{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

MIT6_003S10_lec07

MIT6_003S10_lec07 - 6.003 Signals and Systems Relations b...

Info icon This preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
6.003: Signals and Systems Relations between CT and DT: Insights from Operators and Transforms February 25, 2010
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
ednesday, March 3, o recitations on the day of the exam. overage: Representations of CT and DT Systems Lectures 1–7 Recitations 1–8 Homeworks 1–4 omework 4 will not collected or graded. Solutions will be posted. losed book: 8 1 1 page of notes ( 2 11 inches; front × and back). esigned as 1-hour exam; two hours to complete. eview sessions during open office hours. Mid-term Examination #1 W N C H D R C 7:30-9:30pm.
Image of page 2
˙ x ( t ) x ( t ) X A X Concept Map: Continuous-Time Systems Relations among CT representations. Block Diagram System Functional + Y Y 2 A 2 = X 2 + 3 A + A 2 + 1 1 2 X Impulse Response h ( t ) = 2( e t/ 2 e t ) u ( t ) Differential Equation System Function y ( t ) + 3 ˙ y ( t ) + y ( t ) = 2 x ( t ) Y ( s ) 2 X ( s ) = 2 s 2 + 3 s + 1
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
index shift Delay → R Concept Map: Discrete-Time Systems Relations among DT representations. Block Diagram System Functional Y Y 1 = Delay X 1 − R − R 2 + Delay + X Unit-Sample Response h [ n ]: 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , . . . Difference Equation System Function z y [ n ] = x [ n ] + y [ n 1] + y [ n 2] H ( z ) = Y ( z ) = 2 X ( z ) 1 z z 2
Image of page 4
˙ x ( t ) x ( t ) X A X index shift Delay → R Concept Map Relations between CT and DT representations. Block Diagram System Functional Y Y 1 = X 1 − R − R 2 + Delay + Delay X System Function y [ n ] + y [ n 1] + y [ n 2] H ( z ) = Y ( z ) z 2 System Functional Impulse Response + + Y X = 2 A 2 2 + 3 A + A 2 e t/ 2 e t ) u ( t ) Difference Equation Unit-Sample Response n ] = x [ h [ n ]: 1 , 1 , 2 CT DT CT DT X ( z ) = 1 z z 2 Block Diagram X Y 1 1 2 h ( t ) = 2( , 3 , 5 , 8 , 13 , 21 , 34 , 55 , . . . Differential Equation System Function Y ( s ) 2 y ( t ) + 3 ˙ y ( t ) + y ( t ) = 2 x ( t ) X ( s ) = 2 s 2 + 3 s + 1
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
˙ x ( t ) x ( t ) X A X First-Order CT System Example: leaky tank. r 0 ( t ) r 1 ( t ) h 1 ( t ) Block Diagram System Functional Y A Y = X A + τ + 1 τ X Impulse Response h ( t ) = 1 τ e t/τ u ( t ) Differential Equation System Function τr ˙ 1 ( t ) = r 0 ( t ) r 1 ( t ) H ( s ) = Y ( s ) = 1 X ( s ) 1 + τs
Image of page 6
Check Yourself What is the “step response” of the leaky tank system? Leaky Tank u ( t ) s ( t ) =? t 1 τ 1. t 1 τ 2. t 1 τ 3. t 1 τ 4. 5. none of the above
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Check Yourself What is the “step response” of the leaky tank system? h ( t ) = τ 1 e t/τ u ( t ) H ( s ) δ ( t ) H ( s ) u ( t ) s ( t ) =? 1 s H ( s ) δ ( t ) u ( t ) s ( t ) =? H ( s ) 1 s δ ( t ) h ( t ) t s ( t ) = h ( t ) dt −∞ s ( t ) = t 1 e t u ( t ) dt = t 1 e t dt = (1 e t/τ ) u ( t ) −∞ τ 0 τ Reasoning with systems.
Image of page 8
Check Yourself What is the “step response” of the leaky tank system? 2 Leaky Tank u ( t ) s ( t ) =?
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
This is the end of the preview. Sign up to access the rest of the document.
  • Fall '11
  • Freeman
  • Trigraph, Yd, Numerical differential equations, Runge–Kutta methods, Numerical ordinary differential equations

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern