Homework 4 Solutions

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EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley. b Clearly n2 28 5 rad s and 2 n 1 Therefore n 2 37 0 211 . 4, Ts 8 01 sec, n, , Tp p 1 36 sec, n 1 2, 2, OS e 1 100 50 7 . Tr 1n 1 76 3 0 417 2 1 093 1 0 514 sec, 3 Transfer Function From Unit Step Response. For each of the unit step responses shown below find the transfer function of the system . Solution , K, a This is a first order system of the form G s Using the graph we can estimate the. s a, 1 K, time constant as T 0 0244 sec But a 40 984 and DC gain is 2 Thus 2 Hence . T a, K 81 967 , Thus , 81 967, G s , s 40 984, K, b This is a second order system of the form G s We can estimate the.
s2 2 n s n2, percent overshoot and the settling time from the graph . OS 13 82 11 03 , 11 03 100 25 3 , Ts 2 62sec,Rev 1 0 02 23 2014 2 of 9. EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley. We can now calculate and n from the given information . ln OS 100 , q 0 4, 2 2, ln OS 100 , 4, n 3 82, Ts. K, DC Gain 11 03 Therefore 2 11 03 Hence K 160 95 Substituting all values we get. n, 160 95, G s , s2 3 056s 14 59, K, c This is a second order system of the form G s We can estimate the.
s2 2 n s n2, percent overshoot and the peak time from the graph . OS 1 4 1 0 , 1 0 100 40 , Tp 4, We can now calculate and n from the given information . ln OS 100 , q 0 28, 2 2, ln OS 100 , , n p 0 818, Tp 1 2. K, DC Gain 1 0 Therefore 2 1 0 Hence K 0 669 Substituting all values we get. n, 0 669, G s , s2 0 458s 0 669, 4 State Space Analysis.
A system is represented by the state and output equations that follow Without solving the state. equation find the characteristic equation and the poles of the system . , 0 2 3 0, x 0 6 5 x 1 u t , , 1 4 2 1, h i, y 1 2 0 x. Solution , 1 0 0 0 2 3 s 2 3, sI A s 0 1 0 0 6 5 0 s 6 5 . , 0 0 1 1 4 2 1 4 s 2, Characteristic Equation det sI A s3 8s2 11s 8. Factoring yields poles 9 111 0 5338 and 1 6448,Rev 1 0 02 23 2014 3 of 9. EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley. 5 Block Diagram To Transfer Function, Reduce the system shown below to a single transfer function T s C s R s .
Solution Push G2 s to the left past the summing junction . Collapse the summing junctions and add the parallel transfer functions . Rev 1 0 02 23 2014 4 of 9, EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley. Push G1 s G2 s G3 s to the right past the summing junction . Collapse the summing junctions and add feedback paths . Applying the feedback formula , G3 s G1 s G2 s , T s . 1 H s G3 s G1 s G2 s G2 s G4 s , 6 Block Diagram To Transfer Function. For the system shown below find the poles of the closed loop transfer function T s C s R s . Rev 1 0 02 23 2014 5 of 9, EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley. Solution Push 2s to the left past the pickoff point and combine the parallel combination of. 2 and 1 s , Push 2s 1 s to the right past the summing junction and combine summing junctions .
Hence , 2 2s 1 s 5, T s where Heq 1 , 1 2 2s 1 Heq s 2s 1 2s. 4s2 2s, T s , 6s2 13s 5, 7 Finding Constants To Meet Design Specifications. For the system shown below find the values of K1 and K2 to yield a peak time of 1 5 seconds and. a settling time of 3 2 seconds for the closed loop system s step response . Rev 1 0 02 23 2014 6 of 9, EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley. Solution The closed loop transfer function is given by . 10K1, T s , s2 10K2 2 s 10K1, 4 p , We know n 1 25 And n 1 2 2 09 Therefore poles are at n . Ts Tp, p , n 1 2 1 25 j2 09 Hence n 1 252 2 092 10K1 and 10K2 2 2 1 25 .
Therefore K1 0 593 and K2 0 05 , 8 Signal Flow Graphs. Draw a signal flow graph for the following equation . , 0 1 0 0, x 0 0 1 x 0 r t , , 2 4 6 1, h i, y 1 1 0 x. Solution , x 1 x2, x 2 x3, x 3 2x1 4x2 6x3 r, y x1 x2. 9 Signal Flow Graphs, Using Mason s rule find the transfer function T s C s R s for the system represented by the. following figure ,Rev 1 0 02 23 2014 7 of 9, EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley.
Solution , Closed loop gains G2 G4 G6 G7 H3 G2 G5 G6 G7 H3 G3 G4 G6 G7 H3 G6 H1 G7 H2. Forward path gains T1 G1 G2 G4 G6 G7 T2 G1 G2 G5 G6 G7 T3 G1 G3 G4 G6 G7 T4 G1 G3 G5 G6 G7. Nontouching loops 2 at a time G6 H1 G7 H2, 1 H3 G6 G7 G2 G4 G2 G5 G3 G5 G6 H1 G7 H2 G6 H1 G7 H2 . 1 2 3 4 1, T1 1 T2 2 T3 3 T4 4, T s , , G1 G 2 G4 G 6 G7 G1 G 2 G5 G6 G7 G1 G3 G4 G6 G7 G1 G3 G5 G6 G 7. T s , 1 H3 G6 G7 G2 G4 G2 G5 G3 G4 G3 G5 G6 H1 G7 H2 G6 H1 G7 H2. 10 State Space Representation and Signal Flow Graphs. Represent the system shown below in state space form and draw its signal flow graph . s 3, G s , s2 2s 7, Solution Writing the state equations .
x 1 x2, x 2 7x1 2x2 r, y 3x1 x2, , 0 1 0, x x r, 7 2 1. h i, y 3 1 x,Rev 1 0 02 23 2014 8 of 9, EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley. 11 Canonical Forms, Represent the system given in Problem 10 in controller canonical form and observer canonical form . Solution , Controllable canonical form , 2 7 1, x x r. 1 0 0, h i, y 1 3 x, Observable canonical form , 2 1 1.
x x r, 7 0 3, h i, y 1 0 x,Rev 1 0 02 23 2014 9 of 9. EE C128 ME C134 Spring 2014 HW4 Solutions UC Berkeley Homework 4 Solutions Note Each part of each problem is worth 3 points and the homework is worth a total of 42 points

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