6.003: Signals and Systems Discrete Approximation of Continuous-Time - PowerPoint PPT Presentation

6.003: Signals and Systems Discrete Approximation of Continuous-Time Systems September 29, 2011 1

Mid-term Examination #1 Wednesday, October 5, 7:30-9:30pm, No recitations on the day of the exam. Coverage: CT and DT Systems, Z and Laplace Transforms Lectures 1–7 Recitations 1–7 Homeworks 1–4 Homework 4 will not collected or graded. Solutions will be posted. Closed book: 1 page of notes ( 8 1 2 × 11 inches; front and back). No calculators, computers, cell phones, music players, or other aids. Designed as 1-hour exam; two hours to complete. Review sessions during open office hours. Conflict? Contact before Friday, Sept. 30, 5pm. Prior term midterm exams have been posted on the 6.003 website. 2

Concept Map Today we will look at relations between CT and DT representations. Delay → R Block Diagram System Functional + + X Y 1 Y X = H ( R ) = 1 − R − R 2 Delay Delay DT DT Unit-Sample Response index shift h [ n ]: 1 , 1 , 2 , 3 , 5 , 8 , 13 , 21 , 34 , 55 , . . . Difference Equation System Function CT CT z 2 H ( z ) = Y ( z ) y [ n ] = x [ n ] + y [ n − 1] + y [ n − 2] X ( z ) = z 2 − z − 1 Delay → R Block Diagram System Functional DT DT + � + � X Y 2 A 2 Y − − X = 2 + 3 A + A 2 1 1 2 Impulse Response x ( t ) ˙ � x ( t ) h ( t ) = 2( e − t/ 2 − e − t ) u ( t ) CT CT Differential Equation System Function Y ( s ) 2 2¨ y ( t ) + 3 ˙ y ( t ) + y ( t ) = 2 x ( t ) X ( s ) = 2 s 2 + 3 s + 1 3

Discrete Approximation of CT Systems Example: leaky tank r 0 ( t ) h 1 ( t ) r 1 ( t ) � A X X Block Diagram System Functional Y A 1 + � X Y X = τ − A + τ Impulse Response � x ( t ) ˙ x ( t ) h ( t ) = 1 τ e − t/τ u ( t ) Differential Equation System Function H ( s ) = Y ( s ) 1 τ ˙ r 1 ( t ) = r 0 ( t ) − r 1 ( t ) X ( s ) = 1 + τs Today: compare step responses of leaky tank and DT approximation. 4

Check Yourself (Practice for Exam) What is the “step response” of the leaky tank system? u ( t ) Leaky Tank s ( t ) =? 1 1 1. 2. t t τ τ 1 1 3. 4. t t τ τ 5. none of the above 5

Check Yourself What is the “step response” of the leaky tank system? de: ˙ 1 ( t ) = u ( t ) − r 1 ( t ) τ r t < 0 : r 1 ( t ) = 0 t > 0 : r 1 ( t ) = c 1 + c 2 e − t/τ ˙ 1 ( t ) = − c 2 e − t/τ r τ Substitute into de: τ − c 2 e − t/τ = 1 − c 1 − c 2 e − t/τ � � → c 1 = 1 τ Combine t < 0 and t > 0 : − t/τ u ( t ) r 1 ( t ) = u ( t ) + c 2 e ˙ 1 ( t ) = δ ( t ) + c 2 δ ( t ) − c 2 e − t/τ r u ( t ) τ Substitute into de: τ (1 + c 2 ) δ ( t ) − τ c 2 e − t/τ − t/τ u ( t ) = u ( t ) − u ( t ) − c 2 e u ( t ) c 2 = − 1 → τ − t/τ ) u ( t ) r 1 ( t ) = (1 − e 6

Check Yourself Alternatively, reason with systems! A h ( t ) = 1 τ e − t/τ u ( t ) δ ( t ) A + τ A u ( t ) s ( t ) =? A + τ u ( t ) A δ ( t ) s ( t ) =? A A + τ � t h ( t ) A h ( t ′ ) dt ′ δ ( t ) s ( t ) = A A + τ −∞ � t � t 1 1 − t I /τ u ( t ′ ) dt ′ = − t I /τ dt ′ = (1 − e − t/τ ) u ( t ) s ( t ) = e e τ τ −∞ 0 7

