Chapter2

Chapter2 - CHAPTER 2 2.1 DYNAMIC MODELS AND DYNAMIC...

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CHAPTER 2 DYNAMIC MODELS AND DYNAMIC RESPONSE 2.1 Revisiting Examples 2.4 and 2.5 will be helpful. (a) R ( s ) =A ; Ys KA s () = + t 1 y ( t ) = KA e t t t - / ; y ss = 0 (b) R ( s ) = A s ; Y ( s ) = KA ss τ + 1 y ( t )= KA / 1 - - e t ; y ss = KA (c) R ( s ) = A s 2 ; Y ( s ) = KA 2 1 + y ( t KA / te t -+ - ττ ; y ss = KA t - t (d) R ( s A s w w 22 + ; Y(s) = KA ϖ τϖ ( ) ++ 1 y ( t KA e KA t t ϖτ ϖθ 1 1 + + + + - / sin( ) θ = tan ( ) - - 1 yt t fi¥ = KA t 1 + + sin( ) 2.2 (a) The following result follows from Eqns (2.57 - 2.58) y ( t ) = 1 1 2 1 2 1 2 - - + - F H G I K J - - e t n t d ζϖ ζ sin tan d = ϖζ n 1 2 - The final value theorem is applicable: the function sY ( s ) does not have poles on the j w -axis and right half of s -plane. y = lim ( ) s sY s = 0 1 (b) The following result follows from Review Examples 2.2.
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SOLUTION MANUAL 3 y ( t )= t e t n t d d n -+ + - F H G I K J - - 2 2 1 1 2 ζ ϖϖ ϖ ζϖ sin tan y ss = t n - 2 The final value theorem is not applicable; the function sY ( s ) has a pole on the j w -axis. 2.3 Revisiting Review Example 2.1 will be helpful. Es R RCs i 0 4 10 1 () / = + ; s i = 10 2 et R te R C t 0 4 10 10 ( ) ; / = - tt t t For the output to track the input with a steady-state delay 100 × 10 –6 sec, it is necessary that R 10 1 4 = , and t ==· - RC 100 10 6 This gives R = 10 k W ; C = 0.1 m F Steady-state error = 10 t - (10 t – 10 t ) = 0.001 2.4 Revisiting Review Example 2.2 will be helpful. wz z w nn == = 100 3 2 6 100 ;; / / sec = 60 msec Steady-state error = 25 25 25 2 n -- · F H G I K J z w = 1.5 2.5 My t By t Ky t && & ++ = F ( t ) = 1000 m ( t ) Y ( s 1 10 100 2 s n = 10, = 0.5, w d = 53 Using Eqns (2.57)-(2.58), we obtain
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4 CONTROL SYSTEMS: PRINCIPLES AND DESIGN y ( t ) = 0.01 1 2 3 53 3 51 -+ L N M O Q P -- et t sin( tan 2.6 Ys Rs Gs ss () == ++ = 6 76 6 16 2 |( ) | Gj ϖ = 2 = 3 52 ; G ( j 2) = – 81.87º y = 3 28 1 8 7 sin( .
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Chapter2 - CHAPTER 2 2.1 DYNAMIC MODELS AND DYNAMIC...

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