# sol08 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of...

This preview shows pages 1–4. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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
± ± ± ± ± 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.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

### Page1 / 12

sol08 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online