Problem
Consider the system shown in Fig., where
Dc{s) = K
(a) Prove that if the system is stable, it is capable of tracking a sinusoidal reference input r = sin wof with zero steady-state error. {Hint: Look at the transfer function from R to E and consider the gain at wo.)
(b) Use Routh’s criterion to find the range of K such that the closed-loop system remains stable if wo = 1 and a = 0.25.
Figure Control system
(b) S.S.E=lim sE(s)
lim s’ (s + l)(s ’ +<D,’ )
■-*" (s’ +£0,’ )s (s+ l)+ K (s+ a )’ (s’ +00,’ )
= 0
Step 3 of 3
(c) Characteristic equatioii is s ([s^ + lj ( r H ) +K (s+0.25)^ =0
i.e., s*-h’ +s’ (K+1)+s(1+0.5K)+(0.25)’k=0 Using Routh's criterion for stability.
s* 1 K+1 (0.25)’ K
s’ 1 m ).5 K 0
s’ 0 5 K (0.25)’ K
s‘ O5K+0.875 0
s" (0 2 5 )’ K
For stabili^, as all the first column elements should be greater than 0,
Consider the system shown in Fig., which represents control of the angle of a pendulum that has no damping.
(a) What condition must Dc(s) satisfy so that the system can track a ramp reference input with constant steady-state error?
(b) For a transfer function Dc(s) that stabilizes the system and satisfies the condition in part(a), find the class of disturbances w(t) that the system can reject with zero steady-state error.
Figure Control system
c Control O f Dynamic Systems (7th Edition)
Problem
Consider the system shown in Fig., which represents control of the angle of a pendulum that has no damping.
(a) What condition must Dc(s) satisfy so that the system can track a ramp reference input with constant steady-state error?
(b) For a transfer function Dc(s) that stabilizes the system and satisfies the condition in part(a), find the class of disturbances w(t) that the system can reject with zero steady-state error.
Figure Control system
j* +a: (1)
step 2 of 6
The reduced block diagram is shown in Figure 1.
w
Step 3 of 6
From Figure 1, the closed loop transfer function is.
Substitute — !— for r ( s ) in the above equation.
A M _ ! _
W (s]
step 4 of 6
Determine the expression for error signal.
=*M-
i ’ + A : + o . ( j ) ' ' i '+ A : + z ) . ( j )A Mw ( , )
Problem 4 .1 2PP
A unity feedback system has the overall transfer function
R(s) ' + +
Give the system type and corresponding error constant for tracking polynomial reference inputs in terms of ^ and ojn.
Step-by-step solution
step 1 of 2
Step 1 of 2
t m - ^ ( 0 - V
^ R (s) E ( 0 = E ( 0 - Y ( 0
= j R ( 0
s +2^cd^s+cch^ S .S .E -lim sE (s)
By observing the above function.we can say &at &e system is of [Type l\.
Step 2 of 2
■ S.S.E.=lim
= 2 i
25
Consider the second-order system
d s )
? + 2 f 7 + T *
We would like to add a transfer function of the fonn Dc(s) = in series with G(s) in s unity feedback structure.
(a) Ignoring stability for the moment, what are the constraints on K, a, and b so that the system is Type 1?
(b) What are the constraints placed on K. a, and b so that the system is both stable and Type 1 ?
(c) What are the constraints on a and b so that the system is both Type 1 and remains stable for
(c) What are the constraints on a and b so that the system is both Type 1 and remains stable for every positive value for K?
Step-by-step solution
step 1 of 5
D (s ) = s"+2?s+l
s+b Y (s) _________ K (s+ a) E ( s ) (s“+2?s+l)(s+ b)+ K (s+ a)
( ! “ + 2 ? !+ l)(s + b ) E = R -Y = . , \ --- -R (s)
(s“ + 2? s+ l)(s+ b )+ K (s+ a )
Step 2 of 5
, . sfs+b)fs^+2^s+ll , .
a. S.S.E .=lim sE(s)= ^ R ( 0
(s^+ 2 ^+ l) ( s ^ ) +K (s+a) Fortype -1 systemJb=0, K ^ O , a ^ O
Step 3 of 5
b. For the given system, b=0;
R ( s ) = | . s
and the system shoul d be of type -1.
S.S.E.=lim sEfs) i-»0 ' '
s"(s"+2?s+ l) 1
= lim --- ^--- --- X - r
*-»«(s^+24s+l)s+K(s+a) s^
Char, equation = s^+2^^+(K+l)s+Kat*0
Step 4 of 5
For Stability,
1 K+1
s’ K a
s‘ 0
s" Ka
Ka>0, (K+1) - E i > 0 25
=>K a<2^(K +l)
The constraints are | b=0,Ka>0, K a<2^(K +l)|
Step 5 of 5
c. A sK >0, r^>0| a<2^(K +1)
>1 ^ |0<a<2^(K +l)|
Problem 4.14PP
Consider the system shown in Fig.(a).
(a) What is the system type? Compute the steady-state tracking error due to a ramp input r(t) = rof\{t).
(b) For the modified system with a feed forward path shown in Fig.(b), give the vaiue of H f so the system is Type 2 for reference inputs and compute the Ka in this case.
(c) Is the resulting Type 2 property of this system robust with respect to changes in Hf. that is, will the system remain Type 2 if H f changes slightly?
Figure Control system
Figure Control system
»o— ^ ( i ) - A
V -o r
HfS
Step-by-step solution
step 1 of 4
T ( 0 = -s (t-s + l) +A R ( s ) = ^
E ( . ) = R - Y = ^ ! i ^ s(ts+ 1 )+ A
S.S.E. = lim sE (s)= lim f
*-.0 ' * - * 0 s ( t s + l) 4 A U v
S .S .E .= ^ ____ A System is of [Type -1[
Step 2 of 4
b. From block dis^am , we get (R E , -Y+R Hf s ) A=Ys
A + s(ts+ l)
A +s(ts+1) Niim erator= ts^+s -Hf As+A (1-H,)
Step 3 of 4
Fortype-2sy8tem, itm usthavetw o zeros at s ^ .
= >(l-H fA )=0
a n d [ ^ ^
S. S.E. =lim sE (s) =lim . ,---x f -^1
*-»o »-»<• s(ts+1)+A
A K a
Step 4 of 4 ^
c. system wouldbecome |T]rpe ^
A controller for a satellite attitude control with transfer function G=1/s2 has been designed with c unity feedback structure and has the transfer function
DcW = - ! ^ .
(a) Find the system type for reference tracking and the corresponding error constant for this system.
( b )if a disturbance torque adds to the control so that the input to the process \s u + w. what is the system type and corresponding error constant with respect to disturbance rejection?
Step-by-step solution
Step-by-step solution
step 1 of 8
(a)
The transfer function of the controller for a satellite attitude control is.
c = 4
-The transfer function of unity feedback structure is.
10(1+2)
£ ) ( . ) s+5
In order to determine a system for reference tracking, first identify the poles, and then determine the type of system and the finally evaluate the error constants.
By looking at the transfer function, the system has two poles at 5 = 0 - Thus, the system type is [Type 2 | .
Step 2 of 8
The closed loop transfer function is.
r( s ) C(s)D(s)
« ( s ) 1 + G ( j ) c ( s )
1-h
10(^ + 2)
^ * ( « + 5 ) + I 0 ( f + 2 )
Step 3 of 8
Calculate the steady state error.
s^(s+S)+W(s
+ 2)i’J
f s’ (s + S) l']
™ [j> ( , + 5 ) + 1 0 ( 5 + 2 )5 ’ J
= lim f- 7 ^
---— > |_ * '( i + 5 ) + 1 0 ( i + 2 ) J 0 + 5
( 0 ) ( 0 + 5 ) + 1 0 ( 0 + 2 ) ___ 5 _
°0+20
• 0.25
Step 4 of 8
Determine error constant i f .
j f . - —
0.25
• 4
= 4
Therefore, the error constant. K is 0
Step 5 of 8
(b)
When a disturbance torque adds to the control, it acts as a disturbance input. For a disturbance input, the poles at 5 s 0 ^re usually after the Input. Hence the system type is Type 0.
