Step 3 of 10
The root locus condition is,
* = v , + '5 ', + H « , - ( ^ + A + ^ i ) - i 8 c r . = l a n - ' ^ Y j + W +90‘ - |^ 9 0 '+ 9 (r + 9 0 ‘ + tan-' = 8 4 . 2 8 '- 9 0 '- 3 7 . 5 6 '- 1 8 ( r
= -2 2 3 .2 T
Step 4 of 10
RULE 5: Find the breakaway points.
\-¥K\- j * + 8 l sO U + 1 3 j [ j ’ ( i ' + 100)J j* (s+ 1 3 ) (s* + 1 0 0 ) + A :(j* + 8 1 ) (j + 1) = 0 Solve for X - f * ( j + 1 3 )(j* + 100) ( j ’ +8l) ( s + l) j* + i 3 s * + i o a ! ’ + i3 o a ! ' j* + j* + 8 1 j+ 8 l
Differentiate the equation with respect to s .
( i ’ + 8 U + 8 1 )(5j* + 5 2 i’ +300j’ +2600i)- ( i ’ + 13 « V 100^*+130fts’ )(3*’ + 2 i +81)
ds + s * +8U +8IJ
d K Equate ---to zero.
ds y + » * + 8 1 i + 8 l ) ( 5 / + 52s" + 300s’ + 2 6 0 0 s )- ( i ’ +13 i‘ +1 OOj’ +1 300j" ) (3 * ’ + 2 i + 81) = 0 (s’ + i'+ 8 1 s + 8 l) (j* +»’ +81i+ 8l)(5j*+52s’ +300J*+2600j) - (j’ +13j* + 100j’ + 1300j')(3j*+2i+81) 5 i’ +57s‘ +757j’ + 7517J* +31112j’ +234900*' + 210600*' -(3*’ +4I*‘ + 4 0 7 / + 5153** + 10700*’ +105300*') 2 * '+ 1 6*‘ +350*’ +2364** + 20412*’ + 129600*'+ 210600* = 0 = 0 Step 5 of 10
Use MATLAB to solve the equation.
»solve('x'‘7■^16*x^6+350*x''5■^2364*xM+20412*x''3■^129600*x''2■^210600*x'); ans = 0
-5.6417302144823378589757341357966-2.3112175660279159755982613661971 All the three roots are valid break-away points since the real-axis segment defined by
- 1 3 ^jS - 1 is part of the locus.
Remaining four roots are complex roots which are not valid for breakaway points. The locus of closed loop poles with respect to x shown in Figure 3.
Therefore, root locus has been implemented.
Step 6 of 10
(b)
Draw the root locus for i ( * ) . ( * + ! ) ( * ' +8l) i ( * ) =
’ * '( * + 1 3 ) ( * '+ 1 0 0 )
Enter the following code in MATLAB to draw the root locus and asymptotes at ^ s 0.5 • » num=conv([11],[10 81]);
» den=conv([1 13 00],[10100]); » sys=tf(num,den):
» rltool(sys)
Step 7 of 10
Observe the MATLAB output on the Figure window. Root Locus Edior for Op«n Loop 1(OL1)
From the root locus in Figure 4, it is observed that the asymptote at ^ = 0.5 cuts the root locus at a point of the form,
In under damped system, roots are of form -^6)^ ± Jo>^ ( complex conjugates), where ^ value is in the range 0 < ^ < 1 •
From the value of ^ it is clear that system is under damped.
There exists a value o f X ^bat causes roots to have damping ratio greater than 0.5.
Step 8 of 10
(c)
To find the value of x ^bat yield closed loop poles with damping ratio ^ s 0.707. Consider the magnitude condition
1 i i ( * ) i ■
K =* ' ( * ’ + 1 0 0 )(*+ 1 3 ) (3) ( * + ! ) ( * ' + 8 l)
To find value of x that yields closed loop poles with 0.707, d value is to be found. As Rvalue is known, calculate the angle 0 ,
c o s $ = ^ 9 ■ cos"' ^
= co s"'(0 ,7 0 7 ) = 45'
Draw a line at an angle ^ —45*00 the root locus plot and mark the intersecting point as shown in Figure (4).
