22, 2009 1 6.003: Signals and Systems CT Frequency Response and Bode Plots October 22, 2009 Mid-term Examination #2 Wednesday, October 28, 7:30-9:30pm, Walker Memorial. No recitations on the day of the exam. Coverage: cumulative with more emphasis on recent material lectures 1–12 homeworks 1–7 Homework 7 includes practice problems for mid-term 2. It will not collected or graded. Solutions will be posted. Closed book: 2 pages of notes ( 8 1 2 × 11 inches; front and back). Designed as 1-hour exam; two hours to complete. Review sessions Monday 5-6pm and 8-9pm in 32-044. Conflict? Contact [email protected] by Friday, October 23, 5pm. Last Time Complex exponentials are eigenfunctions of LTI systems. H ( s ) e s t H ( s ) e s t H ( s ) can be determined graphically using vectorial analysis. H ( s ) = K ( s − z )( s − z 1 )( s − z 2 ) ··· ( s − p )( s − p 1 )( s − p 2 ) ··· z z s − z s s-plane s Response of an LTI system to an eternal cosine is an eternal cosine: same frequency, but scaled and shifted. H ( s ) cos( ω t ) | H ( jω ) | cos ( ω t + ⚢ H ( jω ) ) Frequency Response: H ( s ) | s ← jω s-plane σ ω 5 − 5 5 − 5 H ( s ) = s − z 1 − 5 5 5 | H ( jω ) | − 5 5 π/ 2 − π/ 2 ⚢ H ( jω ) Frequency Response: H ( s ) | s ← jω s-plane σ ω 5 − 5 5 − 5 H ( s ) = 9 s − p 1 − 5 5 5 | H ( jω ) | − 5 5 π/ 2 − π/ 2 ⚢ H ( jω ) Frequency Response: H ( s ) | s ← jω s-plane σ ω 5 − 5 5 − 5 H ( s ) = 3 s − z 1 s − p 1 − 5 5 5 | H ( jω ) | − 5 5 π/ 2 − π/ 2 ⚢ H ( jω ) 6.003: Signals and Systems Lecture 12 October 22, 2009 2 Poles and Zeros Thinking about systems as collections of poles and zeros is an im- portant design concept. • simple: just a few numbers characterize entire system • powerful: complete information about frequency response Today: poles, zeros, frequency responses, and Bode plots. Asymptotic Behavior: Isolated Zero The magnitude response is simple at low and high frequencies. σ ω 5 − 5 5 − 5 H ( jω ) = jω − z 1 − 5 5 5 | H ( jω ) | z 1 ω − 5 5 π/ 2 − π/ 2 ⚢ H ( jω ) Asymptotic Behavior: Isolated Zero Two asymptotes provide a good approxmation on log-log axes....
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This note was uploaded on 08/24/2011 for the course EECS 6.003 taught by Professor Dennism.freeman during the Spring '11 term at MIT.

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Lec12-handout-6up - 6.003 Signals and Systems Lecture 12 1 6.003 Signals and Systems CT Frequency Response and Bode Plots Mid-term Examination#2

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