1 (156)

Prévia do material em texto

(a) C(i) = ;p T n ^ rF I= 57'
For the feedback configuration of Fig., use asymptofes, center of asymptofes, angies of
deparfure and arrivai, and fhe Routh array to sketch root ioci for the characteristic equations of
the iisted feedback controi systems versus the parameter K. Use Mattab to verity your resuits.
(») C ( j)= 5 ,
(c ) =
m = '+ 3 * *
Figure Feedback system
m
Figure Feedback system
Step-by-step so lution
S lep t of 30
(a)
Consider the foiiowing equation. 
« , ) = (. 
i ( a ) =
a ( j+ l+ 3 y ) ( s + l - 3 y ) 
s+2
a (a + I+ 3 y )(a + I-3 y )(s + 8 )
Consider the generai torm of characteristics equation. 
l + * i ( a ) = 0 ......(1)
1 + i r -
i(a + I0 ) (a + I+ y ') (a + I-y ) 
a + 2
fyfor ^ 5)in Equation (1). 
= 0 ...... (2)a (a + l+ 3 y )(a + l-3 y ) ( i+ 8 )
Consider the roots of the generai form of an equation by the root iocus method.
l + i C - ^ - O ......(3)
ZKs)
Where.
The roots of ^ ( 5) = 0 are catted the zeros of the probiem.
The roots of O(a) = 0 are the potes.
Consider the number of potes and zeros from the characteristics equation. 
Compare the Equation (2) and the Equation (3).
To find zeros put numerator fV(s)=0 
Thus, the zero is -2.
To find potes put denominator D(_s) = 0. 
j ( j+ l+ 3 y ) ( i+ l- 3 y ) ( » + 8 ) = 0 ......(4)
The roots of the equation (4) are 0 .-8 . —1—3y and —l+ 3 y .
Thus, the tour poies are 0. - 8. —I-3 y and - l+ 3 y .
Step 2 of 30 ^
Consider the formuta for the asymptotes. 
n -m
(sum o f finite poles)-(sum o f finite zeros) 
(numberof finite poles )-(n i]inbero f finite zeros) 
- l - 3 y - l+ 3 y -8 - ( -2 )
3
3
Thus, the asymptotes is |-2.67|.
Consider the formuta for the angte of asymptotes.
180-m]
■ j’ (j+3)
Step 10 of 30
Substitute ) * for £ ( j) in Equation (1).
i ( j+ 3 )
.(9)i ’ (i+3)
Compare the Equation (9) and the Equation (3). 
To find zeros put numerator N{s^=0 
Thus, the zero is -1.
To find poies put denominator D(_s) = 0.
The roots of the equation (4) are 0.0 and -3. 
Thus, the tour poies are 0. 0 and -3.
Step 11 of 30 -rv
Consider the formuta for the asymptotes. 
n -m
(sum o f finite poles)-(sum o f finite zeros) 
(numberof finite poles )-(m unberof finite zeros)
2
~ 2
Thus, the centre of asymptotes is .
Substitute 1 for /. 3 for n and 1 for m in equation (5).
180+ 360(1-1)
f t “ +
180 
” 2 
=90“
Substitute 2 for /. 3 for n and 1 for m in equation (5).
180+ 360(2-1)
540
2
= 270“
= -90“
Thus, the angle of asymptotes are |90“ |and
Step 12 of 30
Consider the following transfer tunction.
T - r 0 (s ) f f (s ) 
l+ G (s )W (s )
X
l+ X
(» + 0
” s’ (s+3)
s’ (s+3)
Consider denominator polynomial to find the K value.
s’ +3s’ + J&+A:
Consider the denominator polynomial is equal to zero.
s’ +3s’ +&+A: = 0 (10)
From equation (10). the characteristics equation is given below. 
A (s )= s ’ +3s’ + A i+ j : 
s’ +3s’ +&+A: = 0 (11)
By applying Routh-Huiwitz criteria to equation (11).
a :
a :
2a :
K
Figure 4
Step 13 of 30 ^
The sysfem is stable if the equation satisties the following condition.
• All the terms in the tirst column of the Routh’s array must be positive sign. 
From Figure 4. the stable condition is JK > Q.
Thus, the value of K is JK = Qand it is satistied the stability condition. 
Substitute 0 for X in equation (11).