Thus, the system type is |t^
Step 6 of 8
Determine the transfer function with disturbance feedback.
I W - I G
i r ( s ) I + G D
=1—
• 1— j + 5
j ' ( i + 5 ) + I O ( j + 2 ) f ^ ( i + 5 ) + 1 0 ( i + 2 ) - » + 5
i ’ ( i + 5 ) + 1 0 ( * + 2 ) ( » '- l ) ( j + 5 ) + I O ( i + 2 )
i ’ ( j + 5 ) + 1 0 ( i + 2 )
Step 7 of 8
Calculate the steady state error.
*lim'( j * - l ) ( s + 5 ) + 1 0 ( j + 2 ) j ’ ( i + 5 ) + 1 0 ( j + 2 ) ( 0 - l ) ( 0 + 5 ) + 1 0 ( 0 + 2 )
( 0 ) ( 0 + 5 ) + 1 0 ( 0 + 2 ) - 5 + 2 0
20 _15 '2 0 3
° 4
Step 8 of 8
Determine error constant K . 1
3 1
4 1 +a: .
'
3 4 - 33
Therefore, the error constant,
I
Problem 4 .1 6PP
A compensated motor position control system is shown in Fig. Assume that the sensor dynamics are H(s) = 1.
Figure Control system
(a) Can the system track a step reference input rw ith zero steady-state error? If yes. give the value of the velocity constant.
(a) Can the system track a step reference input rw ith zero steady-state emor? If yes. give the value of the velocity constant.
(b) Can the system reject a step disturbance wwith zero steady-state en^or? If yes. give the value of the velocity constant.
(c) Compute the sensitivity of the closed-loop transfer function to changes in the plant pole at -2.
(d) In some instances there are dynamics in the sensor. Repeat parts (a) to (c) for |W(*) = and compare the corresponding velocity constants.
Step-by-step solution
s te p 1 of 8
Refer to Figure 4.29 in the textbook.
(a)
From the block diagram, it is clear that the system is Type 1 with
y ( 4 0 ( 4 R (s ) l+ H ( s ) G ( s ) y ( « ) g ( 4 R (s ) 1 + G ( j)
Write the expression for the error signal.
' ' i + a ( s ) '• ’ Now. evaluate the steady-state system error.
e„ - liin
= lin ii f ---
^
'-*• ^^
s(5 + 2)(5+30)+160(
j+4)J
sO
Therefore, the system can track a step input with zero steady-state emor.
Step 3 of 8
Determine the value of velocity constant.
= H m j(7 (5 )
Therefore, the value of the velocity constant, is
|1Q.67|-Step 4 of 8
(b)
The system is Type 0 with respect to the disturbance and has the steady-state emor. Find the transfer function for 7(4) •
Determine the steady-state error to a disturbance input.
y .
*-*® (^4(4 + 2)(4+30)+160(4+4)J
Hence, the system cannot reject a constant disturbance.
Step 5 of 8
(c)
In order to compute the sensitivity of the closed-loop transfer function to changes in the plant pole at _2. Now by determine the transfer function.
r ( ^ ) = - l6 0 ( ^ + 4 )
«(« + >4)(s+ 3 0 )+ I6 0 ( 2 + 4 )
Where. A was inserted for the pole at the nominal value of 2.
Now evaluate the sensitivity of the system.
A S T ... (3) T S A
Step 6 of 8
Apply partial differentiation to the transfer function with respect to A.
160(5+4
Substitute — value in equation (3).