Check Yourself What is the “step response” of the leaky tank system? 2 u ( t ) Leaky Tank s ( t ) =? 1 1 1. 2. t t τ τ 1 1 3. 4. t t τ τ 5. none of the above 8

Forward Euler Approximation Approximate leaky-tank system using forward Euler approach. Approximate continuous signals by discrete signals: x d [ n ] = x c ( nT ) y d [ n ] = y c ( nT ) Approximate derivative at t = nT by looking forward in time: y c ( nT ) = y d [ n +1] − y d [ n ] ˙ T y c ( t ) y d [ n +1] y d [ n ] t nT ( n +1) T 9

Forward Euler Approximation Approximate leaky-tank system using forward Euler approach. Substitute x d [ n ] = x c ( nT ) y d [ n ] = y c ( nT ) � � � � � � � � y c ( n + 1) T − y c nT y d [ n + 1] − y d [ n ] y ˙ c ( nT ) ≈ = T T into the differential equation τy ˙ c ( t ) = x c ( t ) − y c ( t ) to obtain τ y d [ n + 1] − y d [ n ] � � = x d [ n ] − y d [ n ] . T Solve: � � T y d [ n ] = T x d [ n ] y d [ n + 1] − 1 − τ τ 10

Forward Euler Approximation Plot. 1 T τ = 0 . 1 t 1 T τ = 0 . 3 t 1 T τ = 1 t 1 T τ = 1 . 5 t 1 T τ = 2 t τ T Why is this approximation badly behaved for large τ ? 11

Check Yourself DT approximation: y d [ n + 1] − 1 − T � � y d [ n ] = T � � τ x d [ n ] τ Find the DT pole. 1. z = T 2. z = 1 − T τ τ 3. z = τ 4. z = − τ T T 1 5. z = 1 + T τ 12

Check Yourself DT approximation: � � T � � y d [ n ] = T x d [ n ] y d [ n + 1] − 1 − τ τ Take the Z transform: � � � � T Y d ( z ) = T X d ( z ) zY d ( z ) − 1 − τ τ Solve for the system function: T H ( z ) = Y d ( z ) = τ � � � � X d ( z ) z − 1 − T τ Pole at z = 1 − T . . τ 13

Check Yourself DT approximation: y d [ n + 1] − 1 − T � � � � y d [ n ] = T τ x d [ n ] τ Find the DT pole. 2 1. z = T 2. z = 1 − T τ τ 3. z = τ 4. z = − τ T T 1 5. z = 1 + T τ 14

Dependence of DT pole on Stepsize z 1 T τ = 0 . 1 t z 1 T τ = 0 . 3 t z 1 T τ = 1 t z 1 T τ = 1 . 5 t z 1 T τ = 2 t τ The CT pole was fixed ( s = − 1 τ ). Why is the DT pole changing? 15

Dependence of DT pole on Stepsize Dependence of DT pole on T is generic property of forward Euler. Approach: make a systems model of forward Euler method. CT block diagrams: adders, gains, and integrators: X A Y ˙( t ) = x ( t ) y Forward Euler approximation: y [ n + 1] − y [ n ] = x [ n ] T Equivalent system: + X T R Y Forward Euler: substitute equivalent system for all integrators. 16

Example: leaky tank system Started with leaky tank system: 1 � + X Y τ − Replace integrator with forward Euler rule: 1 + + X T R Y τ − Write system functional: T R T T R R Y τ 1 −R τ τ = = = 1 + T T R � � � � X 1 − R + R 1 − 1 − T R τ τ 1 −R τ Equivalent to system we previously developed: � � � � T y d [ n ] = T x d [ n ] y d [ n + 1] − 1 − τ τ 17

