Solution Code: 1AFBC
This assignment falls under Control Systems which was successfully solved by the assignment writing experts at My Assignment Services AU under assignment help service.
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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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
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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