SA
Therefore, the sensitivity of the closed-loop transfer function is
25(5+30) 1 with respect to the reference input.
The new transfer function is.
f f ( 4 ) . l -4 + 20
Step 8 of 8
The velocity feedback J ^ w ill also change. Hence, evaluate the new expression for the error.
*W = [i-r(4 > (4
-Evaluate the velocity constant from the emor expression.
3 0 x2 2.63 - 22.86
Therefore, the velocity constant. is 122.86! •
The general unity feedback system shown in Fig. has disturbance inputs w1, w2, and w3 and is asymptotically stable. Also,
* ' ' '' j'> n 2 i( * + p ii) ’ Vi n f * . r - ■ ~ ' « ^ n rJ i(» + w )'
Show that the system is of Type 0, Type /1. and Type (/1 + 12) with respect to disturbance inputs w1, w2. and vv3 respectively.
Figure Single input-single output unity feedback system with disturbance inputs
Step-by-step solution
step 1 of 5
Draw the diagram of single input-single output unity feedback system with disturbance inputs.
w . w .
Figure 1
Step 2 of 5
Consider the following data;
Step 3 of 5 ^
Write the expression for output from the block diagram shown in Figure 1.
y / A _ y / j X
Substitute for G ,( i) a n d for a , ( s )
1 + G , W G , W '^ ^ ^
) +t U ' n 3 i ( * + / > i i ) J U ^ n ; , i
(*+Pa)JJ
Here. ^ are the poles and zeros of the and ^ not at the origin.
Step 4 of 5
Now evaluate the steady-state error.
( g , g , [ n , ( z 4 - z , ) ] » ^ ( z ) y
[ j * + i ' ’ I l ,( s + p ,) + / :,A :, n ,( i + z ,) J
= Iiin
= liin .
Analyzing this system, we see that there are no poles at the origin, hence its Type 0 system.
Step 5 of 5
Write the expression for output for disturbance ■
l + G | ( j ) G j ( j
r A r , n ; , ( z + Z a ) '|
i . i ‘ r c , ( i + P a ) J
1, n;,(z+z„)Y
K,n;,
W ,{s){k,
n:, (
j+
z„))(
a:, nj, (»+zj,))
(s* n ; , ( z + p „ ) ) j ^ n 2 , ( z + P a ) A T ,r n ,(» + ? .,)V n ,(z + f» „ )„ , , ,
In this expression, in the denominator A (z) is the characteristic polynomial. Now evaluating the steady-state error.
. ( i f L O i .
By analyzing this system, we see that the system is of Type Write the expression for output for disturbance ■
I ' M -i^ A s ) l + G , ( i ) G , ( i )
( F .W
l- ff f ^ , n : , ( » + r , , ) Y A : , n ; , ( z t r , , ) ^^
TC, (j+p„) Ji*' nj,
( s + P u ) } )(» *n ;i(j+ p „ ))(» * -n ;,(z + p „ ))+ (A :,n ;,(j+ z „ ))(A :,n 5 i(z + z a )) ( j ' n ; , ( j + p „ ) ) ( i ' ’ n ; , ( » + P j , ) )
---A W
In this expression, in the denominator A (z) is the characteristic polynomial. Now evaluate the steady-state error.
i n , A
By analyzing this system, we see that the system is of Type k * k
Problem 4 .1 8PP
One possible representation of an automobile speed-control system with integral control is shown in Fig.
(a) With a zero reference velocity input vc = 0, find the transfer function relating the output speed v to the wind disturbance w.
(b) What is the steady-state response of v if w is a unit-ramp function?.
(c) What type is this system in relation to reference inputs? What is the value of the corresponding error constant?
WVWtjat.la.thj?iv/ie aodJ^rreRnnnriinn prrnr mnstant nf this sustpm in rplptinn tn trankinn thp
(d) What is the type and corresponding error constant of this system in relation to tracking the disturbance w?
