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MIT6_003S10_lec07

# MIT6_003S10_lec07 - 6.003 Signals and Systems Relations b...

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6.003: Signals and Systems Relations between CT and DT: Insights from Operators and Transforms February 25, 2010

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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.
˙ 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

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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
˙ 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

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˙ 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
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

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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.
Check Yourself What is the “step response” of the leaky tank system? 2 Leaky Tank u ( t ) s ( t ) =?

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• Fall '11
• Freeman
• Trigraph, Yd, Numerical differential equations, Runge–Kutta methods, Numerical ordinary differential equations

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