sol08 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.004 Dynamics and Control II Fall 2007 Problem Set #8 Solution Posted: Problems 1–3: Friday, Nov. 9, ’07 In problems 1–3, you will explore the characteristics of the root locus. The root lo- cus is the trajectory of the closed–loop pole as a gain K increases. The closed–loop transfer function is KG ( s ) / (1 + ( s ) H ( s )), and the closed–loop poles are the roots of 1 + ( s ) H ( s )=0( i.e. , the roots of the denominator of the closed–loop trans- fer function). In this problem set, we deal with unity–feedback only, which implies H ( s ) = 1. Thus every pole on the root locus should satisfy 1 + ( s ) = 0. Because s is a complex number, ( s )= 1 leads to two requirements: K =1 / | G ( s ) | , ± ( s )=( 2 n +1) × 180 (n is an arbitrary integer) . You can use these relations either geometrically or algebraically. 1. For the complex number s 1 = 1+ j , a. The phase of the complex number ( s 1 + 2)( s 1 + 0). Answer: From the above figure, ± ( s 1 + 2)( s 1 +0) = ± ( s 1 +2)+ ± ( s 1 π 3 π = + = π. 4 4 Algebraically, ( s 1 + 2)( s 1 +0)=( j + 2)( j ) = (1 + j )( j 2 , 1
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and ± ( 2) = π. b. The value of the real number K such that K | s 1 +2 || s 1 +0 | =1 . Answer: From the above figure, K · 2 · 2=1 K / 2 . Algebraically, | s 1 | = | 1+ j | = 1 2 +1 2 = 2 , | s 1 | = |− j | = 1 2 2 = 2 . Hence, K · 2 · K / 2 . c. Does s 1 belong to the root locus? Answer: the problem statement, the open–loop transfer function is given by 1 G ( s )= . s ( s +2) result (a) we determine that ± G ( s 1 π . Therefore, s 1 is on the root locus. To find the value of gain that would drive the closed–loop pole to location s 1 on the complex plane, we must satisfy 1 1 K K √ √ K =2 . ( s 1 +0)( s 1 2 2 Note that result (b) is not directly applicable! d. Matlab result Answer: Root Locus −3 −2 −1 0 1 Real Axis 2. The open loop transfer function G ( s )=1 / { s ( s + 1)( s } . 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 Imaginary Axis
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± ± ± ± ± ² ³² ³ a. Show that ± j 2 belongs to the root locus. Answer: For s = ± j 2 to be on the root locus, it must satisfy ± { s ( s + 1)( s +2) } ± ± = s = ² j 2 π . ± { s ( s + 1)( s } ± ± = s = ² j 2 ± ± ± π ± s ± + ± ( s +1) ± + ± ( s ± = + θ 1 + θ 2 = π, s + ² j 2 s = ² j 2 s = ² j 2 2 π θ 1 + θ 2 = , 2 where θ 1 = tan 1 ( 2) and θ 2 = tan 1 ( 2 / 2). Hence, if θ 1 + θ 2 = π/ 2so that cot( θ 1 + θ 2 ) = 0, then s = ± j 2 is on the root locus. 1 cos( θ 1 + θ 2 ) cos θ 1 cos θ 2 sin θ 1 sin θ 2 cot( θ 1 + θ 2 )= = = =0 , tan( θ 1 + θ 2 ) sin( θ 1 + θ 2 ) sin θ 1 cos θ 2 + cos θ 1 sin θ 2 which requires that cos θ 1 cos θ 2 = sin θ 1 sin θ 2 . From the geometric relation, ´ µ´ µ 1 2 2 2 cos θ 1 cos θ 2 = = = sin θ 1 sin θ 2 , 3 6 3 6 which is true in this case. Therefore s = ± j 2 belongs to the root locus.
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sol08 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of...

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