ps2sol

# ps2sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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Unformatted text preview: MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Aeronautics and Astronautics 16.060: Principles of Automatic Control Fall 2003 PROBLEM SET 2 Solutions Problem 1 P3.27(a): R 6 +-- 6 +- +- 6 G 1 G 2---- C- H 2 H 1 R- + 6-- +- G 2 1+ G 2 H 1 6-- G 1 C- H 1 H 2 R- 6 +- G 1 G 2 1+ G 2 H 1 + G 1 G 2 H 1 H 2-- C T ( s ) = G 1 G 2 ⇒ 1 + G 1 G 2 + G 2 H 1 + G 1 G 2 H 1 H 2 1-- P3.27(b): R + C-- G 1 G 2- 6 + ? Y H 1 X- 6 Y This one is a little trickier. Label the input to H 1 as X, and the output of H 1 as Y, as shown. Therefore, we have: X = C − Y Y = H 1 X Solve for X and Y to get: C X = 1 + H 1 H 1 Y = C 1 + H 1 So the block diagram can be rewritten as: R +-- G 1- G 2 C-- 6 H 1 1+ H 1 And now we can write down the closed-loop transfer function: G 1 G 2 (1 + H 1 ) T ( s ) = 1 + H 1 (1 + G 1 G 2 ) Problem 2 1. f ( t ) is the input force to a standard spring-mass-damper system, so the equation of motion is: m ¨ x + cx ˙ + kx 2 / 3 = f (1) 2. We need to linearize the non-linear spring force f k ( x ) about the point x = x . First, write down the Taylor Series expansion for f k ( x ) about x = x : 2 1 f k ( x − x ) = f k ( x ) + f k ( x )( x − x ) + f k ( x )( x − x ) 2 + . . . (2) 2! 2 / 3 2 = kx + kx − 1 / 3 ( x − x ) − 1 kx − 4 / 3 ( x − x ) 2 + . . . (3) 3 9 Now we drop all the higher order terms so we’re left with only two terms: a constant term and a linear term. 2 f k ( x − x ) ≈ kx 2 / 3 + kx − 1 / 3 ( x − x ) (4) 3 Next, we rewrite the original differential equation in terms of deviations from the operating point. Define: ˆ x = x − x f ˆ = f − f In this case, f represents the input force that will keep the system in equilibrium at the operating ¨ point x = x . To find this force, go back...
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ps2sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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