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Problem 6.16PP
Determine the range of K for which the ciosed-ioop systems (see Fig) are stabie for each of the 
cases beiow by making a Bode piot for K = 1 and imagining the magnitude plot sliding up or 
down until instability results. Verity your answers by using a very rough sketch of a root-locus 
plot.
(a) js:g (s) = ' ^
(b) KG (s) -
v r * t ^ \_ ^(a-F IO)(j-F 1)
(0 * G ( * ) - ( , + , oo)(,+5 )1
Fiaure Block diaoram for YfsVRfst = KGfst/tl + KGtstl
r#-r vr> t-\ ATfa-F I 0 ) ( J - F I )
(0 « G (* * ) - ( . + , oo,(,+ 5 )3
Figure Block diagram for Y{s)/R(s) = KG{s)/[1 + KG(s)]
—■ act*) I or
Step-by-step solution
step 1 of 31
(a)
Consider the given transfers function.
...... (1)' ’ a+30
Rewrite equation (1).
a ( j * 3 i K [ ? * ' ]
Substitute 1 for K in Equation (1).
' ’ ■ • f e - ' )
ii:) = 0 . \ ^ ----- 4
( f * ' )
a v , r . , - y -(4)
Step 3 of 31 'v
Find the magnitude of K G [ji i i) in dB using Equation (4) as listed in table (1). 
Table 1
Term
Corner Frequency
(S)
Slope
©
Change in slope 
©
®.,=30 20 -
(5-) ax,, =30 -20 0
Calculate gain A using table 1.
Assign lower frequency = 1^ ^ and higher frequency =1000^^- ' sec sec
Calculate gain A,, at
A„,=|G(ŷ ■
A ^ , =|G(y Q |.
Step 7 of 31 'v
Consider the number of poles and zeros from Equation (1). 
To find zeros put numerator JV(j) = 0 
Thus, the zero is -3.
To find poles put denominator D(s) = 0 ■
Thus, the pole is - 30.
Step 8 of 31 'v
Consider the formula for the asymptotes. 
n - m
(sum of finite poles)-(sum of finite zeros) 
(numberof finite poles )-(niunber of finite zeros)
- 3 0 - ( - 3 )
1-1
= co
Thus, the asymptotes is .
Step 9 of 31
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.
• Draw the root locus.
The root locus plot is shown in Figure 2.
Step 10 of 31 A
Thus, the root locus is plotted for the given transfer function and it is shown in Figure2. 
From Figure 2, the system is stable for |J(f >Q|.
Step 11 of 31
Hence, the results are verified from root locus plot.
Step 12 of 31
(b)
Consider the transfer function. 
K
( j + 10)(s + l)
Rewrite equation (5).
K
K G (s) = -̂
Rewrite equ 
JCG(4)=7 
Substitute 
ATG(j) =
■(5)
( j + 10)(j+ l)(s + l)
Substitute 1 for K in above equation. 
_1_^____ 1
l o r ^ T T v - ^ . v 
Plot the straight Bode diagram for the transfer function KG{^s) ■ 
Consider the standard form of transfer function up to third order system. 
3r(l-F4/r».)
G(z) -(7)
0 + 4 M ) 0 + 4 M ) 0 + V )
Compare Equations (6), and (7).
Number of zeros present in the system is 0 
Number of poles present in the system is 3.
The corner frequencies are =1, 2421 ■
Step 17 of 31 ^
Consider the number of poles and zeros from Equation (5). 
To find zeros put numerator JV(j) = 0 
Thus, there is no zero.
To find poles put denominator D(s) = 0 ■
Thus, the pole is - 10, - 1 and - 1.
Consider the formula for the asymptotes.
n - m
(sum of finite poles)-(sum of finite zeros) 
(numberof finite poles )-(number of finite zeros)
.zlizM
3-0 
= -4
Thus, the asymptotes is a
step 18 of 31
Consider the formula for the angle of asymptotes.
-(9)2 180°+360«(/-l) 
w — n —m
Where.