Model of Forward Euler Method Replace every integrator in the CT system X A Y with the forward Euler model: + X T R Y Substitute the DT operator for A : T 1 T R T z A = → = = 1 − 1 s 1 − R z − 1 z z − 1 Forward Euler maps s → . T Or equivalently: z = 1 + sT . 18

Dependence of DT pole on Stepsize Pole at z = 1 − T τ = 1 + sT . z 1 T τ = 0 . 1 t z 1 T τ = 0 . 3 t z 1 T τ = 1 t z 1 T τ = 1 . 5 t z 1 T τ = 2 t τ 19

Forward Euler: Mapping CT poles to DT poles Forward Euler Map: z = 1 + sT s → 0 1 1 − 0 T − 2 − 1 T s z 1 T 1 − 1 z → 1 + sT − 1 − 2 T T DT stability: CT pole must be inside circle of radius 1 T at s = − 1 T . − 2 1 T T < − < 0 τ < 2 → τ 20

Backward Euler Approximation We can do a similar analysis of the backward Euler method. Approximate continuous signals by discrete signals: x d [ n ] = x c ( nT ) y d [ n ] = y c ( nT ) Approximate derivative at t = nT by looking backward in time: y c ( nT ) = y d [ n ] − y d [ n − 1] ˙ T y c ( t ) y d [ n ] y d [ n − 1] t nT ( n − 1) T 21

Backward Euler Approximation We can do a similar analysis of the backward Euler method. Substitute x d [ n ] = x c ( nT ) y d [ n ] = y c ( nT ) � � � � ˙ c ( nT ) ≈ y c nT − y c ( n − 1) T = y d [ n ] − y d [ n − 1] y T T into the differential equation τy ˙ c ( t ) = x c ( t ) − y c ( t ) to obtain τ y d [ n ] − y d [ n − 1] = x d [ n ] − y d [ n ] . � � � � T Solve: � � � � 1 + T y d [ n ] − y d [ n − 1] = T x d [ n ] τ τ 22

Backward Euler Approximation Plot. 1 T τ = 0 . 1 t 1 T τ = 0 . 3 t 1 T τ = 1 t 1 T τ = 1 . 5 t 1 T τ = 2 t τ This approximation is better behaved. Why? 23

Check Yourself DT approximation: � � 1 + T � � y d [ n ] − y d [ n − 1] = T τ x d [ n ] τ Find the DT pole. 1. z = T 2. z = 1 − T τ τ 3. z = τ 4. z = − τ T T 1 5. z = 1 + T τ 24

Check Yourself DT approximation: � � 1 + T � � y d [ n ] − y d [ n − 1] = T x d [ n ] τ τ Take the Z transform: � � � � 1 + T − 1 Y d ( z ) = T X d ( z ) Y d ( z ) − z τ τ Find the system function: T H ( z ) = Y d ( z ) = τ z � � � � X d ( z ) 1 + T z − 1 τ 1 Pole at z = . 1 + T τ 25

Check Yourself DT approximation: y d [ n + 1] − 1 − T � � � � y d [ n ] = T τ x d [ n ] τ Find the DT pole. 5 1. z = T 2. z = 1 − T τ τ 3. z = τ 4. z = − τ T T 1 5. z = 1 + T τ 26

Dependence of DT pole on Stepsize z 1 T τ = 0 . 1 t z 1 T τ = 0 . 3 t z 1 T τ = 1 t z 1 T τ = 1 . 5 t z 1 T τ = 2 t τ Why is this approximation better behaved? 27

Dependence of DT pole on Stepsize Make a systems model of backward Euler method. CT block diagrams: adders, gains, and integrators: X A Y y ˙( t ) = x ( t ) Backward Euler approximation: y [ n ] − y [ n − 1] = x [ n ] T Equivalent system: + X T Y R Backward Euler: substitute equivalent system for all integrators. 28

Model of Backward Euler Method Replace every integrator in the CT system X A Y with the backward Euler model: + X T Y R Substitute the DT operator for A : 1 T T A = s → 1 − R = 1 − 1 z 1 Backward Euler maps z → . 1 − sT 29