Figure System using integral control
Step-by-step solution
step 1 of 7
Refer to Figure 4.31 in the textbook.
From the block diagram in Figure 4.31,
y h . v * w = { - + ^ + i C \ v
s \ m s )
V = , . . K + - , --- --- W ...(1) +kyms+mkfk2 * •¥mk^k^-¥kjnis
Step 2 of 7
Thus, the transfer function relating the output speed to the wind disturbance is
f P ( j) + ky/ns+ k^k^m (2)
Step 3 of 7
(b)
The wind disturbance is a ramp function, ' s The steady state response of the velocity v is.
Substitute the value of K (f)fro m equation (2) in this equation.
= lim
Thus, the steady state value of the velocity v is V
step 4 of 7
(c)
Calculate the emor in system with respect to reference inputs.
£ W = n ( i ) - f ' W
The error of the system can be compared with a system of unit gain feedback and having e nk^k^
fonward gain of <7(j) * - 5— — .A s the gain <7(4) has a pole at origin.
s ^ k^frts The positional constant of the system is,
J S r,.B m G W mk^k2 s lim
Step 5 of 7 ^
The steady state error of the system for the unit step input.
e„ = lim — ^-r-r y. (5)
The velocity constant of the system is.
JS:, = limiG(j)
s l i m j - n k ik j
»-»• s +kjm s
Now. the steady state error of the system for the unit ramp input.
s l i m — —
As steady state error for unit step input is zero and for the ramp input is — . the system is a
Type 1 system.
Therefore, the velocity emor constant is _____
Step 7 of 7
(d)
Calculate the emor in system with respect to disturbance w . E ( ,) = W { s )-V (s )
The steady state error for a unit step disturbance input is,
K lim f £ ( 5 )
As the steady state emor for a unit step disturbance is non-zero this is a Type 0 system.
Therefore, the value of the error constant is Q].
For the feedback system shown in Fig., find the value of a that will make the system Type 1 for K
= 5. Give the corresponding velocity constant. Show that the system is not robust by using this value of a and computing the tracking error e = r - y to a step reference for /C = 4 and K = 6.
Figure Control system
Step-by-step solution
5tep^y-step~so1ution
step 1 of 3
Y (s) = - 5 ^ R (s) S+2+K AsK=5,
S.S.E.=lim sE (s)=lim s(R -y ) c^O ' ' *-»0 ' ' ix R ( s ) (s+7) for type-1 system, ( 7 - 5 a ) ^
7
O F —
5
— = S .S .E .= -
K . 7
=7
Step 2 of 3
for :
E(s)=R-y=
s+2— K 2 s + -t____ 5 L __ 5
S + 2 + K F i _5 s+6 s f s + l j S.S.E.=lim sE (s)=lim ^ (s)
c-kO ' ^ c-kO -jj: '• ^
*-»0 ' • *-»0 g+6 system becomes type-0 with S.S.E .=-1
15
Step 3 of 3
for K=6 :
S + 2 - - K
E (s)= R -Y = --- 5 _ | _ 5
‘ S+2+K s+8
S.S.E.=lim s£ fs)= lim
<-kd ' • <-k0
■ H )
R ( 0> System becomes type-0 witti S.S.E.=—
System is not robust.
Problem 4.20PP
Suppose you are given the system depicted in Fig.(a) where the plant parameter a is subject to variations.
(a) Find G(s) so that the system shown in Fig.(b) has the same transfer function from r to y as the system in Fig.(a).
(b) Assume that a = 1 is the nominal value of the plant parameter. What are the system type and the error constant in this case?
(c) Now assume that a = ^ + 5a, where 6a is some perturbation to the plant parameter. What are the system type and the error constant for the perturbed system?
Figure Control system
the system type and the error constant for the perturbed system?
Figure Control system
Step-by-step solution
step 1 of 8
(a)
Refer to the system in Figure 4.35(a) in the textbook.