Figure 5; Graph to find 5value
Step 9 of 10
The intersecting point in Figure 4 is the value of s . Substitute value of 5 In equation (3).
( -5.6+6.5 i f ( ( -5.6+ 6.5y)’ + 1 0 0 )(-5 .6 + 6.5 y+13) K = ( - 5 .6 + 6.5y+ 1 )((-5 .6 +6.57) ' +8I) [V(-5.6)'+(6.5)' ] ’ [V(89.1 D' +(72!iF][>/(7.4)' +(6.5)' [ V ( ^ . 6 ) ' + ( 6 .5 ) '][ V ( 7 0 .1 D ' + ( 7 2 .8 7 ] ” = 103.685
Thevalueof jfth a t yields closed loop poles with the damping ratio ^ = 0.707 is |lQ3.68Sl-
Step 10 of 10
(d)
Enter the following code in MATLAB to plot the response of the closed loop system to a reference input. » num=conv([1 1],[1 0 81]); » den=conv([1 13 0 0],[1 0 100]); » sys=tf(num,den); » sys1=feedback(sys,1); » step(sysl)
The MATLAB output for the step response is shown in Figure 5.
For the feedback system shown in Fig., find the value of the gain K that results in dominant closed-loop poles with a damping ratio ^ = 0.5.
Figure Feedback system
Step-by-step solution
Step-by-step solution
step 1 of 6
Refer to Figure 5.55 in the textbook.
Simplify the feedback system {Figure 5.55) as shown in Figure 1.
Figure 1
Step 2 of 6
Calculate the transfer function of the feedback system.
y w G W
R(s)
1+G(i)ff(j)
1+
K M
The characteristic equation for the feedback system is.
U ( j + i ) J L i - i & J luation (1) is of the fonn
Equation (1) is of the fonn
1 +
KL(s) -0
. Where10 ( 5 + l) ( l- A : r )
L ( j) has two poles, one on real axis and other pole value depends on value . To make system stable consider the ^ value as -1.
Then, 10 i ( s ) .
(s + l)(i + l)
step 3 of 6
Procedure to draw root locus:
RULE 1 : There are two branches to the locus, both of which approach asymptotes. RULE 2 : There is no part of root locus.
RULE 3: Calculate the centre of asymptotes. -1-1
2 - 0 - 2 < r * -
The angles of asymptotes are at ^9q* .
Step 4 of 6
To find the value of fc that yield closed loop poles with damping ratio ^ =
0.5 •
Consider the magnitude condition.^ I | i( * ) | •
K =
(j + l)(i+l)
10 (1)
To find value of ^ that yields closed loop poles with ^ = 0.5. rvalue is to be found. As Rvalue is known, calculate the angle $ ,
c o s $ = ^ 6
■ cos"'
C-
cos"'(0.5)
*60*
Draw the root locus using sisotool function in MATLAB and add design parameter as 0.5 damping ratio to get asymptotes that cut the root locus as shown in Figure 2.
» num=10; » den=[1 2 1]; » sys=tf(num,den): » sisotool(sys) Step 5 of 6 MATLAB output:
Root Locus EdHorfor Open Loop 1(OL1)
Figure 2
Step 6 of 6
From Figure 3, observe that the 0.5 damping ratio asymptote cuts the root locus at
5 = -l±yT.732.
Substitute value of s in equation (1).
(-1+yi.732+l)(-l + ;i.732+l)|
10
-2.991
10 I
= 0.299
Problem 5.15PP
A simplified model of the longitudinal motion of a certain helicopter near hover has the transfer function
<H*)- («+0.66)(«2-a24c + 0.IS>
and the characteristic equation 1 + Dc(s)G(s) = 0. Let Dc(s) = kp at first. (a) Compute the departure and arrival angles at the complex poles and zeros.
