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

March 04, 2018
Author : Ashley Simons

Solution Code: 1AFBC

Question: Control Systems

This assignment falls under Control Systems which was successfully solved by the assignment writing experts at My Assignment Services AU under assignment help through guided sessions service.

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 +System  Y(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=2 As 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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