Number of poles is n 
Number of zeros is m 
/ = l,2,..ji-m
Substitute 1 for /, 3 for rx and 0 for m in equation (9). 
180+ 360(1-1)
3
180 s -----3
-60»
Substitute 2 for /, 3 for rx and 0 for m in equation (9). 
180 + 360(2-1)
3
540
3
= 180*
Substitute 3 for /, 3 fOr rx and 0 for m in equation (9).
, 180+ 360(3-1)
3
300
3
—60*
Thus, the angle of asymptotes are |60v|, |180°|and | - 60**l-
Step19of31
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.
• Draw the root locus.
Step 20 of 31
The root locus plot is shown in Figure 4.
Step 21 of 31 A
Thus, the root locus is plotted for the given transfer function and it is shown in Figure 4.
Step 22 of 31 A
From Figure 4, the system is stable for |j(f 2 4 2 l ■ 
Step 23 of 31 ^
Hence, the results are verified from root locus plot.
Step 24 of 31 ^
(c)
Consider the transfer function.
^ ,^ ( 4 + 10X4 + 11 
' ̂ (4+100 )(4+5)
Rewrite equation (10).
K
.(10)
.(11)
( 4 + 1 0 ) ( 4 + l ) ( 4 + l )
Substitute 1 for K in above equation.
x e |.).-L
Plot the straight Bode diagram for the transfer function KG{^s) ■ 
Substitute J a fOr s in Equation (11).
K G {J a )= .(12)
Find the magnitude of K G [ ja ) In dB using Equation (12) as listed In table (3). 
Table 3
Term
Corner Frequency Slope Change in slope
(y-AX+l) AX„=1 -20 -
(f-) ax„ = 5 -20 ■40
(if-) AX„=10 -20 -60
(£*■) AX„ = 100 -20 -80
Calculate gain A using table 3.
Assign lower frequency ax = 0 .1 ^ ^ ond higher frequency ax = 1 0 0 0 ^ ^ - sec sec
Calculate gain (A) at a —a, ■
= -62db
Calculate gain A at ax = a ^ ■
A^=-59db 
Calculate gain (A) at ax= ax̂ , .
A,̂ =-S6db
Calculate gain (A) at ax = AX., -
Step 25 of 31
Calculate gain (A) at ax= ax̂ , . 
A ^ = - 8 3 d B 
Calculate gain (A) at a —a,,- 
A ^ = -120dB
Find the phase plot for equation (4).
The Bode diagram is shown in Figure 5.
Step 26 of 31
Thus, the Bode diagram is plotted for the given transfer function and it is shown in Figure 5. 
From Figure 5, the phase angle is never crosses —180®.
Step 27 of 31
Hence, the system is stable for |jc > Q |.
Step 28 of 31 ^
Consider the number of poles and zeros from Equation (1). 
To find zeros put numerator JV(4) = 0 
Thus, the zeros are -10and -1.
To find poles put denominator ^ ( 4) = 0 - 
Thus, the pole is - 100, - 5, - 5 and - 5.
Consider the formula for the asymptotes.
n -m
(sum of finite poles)-(sum of finite zeros) 
(numberof finite poles )-(number of finite zeros)
. z i i H i m
4 - 2
= -52
Thus, the asymptotes is □ u
Step 29 of 31
Substitute 1 for /, 4 fOr rx and 2 for m in equation (9).
180+ 360(1-1) 
" ■ 2
^ 180
2
= 90®
Substitute 2 for /, 4 fOr rx and 2 for m in equation (9).
180 + 360(2-1) 
2
540B-----
2
= 270®
Thus, the angle of asymptotes are |9Q®|and |270®l-
Step 30 of 31
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.
• Draw the root locus.
The root locus plot is shown in Figure 6.
Root Locus
Step 31 of 31
Thus, the root locus is plotted for the given transfer function and it is shown in Figure 6. 
From Figure 6, the system is stable for |jc > Q |.
Hence, the results are verified from root locus plot.