(b) Sketch the root locus for this system for parameter K = 9.6kp. Use axes -4 < x < 4; -3 < y < 3.
(c) Verify your answer using Matlab. Use the command axis([-4 4 - 3 3]) to get the right scales.
(c) Verify your answer using Matlab. Use the command axis([-4 4 - 3 3]) to get the right scales.
(d) Suggest a practical (at least as many poles as zeros) alternative compensation Dc(s) that will at least result in a stable system.
S te p -b y -s te p s o lu tio n step 1 of 6 a { s ) : 9.8(s’ -0 .5 s + 6 .3 ) (s+ 0 .6 6 )(s^-0 .2 4 s+ 0 .1 5 ) Step 2 of 6 w Step 3 of 6 Angles o f departure: = 1 8 0 ° -« t> Where = 2 < b > - Z 'f e 2 4 ^ = 90®+25.25'> = 115.25" 2<tir = 106.38"+253.61" = 360" ^ = 180"+244.75" = 424.75" = 64.75" And, ^ « -64 .7 5 " Step 4 of 6 Angles o f arrival: ^ = 180"+<t» ■Where if = S f e - Z ' f e = 70°+86.5°+87.4” = 244° Z f e = 90° If, = 334° And, <bi, = -3 3 4 " . Step 5 of 6 , , A r(5^-0.5ff+6 .3) (b) a f s ) = --- V r . ; ---i ^ where K = 9 .8 ^ , (s+0.66)(ff^-0.245+ 0.15) ' K = 0 points: -0 .6 6 . 0.12±j0.368 iT s a points: 0.25±j2.5 As <ti, = 64.75" ^ = -64.75" = 334" And, ^ = -3 3 4 "
We can plot the root locus o f the given transfer fimction as shown below.
(a) For the system given in Fig., plot the root locus of the characteristic equation as the parameter K^ is varied from 0 t o w i t h A = 2. Give the corresponding L(s), a(s), and b(s).
(b) Repeat part (a) with /\ = 5. Is there anything special about this value?
(c) Repeat part (a) for fixed K^ = 2, with the parameter K = A varying from 0 to Figure Control system
Step-by-step solution
step 1 of 15
Refer to Figure 5.56 in the textbook. Draw the modified block diagram.
Figure 1
Move the summing point
R
Step 2 of 15
of a block as shown in Figure 2.
Figure 2
Step 3 of 15
Draw the reduced block diagram of Figure 2.
Figure 3
Step 4 of 15 ^
Move the summing point ahead of a block as shown in Figure 4.
Figure 4
Step 5 of 15
Draw the reduced block diagram.
Step 6 of 15 ^
(a)
Calculate the characteristic equation. I + G ( s ) f f ( j ) = 0 f lOiC, Y 0.U(^-fA)-fg,(0.2»-Mn . K , ) - j ( j + IO )(j + A) j(j + 10)(5 + A) (1) step 7 of 15 Simplify further. i ’ + ( 1 0 + A ) i’ + 10.1i i r g f 1 - 0 Substitute 2 f o r A in equation (1). ‘ ■ ^ ^ { j ’ + { l l +2) j ‘ + l l (2) j ] “ “
From the characteristic equation, the loop transfer function is. r ( 2 £ + I 2 _ j’ +13j‘ + 22s o ( i ) = »* + lJ s ’ + 2 2 i And, i ( j ) = 2 i+ 1 0 Step 8 of 15
Write the MATLAB code to draw the root locus. » num=[2 10]: » d e n = [1 13 22 Oj; » sys=tf(num,den) sys = 2 S + 10 s ^ 3 + 1 3 s^ 2 + 2 2s
Continuous-time transfer function. » rlocus(sys)
Step 9 of 15
Draw the root locus plot using MATLAB.