Consider the expression for left side loop.
1
Consider the expression for right side loop.
[ j r + ; i r ( s + o ) ] = i y X { s + a + l) = sY
„ sY
( s + a + I ) (2)
Step 3 of 8
Equate both the equations (1) and (2).
4 { R - Y ) sY
The closed-loop transfer function of the system is.
r ( j ) 4 (5 - t- g + i)
R (s) 4 (* + f l - l ) + 4 ( 5 + g + l ) (3)
step 4 of 8
The general fonri of the closed-loop transfer function is.
y ( 4 g ( ^ )
« ( s ) 1 + G ( j ) (4)
Here C {4 ) is the open-loop transfer function.
Find the open-loop transfer function <7(j)- From equations (3) and (4),
5 ( i + f l - l ) + 4 ( 5 + g + l ) 1 + G ( j )
-To define the system type open loop transfer function is considered.
Recall equation (5).
Substitute 1 for a in the equation.
' ' s ( * + l - l ) 4 ( i + 2 )
i ( s ) 4 ( i + 2 )
There are two poles at origin, hence the system is a type-2 system.
Thus, the system type is .
Step 6 of 8
Find the error constant.
For a type-2 system, the error constant exists for parabolic input only and the error constant is zero for step and ramp inputs.
The acceleration error constant is.
it. ■U ni5 ^G (5 )
* j-»0 ' '
s U m [4 ^ 'i'8 ]
= [ 4 ( 0 ) + 8 ]
= 8
Thus, the error constant is
Step 7 of 8
(c)
Assume that j s U d a
Here Sa is the some perturbation to the plant parameter.
To define the system type open loop transfer function is considered.
Recall equation (5).
Substitute 1 + for a in the equation.
s{s+ Sa)
There is only one pole at origin, hence the system is a type-1 system.
Thus, the system type is [ | ] .
Step 8 of 8
Find the error constant.
For a type-1 system, the error constant exists for ramp input only and the error constant is zero for step input and infinite parabola input.
The acceleration error constant is.
f 4 ( 5 + 2 + ^ g )
(a) Determine values for K^. K2, and K3 so that
(i) both systems exhibit zero steady-state error to step inputs (that is. both are Type 1). and (ii) their static velocity constant /Cv = 1 when KO = 1.
Two feedback systems are shown in Fig.
(b) Suppose KQ undergoes a small perturbation: K Q -* + 5KQ. What effect does this have on the system type in each case? Which system has a type which is robust? Which system do you think would be preferred?
Refer from Figure 4.36 (a) in the textbook and write the emor detector output equation.
E ^ R - Y
E ( s U - ^ (1)
^ ^ 1 + G (s )
Substitute the gain of the system from Figure 4.36 (a) in the textbook for G (« ) in equation (1).
^ ' A s ^ + s + K ^ , ' ' (2)
Determine the value of velocity error coefficient K , ■ Write the general formula for velocity error coefficient.
Calculate the ^ . ^ ( o o ) = U in s £ ( j) from equation (2).
Substitute equations (5) in equation (3).
= ... (6)
Hence, from equation (6), the value of is equal to |pnel •
Step 2 of 4
Refer from Figure 4.36 (a) in the textbook and write the emor detector output equation.
E s R — Y
Substitute the gain of the system from Figure 4.36 (b) in the textbook for in equation (7).
---i 4 £ ± i £ _ i d £ ± y j t ( , )
---R(s)
(8) 4J + 1 + C A
4s + l
Calculate the from equation (8).
4s+i+a:,a:„ Equate equations (9) and (10),
U K , K , ( \ - K , )
Determine the value of velocity error coefficient K ,
Calculate the from equation (8).
' ’ ,-M 4s+i+a:,k, *’ Substitute equations (14) in equation (3).
a: . *
\ U K , \ U K ,
For AT, a 1. find the value of K ,- a: , = 3 ... (15)
Substitute equations (15) in equation (12).
i + 3 ( i - * r j ) = o
.(16) K - 1
« p - 3
Hence, from equations (15), and (16). the value
(b)
Consider the value of A^.
