SEE344: Control Systems - Routh-Hurwitz Stability Criterion - Assessment Answer

March 04, 2018
Author : Ashley Simons

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)

Control Systems

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.


Control Systems


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)

Control Systems

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.


A control system as shown in figure (4) has a transfer function of sG )( = ss )1(

Figure (4)

Control Sysytems

(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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Ans Q. No 1

The closed loop system is shown in figure1

Control System


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

Control System

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

Control System

G(s) = 2s-2

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

  1. 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

Control Sysytem

Y(s)Rs= ks*s-s1+ ks*s-s = ks*s-s+k

From charectristic equation s*s - s+k=0

s= 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


Now for k>1 , roots goes to left side and closed loop system gets stabilised.

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