EECE 340 Homework 5 Solution, Spring 2016

AMERICAN UNIVERSITY OF BEIRUT ELECTRICAL AND COMPUTER ENGINEERING DEPARTMENT EECE 340 Homework V – Routh Hurwitz Stability Criterion Solution Problem 1 Consider the unit feedback system shown below a. Determine the error signal E(s). ) K( s ) )(s )(s s ( )X( s ) )(s )(s ( s K ( s ) G X( s ) E( s ) 7 6 4 2 6 4 2 1            b. Determine the range of K for the system to be stable K K s) ( s s ) ( s K ) K( s ) )( s )( s s ( ) ( s K ( s ) X ( s ) Y 7 48 44 12 2 3 7 7 6 4 2 7               3 s 1 44+K 2 s 12 48+7K 1 s 12 5 528  K 0 s 48+7K System is stable for all K>0. K - X(t) Y(t) Problem 2 The open loop transfer function of a unity feedback control is given by: 2 3 4 3 2 3 3 1 ( ) ( 3 3 1) s s G s s s s s s        Check for system stability by determining the number of poles in each half of the complex plane. The closed-loop transfer function is given by 2 7 6 5 4 3 2 ( ) 3 3 1 ( ) 3 3 3 3 1 Y s s s X s s s s s s s s           Constructing the RH table of this system, we obtain 1 s 4 s 1 s 4s )s( A 0 4 s 4 s A )s( 0 0 s 1 0 1 s 1 0 1 s 1 3 1 3 s 3 1 3 1 s 0 2 3 3 4 3 4 5 6 7        According to RH table, 2 sign changes which means two roots in the rhp. Moreover, roots of A(s) are roots of the original polynomial. As no roots of A(s) are on the jw axis, therefore the system has no poles on the jw axis. The remaining roots (5) are in the lhp. (5,0,2) Problem 3 The open-loop transfer function of a unity-feedback control system is given by: 4) 2)(S S 1)(S ( K G )s(     Determine the range of K for which this system is stable. The closed-loop transfer function of the system is given by: K 8 14s 7s s K )s( R )s( C 2 3      The RH table 3 s 1 14 2 s 7 8+K 1s 90-K 0 s 8 +K For stability: -8<K<90 Problem 4 The transfer function of an LTI system is given by 𝑌(𝑠) 𝑋(𝑠) = 𝑠 + 1 𝑠6 + 2𝑠5 + 3𝑠4 + 6𝑠3 + 2𝑠2 + 4𝑠 + 1 Where x(t) is the input and y(t) is the output. Determine the stability of the system by determining the number of poles of the system in each half of the complex plane. Problem 5 The transfer function of a linear time-invariant system, whose input is x(t) and output y(t), is given by 10 5s 3s s s 2 10 s 3 )s( X )s( Y 2 3 4       Check the stability of the system by determining the number of poles of the system in each half of the complex-plane. Problem 6 The transfer function of a linear time-invariant system, whose input is x(t) and output y(t), is given by 1 s s s 10 s 3 )s( X )s( Y 3 4      Check the stability of the system by determining the number of poles of the system in each half of the complex-plane. Problem 7 The forward transfer function of a unity feedback control system is given by: 4) 2s 3s s s(s )1 5 s( )s( G 2 3 4       Check the stability of the system by determining the number of poles in each half of the complex plane. Solution System transfer function 5 s 2s 3s s s )1 s( 5 G )s( 1 )s( G )s( X )s( Y 2 3 4 5          Using RH stability criterion S5 1 3 1 S4 1 2 5 S3 1 -4 S2 6 5 S1 -29/6 S0 5 System is not stable with 2 poles in the RHP; (3, 0, 2) Problem 8 The closed-loop transfer function of a unity feedback control system, whose input is Y(s) and output is R(s), is given by 56 8s 42s 6s 7s s 10 )s( R )s( Y 2 3 4 5       Check the stability of the above system by determining the number of poles in each half of the complex plane. S5 1 6 8 S4 1 6 8 S3 0 0 A(s)=s4+6s2+8 S3 4 12 A/(s)=4s3+12s S2 3 8 S1 4 S0 8 2j 2 ,j and s s 0 A )s(       Root Distributions: (1, 4, 0) Problem 9 The block Diagram of a unity feedback control system is shown below Find the number of poles in the left-half-plane, the right-half-plane, and on the jω- axis. Solution 1 2s 3s 2s 3s s 2 1 )s( R C )s( TF 2 3 4 5        Using RH table, we obtain From the above table, we conclude (3, 0, 2)