SEE344: Control Systems - Routh-Hurwitz Stability Criterion

Solution Code: 1AFBC

Question:Control Systems

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Control Systems Assignment

Assignment Task

Q1:A system with unity feedback is shown in figure (1), Figure (1)

Use the Routh-Hurwitz stability criterion, determine if closed-loop system with G (s)(3( 15 )1)(10 sG )( = ss +)8(100 2 + + + s + s s is stable or not. Q2:

QUESTION 3 A closed-loop control system is shown in figure (3), where )(sG is the transfer function of the system, G c( s ) is the transfer function of the controller, D s( ) is a disturbance, R s( ) and Y s( ) are the input and output, respectively. The open-loop system is unstable with a transfer function of sG )( = s - 2 2. D(s) R(s) + Controller +SystemY(s) Figure (3)

The objectives of the controller, G c( s ) , are to make the closed-loop system stable and at the same time to minimize the effect of the disturbance. Consider a proportional-integral (PI) controller with a transfer function s Y(s) ? )(sGc + ? )(sGc ? )(sG sGc )( = Ks + 1 . Answer the following questions: (a) Determine the value of K so that the closed-loop system is stable and has a critically damped response (b) Find the steady-state error for the case where sR )( = 1 s and sD 0)( = . (c) Find the steady-state output of the system when sR 0)( = and sD )( = 1 s . (d) Base on the above analysis, explain whether the objectives of the controller have been met. QUESTION 4 A control system as shown in figure (4) has a transfer function of sG )( = ss )1( Figure (4)

(a) Let KsGc = )( . By sketching the root-locus, show that the closed-loop system is always unstable. (b) Let )2()( + = sKsGc . Sketch the root-locus and show that the closed-loop system can be stabilized. Determine the range of K for the closed-loop system to remain stable.

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Solution:

Ans Q. No 1 The closed loop system is shown in figure1

here The denominator of closed loopm transfer function is Y(s)R(s)= G(s)1+G(s) = 100(s+8)s+3s*s+15s+1(s+1)1+ 100(s+8)s+3s*s+15s+1(s+1)=100(s+8)s+3s*s+15s+1s+1+ 100(s+8) (s+3)*(s*s+15*s+10)*(s+1)+ 100(s+8)=0 a0*S4 +a1*S3 + a2 * S2 +a3* S +a4 = 0 Here a0 = 1 , a1 = 19, a2=64, a3=149, and a4=803 S4 1 64 803 S3 19 149 S2 b1=56.16 b2=803 Here b1= (a1*a2 - a0*a3)/a1 = 56.16 b2=(a1*a4 - a0*a5)/a1 = 803 S1 c1= -122.67 c1=(b1*a3-a1*b2)/b1 = 149 - (19*803)/56.16= -122.67 Since there is sign change , system is unstable. Ans Q.No 2 The closed loop T.F. is having K and p unknown

Y(s)/R(s) = k/[ s*s*s + p*s*s +k] Charactristic equation of closed look T.F. is a0*S3 + a1 *S2 +a2* S +a3 =0 here a0 = 1 , a1 = p , a2=0, a3=k As per Rouths stability criteria S3 1 0 S2 p k S1 b1= -k/p 0 b1=(a1*a2 - a0*a3)/a1 = -1*k/p , b2=(a1*a4 - a0*a5)/a1 =0 Here for stability must be k/p < 0 , since k is always a +ve number , p will also be -ve number So system is stable for k>= 0 , and p<0 For example K can be 2 and p = -2 Ans Q.No 3 The control system is given in Figure 3

G(s) = 2s-2 Gc(s) = Ks+1s = k + 1/s This prportional and integral controller

a)Closed loop transfer function = 2s-2 * Ks+1s 1+ 2s-2 * Ks+1s = 2(ks+1)s-2s+2(ks+1)Charactristic Equation (s-2)*s +2(ks+1)=0 or 1*s*s + (2k -2)s +2 = 0

here a0 = 1 , a1 = 2k-2 , a2=2As per Rouths stability criteria S2 1 2 S 2k-2 For stability 2k-2 >=0 or K>=1.0 For critically damped case K=1.0 b)Closed loop transfer function = 2s-2 * Ks+1s 1+ 2s-2 * Ks+1s = 2(ks+1)s-2s+2(ks+1) =2s+2s*s-2s+2s+2=2s+2s*s+2 Y(s) = 1s*2s+2s*s+2 = 1s+-s+2s*s+2 Using final value theorem Y(Inf)= Lims->inf s*(1s+-s+2s*s+2) = Lims->inf (11+-1+(2s)1+(2/s)) = 1 - 1 = 0 c)Y(s) = 1s*2ss*s+2 Using final value theorem Y(Inf)= Lims->inf s*(1s*2ss*s+2) = Lims->inf (2/s1+(2/s)) = 0 d) Yes it is met, as error gets fully rejected and steady state error is zero Ans 4: a) Given G(s) = 1s(s-1) Gc(s)= K

Y(s)Rs= ks*s-s1+ ks*s-s = ks*s-s+k From charectristic equation s*s - s+k=0 s= 1±1-4k2 The roots are always in right side, so system is always unstable From 0< k<0.25 roots are realk and for k>0.25 complex. b)Let Gc(s) = k(s+2) Y(s)Rs= k(s+2)s*s-s1+ k(s+2)s*s-s = k(s+2)s*s-s+ks+2k Now the charactristic equation is s*s +(k-1)s+2k=0 s=1-k±k*k-2k+1-8k2 Now for k>1 , roots goes to left side and closed loop system gets stabilised.

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