K ^ ^ K ^ + 8 K ^ ... (17)
Calculate the ^ *pH = ' ™ » ^ W from equation (2).
of ATjis and the value of AT, is [ ^ .
.(18)
‘ i . * p l ) ‘™ * 4 i ’ + i+ ( A r .+ S A r ,) iir , j Substitute AT| value in equation (18).
• b » p v I « P 4 i ' + * + ( * ; , + 8a:„) e . « p H = 0 (19)
Hence, the Figure 4.36 (a) is regardless of K^ value.
Step 4 of 4
Substitute equations (17) in equation (9).
, . i+a: , (a: . + 5a: . ) (i-a: , )
Hence, the Figure 4.36 (b) is robust. So, the designers favor to [choosesystem (a ) than because of output variation is huge with small variation in input changes.
Problem 4.22PP
You are given the system shown in Fig., where the feedback gain jS is subject to variations. You are to design a controller for this system so that the output y(t) accurately tracks the reference input r(t).
(a) Let f3 =1. You are given the following three options for the controller Dci(s):
Choose the controller (including particular values for the controller constants) that will result in e Type 1 system with a steady-state error to a unit reference ramp of less than .
(b) Next, suppose that there is some attenuation in the feedback path that is modeled by p = 0.9.
Find the steady-state error due to a ramp input for your choice of Dei = (s) in part (a).
(d; iNexi, suppose inai iiieie is some uuenuauun in uie leeuuauK paui inai is iiiuueieu oy p = u.s.
Find the steady-state error due to a ramp input for your choice of Dei = (s) in part (a).
(c) If = 0.9, what is the system type for part (b)? What are the values of the appropriate error constant?
Figure Control system
X
S te p -b y -s te p s o lu tio n
step 1 of 8
Refer to Figure 4.35 in the textbook.
(a)
In order to choose a controller that will result in Type 1 system, first evaluate the transfer function and then determine the steady-state error.
First choose as an integrator in the loop and evaluate the transfer function.
10____ Y * » ^ * * ' j
Now evaluate the system error.
io(*^+*,)
Now apply final value theorem to find the steady state emor.
= ---] j .
From equation (2), it is clear that 2 1 0 which meets the steady state specifications. The next step is to determine the stability, and see if all the poles are on the open left half plane, use the Routh criterion
The characteristic equation of the closed-loop poles is, t i l s ’ 4 - i a s 4 - l o ( i ^ + 1^) = 0
Step 5 of 8
To determine the Routh array, we first arrange the coefficients of the characteristic polynomial in two rows, beginning with first and second coefficients and followed by the even numbered and odd-numbered coefficients.
Apply final value theorem to find the steady state error.
e . = l i m s [ £ ( j ) ]
|^5 (5 + l)(j+ 1 0 )+ 9 (A :^ 5 + ifc ,)J L 5 *y
Therefore, the steady-state emor due to ramp input is 0
Since the steady-state error due to ramp input is infinite, the system is no longer Type 1.
Step 8 of 8
(c)
The value of feedback gain, f i = 0.9
The input is, —
s
Calculate the steady state error.
e . - t o s [ £ ( j ) ]
From error formula we see the system is Type 0. For the system Type 0 the error for step position
1
(a) Find the transfer function from the reference input to the tracking error.
Consider the system shown in Fig.
(b) For this system to respond to inputs of the form r(t) = fn1 {t) (Where n < q ) with zero steady- state error, what constraint is piaced on the open-loop poles p1, p2. • • •. pg?