* ’ +3»’ + k »+a: = o
j = 0
Thus, the imaginary axis crossing is .
Step 14 of 30
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid at an angle of
9(r
• Draw the root locus.
The root locus plot is shown in Figure 5.
Root Locus
Hence, the root locus is plotted for the given transfer function and it is shown in Figure 5.
Step 15 of 30
Consider the following given function. 
( j+ 1 )
Write the MATLAB program to obtain root locus. 
s=tf('s');
sysL=(s+1)/(s“2*(s+3));
riocus(sysL)
The root locus plot is shown in Figure 6:
Step 16 of 30
Thus, the root locus is verified from MATLAB output.
Step 17 of 30
(0
Consider the following equation.
Substitute Ifo r L ( j) in Equation (1).
U + lA s + 3 j
■•'(Srlef)-".. '■»
step 18 of 30
Compare the Equation (12) and the Equation (3).
To find zeros put numerator N{s^=0 
Thus, the zero is -5 and -7.
To find poles put denominator D(s) =0.
The roots are -1 and -3.
Thus, the tour poles are -1 and -3.
Step 19 of 30 ^
Consider the formula for the asymptotes.
Asymptodes=(numberof Unite poles )-(numberof finltezeros)
= 2 -2 
=0
Thus, the asymptotes is .
Consider s 0 and simplity equation is given below.
ds
8s’ +64s+I04 = 0
From the above equation, the break in and break away points is -2.27 and -5.73. 
Thus, the break in and break away points is -2.27 and -5.73.
Step 20 of 30
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid.
• Draw the root locus.
The root locus plot is shown in Figure 7.
Root Locus
Hence, the root locus is plotted for the given transfer function and it is shown in Figure 7.
Step 21 of 30
Consider the following given function.
i ( i ) =M S I )
MATLAB program to obtain root locus: 
s=tf('s');
sysL=((s+5)*(s+7))/((s+1)*(s+3)):
riocus(sysL)
The root locus plot is shown in Figure 8.
Figure 8
Step 22 of 30
Thus, the root locus is verified from MATLAB output.
Step 23 of 30
( d )
Consider the following equation.
w A - (, = 180“-[61 +6l,]+[tf, +6I4 +6I5]
Substitute all 0 values in equation.
6>, = 180“-[116.56“+90“]+[108.26“+71.56“-45“]
angleof depature 1 
from a complex A j
Thus, the departure angle in the transfer tunction is \0^ =108.26“| at s = —1+2/.
Step 26 of 30
Consider the following equation. 
d l- lO O -ta n j^ jj 
= 126.86“
6 ^ .1 8 0 - ta n [Y ^ ]
= 123.62“
6( =lS0-tan|^Y 
= 135“
6)4 = 180-tan
= 108.43“
61,-90“
Consider the arrival angle in the transfer function.
sum o f angleof vector to' 
= 180“ — the complex zero A 
from other poles zeros 
6>. = 180“ - [ 6i + 6) ]+ [6l + 6) + ^ 4]
angleof arrival 1 
from a complex A J {sum o f angleof vectors to the] 
complex zero A from poles J
step 27 of 30
Substitute all 0 values in equation. 
6!.-180“-[l23.62“+90“]+[l26.86“+135“+108.43“] 
= 336.67“
Thus, the arrival angle in the transfer tunction is |6t̂ = -22.33“| at s = -3+ 4 /.
Step 28 of 30 ^
Procedure to draw root locus plot:
• Take real and imaginary lines on X axis and Y axis respectively.
• Mark the poles on the real axis.
• Locate the asymptotes on the real axis, and draw the asymptotes from centroid.
• Draw the root locus.
The root locus plot is shown in Figure 9.
Root Locus
Hence, the root locus is plotted for the given transfer function and it is shown in Figure 9.
Step 29 of 30
Consider the following given function.
w A - 0^3s)(s+ 3+ 4y)(s+ 3-4 ;) 
s(s + l+ 2 /) (s + l-2 /)
MATLAB program to obtain root locus: 
s=tf('s');
sysL=((1+(3*s))*(s+3+4i)*(s+3^j)V(s*(s+1+2i)*(s+1-2j));
riocus(sysL)
The root locus plot is shown in Figure 10.
Root Loon
Step 30 of 30
Thus, the root locus is verified from MATLAB output.