Figure Control system
(«+;»iXi+ f t ) " ' C*+/>«)
Step-by-step solution
step 1 of 2
Y ( 0 _ G R (s ) 1+G f E ( s ) = R ( s ) - Y ( s )
, E ( s ) _
n ( ^ ‘)
step 2 of 2
S.S.E.=lim sEfs) t-»o ' ''
n ( ' + P i ) „ i
sn(s+pi)
» i_______
A nd Hm--- is required to be equal to zero.
For this condition to be True, at least (n + 1) zeros out o f *q’ should present at the origia
Problem 4.24PP
(a) Compute the transfer function from R(s) to E(s) and determine the steady-state error {ess) for a unit-step reference input signal, and a unit-ramp reference input signal.
Consider the system shown in Fig.
(b) Determine the locations of the closed-loop poles of the system..
(c) Select the system parameters {k, kP, kl) such that the closed-loop system has damping coefficient ^ = 0.707 and ojn= What percent overshoot would you expect in y(t) for unit-step reference input?
fdXJFinrlthft,trackinn prmr <;innpl a.<? a fiinrtinn nf timp if thp rpfprpnnp inniit tn thp svRfpm
(d) Find the tracking error signal as a function of time, e(t), if the reference input to the system, r(t), is a unit-ramp.
(e) How can we select the PI controller parameters {kP, kl) to ensure that the amplitude of the transient tracking error, |effj|, from part (d) is small?.
(f) What is the transient behavior of the tracking error, e(t), for a unit-ramp reference input if the magnitude of the integral gain, kl. is very large? Does the unit-ramp response have an overshoot in that case?
Refer from Figure 4.39 in the textbook and write the error detector output equation.
E = R -Y
E{s) = - I > « ( * ) (1) 1 + G(5)Dc(»)
Substitute the plant gain o f the system and controller gain from Figure 4.39 in the textbook for G { .) in equation (1).
Calculate the from equation (2).
= lim
Write the denominator part of the equation (2).
s ^ + k ^ + k , k » 0 ... (5) Find the roots of the above equation.
■ ■ 2
Hence, the roots of the closed loop poles from the transfer function from J?(5)to is - k ^ ± y l { k ^ y - 4 ( k , k )
Step 5 of 10
(c)
Write the characteristic equation of the second order system.
5’ -f2Co^ - Ki> / » 0 ... (6) Equate the equations (5) and (6).
s * + k f k s + k fk ^ s ^ + 2 f y a ^ + t o * ... (7) Substitute 0.707 for ^ and 1 for <o^ in equation (7).
j ^ + * ^ + * , * = j '+ 2 ( 0 . 7 0 7 ) ( l ) f + ( l ) ’
* = »’ + 2 ( 0 . 7 0 7 ) i + I ...(8) Write the coefficients of the equation.
* , * = ! ... (9)
* ,Jt = 2 (0 .7 0 7 )
*,Jt = 1.414...(10)
Write the transfer function of the entire system.
K ( , ) _ 0 { s ) D A s ) R(s) 1+G (*)Dc( 4
Step 6 of 10 ^
Substitute the plant gain o f the system and controller gain from Figure 4.39 in the textbook for in equation (11).
Substitute equations (9) and (10) in equation (12).
... ,t3 ) i* + l. 4 1 4 . f + l
Determine the peak overshoot value from damping ratio value and natural frequency value («>.)
Write the general formula for percent of peak overshoot.
W , = e”* ''^ x l 0 0 (14)
Determine the tracking emor signal as a function of time e ( /) at unit ramp signal.
Write £ ( j) fr o m equation (2).
Consider the general form of inverse Laplace transform.
r ' f - --- 1 — - | = « '* 's i i i i < (18)
Apply inverse Laplace transform in equation (17).
e(,)=
V 4 t t , - ( t t , y
^ Sin (19)
Hence, the value of tracking error signal as a function of time e ( /) a t unit ramp signal is
^ 4 t t , - ( « , ) ’ 2 '
^ /
Step 8 of 10
(e)
Write the small value of |e(/)| from the equation (19).
Write the small value of |e(/)| from the equation (19).